% ****** Start of file DMAA.tex ******
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%   Copyright (c) 2003-2004 Carl Brannen
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% Poetry:
%
% Chapter               ck  Kipling
% ------------------    --  ----------------
% Preface:               x  When 'Omer Smote... (1899)
% Introduction:          x  The Virginity (1919)
% Foundations:           x  The Pro-consuls (1919)
% Density Operators:     x  The Lesson (1903)
% Geomery:               x  Sappers (1896)
% Primitive Idempotents: x  In the Neolithic Age (1897)
% Algebra tricks:        x  Pan in Vermont (1919)
% Representations:       x  The Conundrum of the Workshops (1897)
% Measurement:           x  Arithmetic on the Frontier (1900)
% Particle Zoo:          x  Mulholland's Contract (1919)
% Force:                 x  The Inventor (1919)
% Mass:                  x  The Explorer (1903)
% Cosmic haze:           x  Tomlinson (1919)
% Conclusion             x  A Dedication (1919) 


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\frontmatter %


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$ $\\[120pt] \Huge{ Operator Guide} \\ \LARGE{$ $} \\
\LARGE{to the}\\$ $ \\ \Huge{Standard Model}
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{\LARGE{ Operator Guide} \\$ $ \\
\LARGE{to the}\\$ $ \\ \LARGE{Standard Model} \\ $ $ \\ $ $ \\
\Large{Snuarks and Anions,\\ $ $ \\for the Masses} \\$ $ \\$ $ \\$ $
\\$ $
Carl Brannen\\Liquafaction Corporation \\
Redmond, Washington, USA\\$ $\\$ $ \today $ $ \\ $ $ \\
( Draft )}
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$ $ \\[20pt]
This edition published on the World Wide Web at \\[10pt]
www.brannenworks.com/dmaa.pdf \\[200pt]
\date{\today} \\[200pt]
Copyright \copyright $\; 2008$ by Carl A. Brannen.  All rights
reserved.
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$ $ \\[200pt]
\emph{For Mom and Dad}
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$ $ %
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%Foreword

%Preface
\chapter{Preface}

\begin{quote}
The chance is high that the truth lies in the fashionable direction.
But, on the off-chance that it is in another direction---a direction
obvious from an unfashionable view of field theory--who will find
it?  Only someone who has sacrificed himself by teaching himself
quantum electrodynamics from a peculiar and unfashionable point of
view; one that he may have to invent for himself.\end{quote} %
\begin{flushright}
{\textrm {Richard Feynman,\\}} %
{\textrm {Stockholm, Sweden,\\}} %
{\textrm {December 11, 1965}}
\end{flushright}
$ $

\lettrine{A}{s I write this} in October 2006 particle physics is in trouble.  Two
books stand out on the physics best sellers list: Lee Smolin's
\emph{The Trouble With Physics}, and Peter Woit's \emph{Not Even
Wrong}. These books show that mankind's centuries long effort to
understand the nature of the world has come to a quarter century
long pause. The methods that worked between 1925 and 1980 have not
succeeded in pushing back the frontier since then.  This book
defines an alternative path, a path that may once again allow
nature's mysteries to unfold to us.


Lee Smolin writes, ``When the ancients declared the circle the most
perfect shape, they meant that it was the most symmetric: Each point
on the orbit is the same as any other.  The principles that are the
hardest to give up are those that appeal to our need for symmetry
and elevate an observed symmetry to a necessity.  Modern physics is
based on a collection of symmetries, which are believed to enshrine
the most basic principles.''  In this book we will reject symmetries
as the most basic principle and instead look to geometry, more
specifically, the geometric algebra\cite{HestenesSpacetimeGeometry}
of David Hestenes.  But instead of applying geometric algebra (a
type of Clifford algebra) to spinors, we will be applying it to
density operator states.

The geometric algebra is elegant and attractive and several authors
have applied it to the internal symmetries of the elementary
particles. \cite{LewisElectroWeakGA_1998}  This book has the
advantage over previous efforts in that it derives the relationship
between the quarks and leptons, the structure of the generations,
and provides exact formulas for the lepton masses.  On the other
hand, this book suffers from the disadvantage of requiring a hidden
dimension and that the geometric algebra be complex.  We will
attempt to justify these extensions; in short, they are required
because the usual spacetime algebra is insufficiently complicated to
support the observed standard model particles.


This book is intended as a textbook for graduate students and
working physicists who wish to understand the density operator
foundation for quantum mechanics.  The density operator formalism is
presented as an alternative to the usual Hilbert space, or state
vector, formalism.  In the usual quantum mechanics textbooks,
density operators (or density matrices) are derived from spinors. We
reverse this, and derive spinors from the density operators.  Thus
density operators are at least equal to spinors as candidate
foundations for quantum mechanics.  But we intend on showing more;
that the density operator formalism is vastly superior.


In the state vector formalism, one obtains a density operator by
multiplying a ket by a bra: $\rho = |A\ra \la A|$.  Thus a function
that is linear in spinors becomes bilinear in density operators. And
a function that is linear in density operators becomes non linear
when translated into spinor language.  This means that some problems
that are simple in one of these formalisms will become nonlinear
problems, difficult to solve, in the other.  To take advantage of
both these sorts of problem solving, we must have tools to move back
and forth between density operator and state vector form.  While
most quantum textbooks provide no method of obtaining spinors from
density operators, we will, and we will show how to use these
methods.


The standard model of the elementary particles has been very good at
predicting the results of particle experiments but it has such a
large number of arbitrary constants that it has long been expected
that it would be eventually replaced with a deeper theory. Dr.
Woit's book describes the attraction and ultimate disappointment of
string theory.  The attraction was the promise of a theory with no
need for all those constants; the disappointment was that there were
$10^{500}$ possible quantum vacua with no method to pick out the
right one.  This fits well with the density operator formalism
which, taken literally, suggests that the vacuum is not a part of
physics but instead is simply an artifact of the mathematics.


And the large number of arbitrary constants do not appear so
completely arbitrary.  For example, experimental measurements of
some of the neutrino mixing angles turn out to be small rational
fractions of pi.  And the masses of the leptons are related (to
within experimental error) by the formula discovered by Yoshio Koide
in 1982: $3(m_e + m_\mu + m_\tau)$ $= 2(\sqrt{m_e} + \sqrt{m_\mu} +
\sqrt{m_\tau})^2$, a $5-$digit coincidence.  As Dr. Koide
writes\cite{KoideChallengeOfMassFormula_0506247}, the presence of
the square root ``suggests that the charged lepton mass spectrum is
not originated in the Yukawa coupling structure at the tree level,
but it is given by a bilinear form on the basis of some mass
generation mechanism.''  Density operators provide that bilinear
form.


Any time a new formalism is found, it is natural to first pick the
low hanging fruit.  We will treat mass as if it were a force that
converts left handed particles to right handed particles and vice
versa.  As a ``force'', mass is particularly simple because, in this
model, the interactions correspond to Feynman diagrams with only two
legs.  For the density operator theory, these are the low hanging
fruit, and we will apply it to the Koide relation, extending it to
the neutrinos.  And we will finish with other, more speculative
applications.


A word on the poetry that begins each chapter. These are works by a famous author. They were published sufficiently long ago that their copyright has expired. I quote them without attribution in the knowledge that, so long as our civilization survives, you will be able to quickly locate the author using the internet. Perhaps, after such a search, you will find that the unquoted portions of the poetry reads on the physics topic.  And if you are insufficiently interested to make this small effort, a proper citation will provide you no advantage.

Regarding citations of other's work, this text is intended as a practical work, a training tool for graduate students more interested in the methods than the authors. A bit of stolen doggerel:
$ $\\ $ $\\

\settowidth{\versewidth}{$\;\;\;\;\;$    An' what he thought 'e might require,}
\begin{verse}[\versewidth]
When 'Omer smote 'is bloomin' lyre,\\
\vin  He'd 'eard men sing by land an' sea;\\
An' what he thought 'e might require,\\
\vin  'E went an' took -- the same as me!\\
\end{verse}

% Test of Kanji here: ``$>>\;$ \unichar{29992} $\;<<$ '' % (didn't work)
% Test of Kanji here: ``$>>\;$ \unichar{7528} $\;<<$ ''



$ $\\ $ $\\

{\raggedleft{\scshape Carl Brannen} \\ Redmond, Washington, USA \\
January 19, 2008 \par}

 %
\clearpage


% Acknowledgements

% Introduction
\chapter{Introduction}

%\begin{quote}
\settowidth{\versewidth}{$\;\;\;\;\;$    From his first love, no
matter who she be.}
\begin{verse}[\versewidth]
%\em
TRY as he will, no man breaks wholly loose\\
$\;\;\;\;\;$    From his first love, no matter who she be.\\
Oh, was there ever sailor free to choose,\\
$\;\;\;\;\;$    That didn't settle somewhere near the sea? \\
\end{verse}
%\end{quote}
$ $

\section{\label{sec:ThiefsLover}  The Thief's Lover}
\lettrine{T}{$\;$o be added} later.  At the moment I can't get
myself to actually put on paper what I am thinking of for this, but
it amounts to a short story.






% Terminology




\mainmatter



\chapter*{\label{sec:Foundations} Foundations}
%\addtocontents{toc}{\contentsline {chapter}{\chapternumberline {0} Fundamentals}{1}}
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%need: \contentsline {chapter}{\chapternumberline {0} Fundamentals}{1}


%\begin{quote}
\settowidth{\versewidth}{Lesser men feign greater goals}
\begin{verse}[\versewidth]
%\em
Lesser men feign greater goals,\\
\vin Failing whereof they may sit\\
Scholarly to judge the souls\\
\vin That go down into the pit,\\
And, despite its certain clay,\\
Heave a new world toward the day.
\end{verse}
%\end{quote}
$ $


%\setcounter{DefaultLines}{3}
\lettrine{T}{$\;$his text differs from} the usual introductions to elementary particles in that it assumes that elementary particles are best represented in density matrix form rather than state vector form. In quantum mechanics it is often said that density matrices are an alternative, equivalent, method of representing quantum states and this is true. Where density matrices differ from state vectors is in quantum field theory. Since the standard model of elementary particle physics is a quantum field theory, the difference between the theories gives a different persepctive on elementary particles. In this chapter we review the foundations of quantum mechanics with an eye to justifying the use of density matrices instead of state vectors.

% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:HestenesFactorization})
\section{\label{sec:QWs}  Particle Waves}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The first surprise that quantum mechanics brought to physics was the fact that matter can interfere with itself. Let a beam of particles impinge on a barrier with two slits. This produces two beams of spreading particles on the far (right) side of the barrier. These two beams are caught on a screen. One finds that the particles produce an interference pattern on the screen. See Fig.~(\ref{fig:TwoSlitExp}).


% Fig.~(\ref{fig:TwoSlitExp})
\begin{figure}[!htp]
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% Particle Source
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% Source Scattering
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% Slit Scattering Low
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% Interference Pattern
\qbezier(191,31)(195,31)(195,33) %
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\put(210, 5){\vector(-1,4){6}} %
\put(114, 2){Interference Pattern} %
%\put(120, 5){\vector(-1,4){8}} %
\end{picture}
} %
\caption{\label{fig:TwoSlitExp} The Two Slit Experiment: Quantum particles impinge on a barrier with two slits. Particles that pass the barrier form an image on the screen that exhibits interference.}
\end{figure}

The intereference patterns are reminiscent of those resulting from classical waves, however, the image is formed by a large number of individual particle hits on the screen, and the image will form even when the experiment is run at such low particle production rates that only a single particle is present in the apparatus at a time.

The pattern only forms when a particle is allowed to go through either of the two slits. If, for example, one arranged for two different particles sources to feed the slots, one feeding only the upper slit and the other feeding only the lower slit, the interference pattern disappears. The implication is that the particle somehow interferes with itself by passing through both slits simultaneously.

For any single particle, quantum mechanics assumes that it is impossible to predict exactly where the particle will end up. Instead, quantum mechanics allows one to compute a probability density. Given a wave function $\psi(x;t)$,\footnote{We will use $x$ to mean the $x$-coordinate or all the spatial coordinates, but we will try to carefully use the semicolon to separate the space and time parameters, for example, $\psi(x;t) = \psi(x,y,z;t)$.} the probability density $\rho(x;t)$ is given by $\rho(x;t) = |\psi|^2 = \psi^*\psi$ where $a^*$ indicates the complex conjugate of $a$.

In classical wave interference, one supposes that two wave sources, say $A$ and $B$ are present in the same region. These two wave sources produces two waves $F_A(x;t)$ and $F_B(x;t)$ which we will take, for simplicity to be real valued functions of position and time. In order for interfere to occur, $F_A$ and $F_B$ must be able to take both positive and negative values. The interference occurs because their signs cause them to partially cancel when they are added together to give the total wave, $F = F_A(x;t) + F_B(x;t)$.

In summing two classical (or quantum) waves, we are making the assumption of the ``law of superposition''. Classically, superposition tends to work for waves of sufficiently small amplitude. In quantum mechanics, superposition always works, but is of a nature somewhat different from classical superposition as will discuss in the next chapter.

Quantum mechanics is a probabilistic theory and the results of a calculation is a probability. Since probabilities cannot be negative, and yet matter waves must be able to interfere, one cannot take the quantum wave function to be a probability. Instead, one takes the squared magnitude, $|\psi|^2$, of the wave function as the probability. In the classical case, the squared magnitude of a wave function is the energy.

In a classical wave, both the amplitude and phase are observable. For example, one can measure the height and arrival times of waves on the ocean. With the right instruments, one can similarly measure the amplitude and phase of an electromagnetic waves. One can do this without significantly modifying the classical wave. One can therefore arrange for experiments with waves with known amplitude and phase.

The situation in quantum mechanics is more difficult. The squared amplitude of the probability wave gives the probability density for position $|\psi(x;t)|^2$, while the momentum operator $i\hbar\nabla$ applied to the wave function gives the wave function for momentum, $\psi_p$:
\begin{equation}\label{eq:WaveFncMD}
\psi_p(x;t) = i\hbar \nabla \psi(x;t).
\end{equation}
The probability density for momentum is then $|\psi_p(x;t)|^2$.

If we multiply the wave function $\psi(x;t)$ by a complex phase $\exp(i\kappa)$ where $\kappa$ is real, neither of the probability densities, that for momentum or position, is changed. More generally, let $Q$ be an operator, for which we wish to calculate an average value. We have:
\begin{equation}\label{eq:AverageOfQ}
\la Q_{\psi} \ra (t) = \int \psi^*(x;t) Q \psi(x;t)\; d^3x,
\end{equation}
where the subscript on $Q$ denotes that this is the average for the quantum state $\psi$, which average in general depends on the time $t$. Define $\psi'(x;t) = \exp(i\kappa)\psi(x;t)$ with $\kappa$ real so that $\psi'$ is related to $\psi$ by an overall (global) phase change. Since $\psi$ contributes bilinearly in the formula for average value, Eq.~(\ref{eq:AverageOfQ}), the average value for $Q$ will be the same for $\psi$ and $\psi'$.

We are left in the somewhat contradictory situation that the phase of a quantum state is apparently unneeded in physical measurements of the system it represents, but is necessary for interference to occur. The contradiction will be resolved by going to the density matrix representation.

% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:HestenesFactorization})
\section{\label{sec:IDMs}  Wave Function Density Matrices}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In the usual state vector formalism, the fundamental object is the wave function $\psi(x;t)$. This must be modified to produce a probability density by multiplying by $\psi^*(x;t)$. However, if one so multiplies a wave function, one loses the phase information. A method that allows both the probability and phase information to be directly encoded in the same mathematical object is to instead multiply by $\psi^*(x';t)$ where $x'$ is allowed to vary independently of $x$. The result is the ``density matrix'' wave function:
\begin{equation}\label{eq:WDMF}
\rho(x,x';t) = \psi^*(x';t)\;\psi(x;t).
\end{equation}

Since the density matrix wave function depends bilinearly on $\psi$, multiplying the state $\psi$ by an arbitrary complex phase results in no change to the density matrix wave function. The unphysical arbitrary complex phase has been removed. In addition, the probability density for position is now given by the ``diagonal elements'' of the density matrix: 
\begin{equation}\label{eq:ProbDensInRho}
|\psi(x;t)|^2 = \rho(x,x;t).
\end{equation}
To get the probability density for momentum, we 

% ???????????????



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:HestenesFactorization})
\section{\label{sec:CHI}  Consistent Histories Interpretation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Consistent Histories section content goes here


\chapter{\label{sec:DensityOperators} Finite Density Operators}

%\begin{quote}
\settowidth{\versewidth}{Have felt the effects of the lesson we got---an advantage no money could buy us!}
\begin{verse}[\versewidth]
%\em
For remember (this our children shall know: we are too near for that knowledge)\\
Not our mere astonied camps, but Council and Creed and College---\\
All the obese, unchallenged old things that stifle and overlie us---\\
Have felt the effects of the lesson we got---an advantage no money could buy us!
\end{verse}
%\end{quote}
$ $

%\setcounter{DefaultLines}{3}
\lettrine{T}{$\;$he standard practice} in quantum mechanics has been to
treat the state vector as the fundamental description of a quantum
state and the density operator as a derived object.  In this chapter
we reverse this relationship and treat the density operator as the
fundamental description of a quantum state.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:HestenesFactorization})
\section{\label{sec:TDMs}  Traditional Density Operators}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section we introduce density operators as they are commonly
taught, with a bit of an emphasis on the fundamental nature of them.
We will loosely skim the excellent class notes of Frank C.
Porter\index{Porter, Frank C.}\cite{PorterDensityMatrixClassNotes},
to which the reader is directed.

We begin with a state space, with a countable orthonormal basis
$\{|u_n\rangle,n=1,2,...\}$.  A system in a normalized state
$|\psi(t)\rangle$ at time $t$ can be expanded as:
\begin{equation}\label{eq:DefnExpansion}
|\psi(t)\rangle = \sum_n a_n(t)|u_n\rangle.
\end{equation}
Normalization implies that $\sum_n |a_n(t)|^2 = 1$.

An observable $Q$ can be expanded in this basis as:
\begin{equation}\label{eq:ExpanQ}
Q_{mn} = \langle u_m|Q|u_n\rangle,
\end{equation}
and the expectation value of $Q(t)$ for the
system $|\psi(t)\rangle$ is:
\begin{equation}\label{eq:ExpQ}
\langle Q\rangle = \langle\psi(t)|Q\psi(t)\rangle = \sum_n\sum_m
a_m^*(t)a_n(t) Q_{mn}.
\end{equation}
Note that $\langle Q \rangle$ is quadratic in the coefficients
$a_n$.


Define the density operator $\rho(t)$ as:
\begin{equation} \label{eq:DensityOpDefn}
\rho(t) = |\psi(t)\rangle \langle\psi(t)|.
\end{equation}
Since $|\psi(t)\rangle$ is normalized, the density operator is
idempotent:\marginpar{\em Pure density operators are idempotent.}
\begin{equation}\label{eq:IdempotentRho}
\rho^2(t) = \rho(t).
\end{equation}
Writing $\rho(t)$ in the $u_m$ basis we have:
\begin{equation}\label{eq:RhoInU}
\rho_{mn} = \langle u_m|\psi(t)\rangle\langle\psi(t)|u_n\rangle =
a_m(t) a_n^*(t).
\end{equation}
These matrix elements appear in Eq.~(\ref{eq:ExpQ}) and consequently
we can rewrite the expectation value of $Q$ using the density
operator:
\begin{equation}\label{eq:UsualExpValue}
\begin{array}{rcl}
\langle Q \rangle(t) &=& \sum_m \sum_n a_m(t) a_n^*(t) Q_{mn}\\
&=& \tr(\rho(t)\;Q).
\end{array}
\end{equation}


Let $\{q\}$ be a subset of the eigenvalues of $Q$.
Define $P_{\{q\}}$ as the projection
operator that selects these eigenvalues.  Then the probability that
a measurement will lie in $\{q\}$ is
\begin{equation}\label{eqProbDefn}
P(\{q\}) = \tr(P_{\{q\}}).
\end{equation}
If $\{q\}$ is the whole spectrum of $Q$, then the projection
operator is unity and the probability is one.  Thus:
\begin{equation}\label{eq:TrpEqOne}
\tr(\rho(t)) = 1.
\end{equation}
The time evolution of a state $|\psi(t)\rangle$ is given by
Schroedinger's equation:
\begin{equation}\label{eq:SchEqn}
i\frac{d}{dt}|\psi(t)\rangle = H(t)|\psi(t)\rangle,
\end{equation}
where $H(t)$ is the Hamiltonian operator.  When
put into density operator form, the equation becomes:
\begin{equation}\label{eq:DenOpEqn}
\frac{d}{dt}\rho(t) = \frac{1}{i}[H(t),\rho(t)].
\end{equation}


We\marginpar{\em Density operators are on an equal footing with
state vectors in the foundations of quantum mechanics.} have showed
that the density operator $\rho(t)$ allows computation of
expectation values and probabilities, and we've shown the equation
for time evolution. This is apparently all that can be known about a
quantum state, so the density operator is an alternative formulation
for quantum mechanics on an equal footing with the state vector
formalism from which we derived it.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:HestenesFactorization})
\section{\label{sec:DMaFoP}  An Alternative Foundation for QM}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


With two alternative formulations for quantum
mechanics we have a choice.  The two methods will give the same
answer, but for any particular problem, one or the other is likely
to give an easier calculation. And one or the other might be closer
to the underlying physics.  We now look at the two formulations from
the point of view of which is more likely to be useful in
understanding the foundations of physics.


Both the density operator formalism and the state vector formalism
share the same operators so there is no difference here. But the
state vector formalism also requires states and in this sense the
density operator formalism gets by with fewer mathematical objects.
Since the operators alone are sufficient to describe a quantum
state, the state vectors are only ancillary mathematical devices
used for calculational convenience.


Density operator formalism is particularly well suited to
statistical analysis of quantum mechanical systems.  For example,
the entropy of a quantum ensemble is
defined by the simple equation:
\begin{equation}\label{eq:EntropyAndRho}
S = -k \tr(\rho \; \ln(\rho)).
\end{equation}
This is an advantage for the density operator formalism.


Given two solutions to the Schroedinger equation, $|\psi(t)\rangle$
and $|\phi(t)\rangle$, any linear combination is also a solution.
That is, the solution set is linear.  The same cannot be said of the
density operator formulation.  The advantage is with the state
vector formalism, but this is a calculational advantage only, and
later in this chapter we will show that linear superposition can be
translated advantageously into the density operator formalism. Our
models of reality are not inherently simpler when they are linear,
instead they are simpler to use in calculations.


Calculations in the state vector formalism use an inner product
which is inherently complex valued, while the corresponding
calculations in the usual density operator formalism use the trace
of a matrix. The trace is a complex function defined on the set of
operators.  In later chapters we will give a geometric
interpretation of these complex numbers that will allow us to make
calculations that are difficult or impossible in the state vector
language.

States\marginpar{\em Density operators have no phase ambiguity.}
represented by a state vector carry a phase ambiguity while the
density operator states are completely defined. This is an advantage
for the density operator formulation. We will later show that when
translating a density operator state into state vector form, one
must reintroduce this phase ambiguity in the form of a choice of
spin direction. Thus the origin of the gauge symmetries appears to
be related to a geometric choice.\footnote{One might suppose that
the density operator formulation would be at a disadvantage to the
state vector form on problems associated with gauge forces, but this
was recently shown not to be the case by Brown and
Hiley.\cite{HileyAlgebraicSchroedingerEqn_0005026} Also see
\cite{MaroneyDensityMatrix_0311149}.}




% Fig.~(\ref{fig:FermCube}) and  Eq.~(\ref{eq:HestenesFactorization})
\section{\label{sec:EvcaEva}  Eigenvectors and Eigenmatrices}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


Up to this point we've been discussing density operators in general.
We will now specialize to the pure density operators, that is, the
ones that correspond to state vectors.  If the author mentions
``density operator'' or ``density matrix'' the reader should assume
that he means ``pure density operator''.  Furthermore, we will
almost entirely be dealing with spin and
internal degrees of freedom.


For the remainder of this chapter we will explore the density
operator theory of the Pauli algebra.  The
usual introduction to the Pauli algebra involves the
representation known as the Pauli spin
matrices:
\begin{equation}\label{eq:PauliSpinMatrices}
\sigma_x = \left(\begin{array}{cc}0&1\\1&0\end{array}\right),\; %
\sigma_y = \left(\begin{array}{cc}0&-i\\+i&0\end{array}\right),\; %
\sigma_z = \left(\begin{array}{cc}+1&0\\0&-1\end{array}\right).\; %
\end{equation}
Most physicists would associate the Pauli spin matrices with a
representation of a Lie algebra, but in this book
we will instead associate them with a Clifford
algebra.  The reader is not expected to know
anything of Clifford algebra; we will introduce the necessary
concepts in this chapter and the next.  For now, let us only mention
that Clifford algebras are also Lie algebras, but there are many Lie
algebras that are not Clifford algebras.


In order to illustrate the uses of the pure density operators, we
now discuss the classic eigenvector
problem\cite[\S 54]{Landau} for spin-$1/2$ from the point of view of
state vectors and density operators.

\begin{quote}
A particle with spin 1/2 is in a state with a definite value $s_z =
1/2$.  Determine the probabilities of the possible values of the
components of spin along an axis $z'$ at an angle $\theta$ to the
$z$-axis.
\end{quote}

This problem can be solved in a number of ways.
Landau uses the fact that the mean
spin vector of the particle along the $z'$ directions is
$\cos(\theta)$, along with the fact that this average is given by
$(w_+-w_-)/2$ where $w_\pm$ are the probabilities for the spin value
along $a'$ being measured as $\pm 1/2$.  Then, since $w_++w_- = 1$,
the result, $w_+ = (1+\cos(\theta))/2$, can be deduced by algebra.

A more direct way of solving this problem is to find
eigenvectors corresponding to spin-$1/2$ oriented
in the $z$ and $z'$ directions, and then computing the probability
with the formula $P = |\langle z|z'\rangle|^2$.  This method is
somewhat involved. If the vectors $z$ and $z'$ were more arbitrary,
the problem would be even worse.

In the density operator formalism, the states are operators along
with the particles.  So the solution of the
eigenvector equations are trivial.  As with the
state vector formalism, let us first define the
operator for spin-$1/2$ in an arbitrary direction. To do this, we
first define:
\begin{equation}\label{eq:PauliVector}
\vec{\sigma} = (\sigma_x,\sigma_y,\sigma_z)
\end{equation}
where $\sigma_\chi$ are the usual Pauli spin matrices.  Then the
projection operators for spin in the
$\vec{u}$ direction is given by:
\begin{equation}\label{eq:ProjOpU}
\vec{u}\cdot\vec{\sigma} = u_x\sigma_x + u_y\sigma_y + u_z\sigma_z.
\end{equation}
where $u_\chi$ are the components of the vector $\vec{u}$.

The traditional way of solving this problem with state vectors is to
write out the eigenvector equation with the
operator for spin-$1/2$ in the $\vec{u}$ direction, which is simply
half the projection operator:
\begin{equation}\label{eq:EigenVectorSV}
((1/2)\vec{u}\cdot\vec{\sigma})\;\; |u+\rangle =
(+1/2)\;\;|u+\rangle,
\end{equation}
then substituting the Pauli spin matrices for $\vec{\sigma}$, and
then solving the resulting matrix equation.  This can be a fairly
involved activity for the new student.  In addition, even if the
student uses the technique that requires normalization of the
eigenvectors, their phases are still
arbitrary.

One would think that solving the same problem with density operators
would be more difficult because there are more apparent degrees of
freedom with the state, but this is not the case.  First, one must
know how two projection operators multiply:\marginpar{\em
Multiplication of Pauli matrices.}
\begin{equation}\label{eq:ProdSpinProjOps}
(\vec{u}\cdot\vec{\sigma})(\vec{v}\cdot\vec{\sigma}) =
\vec{u}\cdot\vec{v} + i(\vec{u}\times\vec{v})\cdot\vec{\sigma}.
\end{equation}
In particular, when $\vec{u}=\vec{v}$ in the above equation, one
finds that the right hand side is one.

Thus\marginpar{\em Density operator equations have a simple closed
form solution.} the matrix eigenvector equation is solved trivially
by $(1+\vec{u}\cdot\vec{\sigma})$.  That is:
\begin{equation}\label{eq:MatrixEVeqn}
((1/2)\vec{u}\cdot\vec{\sigma})\;\;(1 + \vec{u}\cdot\vec{\sigma}) =
(+1/2) (1 + \vec{u}\cdot\vec{\sigma}).
\end{equation}
While this is a solution to the matrix
eigenvector equation, it is not normalized as
density operators are supposed to be, that is, $\rho^2 = \rho$ by
Eq.~(\ref{eq:IdempotentRho}). Instead, the square of
$(1+\vec{u}\cdot\vec{\sigma})$ is twice itself, so our
normalization is off by a factor of two.  The
correct density operator corresponding to spin-$1/2$ in the
$\vec{u}$ direction is therefore given by:
\begin{equation}\label{eq:DensityMatrixUSoln}
\rho_u = (1 + \vec{u}\cdot\vec{\sigma})/2,
\end{equation}
and we have a closed form solution for the density operator
eigenmatrix problem in an arbitrary direction
$\vec{u}$.


\SubSectionBreak

When one tries to solve the general spin-$1/2$ eigenvector problem
in the state vector formalism, one discovers that it is not so
simple as the matrix problem.  One does not have the easy option of
writing the answer in terms of the Pauli matrices themselves (and
therefore avoiding any mention of the particular
representation chosen).  One finds that ones
solution fails for certain vectors, which we now illustrate.

Canceling\marginpar{\em The corresponding state vector equation has
no closed form solution.} out the factor of $1/2$ on both sides, the
state vector eigenvector equation is:
\begin{equation}\label{eq:EVeqnGenSV}
\left(\begin{array}{cc}u_z&u_x-iu_y\\u_x+iu_y&-u_z\end{array}\right)
\left(\begin{array}{c}a\\b\end{array}\right) =
\left(\begin{array}{c}a\\b\end{array}\right)
\end{equation}
or
\begin{equation}\label{eq:EVeqnGenSV2}
\left(\begin{array}{cc}u_z-1&u_x-iu_y\\u_x+iu_y&-u_z-1\end{array}\right)
\left(\begin{array}{c}a\\b\end{array}\right) = 0.
\end{equation}
An obvious solution to this problem is
\begin{equation}\label{eq:ObvSolnSVEV}
\left(\begin{array}{c}u_x-iu_y\\1-u_z\end{array}\right),
\end{equation}
however, when $\vec{u} = (0,0,1)$, the above is zero and cannot be
normalized.  Another obvious solution is:
\begin{equation}\label{eq:ObvSolnTwo}
\left(\begin{array}{c}1+u_z\\u_x+iu_y\end{array}\right),
\end{equation}
but this solution is zero when $\vec{u} = (0,0,-1)$.  In addition to
these two solutions, we can choose any linear combination.  As the
astute reader recognizes, any of these more general solutions will
also be zero for some value of $\vec{u}$.

So we see that the problem, when written in terms of the pure
density operators, has a simple general solution, but when written
in the traditional state vector form, the problem is more difficult.

Let\marginpar{\em Density matrix formalism is more powerful.} us
illustrate the power of the density operator formalism by restating
the given problem in more general terms:
\begin{quote}
A particle with spin 1/2 is in a state with a definite value $s_u =
1/2$ for spin measured in the direction $\vec{u}$.  Determine the
probability of a measurement of spin $+1/2$ along the $\vec{v}$
axis.
\end{quote}

By symmetry, we know that the answer to the above problem is
$(1+\cos(\theta))/2$ where $\theta$ is
the angle between $\vec{u}$ and $\vec{v}$.  To solve it with the
state vector formalism, we compute the eigenvectors for the two
directions, then take the probability as the square of their dot
product.  In the density operator formalism, the answer is given by
a trace:
\begin{equation}\label{eq:DMFprobability}
w_+ = \tr(\rho_u \rho_v),
\end{equation}
where $\rho_\chi$ is the density operator for spin in the
$\vec{\chi}$ direction.

Using the formula for the density operator solution of the
eigenvector problem, Eq.~(\ref{eq:DensityMatrixUSoln}), the
probability computes as:
\begin{equation}\label{eq:DMsolnForTProb}
\begin{array}{rcl}
w_+ &=& \tr(\rho_u\;\rho_v)\\
&=&\tr((1 + \vec{u}\cdot\vec{\sigma})/2 (1 +
\vec{v}\cdot\vec{\sigma})/2)\\
&=& \tr(1 + (\vec{u}+\vec{v})\cdot\vec{\sigma}/2 +
(\vec{u}\cdot\vec{\sigma})(\vec{v}\cdot\vec{\sigma}))/4.
\end{array}
\end{equation}
The trace function keeps only the scalar part of its argument, and
since the representation is done with $2\times 2$
matrices, we have that $\tr(1) = 2$.
Applying Eq.~(\ref{eq:ProdSpinProjOps}), we obtain:
\begin{equation}\label{eq:DMsolnForTPcnt}
w_+ =  2(1 + \vec{u}\cdot\vec{v})/4 = (1+\cos(\theta))/2,
\end{equation}
the same as with the state vector
calculation.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ObvSolnTwo})
\section{\label{sec:BaK}  Bras and Kets}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In the previous section we saw that the density operator formulation
allows the spin eigenvector problem to be solved
in closed form while the state vector formulation cannot.
Consequently, when the theory is taught in the state
vector formulation, problems are left in
eigenvector form rather than solved.  The reason for the failure was
that for any given general form solution in the state vector
language, there is a direction where the solution becomes zero.  But
as we saw above, it is possible to find two complementary solutions,
that is, solutions complementary in that one or the other (or both)
provide solutions for any given direction.  In this section we
further discuss this fact in the context of how one obtains a state
vector from a density operator.

One\marginpar{\em Density operator formalism is naturally
independent of the choice of representation of the Pauli algebra.}
of the strengths of the density operator formulation is that it
allowed us to write the normalized solution to the
eigenvector equation without reference to the
representation of the Pauli algebra:
$(1+\vec{u}\cdot\vec{\sigma})/2$.  But to show the connection to
state vectors, let us write this solution explicitly using the
Pauli matrices:
\begin{equation}\label{eq:DMevsolnExplicit}
\begin{array}{rcl}
\rho_u &=& (1+\vec{u}\cdot\vec{\sigma})/2\\
&=&\frac{1}{2} \left(\begin{array}{cc}
1+u_z&u_x-iu_y\\u_x+iu_y&1-u_z
\end{array}\right).
\end{array}
\end{equation}
Comparing the vectors of the above with Eq.~(\ref{eq:ObvSolnSVEV})
and Eq.~(\ref{eq:ObvSolnTwo}), we see that the density operator
eigenmatrix solution is composed of the two
obvious solutions to the state vector eigenvector
problem.

Now the two obvious state vector solutions to the eigenvector
problem can be written in ``square spinor'' \cite{LounestoCAS}
form by filling in
the unneeded columns of the matrices with zero. Then those two
solutions sum to the density operator solution as follows:
\begin{equation}\label{eq:EVsolnsSumTogether}
\frac{1}{2}
\left(\begin{array}{cc}1+u_z&u_x-iu_y\\u_x+iu_y&1-u_z\end{array}\right)
=\frac{1}{2}
\left(\begin{array}{cc}1+u_z&0\\u_x+iu_y&0\end{array}\right)
+\frac{1}{2}
\left(\begin{array}{cc}0&u_x-iu_y\\0&1-u_z\end{array}\right).
\end{equation}
In the above, the left hand side is the density operator solution,
and the right hand side are two (not normalized) solutions to the
state vector problem.

In the context of density operators, a natural method of converting
the density operator solution to one of the square spinor solutions
on the right hand side of Eq.~(\ref{eq:EVsolnsSumTogether}) is to
multiply by another density operator.  In particular, note that the
matrices for spin-$1/2$ in the $\pm z$ direction, that is,
$\rho_{\pm z} = (1\pm \sigma_z)/2$;
\begin{equation}\label{eq:ProjectionOnZs}
\begin{array}{rcl}
\rho_{+z} &=& \left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\
\rho_{-z} &=& \left(\begin{array}{cc}0&0\\0&1\end{array}\right)
\end{array}
\end{equation}
convert the density operator solution to the two complementary
spinor solutions:
\begin{equation}\label{eq:SpinorSolnsFromDM}
\begin{array}{rcl}
\frac{1}{2}\left(\begin{array}{cc}1+u_z&0\\u_x+iu_y&0\end{array}\right)
&=&\frac{1}{2}
\left(\begin{array}{cc}1+u_z&u_x-iu_y\\u_x+iu_y&1-u_z\end{array}\right)
\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\
\frac{1}{2}
\left(\begin{array}{cc}0&u_x-iu_y\\0&1-u_z\end{array}\right)
&=&\frac{1}{2}
\left(\begin{array}{cc}1+u_z&u_x-iu_y\\u_x+iu_y&1-u_z\end{array}\right)
\left(\begin{array}{cc}0&0\\0&1\end{array}\right)
\end{array}
\end{equation}
The\marginpar{\em A column vector can be kept in matrix (``square
spinor'') form.} matrices on the left hand side above are equivalent
to spinors. To see this, notice that matrices that have all but one
column zero act just like vectors. That is, the zeroes are conserved
under multiplication by a constant, under addition with another
square spinor matrix, and under multiplication by an arbitrary
matrix on the left.  These are all the things we require of
kets.

Putting what we have found into density operator language, we find
that the way one converts a density operator into ket form is to
simply multiply on the right by a constant density operator.  In the
examples above, multiply on the right by $\rho_{\pm z}$.

For\marginpar{\em Bras and kets defined using density operators.}
the moment let us choose the positive spin$-1/2$ in the $+z$
direction. The ket is defined as:
\begin{equation}\label{eq:KetsFromDMs}
|u\rangle = \rho_u\;\rho_{+z}
\end{equation}
If one reverses the order:
\begin{equation}\label{eq:BrasFromDMs}
\langle v| = \rho_{+z}\;\rho_v
\end{equation}
one obtains the bra form. These bras and kets
are not normalized. Consequently, if we are to perform calculations,
we must divide by the proper normalization constant, a subject we
will take up after discussing scalars.

In the state vector formalism, multiplying a bra by a ket gives a
scalar.  Let's work out what it does in density operator formalism:
\begin{equation}\label{eq:BraKetFromDM}
\begin{array}{rcl}
\langle v| u\rangle\!\!\!\! &=& \rho_{+z}\;\rho_v\;\rho_u\;\rho_{+z}\\
&=&
\frac{1}{4}\left(\begin{array}{cc}\!1\!\!&\!\!0\!\\\!0\!\!&\!\!0\!\end{array}\right)\!
\left(\begin{array}{cc}1\!+\!u_z&u_x\!-\!iu_y\\u_x\!+\!iu_y&1\!-\!u_z\end{array}\right)\!
\left(\begin{array}{cc}1\!+\!v_z&v_x\!-\!iv_y\\v_x\!+\!iv_y&1\!-\!v_z\end{array}\right)\!
\left(\begin{array}{cc}\!\!1\!\!&\!\!0\!\!\\\!\!0\!\!&\!\!0\!\!\end{array}\right)\\
&=&\frac{1}{4}\left(\begin{array}{cc}1\!+\!u_z&u_x\!-\!iu_y\\0&0\end{array}\right)
\left(\begin{array}{cc}1\!+\!v_z&0\\v_x\!+\!iv_y&0\end{array}\right)\\
&=&\frac{1}{4}(1\!+\!u_z)(1\!+\!v_z)+(u_x\!-\!iu_y)(v_x\!+\!iv_y)\left(\begin{array}{cc}
1&0\\0&0\end{array}\right).
\end{array}
\end{equation}

Note\marginpar{\em Complex numbers as complex multiples of
idempotents.} that the above is a complex multiple of $\rho_{+z}$.
The set of all such matrices act just like the complex
numbers. For example, if $a$ and $b$ are
arbitrary complex numbers, then:
\begin{equation}\label{eq:DefnComplex}
\begin{array}{rcl}
a\rho_{+z} + b\rho_{+z} &=& (a+b)\rho_{+z}, \;\;\;\;\textrm{and}\\
(a\rho_{+z})(b\rho_{+z}) &=& (ab)\rho_{+z},
\end{array}
\end{equation}
so complex multiples of $\rho_{+z}$ act just like complex numbers.
This means that if we define all of our bras and
kets in a consistent manner with $\rho_{+z}$, our bras
and kets will multiply to complex multiples of $\rho_{+z}$, and
these act just like the complex numbers.

It remains to normalize the bras and kets we've defined in
Eq.~(\ref{eq:BrasFromDMs}) and Eq.~(\ref{eq:KetsFromDMs}).  Using
the bra ket multiplication formula in Eq.~(\ref{eq:BraKetFromDM}),
we have that
\begin{equation}\label{eq:UbraTimesUket}
\begin{array}{rcl}
\langle u | u \rangle &=&\frac{1}{4}(1+2u_z+u_z^2 + u_x^2 + u_y^2).
\end{array}
\end{equation}
So the normalization factor for $\langle v| u
\rangle$ is the square root of the above multiplied by the same
thing for $v$. That is, since probabilities are proportional to the
squared magnitude of $\langle v|u\rangle$, the probability of the
spin being measured as $+1/2$ in the $v$ direction is:
\begin{equation}\label{eq:PinUandV}
w_+ = \frac{\langle v|u\rangle \langle u| v\rangle} {\langle
v|v\rangle \langle u|u\rangle}
\end{equation}
We leave it as an exercise for the reader to verify that the above
reduces to $(1+\vec{u}\cdot\vec{v})/2$.

We have shown that density operators can be converted to
spinor form by pre or post multiplying by $\rho_{+z}$,
a constant density operator.  This brings the spinors into density
operator form, but there is a problem.  If $u=-\vec{z}$, then the
product $\rho_{-z}\;\rho_{+z}$ is zero.  This problem is identical
to the issue that spinors had when we tried to write down a general
solution to the eigenvector problem.  We can get around it as we did
before, by choosing a different vacuum density
operator.  Or better, by avoiding the spinor formalism where
possible.

Since\marginpar{\em The ``vacuum'' density operator state is an
arbitrary choice.} the $z$ direction is not in any way special, our
analysis of how to convert density operators to bras and kets using
$\rho_{+z}$ can be redone with any other constant density operator.
The choice of this constant density operator defines the phase of
the bra and ket that is produced.  We will discuss this in greater
detail in a later sections and chapters.  For now, let us choose the
notation $\rho_0$ to specify a constant density operator that need
not be aligned in the $+z$ direction.  For reasons that will become
clear in later chapters, we will follow Julian
Schwinger\cite{QKAD} and call $\rho_0$ the
``vacuum'' state.


So\marginpar{\em Bras and kets from vacuum state.} with $\rho_0$ as
our choice of constant density operator, the conversion from density
operators to bras and kets is:
\begin{equation}\label{eq:KetsFromVac}
\begin{array}{rcl}
| u \rangle &=& \rho_u \; \rho_0\\
\langle v | &=& \rho_0 \; \rho_u.
\end{array}
\end{equation}
We now prove that these definitions give bra-kets that are complex
multiples of $\rho_0$.


Write $\rho_0 = |0\rangle\langle 0|$, and similarly for $u$ and $v$.
Let $M$ be an arbitrary operator.  Then the matrix element of $M$
for $u$ and $v$ is:
\begin{equation}\label{eq:ProofComplexes}
\begin{array}{rcl}
\langle u |M|v\rangle &\to & \rho_0\;\rho_u\;M\;\rho_v\;\rho_0\\
&=&|0\rangle\langle 0||u\rangle\langle u|M|v\rangle\langle v|
|0\rangle\langle 0|\\
&=& \langle 0|u\rangle \langle u|M|v\rangle \langle v|0\rangle
|0\rangle\langle 0|,\\
&=& \langle 0|u\rangle \langle u|M|v\rangle \langle v|0\rangle
\;\rho_0.
\end{array}
\end{equation}
which is seen to be a complex multiple of $\rho_0$.  The result is
that this definition differs from the usual only in the
normalization. From the above, we see that the
normalization can be fixed by dividing bras by $\langle 0|u\rangle$
and dividing kets by $\langle v|0\rangle$. From this normalization
it is clear why it is that this method only works when $|0\rangle$
is not antiparallel with any of the bras and kets which are being
converted.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:PauliSoverCBE})
\section{\label{sec:NSB}  Naughty Spinor Behavior}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


For the Pauli algebra, we convert from density operator formalism to
state vector formalism by choosing a vacuum, a
constant pure density operator $\rho_0$. An arbitrary pure Pauli
density operator has, at any given point in space-time, a spin
orientation. This gives us a geometric interpretation of the source
of the arbitrary complex phase seen in the state vector formalism.
We can apply this interpretation to explain the odd behavior of
spinors from a density operator point of view.

It\marginpar{\em Spinors are naughty.} is well known that when a
spinor is rotated by $2\pi$ it does not return to its original value
but instead is multiplied by $-1$. Since density operators do not
have an arbitrary complex phase, this behavior cannot happen with
the density operators.

The operator that rotates a spinor by an angle $\lambda$ around a
rotation axis defined by the vector $\vec{u}$ is simply:
\begin{equation}\label{eq:RotKetByLambda}
U(\lambda) = e^{i\lambda\vec{u}\cdot\vec{\sigma}/2} = e^{i\lambda
\sigma_u/2}
\end{equation}
Let $|v\rangle$ be an arbitrary ket.  We can write:
\begin{equation}\label{eq:KetSplit}
|v\rangle = (1 + \sigma_u)/2\;|v\rangle + (1 -
\sigma_u)/2\;|v\rangle.
\end{equation}
Applying the rotation operator to this gives:
\begin{equation}\label{eq:KetSplitRotated}
\begin{array}{rcl}
U(\lambda)\;|v\rangle &=& U(\lambda)(1+\sigma_u)/2\;|v\rangle +
U(\lambda)(1-\sigma+u)/2\;|v\rangle\\
&=&e^{+i\lambda/2}(1+\sigma_u)/2\;|v\rangle +
e^{-i\lambda/2}(1-\sigma_u)/2\;|v\rangle,
\end{array}
\end{equation}
where we have taken advantage of the fact that $(1+\sigma_u)/2$ is
both a projection operator and an eigenvector of $\sigma_u$.
Putting $\lambda = 2\pi$ gives
\begin{equation}\label{eq:Rotatebycircule}
\begin{array}{rcl}
U(2\pi)|v\rangle &=& -(1+\sigma_u)/2\;|v\rangle -
(1-\sigma_u)/2\;|v\rangle,\\
&=&-|v\rangle.
\end{array}
\end{equation}
The above showed how one rotates a ket.  To rotate a
bra, one puts the rotation operator on the other side,
and because of the complex conjugate, the spin operator takes a
negative angle, $U(-\lambda)$


If\marginpar{\em Density operators are well behaved.} we replace
$|v\rangle$ with a density operator, or any other operator, the same
mathematics would apply. What is different about density operators
is how they are rotated. For a density operator, the rotation
operator must be applied to both sides of the density operator. This
gives two factors of $-1$. Thus a density operator is unmodified
when rotated through $2\pi$ using the rotation
operators.


Applying the rotation operator to a spinor made from density
operators, we see what the source of the factor of $-1$ is.  When a
spinor made from density operators is to be rotated by spinors, one
must include an extra rotation operator.  Done the density operator
way we have:
\begin{equation}\label{eq:RotateRealSpinor}
\begin{array}{cl}
&U(\lambda)|u\rangle\langle u|U(-\lambda)\;U(\lambda) |
0\rangle\langle0|U(-\lambda)\\
=&U(\lambda)|u\rangle\langle u| 0\rangle\langle0|U(-\lambda).
\end{array}
\end{equation}
Putting $\lambda=2\pi$ leaves the state unchanged, consistent with
the fact that density operators are unchanged by
rotations of $2\pi$. Thus, from the density operator
point of view, the $-1$ that a ket takes on rotation by $2\pi$ is a
consequence of failing to rotate the vacuum
bra, $\langle 0|$.


In\marginpar{\em Density operators are simple in normalization.} the
state vector formulation of QM, there is a conflict between
normalization and
linearity.\footnote{In contrast to classical E\&M,
quantum mechanics, even in the usual state vector formulation, is
not physically linear. Three times a state vector is a state vector
that corresponds to the same physical situation (with the
normalization changed), not a physical situation with three times as
many particles or particles that are three times
stronger.} The linear combination of two normalized
state vectors is generally not a normalized state vector.  If, on
the other hand, we associate the states with the rays
then we retain a sort of linearity, but our formula for
probabilities becomes more complicated as we must
normalize. Since density operators are
essentially non linear, there is no temptation to
sacrifice uniqueness for linear superposition.






% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:PauliSoverCBE})
\section{\label{sec:LinSup}  Linear Superposition} %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

As we saw in the previous section, the lack of arbitrary
complex phase makes density operators a natural way
of representing quantum states.  On the other hand, an advantage of
spinors is that they allow linear superposition.
That is, given two spinors $|A\ra$ and $|B\ra$, and
two complex numbers, $a$ and $b$, we can define the linear
superposition:
\begin{equation}\label{eq:LinearSuperposition}
|aA+bB\ra = a\;|A\ra + b\;|B\ra.
\end{equation}

For\marginpar{\em Linear superposition requires a choice of complex
phase.} any two given spinors, for example, $|A\ra$ and $|B\ra$, the
linear superposition is well defined.  But it is not stressed to
those learning physics that the linear superposition is not well
defined for the quantum states $A$ and $B$.  That is, to define
$|aA+bB\ra$, we must first choose kets to represent $A$ and $B$. And
since the choice of ket is arbitrary up to a complex phase, the
linear superposition is also arbitrary.


If we do not require that the kets be normalized the arbitrariness
of linear superposition becomes extreme.
For example, let $A$ be $+1/2$ spin in the $+z$ direction, and let
$B$ be $+1/2$ spin in the $-z$ direction.  For the ket
representing $A$ and $B$, let $u$ and $v$ be arbitrary non zero real
numbers. Then we can choose:
\begin{equation}\label{eq:ArbitraryKets}
\begin{array}{rclcrcl}
|+z\ra &=&\left(\begin{array}{c}u/a\\  0\end{array}\right)&,& %%
|-z\ra &=&\left(\begin{array}{c}  0\\v/b\end{array}\right),
\end{array}
\end{equation}
and the linear superposition gives almost any quantum state:
\begin{equation}\label{eq:ArbKetLS}
a\;|+z\ra + b\;|-z\ra = %%
\left(\begin{array}{c} u\\v\end{array}\right).
\end{equation}
This is the problem of linear superposition for spinors.


The\marginpar{\em Quantum states are a part of physics, spinors are
only mathematics.} world is presumably composed of quantum states
rather than spinors, so it we would like to have a method of
defining linear superposition on quantum
states rather than on spinors. But the above
demonstrates that in making such a definition we must make some sort
of choice. For spinors, the choice consists of the arbitrary complex
phases of the two (or more) spinors which we wish to use, a rather
inelegant definition.

In Eq.~(\ref{eq:KetsFromDMs}) we defined kets from the
density operator formalism by choosing a ``vacuum''
state and multiplying on the right by this state.  We can therefore
define linear superposition by choosing
a vacuum state, using it to convert the (unique)
density operators to bras and kets:
\begin{equation}\label{eq:LinSupVacuum}
\begin{array}{rcl}
|aA + bB\ra &=& a\;\rho_A\;\rho_0 + b\;\rho_B\;\rho_0,\\
\la aA + bB|a &=& a\;\rho_0\;\rho_A + b\;\rho_0\;\rho_B.
\end{array}
\end{equation}
And\marginpar{\em Linear superposition for density operators
requires a choice of vacuum state.} we can now multiply our ket by
our bra to get what we will show to be a complex multiple of a pure
density operator corresponding to the linear
superposition:
\begin{equation}\label{eq:LinSupDM}
\begin{array}{rcl}
k\;\rho_{aA+bB,0} &=& (a\;\rho_A\;\rho_0 + b\;\rho_B\;\rho_0,) %%
(a\;\rho_0\;\rho_A + b\;\rho_0\;\rho_B)\\
&=&(a\;\rho_A\ + b\;\rho_B,)\;\rho_0\; %
(a\;\rho_A + b\;\rho_B),
\end{array}
\end{equation}
where $k$ is a complex constant (and may be zero).  Note that
choosing $a=1, b=0$ or $a=0, b=1$ gives a result that is
proportional to $\rho_A$ or $\rho_B$, respectively, as in the usual
linear superposition.


Let $X$ and $Y$ be arbitrary operators, not necessarily pure states,
and $\rho_0 = |0\ra \la 0|$ be a pure density operator.  We now show
that the product $X\;\rho_0\;Y$ is a complex multiple of a pure
density operator.  Compute the square:
\begin{equation}\label{eq:XOXproof}
\begin{array}{rcl}
(X\;\rho_0\;Y)^2 &=& X\;\rho_0\;Y\;X\;\rho_0\;Y,\\ %
&=& X\;|0\ra   \la 0| \;Y\;X\;|0\ra   \la 0| \;Y,\\   %
&=& X\;|0\ra \;\;\;( \la 0| \;Y\;X\;|0\ra )\;\;\; \la 0| \;Y.
\end{array}
\end{equation}
The quantity in parentheses in the above is a complex
number and so can be factored out of the operator product to give:
\begin{equation}\label{eq:XOX2Proof}
\begin{array}{rcl}
&=&  \la 0| \;Y\;X\;|0\ra  \; (X\; |0\ra \la 0| \;Y), \\  %
&=&  \la 0| \;Y\;X\;|0\ra  \; (X\;    \rho_0    \;Y).
\end{array}
\end{equation}
And since $X\;\rho_0\;Y$ squares to a complex multiple of itself, it
is therefore an idempotent multiplied by that complex number. It
remains to show that $X\;\rho_0\;Y$ is a primitive idempotent.  To
do this, compute the trace:
\begin{equation}\label{eq:XOXtrace}
\begin{array}{rcl}
\tr(X\;\rho_0\;Y) = \tr(X\;|0\ra\la 0|\;Y) = \la 0|\; Y\; X\;|0\ra.
\end{array}
\end{equation}
Since the trace is precisely the complex multiple of
Eq.~(\ref{eq:XOX2Proof}), $X\;\rho_0\;Y$ divided by this multiple is
a pure density operator.







\chapter{\label{sec:Geometry} Geometry}

%\begin{quote}
\settowidth{\versewidth}{There's only one Corps which is perfect --
that's us;}
\begin{verse}[\versewidth]
%\em
I have stated it plain, an' my argument's thus, \\
\vin (``It's all one,'' says the Sapper),\\
There's only one Corps which is perfect -- that's us;\\
\vin An' they call us Her Majesty's Engineers,\\
\vin Her Majesty's Royal Engineers,\\
\vin With the rank and pay of a Sapper!
\end{verse}
%\end{quote}
$ $


\lettrine{T}{$\;$he program of} contemporary physics is to produce a
unified description of nature by looking for symmetries between the
forces and particles.  Where the forces are not symmetric,
similarities are looked for and ``spontaneous symmetry breaking'' is
assumed.  While this plan has been successful in making great
progress, the forward movement of that progress has been stalled for
some years.  The primary difficulty appear to be that symmetry
principles allow too many different possibilities, and their
application leaves too many arbitrary parameters that must be
supplied by experiment.


This book will break with tradition and instead assume that geometry
is at the foundation of physics.  Our wave functions will be written
in terms of scalars, vectors, pseudo vectors and pseudo scalars.
These correspond to the traditional objects of geometry known to the
ancients, points, lines, planes and volumes.  The reader can suppose
the correspondence between the traditional geometric objects and the
geometry we will use here correspond to stresses induced by
particles in the fabric of space-time, but this is not necessary,
and we will not enlarge on the idea.


Traditional quantum mechanics is written with complex numbers. These
are used in very particular ways in the standard theory.  In this
chapter we take advantage of these peculiarities and show that we
can replace them with a geometric theory.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:PauliSoverCBE})
\section{\label{sec:ComplexNumbers}  Complex Numbers} %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $\rho_0$ be any pure density operator, and let
$M$ be any operator. Then the product
\begin{equation}\label{eq:ProductsOfPauliOne}
\rho_0\; M\; \rho_0
\end{equation}
is a complex multiple of $\rho_0$ as can be seen by
replacing $\rho_0$ with the its spinor representation, $|0\ra\la
0|$.  For example, choosing $0$ to be spin$+1/2$ in the $+z$
direction, we have:
\begin{equation}\label{eq:ExampleVev}
\begin{array}{rcl}
\rho_0 M \rho_0 &=&
\left(\begin{array}{cc}1&0\\0&0\end{array}\right) %
\left(\begin{array}{cc}M_{11}&M_{12}\\M_{21}&M_{22}\end{array}\right)
\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\
&=&M_{11}\left(\begin{array}{cc}1&0\\0&0\end{array}\right).
\end{array}
\end{equation}
Any\marginpar{\em In the density matrix formalism, the complex
numbers are a subset of the operators, and which subset depends on
the choice of vacuum.} product of operators that
begins and ends with the same pure density operator thus provides a
version of the complex numbers. But it needs to be
stressed that the form of these complex numbers depend on the
choice of the vacuum operator.


Since our complex numbers depend on the choice of vacuum, we cannot
follow the usual assumption of the spinor formalism
which interprets the complex number $a + ib$ as a complex multiple
of the unit matrix.  That is, we will distinguish the complex
numbers from the operators:\marginpar{\em Numbers, real or complex,
are not a sort of operator.}
\begin{equation}\label{eq:IisNotI}
a+ib \neq \left(\begin{array}{cc}a+ib&0\\0&a+ib\end{array}\right).
\end{equation}
For us, the complex numbers are only a mathematical
convenience used for calculational purposes.  Our
complex numbers will arise from products that begin
and end with the same pure density operator, and when we refer to
them as complex numbers it is only for the
convenience of not having to haul around the pure
density operator that defines them.  A logical consequence is that
we should distinguish between the unity ``$1$'' of the complex
numbers and the Pauli algebra.  We will do this when convenient, but
the old habits are hard to break.


Given this use of the complex numbers, we need to clear up the
interpretation of the use of complex numbers in the
definitions of the Pauli matrices. First, let us note that there are
real representations of the Pauli algebra, for example:
\begin{equation}\label{eq:RealPauliRep}
\left(\begin{array}{ccc}0&&\\&+1&\\&&-1\end{array}\right),
\left(\begin{array}{ccc}-1&&\\&0&\\&&+1\end{array}\right),
\left(\begin{array}{ccc}+1&&\\&-1&\\&&0\end{array}\right).
\end{equation}
Second, the imaginary unit of the Pauli
matrices can be obtained from the product of the three Pauli
matrices:
\begin{equation}\label{eq:PauliXYZ}
\sigma_x\sigma_y\sigma_z =
\left(\begin{array}{cc}i&0\\0&i\end{array}\right),
\end{equation}
and so avoid the use of complex multiples of
operators. As an example of our use of complex
numbers, consider the commutation relations that
define the Pauli algebra.  Rather than writing $[\sigma_x,\sigma_y]
= 2i\sigma_z$, we will instead write:
\begin{equation}\label{eq:PauliCommutationRelations}
[\sigma_x,\sigma_y] = 2(\sigma_x\sigma_y\sigma_z)\; \sigma_z =
2\sigma_x\sigma_y,
\end{equation}
where the second equality comes from the fact that $\sigma_z\sigma_z
= 1$.


After\marginpar{\em The complex commutation relations of the Pauli
algebra are replaced by a real Clifford algebra.} replacing the
imaginary unit in the commutation relations, we
can break apart the commutator and turn the equations into somewhat
simpler anticommutation relations.  Along with the fact that the
squares of the $\sigma_\chi$ square to unity, this gives a
definition of the Pauli algebra that avoids complex operators:
\begin{equation}\label{eq:PauliAntiCommutation}
\begin{array}{c}
\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = 1 \\
\sigma_x\sigma_y = -\sigma_y\sigma_x \\
\sigma_y\sigma_z = -\sigma_z\sigma_y \\
\sigma_z\sigma_x = -\sigma_x\sigma_z.
\end{array}
\end{equation}
These equations form the definition of a Clifford
algebra. More complicated Clifford
algebras will be the subject of later chapters and will be explained
then.


Using the equations of Eq.~(\ref{eq:PauliAntiCommutation}), any
product of Pauli algebra elements can be reduced in length to $\pm$
a product of at most three different Pauli algebra elements. For
example:
\begin{equation}\label{eq:PauliAlgebraProductReduction}
\begin{array}{rcl}
\sigma_x\sigma_y\sigma_y\sigma_y\sigma_x\sigma_z\sigma_x &=&
\sigma_x(\sigma_y\sigma_y)\sigma_y\sigma_x\sigma_z\sigma_x =
\sigma_x\hat{1}\sigma_y\sigma_x\sigma_z\sigma_x\\
&=&\sigma_x(\sigma_y\sigma_x)\sigma_z\sigma_x = -
\sigma_x(\sigma_x\sigma_y)\sigma_z\sigma_x\\
&=&-(\sigma_x\sigma_x)\sigma_y\sigma_z\sigma_x =
-\sigma_y\sigma_z\sigma_x.
\end{array}
\end{equation}
Furthermore, these products can be rearranged so that $\sigma_x$,
$\sigma_y$ and $\sigma_z$ appear in that order.  For example:
\begin{equation}\label{eq:PauliAlgebraOrdering}
\begin{array}{rcl}
\sigma_z\sigma_y\sigma_x &=& \sigma_z(\sigma_y\sigma_x) =
-\sigma_z(\sigma_x\sigma_y)\\
&=&-(\sigma_z\sigma_x)\sigma_y = +(\sigma_x\sigma_z)\sigma_y\\
&=&\sigma_x(\sigma_z\sigma_y) = -\sigma_x(\sigma_y\sigma_z)\\
&=&-\sigma_x\sigma_y\sigma_z.
\end{array}
\end{equation}
There are eight possible end results for these products, they are
the ``Pauli unit multivectors'':\marginpar{\em Pauli unit multivectors}
\begin{equation}\label{eq:PauliMVs}
\begin{array}{ccccccc}
\hat{1}, &\;\;& \sigma_y, &\;\;&  \sigma_x\sigma_y, &\;\;& \sigma_x\sigma_z,  \\
\sigma_x, && \sigma_z&&\sigma_y\sigma_z,
&\;\;&\sigma_x\sigma_y\sigma_z.
\end{array}
\end{equation}


When operators are represented by $2\times 2$
complex matrices, there are four complex
numbers and therefore four complex degrees of freedom.  Since we are
avoiding the use of complex numbers as operators, it makes sense to
think of the $2\times 2$ complex matrices as having
eight real degrees of freedom. Those degrees of freedom are given by
the Pauli unit multivectors. But the
Pauli unit multivectors are written in geometric form, that is, they
are written in terms of $x$, $y$ and $z$.  Thus the Pauli unit
multivectors give us a geometric interpretation
\marginpar{\em Operators always have a geometric
interpretation.} of the $2\times 2$ complex matrices. For example,
\begin{equation}\label{eq:TwoByTwoToMVs}
\begin{array}{rcl}
\left(\begin{array}{cc}1&0\\0&0\end{array}\right) %
&\equiv& (\hat{1}+\sigma_z)/2\\
\left(\begin{array}{cc}i&0\\0&0\end{array}\right) %
&\equiv& (\sigma_x\sigma_y\sigma_z+\sigma_x\sigma_y)/2\\
\left(\begin{array}{cc}0&1\\0&0\end{array}\right) %
&\equiv& (\sigma_x-\sigma_x\sigma_z)/2
\end{array}
\end{equation}
Since any product of Pauli matrices is $\pm$ one of the Pauli unit
multivectors, the Pauli unit multivectors are sufficient to
represent any operator that is made from Pauli matrices.  That is,
real linear combinations of the Pauli unit multivectors are closed
under multiplication.


The\marginpar{\em The Pauli blades are: scalars, vectors,
pseudovector and pseudoscalar.} eight Pauli unit
multivectors can be divided into
subsets in several useful ways.  Clearly $\hat{1}$ is a scalar.
Following the tradition from Clifford algebra, $\sigma_x$,
$\sigma_y$, and $\sigma_z$ are called vectors, $\sigma_y\sigma_z$,
$\sigma_z\sigma_x$ and $\sigma_x\sigma_y$ are
``pseudovector'', and $\sigma_x\sigma_y\sigma_z$
is a ``pseudoscalar''.  These are the four
``blades'' that we mention here only for completeness.
Blade value is conserved under addition but is not conserved under
multiplication.


The\marginpar{\em Signatures are always +1 or -1} Pauli unit
multivectors give either $+1$ or
$-1$ when squared. This is called the signature, and we can divide
the eight elements according to signature:
\begin{equation}\label{eq:DiracMVsignatures}
\begin{array}{rcrcrcrcl}
\hat{1}^2 &=& \sigma_x^2 &=& \sigma_y^2 &=& \sigma_z^2 &=& +1\\
(\sigma_x\sigma_y\sigma_z)^2 &=& (\sigma_y\sigma_z)^2 &=&
(\sigma_z\sigma_x)^2 &=& (\sigma_x\sigma_y)^2 &=& -1.
\end{array}
\end{equation}
In the Pauli algebra, the signature can be thought of as the
presence or absence of the imaginary unit
$\sigma_x\sigma_y\sigma_z$.  In later chapters we will be working
with more complicated Clifford algebras whose elements have more
interesting signatures.  It turns out that the idempotents of a
Clifford algebra are defined by the elements that square to $+1$.
Signature is conserved under addition, but not conserved under
multiplication.


Finally,\marginpar{\em Orientations are $n$, $x$, $y$, and $z$.} the
Pauli unit multivectors can be
organized according to orientation.  Three of the
four orientations are the $x$, $y$ and $z$ directions. The fourth is
$n$, the neutral direction. Orientation is preserved under addition
and under multiplication follows these rules:
\begin{equation}\label{eq:Orientation}
\begin{array}{ccc|c|cccc|}
&&& \times & n & x & y & z \\
\hline
\hat{1}&\sigma_x\sigma_y\sigma_z &\in& n &n &x &y &z\\
\hat{x}&\sigma_y\sigma_z&\in& x&x&n&z&y\\
\hat{y}&\sigma_z\sigma_x&\in& y&y&z&n&x\\
\hat{z}&\sigma_x\sigma_y&\in& z&z&y&x&n\\
\hline
\end{array}
\end{equation}
As the above table shows, under multiplication,
orientation forms a finite group.  For the Pauli
algebra, the neutral orientation is equivalent to half of the
trace. That is, if we use $n(M)$ to denote the
extraction of the neutral portion of an operator,
\begin{equation}
\begin{array}{rcl}
\tr(M) &=& \tr(M_1\hat{1} + M_x\sigma_x + M_y\sigma_y +
M_z\sigma_z\\
&&+ M_{i}\sigma_x\sigma_y\sigma_z + M_{ix}\sigma_y\sigma_z +
M_{iy}\sigma_z\sigma_x + M_{iz}\sigma_x\sigma_y)\\
&=& \tr\left(\begin{array}{cc} %
M_1+M_z+iM_i+iM_{iz}&M_x-iM_y+iM_{ix}+M_{iy}\\
M_x+iM_y+iM_{ix}-M_{iy}&M_1-M_z+iM_i-iM_{iz}\end{array}\right)\\
&=& 2(M_1 + iM_i)\\
&=& 2\;n(M).
\end{array}
\end{equation}
We have some use of orientation in this chapter, and later on it
will be useful in classifying the primitive idempotents of more
complicated Clifford algebras.


The three Pauli matrices $\sigma_x$, $\sigma_y$, and $\sigma_z$ are
associated with the three dimensions $x$, $y$ and $z$.  Special
relativity requires a time dimension, $t$, and to model this
spacetime requires that we add a sort of $\sigma_t$ to the three
Pauli algebras.  Each time one adds a new dimension to a Clifford
algebra, the number of unit multivectors is multiplied by two, so
this will give us $16$ unit multivectors.
We will cover this algebra, the ``Dirac algebra''
and its ``Dirac unit multivectors''
in the next chapter and later chapters will add
one more (hidden) dimension.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:OperatorProbs})
\section{\label{sec:ExpectValues}  Expectation Values} %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

There are at least three ways of defining expectation values of
operators in the pure density operator formalism.  At the risk of
increasing confusion, we will consider all three in this subsection.


Given an operator $M$ and a state $A$, in the usual bra-ket
notation, one finds the expectation value
of $M$ for the state $A$ by finding a normalized
spinor for $A$, and then computing
\begin{equation}\label{eq:SpinorProbInPauli}
\la M \ra_A = \la A| M |A\ra.
\end{equation}
To\marginpar{\em Vacuum expectation value defined.} convert the
above into an expectation value computed in the density operator
formalism, we replace the spinors with the products of the pure
density operator and the vacuum operator as explained
in Eq.~(\ref{eq:KetsFromVac}).  The result is what we will call the
``vacuum expectation value'':
\begin{equation}\label{eq:OperatorProbs}
\la M \ra_{A,0} = \rho_0\; \rho_A\; M\; \rho_A\; \rho_0,
\end{equation}
which, since it begins and ends with the vacuum operator $\rho_0$,
we can interpret as a complex number. Writing
Eq.~(\ref{eq:OperatorProbs}) in spinor form:
\begin{equation}\label{eq:OperatorSpinorProb}
\begin{array}{rcl}
\la M \ra_{A,0} &=& |0\ra \la 0|A\ra \la A|M|A\ra \la A|0\ra\la 0|,\\
&=& \la 0|A\ra\;\;\la A|0\ra\;\;\la A|M|A\ra\;\;|0\ra\la 0|,
\end{array}
\end{equation}
shows that our definition of expectation value
differs from the usual spinor definition of
Eq.~(\ref{eq:UsualExpValue}) due to the influence of the
vacuum.  Of particular interest is the vacuum
expectation value of the unit operator $\hat{1}$:
\begin{equation}\label{eq:VeVofZero}
\begin{array}{rcl}
\la M \ra_{1,0} &=& \rho_0\; \rho_A\; \hat{1}\; \rho_A\; \rho_0,\\
&=&\rho_0\;\rho_A\;\rho_0\\
&=&\la 0|A\ra\;\;\la A|0\ra\;\;|0\la\ra 0|.
\end{array}
\end{equation}
From this we see that we obtain the usual expectation values by
taking the ratio:
\begin{equation}\label{eq:NoVeVExpValue}
\la M\ra_A = \la M\ra_{A,0} / \la \hat{1}\ra_{A,0},
\end{equation}
where the ratio is to be interpreted in our loose use of complex
numbers.  For example, choosing $\rho_0$ to be spin $+1/2$ in the
$+z$ direction,
\begin{equation}\label{eq:NoVeVExample}
a/b \equiv \left(\begin{array}{cc}a&0\\0&0\end{array}\right) \left/
\left(\begin{array}{cc}b&0\\0&0\end{array}\right)\right. .
\end{equation}


We\marginpar{\em Expectation value without vacuum.} can also
eliminate the vacuum from Eq.~(\ref{eq:OperatorProbs}) and define
the ``expectation value'' as:
\begin{equation}\label{eq:GeneralExpecValues}
\begin{array}{rcl}
\la M \ra_A &=& \rho_A\; M\; \rho_A,\\
&=& |A\ra \la A|M|A\ra\la A|\\
&=&\la A|M|A\ra\;\; |A\ra \la A|,
\end{array}
\end{equation}
which defines the expectation value in terms of the ratio of
$\rho_A\rho_M\rho_A$ to $\rho_A$.  This is the form of expectation
value that we will use most often, but we will use it only in the
context of comparing expectation values for different operators $M$
with respect to the same state $A$.  If we wish to compare
expectation values for two different states, we will have to choose
a vacuum and use the earlier method of Eq.~(\ref{eq:OperatorProbs}).


The third method of defining expectation values is to follow the
spinor tradition and use the trace. For the Pauli algebra, there is
a geometric imaginary unit, $\sigma_x\sigma_y\sigma_z$, which
squares to $-1$ and commutes with all the elements of the algebra.
But not all Clifford algebras have an imaginary unit and for such
algebras defining the trace is more difficult.  Accordingly, we will
avoid the use of the trace.


For the expectation value to be real, we must place the same
restriction on $M$ as in the spinor theory, that is, $M$ must be
Hermitian.  Note that as written, the expectation
value depends on the choice of vacuum. In general, we will be
concerned with operators that are not Hermitian and which therefore
do not have real expectation values. For these operators, we can use
the complex interpretation given above, or alternatively we can
write the complex numbers in terms of
$\sigma_x\sigma_y\sigma_z$.\footnote{In later chapters we will
generalize this appropriately.} For example, let the state be
spin$+1/2$ in the $+z$ direction, and the operator be $M =
3-2\sigma_x\sigma_y+\sigma_x$.  We can compute the expectation value
several ways: In the spinor representation, one converts the
operator into $2\times 2$ complex
matrices by using the Pauli matrices, finds the bra and ket
associated with spin $+1/2$ in the $+z$ direction, and multiplies
them together:
\begin{equation}\label{eq:SpinorExpectation}
\begin{array}{rcl}
\la M\ra_{+z} &=&
\left(\begin{array}{cc}1&0\end{array}\right) %
\left(\begin{array}{cc}3-2i&1\\1&3-2i\end{array}\right) %
\left(\begin{array}{c}1\\0\end{array}\right)\\
&=& 3-2i
\end{array}
\end{equation}
In the density operator formalism, the same calculation can be done
without use of the representation.  The student will find it useful
to work the example out in detail.  It can be done easily if one
uses the anticommutation relations to move the left pure density
operator $\rho_{z+}$ over to the right side.  Doing this will cancel
out some parts of the operator and leave other parts unchanged:
\begin{equation}\label{eq:DOExpectation}
\begin{array}{rcl}
\la M\ra_{+z} &=& (1+\sigma_z)/2\; %
(3 - 2\sigma_x\sigma_y+\sigma_x)\; (1+\sigma_z)/2\\
&=& (3 - 2\sigma_x\sigma_y+\sigma_x)\; (1+\sigma_z)/4\\
&&+ \sigma_z(3 - 2\sigma_x\sigma_y+\sigma_x)\; (1+\sigma_z)/4\\
&=& (3 - 2\sigma_x\sigma_y+\sigma_x)\; (1+\sigma_z)/4\\
&&+ (3 - 2\sigma_x\sigma_y-\sigma_x)\;\sigma_z\;(1+\sigma_z)/4\\
&=& (3 - 2\sigma_x\sigma_y+\sigma_x)\; (1+\sigma_z)/4\\
&&+ (3 - 2\sigma_x\sigma_y-\sigma_x)\; (1+\sigma_z)/4\\
&=& (3 - 2\sigma_x\sigma_y)\;(1+\sigma_z)/2,\\
&=& (3 - 2\sigma_x\sigma_y\sigma_z)\;(1+\sigma_z)/2.
\end{array}
\end{equation}
The last equality was obtained by noting that $(1+\sigma_z)/2 =
\sigma_z(1+\sigma_z)/2$, that is, $\sigma_z$ is an eigenvector of
$\rho_{z+}$ with eigenvalue $1$ so we can introduce factors of it;
that is, $\sigma_z(1+\sigma_z) = (1+\sigma_z)$.


In the above calculation, the $3$ and $2\sigma_x\sigma_y\sigma_z$
components contributed to the expectation value
while the $\sigma_x$ component does not.  Let us write $M_\chi$ as
the operator that is any one of the eight degrees of
freedom of the Pauli algebra.  That is,
$M_1 = \hat{1}$, $M_x = \sigma_x$, ..., $M_{iz} = \sigma_x\sigma_y$,
$M_i = \sigma_x\sigma_y\sigma_z$. In computing $\la M_\chi\ra$, we
should note that for any choice of $\chi$, $M_\chi$ will either
commute or anticommute with $\rho_{+z}$.  And of course $M_\chi$
commutes with $\hat{1}$.  This allows us to compute the expectation
value easily. If $M_\chi$ anticommutes with
$\sigma_z$, then the expectation is zero:
\begin{equation}\label{eq:AntiCommExample}
\begin{array}{rcl}
\la M \ra_{+z} &=& (0.5(1+\sigma_z) M_\chi) 0.5(1+\sigma_z)\\
&=& (M_\chi 0.5(1-\sigma_z)) 0.5(1+\sigma_z)\\
&=& 0
\end{array}
\end{equation}
because $(1+\sigma_z)$ and $(1-\sigma_z)$ annihilate
each other. On the other hand, if $M_\chi$
commutes, then the expectation is nonzero:
\begin{equation}\label{eq:AntiCommExpectation}
\begin{array}{rcl}
\la M \ra_{+z} &=& (0.5(1+\sigma_z) M_\chi) 0.5(1+\sigma_z)\\
&=& (M_\chi 0.5(1+\sigma_z)) 0.5(1+\sigma_z)\\
&=& M_\chi 0.5(1+\sigma_z),
\end{array}
\end{equation}
because $0.5(1+\sigma_z)$ is idempotent.


This method of calculation is very useful and it is worth describing
again why it works. If an idempotent is written with unit
multivectors, one can always factor a unit multivector
to the other side of the idempotent. This will either
leave the idempotent unaltered, or it will change the idempotent to
a different idempotent that will annihilate the
original idempotent.  This behavior will be repeated when we later
study the Dirac and more complicated Clifford
algebras.\marginpar{\em To calculate with primitive idempotents,
factor unit multivectors around the primitive idempotents, which
sometimes changes the primitive idempotents.}


Given the idempotent $0.5(1+\sigma_z)$, the Pauli unit
multivectors that commute with it, and
therefore give a nonzero expectation value, are $\hat{1}$,
$\sigma_z$, $\sigma_x\sigma_y$ and $\sigma_x\sigma_y\sigma_z$.  The
other four Pauli unit multivectors will
convert $\rho_{+z}$ to $\rho_{-z}$ which will
annihilate with the other $\rho_{+z}$ and
therefore give an expectation value of zero. Returning to the
classification of the degrees of freedom of the Pauli unit
multivectors according to orientation, we see
that the neutral and $z$ oriented unit multivectors give nonzero
expectation values while the $x$ and $y$ oriented
unit multivectors give zero expectation values. Thus half the
degrees of freedom give a zero expectation
value and the other half gives a nonzero expectation. When we later
study more complicated Clifford algebras we will find somewhat more
complicated behavior; in addition to different orientations giving
zero expectation, there is also the possibility of changing the
``internal'' quantum states which
will then annihilate.


A\marginpar{\em Hermitian Pauli operators include scalars and
vectors, but not pseudoscalar and pseudovector.}
Hermitian operator can be written as a sum of real
multiples of the scalar and vector blades of the Pauli
algebra. That is, the general Hermitian operator $H$ can be written
as:
\begin{equation}\label{eq:HermitianOps}
H= H_1\hat{1} + H_x\sigma_x + H_y\sigma_y + H_z\sigma_z,
\end{equation}
where $H_\chi$ are real numbers.  An
anti-Hermitian operator can be written
similarly as a sum of real multiples of
psuedovectors and the
psuedoscalar blades.


In the Pauli algebra, the pure density operators are of the form:
\begin{equation}\label{eq:PureDensOp}
\rho_u = (1+\sigma_u)/2 = (1+u_x\sigma_x + u_y\sigma_y +
u_z\sigma_z)/2,
\end{equation}
where $(u_x,u_y,u_z)$ is a unit vector, and are therefore
Hermitian.  Consequently, when we compute the
expectation value of a pure density operator,
\begin{equation}\label{eq:PureDensExpVal}
\la \rho_u \ra_{A} = \rho_A\;\rho_u\;\rho_A,
\end{equation}
the result will be a real number (i.e. a real multiple of $\rho_A$).


When\marginpar{\em Transition probabilities and the probability
postulate.} a system is prepared in a state $A$, and we later
measure it to see if it is in a state $B$, it is a fundamental
postulate of quantum mechanics that the ``transition
probability'' will be given by the
expectation value of $\rho_B$. That is,
\begin{equation}\label{eq:TransProbsAB}
\begin{array}{rcl}
P(A\to B) &=& |\la A| B\ra |^2,\\
&=& \la A| B\ra \la B|A \ra,\\
&=&\la A| \rho_B |A \ra.
\end{array}
\end{equation}
Transition probabilities are a very important type of expectation
value and we will be computing them often in this book.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:PauliSoverCBE})
\section{\label{sec:AmpsAndFDs}  Amplitudes and Feynman Diagram} %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


In this section we will analyze a particularly simple sort of
Feynman diagram, one which begins and ends
with states that can be defined in the Pauli algebra.  While these
are very limited in type, they will see extensive use later in this
book in the context of the origin of mass.  We will begin with the
state $I$ represented by the pure density operator $\rho_I$, and end
with the state $F$ represented by the pure density operator
$\rho_F$.

Feynman devotes a section titled ``Path Integral Formulation of the
Density Matrix'' in his lecture notes
\cite{FeynmanStatisticalMechanics} on statistical mechanics.  We
differ in our presentation in that we are dealing with the internal
degrees of freedom (i.e. spin), while Feynman was dealing with the
external degrees of freedom (i.e. position or momentum).  But the
expansions are otherwise similar.


In the method of Feynman diagrams, one computes the probability of a
transition from an initial state $I$ to a
final state $F$ by first computing the
``amplitude'' of the transition
as a sum over various paths, and then computing the
probability as the square of the absolute value of the amplitude. In
the case of the simple Feynman diagrams we are working with here we
know that the amplitude is simply $\langle
I|F\rangle$ which we will write as:
\begin{equation}\label{eq:AmplitudesDefn}
Amp(I\to F) = \langle I | F \rangle\;\;\;\textrm{(spinor
formalism)}.
\end{equation}
In\marginpar{\em Feynman amplitudes are not to be complex numbers.}
converting this to the density operator formalism, we have several
choices.  We could define geometric complex numbers by computing in
the multiples of the $I$, $F$ or vacuum $0$ primitive idempotents
$\rho_I$, $\rho_F$ or $\rho_0$.  However, Feynman amplitudes require
only the operations of addition and computation of a magnitude, they
do not require complex multiplication.


Let us consider what happens when we insert a new state between the
initial and final states in
the Feynman method. Since the Feynman method requires us to sum over
all possible paths, we must also include the complement
of the state. We can write the interior parts of this calculation in
pure density matrix form. For example, if $A$ is the state with spin
$+1/2$ in the $a$ direction:
\begin{equation}\label{eq:AmpIAF}
\begin{array}{rcl}
Amp(I\to F)  &=& \la I | F\ra,\\
&=& \la I | \hat{1} | F\ra, \\
&=&\la I| 0.5(1 + \sigma_a) |F\ra + %
\la I| 0.5(1 - \sigma_a) |F\ra, \\
&=& \la I| |A\ra \la A| |F\ra + %
\la I| |\bar{A}\ra \la \bar{A}| |F\ra,\\
&=& \la I| |A\ra \la A| |F\ra + %
\la I| |\bar{A}\ra \la \bar{A}| |F\ra,\\
&=& \la I| \rho_A |F\ra + \la I| \bar{\rho_A} |F\ra,
\end{array}
\end{equation}
where we have introduced the notation $\bar{\rho_A}$ to indicate the
complement of the pure density operator $\rho_A$.


This idea can be continued to give a sequence of pure density
operators between the initial bra and final ket.  The various
intermediate states are all treated as pure density operators, only
the initial and final states
are in ket form. This suggests that the natural way of writing an
amplitude in pure density operator form is to
convert the initial and final objects to look the same as the
intermediate ones. In this manner, again
inserting the state $A$ and its complement between
initial and final state we have:\marginpar{\em Definition of
amplitude in density operator formalism.}
\begin{equation}\label{eq:AmpIAinPDOform}
\begin{array}{rcl}
Amp(I\to F) &=& \rho_I\;\rho_F,\\
&=&\rho_I\;\rho_A\;\rho_F + \rho_I\;\bar{\rho_A}\;\rho_F.
\end{array}
\end{equation}
Using this formalism will allow us to connect Feynman
diagrams together without having to distinguish between
internal and external lines.  But to
justify this, we need to show that we can add these sorts of
objects, and that we can compute squared magnitudes
of them.


For definitiveness, let $I$ be spin $+1/2$ in the $+z$ direction,
and let $F$ be spin $+1/2$ in the $+x$ direction.  With Pauli spin
matrices, the various objects are:
\begin{equation}\label{eq:FeynmanAmpExample}
\begin{array}{rcl}
\rho_I &=& \left(\begin{array}{cc}1&0\\0&0\end{array}\right),\\
\rho_F &=& \left(\begin{array}{cc}0.5&0.5\\0.5&0.5\end{array}\right),\\
\langle I| &=& \left(\begin{array}{cc}1&0\end{array}\right),\\
|F\rangle&=& \sqrt{0.5}\left(\begin{array}{c}1\\1\end{array}\right).
\end{array}
\end{equation}
Suppose that we are considering a particular sequence of states that
happens to reduce to the matrix $M = M_{ij}$.  For the spinor
method, we get an amplitude of:
\begin{equation}\label{eq:AmpIMFbySpinors}
\begin{array}{rcl}
Amp(I \to M \to F) &=& \left(\begin{array}{cc}1&0\end{array}\right)
\left(\begin{array}{cc}M_{11}&M_{12}\\M_{21}&M_{22}\end{array}\right)
\sqrt{0.5}\left(\begin{array}{c}1\\1\end{array}\right),\\
&=& \sqrt{0.5}(M_{11} + M_{21}).
\end{array}
\end{equation}
Replacing the spinors with pure density operators gives:
\begin{equation}\label{eq:AmpIMFbyPDOs}
\begin{array}{rcl}
Amp(I \to M \to F) &=&
\left(\begin{array}{cc}1&0\\0&0\end{array}\right)
\left(\begin{array}{cc}M_{11}&M_{12}\\M_{21}&M_{22}\end{array}\right)
\left(\begin{array}{cc}0.5&0.5\\0.5&0.5\end{array}\right),\\
&=&0.5(M_{11}+M_{21})\left(\begin{array}{cc}1&0\\1&0\end{array}\right).
\end{array}
\end{equation}
Instead of getting a complex number, the amplitude
is a more general operator, a product of two primitive
idempotents.\marginpar{\em Amplitudes are, in general, products of
primitive idempotents.} Comparing Eq.~(\ref{eq:AmpIMFbyPDOs}) to
Eq.~(\ref{eq:AmpIMFbySpinors}), we see that summation of
amplitudes will give analogous results.  For
example, if to the above amplitude we add another
amplitude that has an internal
operator of $N_{ij}$, then the two methods will give results that
are proportional to $(M_{11}+M_{21})+(N_{11}+N_{21})$. Therefore, to
get a pure density operator formalism for amplitudes, we need only
define the equivalent of the squared magnitude.


To convert the amplitude into a squared
magnitude, we can use the methods of
Sec.~(\ref{sec:ExpectValues}), that is, work with
vacuum expectation
values and use multiples of $\rho_0$, or work in
expectation values using multiples of either $\rho_I$ or $\rho_F$.
For example, let us compute the squared magnitude for the amplitude of
Eq.~(\ref{eq:AmpIMFbyPDOs}). In the traditional spinor
formalism, the answer is the squared magnitude of
Eq.~(\ref{eq:AmpIMFbySpinors}), that is $0.5|M_{11}+M_{21}|^2$.  We
begin by multiplying the amplitude by its complex conjugate
transpose:
\begin{equation}\label{eq:ExSqMagAmp}
\begin{array}{rcl}
P(I\to F) &=&
0.5(M_{11}+M_{21})\left(\begin{array}{cc}1&0\\1&0\end{array}\right)
0.5(M_{11}^*+M_{21}^*)\left(\begin{array}{cc}1&1\\0&0\end{array}\right)\\
&=&
0.25|M_{11}+M_{21}|^2\left(\begin{array}{cc}1&1\\1&1\end{array}\right)\\
&=&0.5|M_{11}+M_{21}|^2\rho_F.
\end{array}
\end{equation}
If we multiply on the left by the complex conjugate the result is a
multiple of the initial primitive
idempotent:
\begin{equation}\label{eq:ExSqMagAmp2}
\begin{array}{rcl}
P(I\to F) &=&
0.5(M_{11}^*+M_{21}^*)\left(\begin{array}{cc}1&1\\0&0\end{array}\right)\\
0.5(M_{11}+M_{21})\left(\begin{array}{cc}1&0\\1&0\end{array}\right)
&=&
0.25|M_{11}+M_{21}|^2\left(\begin{array}{cc}2&0\\0&0\end{array}\right)\\
&=&0.5|M_{11}+M_{21}|^2\rho_I.
\end{array}
\end{equation}
We can also define the squared magnitude
using an arbitrarily chosen vacuum state.


As before, these methods allow us to compute
amplitudes without the need for specifying a
representation of the Pauli
algebra. In the above, we have used the Pauli
spin matrices because of their familiarity.  Let us now redo the
calculation in geometric form. The amplitude is:
\begin{equation}\label{eq:SqrdMagGeometric}
Amp(I \to M \to F) = 0.5(1+\sigma_z) M 0.5(1+\sigma_x).
\end{equation}
We will compute the squared magnitude in
terms of the initial idempotent,
$0.5(1+\sigma_z)$. Then the squared magnitude is:
\begin{equation}\label{eq:SqrdMag1Geom}
P(I \to M \to F) = 0.5(1+\sigma_z) M 0.5(1+\sigma_x) 0.5(1+\sigma_x)
M (1+\sigma_z).
\end{equation}
To compute this, we need to commute $M$ around the
idempotents. Therefore, we divide $M$ into its
orientations and write $M = M_n+M_x+M_y+M_z$ and
compute:
\begin{equation}\label{eq:SqrdMag2Geom}
\begin{array}{rcl}
P(I \to M \to F) &=& 0.5(1+\sigma_z)(M_n+M_x+M_y+M_z)0.5(1+\sigma_x)\\
&&0.5(1+\sigma_x) (M_n^*+M_x^*+M_y^*+M_z^*) 0.5(1+\sigma_z).
\end{array}
\end{equation}
In our previous analysis of expectation
values, we found that the $x$ and $y$
orientations give zero expectation values.
Therefore, to perform the above computation we can multiply out the
terms between $0.5(1+\sigma_z)$ and group terms according to their
orientations.  Only the ``$n$'' and ``$z$'' orientations survive
giving:
\begin{equation}\label{eq:GutsOfPrevEqn}
\begin{array}{rcl}
&&0.5(M_n+M_z+M_x+M_y)(1+\sigma_x)(M_n^*+M_x^*+M_y^*+M_z^*)\\
&=&0.5(M_nM_n^*+M_xM_x^*+M_yM_y^*+M_zM_z^*)\;\;\textrm{``$n$''}\\
&+&0.5(M_n\sigma_xM_x^*+M_x\sigma_xM_n^*+M_y\sigma_xM_z^*+M_z\sigma_xM_y^*)\;\;\textrm{``$n$''}\\
&+&0.5(M_nM_z^*+M_zM_n^*+M_yM_x^*+M_zM_n^*)\;\;\textrm{``$z$''}\\
&+&0.5(M_n\sigma_xM_y^*+M_x\sigma_xM_z^*+M_y\sigma_xM_n^*+M_z\sigma_xM_x^*)\;\;\textrm{``$z$''}.
\end{array}
\end{equation}
Since $\sigma_z(1+\sigma_z) = (1+\sigma_z)$, we can turn the ``$z$''
terms into ``$n$'' form by bringing in an extra factor of $\sigma_z$
on the right:
\begin{equation}\label{eq:Guts3OfPrevEqn}
\begin{array}{rcl}
&=&0.5(M_nM_n^*+M_xM_x^*+M_yM_y^*+M_zM_z^*)\\
&+&0.5(M_n\sigma_xM_x^*+M_x\sigma_xM_n^*+M_y\sigma_xM_z^*+M_z\sigma_xM_y^*)\\
&+&0.5(M_nM_z^*+M_zM_n^*+M_yM_x^*+M_xM_y^*)\sigma_z\\
&+&0.5(M_n\sigma_xM_y^*+M_x\sigma_xM_z^*+M_y\sigma_xM_n^*+M_z\sigma_xM_x^*)\sigma_z.
\end{array}
\end{equation}
The above is all in the ``$n$'' orientation; the
only degrees of freedom are $\hat{1}$ and
$\sigma_x\sigma_y\sigma_z$. Writing $M_n = m_n$, $M_x =
m_x\sigma_x$, $M_y = m_y\sigma_y$, and $M_z = m_z\sigma_z$ allows us
to simplify.  In computing the result, we need to keep in mind the
anticommutation relations so that, for
instance, $\sigma_x\sigma_z\sigma_y = -\sigma_x\sigma_y\sigma_z$.
The result is:
\begin{equation}\label{eq:Guts4OfPrevEqn}
\begin{array}{cl}
=&0.5(m_nm_n^*+m_xm_x^*+m_ym_y^*+m_zm_z^*+m_nm_x^*+m_xm_n^*-im_ym_z^*\\
&+im_zm_y^*+m_nm_z^*+m_zm_n^*-im_ym_x^*+im_xm_y^*+im_nm_y^*+m_xm_z^*\\
&-im_ym_n^*+m_zm_x^*)\\
=&0.5(m_n+m_x-im_y+m_z)(m_n^*+m_x^*+im_y^*+m_z^*)\\
=&0.5|m_n+m_x-im_y+m_z|^2
\end{array}
\end{equation}
where ``$i$'' indicates multiplication by
$\sigma_x\sigma_y\sigma_z$.


The use of $\sigma_x\sigma_y\sigma_z$ as $i$ in the above
calculation suggests that we should use it in defining complex
multiples of elements of the Pauli algebra. For
example, let $I$ be the initial state and $F$
the final state in a Feynman. We insert the state $A$ between $I$ and
$F$ as follows:
\begin{equation}\label{eq:FAIandFBIex}
\rho_F\rho_A\rho_I,
\end{equation}
and using the methods of this section we can treat these as complex
numbers to the extent that we can add them together and compute
squared magnitudes.  But we can go a little
farther than this, and we can write this object as a complex
multiple of $\rho_F\rho_I$ as follows:
\begin{equation}\label{eq:FAIasComplexProduct}
\rho_F\rho_A\rho_I = %
(a_R + a_I\sigma_x\sigma_y\sigma_z)\rho_F\rho_I,
\end{equation}
where $a_R$ and $a_I$ are real numbers.


This\marginpar{\em Amplitudes are complex numbers times the product
of two primitive idempotents.} is easiest to show using the Pauli
spin matrices. Let us assume that $\rho_F = 0.5(1+\sigma_z)$ and
that $\rho_I = 0.5(1+\cos(\theta)\sigma_z + \sin(\theta)\sigma_x)$
so that the angle between the spin axes of $I$ and $F$ is $\theta$.
Let $M$ be an arbitrary $2\times 2$
matrix. Then
\begin{equation}\label{eq:FMI}
\begin{array}{rcl}
\rho_FM\rho_I \!\!&=&
\left(\begin{array}{cc}1&0\\0&0\end{array}\right)
\left(\begin{array}{cc}m_{11}&m_{12}\\m_{21}&m_{22}\end{array}\right)
0.5\left(\begin{array}{cc}1+\cos(\theta)&\sin(\theta)\\
\sin(\theta)&1-\cos(\theta)\end{array}\right),\\
&=&0.5 \left(\begin{array}{cc}m_{11}&m_{12}\\0&0\end{array}\right)
\left(\begin{array}{cc}1+\cos(\theta)&\sin(\theta)\\
\sin(\theta)&1-\cos(\theta)\end{array}\right),\\
&=&0.5 m_{11}\left(\begin{array}{cc}
\!1\!+\!\cos(\theta)&\sin(\theta)\!\\0&0\end{array}\right)
+0.5m_{12}
\left(\begin{array}{cc}\!\sin(\theta)&1\!-\!\cos(\theta)\!\\0&0
\end{array}\right)\\
&=&0.5(m_{11}+m_{12}(1-\cos(\theta))/\sin(\theta))\left(\begin{array}{cc}
1+\cos(\theta)&\sin(\theta)\\0&0\end{array}\right).
\end{array}
\end{equation}
since $m_{11}$ and $m_{12}$ are
complex, the result is as desired. In the state vector formalism,
amplitudes are also complex numbers, but they
depend on the phases chosen for the spinors.  In
the\marginpar{\em Complex phases of density operator amplitudes are
uniquely defined.} density operator formalism, there is no arbitrary
complex phase and the complex phase of the amplitude defined here is
not arbitrary.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:FAIasComplexProduct})
\section{\label{sec:ProdsDensOps}  Products of Density Operators} %

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The calculation of Feynman  diagram
amplitudes using products of density operators as
discussed in the previous section is the primary subject of this
book.  In this section, we derive some useful methods of reducing
these sorts of calculations.

The first thing to note is that if we wish to expand an
amplitude by inserting states between and
initial and final state,
there is no reason to do this if the initial and final states
annihilate. Such a transition has zero probability
and is not of interest.  So we will assume that the initial and
final states do not annihilate.


In the Pauli algebra, the pure density operator states correspond to
primitive idempotents in various
directions. Given three unit vectors, $\vec{u}$, $\vec{v}$, and
$\vec{w}$, the corresponding pure density operator states are:
\begin{equation}\label{eq:PureDensityOpStatesUVW}
\begin{array}{rcl}
\rho_u &=& 0.5(1 + \sigma_u) = 0.5(1+u_x\sigma_x + u_y\sigma_y + u_z\sigma_z),\\
\rho_v &=& 0.5(1 + \sigma_v),\\
\rho_w &=& 0.5(1 + \sigma_w).
\end{array}
\end{equation}
We will interpret the $\rho_w$ as the initial state,
$\rho_u$ as the final state, and $\rho_v$
as the intermediate state that is inserted
between them in a Feynman amplitude
calculation. Since antiparallel projection
operators annihilate, under this assumption, we
suppose that $\vec{u}$ and $\vec{w}$ are not antiparallel.


In the previous section, we showed that the product of $\rho_u$,
$\rho_v$ and $\rho_w$ can be written as a ``complex'' multiple of
the product of $\rho_u$ and $\rho_w$.  Let us write this complex
number as a magnitude and phase in the
following fashion:
\begin{equation}\label{eq:RuvwSuvwDefn}
\rho_u\rho_v\rho_w = \sqrt{R_{uvw}\exp(iS_{uvw})}\;\rho_u\rho_w,
\end{equation}
where $R_{uvw}$ and $S_{uvw}$ are real functions of the three
vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$.  The factors of two are
included for later convenience. Rewriting the above in
spinor form we have:
\begin{equation}\label{eq:RuvwSpinor}
\begin{array}{rcl}
\rho_u\rho_v\rho_w &=& |u\ra\la u|v\ra\la v|w\ra\la w|,\\
&=&\la u|v\ra\la v|w\ra \;|u\ra\la w|,\\
&=&(\la u|v\ra\la v|w\ra / \la u|w\ra)\;|u\ra\la u|w\ra\la w|,\\
\end{array}
\end{equation}
and therefore
\begin{equation}\label{eq:RuvwSuvwInSpinors}
\sqrt{R_{uvw}\exp(iS_{uvw})} = \la u|v\ra\la v|w\ra / \la u|w\ra.
\end{equation}
In the above, the left hand side is completely defined within the
density matrix formalism and therefore does not depend on choice of
phase.  The right hand side is therefore invariant with
respect to the arbitrary complex phases chosen for the
spinors; while the individual
amplitudes depend on the choice of
phase, the above product does not.


Eq.~(\ref{eq:RuvwSuvwInSpinors}) includes three spinor
amplitudes; a typical one is $\la u|v\ra$.  The
squared magnitude of each of these can be
written as $0.5(1+\cos(\theta_{uv}))$
where $\theta_{uv}$ is the angle between $\vec{u}$ and $\vec{v}$.
Similarly for the other three amplitudes, and this
gives the formula for $R_{uvw}$ as:
\begin{equation}\label{eq:RuvwDefn}
R_{uvw} =
0.5(1+\cos(\theta_{uv}))(1+\cos(\theta_{vw}))/(1+\cos(\theta_{uw})).
\end{equation}
It remains to find the equation for $S_{uvw}$.


The three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ are on the
surface of the unit sphere and define a
spherical triangle, which we will call
$\triangle uvw$. We next show that $S_{uvw}$ is proportional to the
oriented\footnote{Oriented here means that order
matters, for example, $S_{uwv} = -S_{uvw}$.} area of that spherical
triangle. To see this, pick a vector $\vec{x}$ inside the triangle
$\triangle uvw$ that is not opposite to any of the three vectors
(which would result in an inconvenient zero
amplitude). Since $\la w|x\ra \la x|w\ra = |\la
w|x\ra|^2$ is real, we can multiply by it without changing the
phase.  Accordingly, we can write $S_{uvw}$ as:
\begin{equation}\label{eq:Suvwx}
\begin{array}{rcl}
S_{uvw} &=& 2\arg(\la u|v\ra \la v|\ra / \la u|w\ra),\\
&=& \arg((\la u|x\ra \la x|w\ra / \la u|w\ra) %
(\la u|v\ra \la v|x\ra /\la u|x\ra) %
(\la v|w\ra \la w|x\ra / \la v|x\ra)),\\
&=&S_{uxw} + S_{uvx} + S_{vwx},
\end{array}
\end{equation}
where in the last equality we have used that $\arg(\alpha\beta)
=\arg(\alpha) + \arg(\beta)$.  It remains to find the constant of
proportionality. Reversing two points on an oriented
spherical triangle defines a new spherical
triangle that is complementary to the old one.
The area of the triangle and its complement must add to the total
area of the unit sphere, $4\pi$, but the triangle and its complement
must carry opposite phases. That is, since $S_{uvw}$ is defined as
twice the complex phase,
\begin{equation}\label{eq:SuvwConsistency}
\begin{array}{rcl}
0.5\; S_{uvw}  &=& - 0.5\; S_{uwv} + 2n\pi,\\
0.5\; S_{uvw} + 0.5\; S_{uwv} &=& 2n\pi,\\
\textrm{Area}(\triangle uvw)\;\; +\;\; \textrm{Area}(\triangle uwv)
&=& 4\pi,\\
\end{array}
\end{equation}
where $n$ is an integer, and the simplest solution is the one
claimed:
\begin{equation}\label{eq:SuvwDefn}
S_{uvw} = \textrm{Area}(\triangle uvw),
\end{equation}
the verification of which we leave as an exercise for the reader.


In order to solidify this method of reducing products of projection
operators, we will now compute several products using
Eq.~(\ref{eq:RuvwDefn}) and Eq.~(\ref{eq:Suvwx}).  Let $\rho_x$,
$\rho_y$, and $\rho_z$ be spin $+1/2$ in the $x, y$, and $z$
directions.  Then
\begin{equation}\label{eq:rhoxyzdefns}
\rho_x=\left(\begin{array}{cc} 0.5&0.5\\0.5&0.5\end{array}\right), %
\rho_y=\left(\begin{array}{cc} 0.5&-0.5i\\+0.5i&0.5\end{array}\right), %
\rho_z=\left(\begin{array}{cc} 1&0\\0&0\end{array}\right).
\end{equation}
For this example, $\theta_{xy} = \theta_{xz}=\theta_{yz} = \pi/2$ so
the $0.5(1+\cos(\theta))$ factors will
each give $0.5$, and so $R_{xyz} = 0.5(0.5)^2/0.5 = 0.25$.  The
spherical triangle is an octant, and so its area, $4\pi/8 = \pi/2$
is $S_{xyz}$.  Multiplying out the matrices gives:
\begin{equation}\label{eq:rhoxyzprod}
\begin{array}{rcl}
\rho_x\rho_y\rho_z &=& \frac{1+i}{2}\left(\begin{array}{cc}
0.5&0\\0.5&0\end{array}\right),\\
&=& \sqrt{0.5 e^{i\pi/2}}\rho_x\rho_z,
\end{array}
\end{equation}
as claimed.  Redoing the same problem in the geometric notation,
$\rho_x = 0.5(1+\sigma_x)$, etc., gives:
\begin{equation}\label{eq:GeometricRhoxyz}
\begin{array}{rcl}
\rho_x\rho_y\rho_z
&=&(0.5)^3(1+\sigma_x)(1+\sigma_y)(1+\sigma_z)\\
&=&0.125((1+\sigma_x)(1+\sigma_z)+(1+\sigma_x)\sigma_y(1+\sigma_z))\\
&=&0.125(1+\sigma_x)(1+\sigma_z)\\
&&+0.125(1+\sigma_x)\sigma_x\sigma_y\sigma_z(1+\sigma_z))\\
&=&0.5(1+\sigma_x\sigma_y\sigma_z)0.5(1+\sigma_x)0.5(1+\sigma_z)\\
&=&0.5(1+i)\rho_x\rho_z,
\end{array}
\end{equation}
gives the same answer without need to choose a representation.


Products of more than three projection operators can be reduced
using the above formula, provided one avoids a division by zero in
Eq.~(\ref{eq:RuvwDefn}), the definition of $R_{uvw}$.  Earlier we
saw that products of three projection operators that begin and end
with the same projection operator are always real multiples of that
same projection operator.  Products of four projection operators
are, in general, complex:
\begin{equation}\label{eq:Ruvwu}
\begin{array}{rcl}
\rho_u\rho_v\rho_w\rho_u
&=&\sqrt{R_{uvw}\exp(iS_{uvw})}\;\rho_u\rho_w\rho_u,\\
&=&\sqrt{R_{uvw}R_{uwu}\exp(iS_{uvw})\exp(iS_{uwu})}\rho_u\rho_u,\\
&=&\sqrt{R_{uvw}R_{uwu}\exp(iS_{uvw})}\rho_u,\\
\end{array}
\end{equation}
and the phase, $S_{uvw}/2$, is the same as for the product of the
first three projection operators, $\rho_u\rho_v\rho_w$.  In
addition, the product $R_{uvw}R_{uwu}$ simplifies:
\begin{equation}\label{eq:Ruvwuwu}
\begin{array}{rcl}
R_{uvw}R_{uwu} &=& 0.5
\frac{(1+\cos(\theta_{uv}))(1+\cos(\theta_{vw}))}/(1+\cos(\theta_{uw}))\\
&&\times 0.5
\frac{(1+\cos(\theta_{uw}))(1+\cos(\theta_{wu}))}/(1+\cos(\theta_{uu}))\\
&=&
0.5(1+\cos(\theta_{uv}))0.5(1+\cos(\theta_{vw}))0.5(1+\cos(\theta_{wu})),
\end{array}
\end{equation}
to the product of three $0.5(1+\cos(\theta))$
factors, as expected.  This product is
unchanged when $u$, $v$, and $w$ are pairwise swapped.  On the other
hand, since the area is oriented, swapping any two of $u$, $v$, and
$w$ negates $S_{uvw}$:
\begin{equation}\label{eq:Swappinguvw}
S_{uvw}=S_{vwu}=S_{wuv}=-S_{uwv}=-S_{vuw}=-S_{wvu}.
\end{equation}


If we interpret the product $\rho_u\rho_v\rho_w\rho_u$ as a Feynman
amplitude, then physically, the path that the particle takes is to
visit the states in the order $u$, $w$, $v$, and then back to $u$.
When we swap $v$ and $w$, we reverse the sequence of states. As we
saw above, this causes the phase to be negated.  In other words, the
amplitude of the reversed sequence is the complex conjugate of the
amplitude of the forward sequence.  It is left as an exercise for
the reader to show that this is generally true for more complicated
paths.  Taking the continuous limit, we have that paths that begin
and end on the same point take a complex phase equal to half the
area of the region encircled (possibly circled multiple times).


% Fig.~(\ref{fig:AdjoiningTriangles})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,140)
% Vectors
\linethickness{0.4pt}
\qbezier( 45,110)( 45,110)(145, 30)  % A
\qbezier( 45,110)( 45,110)(167,118)  % B
\qbezier( 45,110)( 45,110)(122,100)  % C
\qbezier( 45,110)( 45,110)(110, 24)  % D
\linethickness{1.0pt}
% Great Circles
\qbezier(145, 30)(162, 74)(167,118)  % AB 156+6
\qbezier(167,118)(145,107)(122,100)  % BC 109-2
\qbezier(122,100)(138, 65)(145, 30)  % CA 133+5
\qbezier(145, 30)(120, 25)(110, 24)  % AD 27-2
\qbezier(110, 24)(119, 65)(122,100)  % DC 116+3
% Labels
\put( 40,112){$O$} \put(146, 25){$A$} \put(168,120){$B$}
\put(120,104){$C$} \put(102, 16){$D$}
% Arrows
\put(155, 70){\vector( 1, 4){ 4}} %AB
\put(150,107){\vector(-2,-1){10}} %BC
\put(138, 70){\vector( 1,-3){ 5}} %CA
\put(135, 59){\vector(-1, 3){ 5}} %AC
\put(123, 67){\vector(-1,-4){ 4}} %CD
\put(123, 29){\vector( 4, 1){10}} %DA
\end{picture}
} %
\caption{\label{fig:AdjoiningTriangles} Two adjoining spherical
triangles on the unit sphere. $S_{uvw}$ is additive, that is,
$S_{ABCD} = S_{ABC}+S_{ABD}$, and therefore the complex phase of a
series of projection operators can be determined by the spherical
area they encompass.}
\end{figure}



One last comment on products of four projection operators.  In the
spinor formulation, each of the legs corresponds to a complex
number, and the final amplitude is the product of the three complex
numbers.  As noted before, the actual complex phase for each leg
will depend on the choices of arbitrary complex phases for the
spinors, but the overall product will not depend on these choices.


Within the spinor formalism, it makes sense to attribute a phase
contribution to each leg, that is, $S_{uvw} = S_{uv} + S_{vw} +
S_{wu}$.  This is reminiscent of  Eq.~(\ref{eq:Suvwx}) which gives
$S_{uvw}$ in terms of the three phases $S_{uxw}$, $S_{uvx}$,
$S_{vwx}$.  To make the equivalence more exact, we can choose the
point $x$ to be the vacuum, that is the constant projection operator
$\rho_0$. Each of the three sides of the spherical triangle gives a
contribution to $S_{uvw}$, that is,
\begin{equation}\label{eq:Suvwx2}
S_{uvw} = S_{u0w} + S_{uv0} + S_{vw0}.
\end{equation}
Since $S_{u0w} = -S_{uw0} = +S_{wu0}$, we can bring the vacuum to
the right most position in the indices and the result is a symmetric
form for the complex phase:
\begin{equation}\label{eq:Suvwx3}
S_{uvw} = S_{wu0} + S_{uv0} + S_{vw0}.
\end{equation}
The fact that $S_{uw0} = -S_{wu0}$ suggests that we consider the
legs of the spherical triangle as ordered paths.  That is, they can
be traversed in either of two directions and the complex phase they
contribute will take a sign depending on which direction is taken.
This gives a natural way of interpreting the fact that $S_{uvw}$ is
proportional to the area.  See Fig.~(\ref{fig:AdjoiningTriangles}).
Consider a spherical quadrilateral, $ABCD$.  We can split the region
into two spherical triangles by adding line $AC$.  Letting the two
triangles be traversed in the same direction starting at $A$, the
left $AC$ is traversed in opposite directions by the two triangles.
Therefore the contribution to the complex phase by the leg $AC$ is
canceled and the complex phase of $ABCD$ is the sum of the complex
phases of the two triangles.








% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




\chapter{\label{sec:PrimitiveIdempotents} Primitive Idempotents}


\settowidth{\versewidth}{Then I stripped them, scalp from skull, and
my hunting dogs fed full,}
\begin{verse}[\versewidth]
%\em
Then I stripped them, scalp from skull, and my hunting dogs fed full, \\
And their teeth I threaded neatly on a thong; \\
And I wiped my mouth and said, ``It is well that they are dead, \\
For I know my work is right and theirs was wrong.''\\
$ $\\
But my Totem saw the shame; from his ridgepole shrine he came, \\
And he told me in a vision of the night: -- \\
``There are nine and sixty ways of constructing tribal lays, \\
And every single one of them is right!''
\end{verse}
%\end{quote}
$ $




\lettrine{T}{$\;$his chapter will discuss} primitive
idempotents for Clifford algebras in
more depth than the Pauli algebra can provide.
We will also cover idempotents in general, and ``roots of
unity'', the elements that square to
$\hat{1}$. Of particular interest are groups of
idempotents that commute.  In quantum
mechanics, these correspond to measurements that
can be made in any order, and are therefore
``compatible''. Also of interest are groups of
commuting roots of unity;  these we will use to define the quantum
numbers of states.







% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ObvSolnTwo})
\section{\label{sec:PAPI}  The Pauli Algebra Idempotents}


An ``idempotent'' is an element of an algebra that
is unchanged when squared:
\begin{equation}\label{eq:IdemDefn}
\rho^2 = \rho.
\end{equation}
The above is a simple equation, but it is a
nonlinear equation, and as such it is not
necessarily very easy to solve.


A ``primitive idempotent'' is a nonzero
idempotent which cannot be written as the sum of two nonzero
idempotents. This definition can be confusing to the student.  It
seems circular. Let us apply it to several simple
algebras and see how to understand it on an intuitive and physical
level.


For\marginpar{\em Complex numbers have only one primitive
idempotent: 1.} the complex numbers, the
solutions to Eq.~(\ref{eq:IdemDefn}), that is, the
idempotents, are $0$ and $1$. Of these, only $1$
is nonzero.  The only possible sum of two nonzero idempotents is
$1+1=2$, so $1$ is a primitive idempotent of the complex (or real)
numbers.


For the complex $2x2$ matrices, the
idempotents are obtained by solving:
\begin{equation}\label{eq:Real22Id}
\begin{array}{rcl}
\left(\begin{array}{cc}a&c\\b&d\end{array}\right) &=&
\left(\begin{array}{cc}a&c\\b&d\end{array}\right)^2\\
&=&\left(\begin{array}{cc}a^2+bc&c(a+d)\\b(a+d)&d^2+bc\end{array}\right)
\end{array}
\end{equation}
There are four resulting equations:
\begin{equation}\label{eq:Idem22eqns}
\begin{array}{rcl}
a &=& a^2 + bc, \\
d &=& d^2 + bc, \\
b &=& b(a+d), \\
c &=& c(a+d).
\end{array}
\end{equation}
If $a+d \neq 1$, then, by the last two of the above, $b=c=0$.  The
other two equations then give $a^2=a$ and $d^2=d$.  There are four
solutions of this sort, according as $a$ and $d$ are independently
$0$ or $1$. Two of these have $a+d=1$, the other two are:
\begin{equation}\label{eq:First22SolnSet}
\left(\begin{array}{cc}0&0\\0&0\end{array}\right),
\left(\begin{array}{cc}1&0\\0&1\end{array}\right),
\end{equation}
the $0$ and $\hat{1}$ of the algebra.


The solutions with $a+d=1$, are more interesting -- they are simply
the primitive idempotents of the Pauli algebra -- but they are more
difficult to find. To solve these equations, let us rewrite $a$,
$b$, $c$, and $d$ in geometric form:
\begin{equation}\label{eq:Geometricabcd}
\left(\begin{array}{cc}a&c\\b&d\end{array}\right) =
\left(\begin{array}{cc}a_1+a_z&a_x-ia_y\\a_x+ia_y&a_1-a_z
\end{array}\right),
\end{equation}
where $a_1$, $a_x$, $a_y$, and $a_z$ are complex numbers.  Then $a+d
= (a_1+a_z) + (a_1-a_z) = 2a_1 = 1$, so $a_1 = 0.5$.


In solving $\rho^2=\rho$ for the case $a+d=1$, we can use either
matrix arithmetic and square the right hand side of
Eq.~(\ref{eq:Geometricabcd}), or we can do the calculation directly
in geometric notation:
\begin{equation}\label{eq:SolveIdempotencyCalc}
\begin{array}{rcl}
0.5+a_x\sigma_x+a_y\sigma_y+a_z\sigma_z &=&
(0.5+a_x\sigma_x+a_y\sigma_y+a_z\sigma_z)^2\\
&=&(0.25 + a_x^2 + a_y^2 + a_z^2)\\
&& + a_x\sigma_x + a_y\sigma_y+a_z\sigma_z,
\end{array}
\end{equation}
where the remaining terms cancel by anticommutation.  The solution
is $a_x^2 + a_y^2 + a_z^2 = 0.25$, and factoring $0.5$ out gives the
general solution as:
\begin{equation}\label{eq:SolnIdempotencyPauli}
\rho = 0.5(\hat{1} + u_x\sigma_x + u_y\sigma_y + u_z\sigma_z),
\end{equation}
where $u_x^2+u_y^2+u_z^2 =1$.  The full set of nonzero
idempotents are the above, plus $\hat{1}$.  These
elements have scalar parts of $0.5$ or $1$. If we add any two of
these idempotents together, the sum will have a scalar part of at
least $1$, so the elements with scalar part $0.5$, that is, the
idempotents given in Eq.~(\ref{eq:SolnIdempotencyPauli}), are all
primitive idempotents. Since $\hat{1} =$
$0.5(1+\sigma_z) + 0.5(1-\sigma_z)$, we have that $\hat{1}$ is not
primitive.

If $u_x$, $u_y$, $u_z$ are all real, then the primitive idempotent
defined in Eq.~(\ref{eq:SolnIdempotencyPauli}) is one of the pure
density operators of the Pauli algebra
discussed in the first chapter. If one or more of them are
complex, then we still have a primitive
idempotent, but it is no longer Hermitian. For
example, $u_x=2.4i$, $u_y=0$, $u_z=2.6$ gives the following
primitive\marginpar{\em Pure density operators of spin$-1/2$ are
Hermitian primitive idempotents.} idempotent:
\begin{equation}\label{eq:NonHermitianPI}
0.5+1.2i\sigma_x+1.3\sigma_z = %
\left(\begin{array}{cc}1.8&1.2i\\1.2i&-0.8\end{array}\right)
\end{equation}
This completes the solution of the primitive idempotents of the
$2\times 2$ complex matrices, and also
the Pauli algebra. It should be clear that the
geometric notation is very useful in
solving the system of equations generated by $\rho^2 = \rho$.







% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ObvSolnTwo})
\section{\label{sec:CRoU}  Commuting Roots of Unity}

All of the primitive
idempotents of the Pauli
algebra have the same scalar part, $0.5$; they
differ in the remaining part. This variable part squares to one:
\begin{equation}\label{eq:RemPartOfPPI}
\begin{array}{rcl}
\iota_u &=& u_x\sigma_x + u_y\sigma_y + u_z\sigma_z,\\
\iota_u^2 &=& u_x^2 + u_y^2 + u_z^2 = \hat{1}.
\end{array}
\end{equation}
We will call elements of a Clifford algebra or matrix algebra that
square to $\hat{1}$, ``roots of unity''.  That
they are square roots is assumed.


The equations can be reversed.  Suppose that $\iota$ is not a scalar
and that $\iota^2 = \hat{1}$.  Then $0.5(1\pm\iota)$ is an
idempotent:
\begin{equation}\label{eq:IdemFromRoU}
\begin{array}{rcl}
(0.5(\hat{1}\pm \iota))^2 &=& 0.25( \hat{1} \pm 2\iota + \iota^2)\\
&=&0.25(\hat{1} \pm 2\iota + \hat{1})\\
&=&0.5(\hat{1}\pm\iota).
\end{array}
\end{equation}
In the first chapter, we associated a product of two different
primitive idempotents of the Pauli
algebra with the complex numbers of a Feynman
amplitude. We saw that these products do
not commute, that is, $\rho_a\rho_b \neq \rho_b\rho_a$. This is a
general property of \emph{primitive} idempotents
in matrix or Clifford algebras, that is, they either are
identical, annihilate, or they don't commute.


But the products of distinct idempotents that are
neither primitive nor equal to $\hat{1}$, even in matrix and
Clifford algebras, need not be zero.  For example, the following two
$3\times 3$ matrices are idempotent and commute, but
their product is neither zero nor either of the two:
\begin{equation}\label{eq:ExamplePIs}
\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right)
\left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&1\end{array}\right) =
\left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&0\end{array}\right).
\end{equation}
Note that the product is also idempotent.  This is true in general:
if $\rho_1$ and $\rho_2$ commute and are idempotent, then
$\rho_1\rho_2$ is also idempotent and
commutes with $\rho_1$ and $\rho_2$:
\begin{equation}\label{eq:ProductsCIs}
\begin{array}{rcl}
(\rho_1\rho_2)^2 &=& = \rho_1^2\;\rho_2^2 = \rho_1\rho_2,\\
\rho_1(\rho_1\rho_2) &=& (\rho_1\rho_2)\rho_1,\\
\rho_2(\rho_1\rho_2) &=& (\rho_1\rho_2)\rho_2.
\end{array}
\end{equation}
This\marginpar{\em Products of commuting idempotents commute.} fact
generalizes. A set of commuting idempotents generate a group under
multiplication that consists entirely of commuting
idempotents.


Let $\rho_1 = 0.5(1+\iota_1)$ and $\rho_2 = 0.5(1+\iota_2)$ be two
commuting idempotents.  Then $\iota_1$
and $\iota_2$ commute:
\begin{equation}\label{eq:CommIdemExample}
\begin{array}{rcl}
\rho_1\;\rho_2 &=& \rho_2\;\rho_1,\\
0.5(1+\iota_1)0.5(1+\iota_2) &=& 0.5(1+\iota_2)0.5(1+\iota_1),\\
1+\iota_1 +\iota_2+\iota_1\iota_2
&=&1+\iota_1+\iota_2+\iota_2\iota_1,\\
\iota_1\iota_2 &=& \iota_2\iota_1.
\end{array}
\end{equation}
The above equations can be reversed:  If $\iota_1$ and $\iota_2$
commute and square to $+1$, then $\rho_1 = 0.5(1+\iota_1)$ and
$\rho_2 = 0.5(1+\iota_2)$ are commuting idempotents.  Thus the sets
of commuting roots of unity match
the sets of commuting idempotents.


An idempotent is primitive only when it
cannot be written as the sum of two other non zero idempotents.
Let's suppose that an idempotent $\rho$ is not
primitive so we can write:
\begin{equation}\label{eq:NotPrimitive}
\begin{array}{rcl}
\rho &=& \rho_1 + \rho_2,\\
\rho^2 &=& \rho_1^2 + \rho_2^2 + \rho_1\rho_2 +
\rho_2\rho_1,\\
\rho &=& \rho_1 + \rho_2 + \rho_1\rho_2 +
\rho_2\rho_1,\;\;\textrm{and so}\\
0 &=& \rho_1\rho_2 + \rho_2\rho_1.
\end{array}
\end{equation}
This says that $\rho_1$ and $\rho_2$ anticommute.
Let us reduce the product $\rho_2\rho_1\rho_2$ using anticommutation
two different ways:
\begin{equation}\label{eq:Rho1andRho2Arezero}
\begin{array}{rcl}
\rho_2\rho_1\rho_2 &=& (\rho_2\rho_1)\rho_2 = -\rho_1\rho_2\rho_2 =
-\rho_1\rho_2,\\
&=&\rho_2(\rho_1\rho_2) =-\rho_2\rho_2\rho_1 =
-\rho_2\rho_1.\\
\end{array}
\end{equation}
So\marginpar{\em To prove an idempotent is not primitive, look for
it among the sums of the idempotents that commute with it.} $
-\rho_2\rho_1\rho_2 = \rho_1\rho_2 = \rhoo_2\rho_1$, and we have that
$\rho_1$ and $\rho_2$ commute. But since they commute with each
other, they also must commute with their sum, $\rho$. This suggests
that a way of finding primitive idempotents is to look at sets of 
commuting idempotents
and eliminate all the ones that can be written as sums of the
others. This is a finite problem.


We can make this problem into a ``lattice''.  Given a
set of commuting idempotents, define an
inequality as follows:
\begin{equation}\label{eq:DefineIneq}
\rho_L < \rho_G \;\;\textrm{iff}\;\;\; (\rho_G-\rho_L)\;\;
\textrm{is a non zero idempotent}.
\end{equation}
Equality is defined as usual.  This definition is a ``partial
ordering'', which means that it acts like
the usual inequality does on the real numbers except that some
elements cannot be compared.


% Fig.~(\ref{fig:PauliPIs})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,140)
% Labels
\put( 97, 16){$0$}            % bottom
\put( 97,110){$\hat{1}$}      % top
\put( 63, 65){$\rho_z$}       % left
\put(126, 65){$\bar{\rho_z}$} % right
% Vectors
\linethickness{0.4pt} %
\qbezier(100, 25)(100, 25)( 75, 65) %b l
\qbezier(100, 25)(100, 25)(125, 65) %b r
\qbezier(100,105)(100,105)( 75, 65) %t l
\qbezier(100,105)(100,105)(125, 65) %t r
\end{picture}
} %
\caption{\label{fig:PauliPIs} Drawing of the lattice of a complete
set of commuting idempotents of the Pauli algebra, see text,
Eq.~(\ref{eq:PauliPIineq}).}
\end{figure}


As an example of this partial ordering, consider the set of four
commuting idempotents of the Pauli algebra, $0$, $\rho_z =
0.5(1+\sigma_z)$, $\bar{\rho_z} = 0.5(1-\sigma_z)$, and $\hat{1}$.
These elements satisfy the following relationships:
\begin{equation}\label{eq:PauliPIineq}
\begin{array}{c|cccc|}
& 0 & \rho_z & \bar{\rho_z} & \hat{1} \\ \hline
0            & = & < & < & < \\
\rho_z       & > & = &   & < \\
\bar{\rho_z} & > &   & = & < \\
\hat{1}      & > & > & > & = \\
\hline
\end{array}
\end{equation}
Any two elements can be compared except for $\rho_z$ and
$\bar{\rho_z}$.  To make this into a lattice, put $\hat{1}$ at the
top, $0$ at the bottom, and the rest of the elements in between.  If
$\rho_L < \rho_G$, then arrange $\rho_L$ to be closer to $0$ than
$\rho_G$. Add a line between two elements if they are
ordered, and there is no other element between them.
See Fig.~(\ref{fig:PauliPIs}).  In general, if there is a relation
between two elements, then there is a way of climbing from the lower
one to the upper one.  The primitive idempotents are the elements
closest to the bottom.  In quantum mechanics theory, two
measurements are compatible
(i.e. commute) if their idempotents
are related by an inequality.  For this reason,
the lattice is sometimes called the ``lattice of
propositions'' in the literature.







% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ObvSolnTwo})
\section{\label{sec:PIsOf33}  $3\times 3$ Matrices}




We now consider the problem of the primitive idempotents of $3\times
3$ matrices.  As before, we will make
these from commuting roots of unity.
The diagonal matrices commute, so
let's consider the diagonal roots of
unity:
\begin{equation}\label{eq:CommRootsUnity33}
\left(\begin{array}{ccc}a_1&0&0\\0&a_2&0\\0&0&a_3
\end{array}\right)^2 =
\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1
\end{array}\right)
\end{equation}
that is,
\begin{equation}\label{eq:CommRootsUnity3eqns}
\begin{array}{rcl}
a_1 &=& \pm 1,\\
a_2 &=& \pm 1,\\
a_3 &=& \pm 1.
\end{array}
\end{equation}
The $\pm 1$ in the above three equations are independent; there are
thus $2^3 = 8$ roots of unity.  When one takes
all three right hand sides as $+1$, the root of unity is just
$\hat{1}$. We can label these roots by a suffix that gives the
signs.  The eight roots are then:
\begin{equation}\label{eq:EightRootsofUnityso}
\begin{array}{rcl}
\iota_{---} &=&
\left(\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&-1\end{array}\right),\\
&$ $&\\
 \iota_{--+} &=&
\left(\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&+1\end{array}\right),\\
...&&...\\
\iota_{+++} &=& \hat{1}.
\end{array}
\end{equation}
Each of these eight roots of unity defines an
idempotent (which may or may not be primitive) by
$\rho_\chi = 0.5(1+\iota_\chi)$. The $+1$s on the diagonal of
$\iota_\chi$ produce $+1$s on the diagonal of $\rho_\chi$, but the
$-1$s become $0$s.  There are two idempotents that are particularly
simple: $\rho_{+++} = \hat{1}$ and $\rho_{---} = 0$.  The remaining
six idempotents have mixtures of $0$s and $1$s on their diagonals in
various combinations.


% Fig.~(\ref{fig:ThreeNPIs})
\begin{figure}[!htp]

\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,140)
% Labels
\put( 97, 16){$0=\rho_{---}$}       % bottom
\put( 33, 45){$\rho_{--+}$}         % left
\put(103, 45){$\rho_{-+-}$}         % center
\put(147, 45){$\rho_{+--}$}         % right
\put( 33, 95){$\rho_{++-}$}         % left
\put(103, 95){$\rho_{+-+}$}         % center
\put(147, 95){$\rho_{-++}$}         % right
\put( 97,117){$\hat{1}=\rho_{+++}$} % top
% Vectors
\linethickness{0.4pt} %
\qbezier(100, 25)(100, 25)( 56, 45) %b l
\qbezier(100, 25)(100, 25)(100, 45) %b c
\qbezier(100, 25)(100, 25)(144, 45) %b r
\qbezier( 56, 45)( 56, 45)(100, 95) %l c
\qbezier( 56, 45)( 56, 45)(144, 95) %l r
\qbezier(100, 45)(100, 45)( 56, 95) %c l
\qbezier(100, 45)(100, 45)(144, 95) %c r
\qbezier(144, 45)(144, 45)( 56, 95) %r l
\qbezier(144, 45)(144, 45)(100, 95) %r c
\qbezier(100,115)(100,115)( 56, 95) %t l
\qbezier(100,115)(100,115)(100, 95) %t c
\qbezier(100,115)(100,115)(144, 95) %t r
\end{picture}
} %
\caption{\label{fig:ThreeNPIs} Drawing of the lattice of a complete
set of commuting idempotents of the algebra of $3\times 3$ matrices,
see text, Eq.~(\ref{eq:ThreeMatrixPIineq}). The primitive
idempotents are $\rho_{--+}$, $\rho_{-+-}$, and $\rho_{+--}$.}
\setlength{\unitlength}{1pt}
\end{figure}




The partial ordering of the diagonal
idempotents is as follows:
\begin{equation}\label{eq:ThreeMatrixPIineq}
\begin{array}{r|cccccccc|}
& 0 & \rho_{--+} & \rho_{-+-} & \rho_{+--}  & \rho_{++-} &
\rho_{+-+} & \rho_{-++} & \hat{1} \\ \hline
0=\rho_{---}       & = & < & < & < & < & < & < & < \\
\rho_{--+}         & > & = &   &   &   & < & < & < \\
\rho_{-+-}         & > &   & = &   & < &   & < & < \\
\rho_{+--}         & > &   &   & = & < & < &   & < \\
\rho_{++-}         & > &   & > & > & = &   &   & < \\
\rho_{+-+}         & > & > &   & > &   & = &   & < \\
\rho_{-++}         & > & > & > &   &   &   & = & < \\
\hat{1}=\rho_{+++} & > & > & > & > & > & > & > & = \\
\hline
\end{array}
\end{equation}
These can be arranged into a lattice, see
Fig.~(\ref{fig:ThreeNPIs}). It may be intuitively obvious that the
idempotents are arranged according to the binomial
theorem. That is, with $N\times N$ matrices
there will be $N+1$ rows, and the number of idempotents in the $n$th
row will be the coefficient of $x^n$ in the expansion of the
polynomial $(1+x)^N$.  In the case of Fig.~(\ref{fig:ThreeNPIs}),
$N=3$ and the four rows of the lattice have $3!/(n!(3-n)! = $ $1$,
$3$, $3$, and $1$ element each.  We will see this pattern in the
structure of the elementary particles in that there is one electron,
three anti-up quarks, three down quarks, and one neutrino.


We have found the idempotent structure of the diagonal $3\times 3$
matrices; what does this tell us about the structure of more general
$3\times 3$ matrices? Let $S$ be any
invertible matrix. Then the following transformation:
\begin{equation}\label{eq:TransformMatrices}
A \to SAS^{-1}.
\end{equation}
preserves multiplication:
\begin{equation}\label{eq:PreservesMult}
(SAS^{-1})(SBS^{-1}) = S(AB)S^{-1},
\end{equation}
and therefore preserves idempotency and roots of unity,
and preserves addition:
\begin{equation}\label{eq:PreservesAddition}
SAS^{-1} + SBS^{-1} = S(A+B)S^{-1},
\end{equation}
and therefore preserves primitive idempotency and the partial
ordering.  We will leave it as an exercise
for the reader to prove that the relation is general; that is, any
set of complete primitive idempotents in $3\times 3$ matrices
generates an
idempotency partial ordering that is
equivalent to that of the group of diagonal matrices.\footnote{The
author figures he could prove this but only inelegantly.  If you
happen to know how to do it in an elegant fashion, please send me a
line at carl@brannenworks.com .}


Quarks come in three ``color'' states
which we will label as $R$, $G$, and $B$ which stand for red, green
and blue. Color is completely
symmetric under rotation; that is, if we make the replacement:
\begin{equation}\label{eq:ColorSymmetry}
\begin{array}{rcl}
R &\to& G'\\
G &\to& B'\\
B &\to& R'
\end{array}
\end{equation}
we expect that our physics will remain the same.  The above is an
even permutation on the colors.  We will
not require that color symmetry be preserved on odd
permutations such as $R\to R'$, $G\to B'$,
$B\to G'$.



Suppose there is a process that can take a red particle of type $I$
and convert it to a red particle of type $F$ (which stand for
initial and final states), and this process can be modeled with an
amplitude of $\alpha$. By color symmetry, we
expect the process to use the same amplitude when
the initial state and final states
are both green, or when they are both blue. This is needed in
order to preserve the above symmetry.


The amplitude $\alpha$ need not be at all similar to the
amplitude for the process that takes a red particle
in the initial state $I$ and turns it into a green particle in the
final state $F$.  Let us suppose that the amplitude $\beta$ applies
to such an interaction. By color symmetry, $\beta$ must also apply
to the process which takes a green particle in the initial state and
produces a blue particle.
Similarly for a process that takes a blue particle in the initial
state and produces a red particle in the final state. Finally, let
an amplitude of $\gamma$ apply to the three
remaining conversions. We have defined $9$ amplitudes. They can be
arranged in a $3 \times 3$ table labeled horizontally and vertically
by the particle types:
\begin{equation}\label{eq:ColorSymmTable}
\begin{array}{r|ccc|}
   &I_R   &I_G&I_B\\ \hline
F_R&\alpha&\gamma&\beta\\
F_G&\beta&\alpha&\gamma\\
F_B&\gamma&\beta&\alpha\\ \hline
\end{array}
\end{equation}
In the above table, initial states run along
the column and final states run along the rows.
With this we can easily read off the results when a particle is in a
given initial state. For example, if the particle begins in the
green state, then the middle column applies, and the process will
result in a final red particle with amplitude $\gamma$, a
green particle with amplitude
$\alpha$, and a blue particle with
amplitude $\beta$.


Suppose we had a particle that was initially in some
superposition of color and we were
representing it with a vector with $R$, $G$ and $B$ components:
\begin{equation}\label{eq:ColorVector}
|I\ra  = \left(\begin{array}{c} I_R\\
I_G\\ I_B\end{array}\right).
\end{equation}
To figure out how this particle is modified by our process, we have
to sum the amplitudes for the various
possibilities, and add them together.  For example, what is the
final amplitude for the blue state?  There are three
contributing processes, $I_R \to F_B$, $I_G \to F_B$, and $I_B \to
F_B$.  The amplitudes for these three processes are $\gamma$,
$\beta$, and $\alpha$, respectively. The incoming vector has $I_R$,
$I_G$, and $I_B$ components, so the total final blue
component is:
\begin{equation}\label{eq:FinalBlue}
F_B = \gamma I_R + \beta I_G + \alpha I_B.
\end{equation}
This\marginpar{\em Circulant matrices model color processes.} is
just matrix multiplication. Therefore
we rewrite our table of amplitudes as a $3\times 3$ matrix:
\begin{equation}\label{eq:TableIntoState}
M = \left(\begin{array}{ccc}\alpha&\gamma&\beta\\
\beta&\alpha&\gamma\\
\gamma&\beta&\alpha\end{array}\right).
\end{equation}
and write $|F\ra = M\;|I\ra$.  The matrix in the above equation is a
``circulant'' $3 \times 3$ matrix. Circulant matrices have
the convenient properties that
sums and products of circulant matrices are also circulant -- they
form a subalgebra.


The\marginpar{\em States and operators are the same thing.}
philosophy of the state vector formalism is that the fundamental
objects in quantum mechanics are
states, and these states can be separated, in the
above, into initial and final
states. This is in contrast to the philosophy of
the density operator formalism which treats the density
operators as the fundamental objects which
share the same algebra as the operators on those
objects. This goes both ways.  In addition to treating our states as
operators, we can also treat our
operators as states. This is not at all a
trivial extension and we will discuss it in much greater detail in
later chapters.


For the moment, let us find the idempotent structure of the
circulant $3\times 3$ matrices. Accordingly, suppose that the operator $M$ of
Eq.~(\ref{eq:TableIntoState}) is a density operator, and is
therefore idempotent:
\begin{equation}\label{eq:MisIdemp}
\left(\begin{array}{ccc}\alpha&\gamma&\beta\\
\beta&\alpha&\gamma\\
\gamma&\beta&\alpha\end{array}\right)^2
= \left(\begin{array}{ccc}\alpha&\gamma&\beta\\
\beta&\alpha&\gamma\\
\gamma&\beta&\alpha\end{array}\right).
\end{equation}
This gives three complex equations in three complex unknowns:
\begin{equation}\label{eq:CirculantEqns}
\begin{array}{rcl}
\alpha &=& \alpha^2 + 2\beta\gamma,\\
\beta  &=& \gamma^2 + 2\alpha\beta,\\
\gamma &=& \beta^2  + 2\alpha\gamma.
\end{array}
\end{equation}
Adding all three of the above gives:
\begin{equation}\label{eq:SolvCircOne}
\alpha+\beta+\gamma = (\alpha+\beta+\gamma)^2,
\end{equation}
and therefore $\alpha+\beta+\gamma = 0$ or $= 1$.  For any algebra,
if $\rho$ is an idempotent, then so is $1-\rho$.  Translating this
into circulant form, this means that if
$(\alpha,\beta,\gamma)$ gives an idempotent, then
so does $(1-\alpha,-\beta,-\gamma)$.  Thus we need only consider
values of $\alpha$, $\beta$, and $\gamma$ that satisfy
$\alpha+\beta+\gamma=0$; we can get the other solutions by computing
$1-\rho$.  Accordingly, we assume that
\begin{equation}\label{eq:AlphaLimitCirc}
\alpha = -(\beta+\gamma).
\end{equation}


Multiplying the last two equations of Eq.~(\ref{eq:CirculantEqns})
by $\gamma$ and $\beta$, respectively, and subtracting gives:
\begin{equation}\label{eq:SolvCirTp}
\gamma^3 = \beta^3.
\end{equation}
The complex cubed root of one,
$e^{2i\pi/3}$, will appear over and over in regard to circulant
matrices.  Let us have pity on both author and reader and abbreviate
this number as $\nu$:
\begin{equation}\label{eq:CubedRootOfUnity}
\begin{array}{lclcl}
\nu   &=& e^{+2i\pi/3} &=& -0.5 + i\sqrt{0.75},\\
\nu^* &=& e^{-2i\pi/3} &=& -0.5 - i\sqrt{0.75}.
\end{array}
\end{equation}
With the $\nu$ notation, we can write $\gamma$ in terms of $\beta$
as:
\begin{equation}\label{eq:SolvCirTwo}
\gamma = \beta\;\nu^k.
\end{equation}
We can substitute the above equation and
Eq.~(\ref{eq:AlphaLimitCirc}) into the second equation of
Eq.~(\ref{eq:CirculantEqns}) to get the following equation for
$\beta$:
\begin{equation}\label{eq:SolvCirThree}
\beta = \beta^2\;\nu^{-k} + 2(-\beta-\beta \nu^k)\beta,
\end{equation}
which is solved by either $\beta=0$ or
\begin{equation}\label{eq:SolvCirFour}
\beta = \nu^k/(1-2\nu^{-k}-2\nu^k).
\end{equation}
If $\beta=0$, then $\gamma=0$ and $\alpha=0$.  Thus we have a
complete set of solutions to Eq.~(\ref{eq:CirculantEqns}) subject to
$\alpha+\beta+\gamma=0$.  Adding back in the $\alpha+\beta+\gamma=1$
solutions we have the complete solution set as:
\begin{equation}\label{eq:SolvCirDone}
\begin{array}{ccc}
\alpha&\beta&\gamma \\ \hline %
0&0&0\\
1/3&1/3&1/3\\
1/3&\nu/3 & +\nu^*/3   \\
1/3&\nu^*/3 & +\nu/3   \\
2/3&-\nu/3 & -\nu^*/3   \\
2/3&-\nu^*/3 & -\nu/3   \\
2/3&-1/3&-1/3\\
1&0&0 \\  \hline %
\end{array}
\end{equation}
There\marginpar{\em Circulant $3\times 3$ matrices have the same
lattice as $3\times 3$ diagonal matrices.} are eight solutions. If
we write them out in matrix form so that we can perform addition on
them, we will see that they define a partial ordering
of idempotents that is identical to the
one we computed for the diagonal $3\times 3$
matrices,see
Fig.~(\ref{fig:ThreeNPIs}). In fact, the above table is organized in
the same order as in Eq.~(\ref{eq:ThreeMatrixPIineq}).  The reader
is invited to derive a transformation $S$ that diagonalizes the
circulant matrices; that is, defines a translation of the form $\chi
\to S\chi S^{-1}$ that takes the diagonal idempotents to the
circulant idempotents.


If a set of matrices commute, then they share a set
of eigenvectors. This is the case for any set of
commuting idempotents, so it certainly applies to both of our
$3\times 3$ examples.  But circulant matrices in
general share common eigenvectors.
No matter the values of $\alpha$, $\beta$ and $\gamma$, the $3\times
3$ circulant matrices have three eigenvectors that can be written
as:
\begin{equation}\label{eq:EigenvectorsOfCircs}
\left(\begin{array}{c}1\\1\\1\end{array}\right), %
\left(\begin{array}{c}1\\\nu\\\nu^*\end{array}\right), %
\left(\begin{array}{c}1\\\nu^*\\\nu\end{array}\right).
\end{equation}
Since all circulant matrices share the same
eigenvectors, the corresponding
eigenvalues can be used to classify the
circulant matrices. For the eight
idempotents defined in
Eq.~(\ref{eq:ThreeMatrixPIineq}), the eigenvalues are as follows:
\begin{equation}\label{eq:EvaluesCircIdempotents}
\begin{array}{ccc|ccc|}
\alpha&\beta&\gamma&
\left(\begin{array}{c}1\\1\\1\end{array}\right)&
\left(\begin{array}{c}1\\\nu^*\\\nu\end{array}\right)&
\left(\begin{array}{c}1\\\nu\\\nu^*\end{array}\right)\\ \hline %
0&0&0                  &0&0&0 \\
1/3&1/3&1/3            &1&0&0 \\
1/3&\nu/3 & +\nu^*/3   &0&1&0 \\
1/3&\nu^*/3 & +\nu/3   &0&0&1 \\
2/3&-\nu/3 & -\nu^*/3  &1&0&1 \\
2/3&-\nu^*/3 & -\nu/3  &1&1&0 \\
2/3&-1/3&-1/3          &0&1&1 \\
1&0&0                  &1&1&1 \\  \hline %
\end{array}
\end{equation}
This is another way of writing the structure of the
lattice of the circulant matrices.





% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




\chapter{\label{sec:Representations} Representations}

%\begin{quote}
\settowidth{\versewidth}{For each man knows ere his lip-thatch grows
he is master of Art and Truth;}
\begin{verse}[\versewidth]
%\em
The tale is as old as the Eden Tree---and new as the new-cut
tooth---\\
For each man knows ere his lip-thatch grows he is master of Art and
Truth;\\
And each man hears as the twilight nears, to the beat of his dying
heart,\\
The Devil drum on the darkened pane: ``You did it, but was it Art?''
\end{verse}
%\end{quote}
$ $


\lettrine{S}{$\;$ince this book is} describing a geometric formulation of
quantum mechanics, it is not strictly necessary for us to delve into
the subject of representations of Clifford algebras.  However, we
are very much concerned with the elementary particles, and it turns
out that there is a very strong connection between elementary
particles and representations.

Accordingly, we will now turn to the task of defining
representations from a particle point of view.  In doing this, we
will find that there are more elegant ways of describing a
representation than are typically seen in the literature.


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ThreeMatrixPIineq})
\section{\label{sec:CliffordAlgebras}  Clifford Algebras}




The number of unit multivectors in a real
Clifford algebra with $M$ vectors is $2^M$ as each vector as each
unit multivector can either have zero or one copies of a given
vector. Each of these unit multivectors is a degree of
freedom for the Clifford algebra, so if the
Clifford algebra is real there will be $2^M$ real degrees of freedom
in the algebra, and if the Clifford algebra is complex there will be
$2^M$ complex degrees of freedom or twice that number of real
degrees of freedom.


A real $N\times N$ matrix algebra has
$N^2$ real degrees of freedom and a complex $N\times N$ matrix
algebra has $2N^2$ real degrees of freedom.  Consequently, in making
a faithful\footnote{Faithful here means $1$ to $1$
and onto.} representation of a Clifford
algebra out of matrices, we will choose real or complex matrices
according as the number of vectors, $M$, is even or odd:
\begin{equation}\label{eq:CAtoMatrix}
\begin{array}{l|c|c|}
\textrm{Clifford Algebra}     & M\;\; \textrm{is even}                & M\;\; \textrm{is odd} \\
\hline %
\textrm{real}    & 2^{(M/2)}\;\; \textrm{real matrices}    & 2^{(M-1)/2}\;\; \textrm{complex} \\
%
\textrm{complex} & 2^{(M/2)}\;\;\textrm{complex} & 2^{(M+1)/2}\;\; \textrm{real matrices} \\
\hline
\end{array}
\end{equation}
For example, in the first chapter we treated the Pauli
algebra as a real Clifford algebra with $3$
vectors, $\sigma_x$, $\sigma_y$ and $\sigma_z$; so $M=3$. This is an
odd number of vectors and a real Clifford algebra, so taking the
upper right entry in Eq.~(\ref{eq:CAtoMatrix}), the matrix
representation uses complex matrices of size
$2^{(3-1)/2} = 2$. Sure enough the Pauli spin matrices
are $2\times 2$ complex
matrices. The Dirac algebra is usually treated as a complex Clifford
algebra with $4$ vectors, $\gamma_0$, $\gamma_1$, $\gamma_2$, and
$\gamma_3$, so $M=4$. Examining the lower left entry in
Eq.~(\ref{eq:CAtoMatrix}), the matrix representation uses complex
matrices of size $2^{(4/2)} = 4$. Sure enough the Dirac gamma
matrices are $4\times 4$ complex
matrices.





The next least complicated matrix algebra after the $3\times 3$
matricesare the $4\times
4$ matrices. Four is a power of two, and
so we can give these matrices a geometric interpretation by
considering them as a representation of a
Clifford algebra, the Dirac algebra. Physicists
are more used to working with representations
of the Dirac algebra rather than the algebra itself and the
representations are called the ``Dirac gamma matrices.''


The Dirac gamma matrices are usually
designated as $\gamma^0$, $\gamma^1$, $\gamma^2$, and
$\gamma^3$. These are associated with the four
dimensions of space-time, respectively, $t$, $x$, $y$, and $z$.  The
numeric designation is convenient for the summation convention. This
is particularly convenient when considering the Dirac
gradient operator,
$\gamma^\mu\partial_\mu$, which is explicitly
covariant. This book is mostly concerned with the
internal states of particles; for this we
will need neither covariance nor that much use of the Dirac
operator. On the other hand, we are very concerned with the
geometric meaning of quantum mechanics and we will be doing a lot of
algebra with these objects.


As a geometric improvement on the usual Dirac notation, we can
replace the numbers $0-3$ by letters corresponding to the given
direction in the manner of the Pauli algebra.  For example instead
of $\gamma^3$ we could write $\gamma_z$.  The Dirac algebra would
then be distinguished from the Pauli algebra by the use of $\gamma$
instead of $\sigma$.  But from the point of view of Clifford
algebra, the designation of the vector should be sufficient.  That
is, it is the vectors that generate and define the Clifford algebra;
the Clifford algebra does not define the vectors.


Accordingly,\marginpar{\em Hat notation: $\hat{x}$, $\hat{y}$,
$\hat{z}$, and $\hat{t}$ are an abbreviation for the basis vectors
of the Dirac algebra, $\gamma^0$, $\gamma^1$, $\gamma^2$, and
$\gamma^3$.} we will use a notation that
abbreviates the usual gamma matrix notation and covers the Pauli
algebra too:

\begin{equation}\label{eq:ShortNotation}
\begin{array}{ccc}
\textrm{Pauli}&\textrm{Dirac}&\textrm{Clifford}\\ \hline %
\times   & \gamma^0 & \hat{t}\\
\sigma_x & \gamma^1 & \hat{x}\\
\sigma_y & \gamma^2 & \hat{y}\\
\sigma_z & \gamma^3 & \hat{z}\\ \hline %
\end{array}
\end{equation}
The reader who takes up this notation will find
that it speeds up work considerably.  When we need to distinguish
between coordinates and Clifford operator
elements, the ``hat'' will distinguish the Clifford
algebra unit vectors.  In writing Clifford unit
multivectors, we can simply abut the vectors
as in $\hat{x}\hat{y}$, or better, we can draw a single hat over the
product as in $\at{xy}$.  If we do draw them with a single hat, we
will arrange the elements in a standard order.
\footnote{Or send author a nasty note:
carl@brannenworks.com .}


In general, an element of an algebra is a sum of scalar multiples of
its unit multivectors.  For example, any
element of the Pauli algebra can be written in
the form:
\begin{equation}\label{eq:PauliElementScalar}
\alpha_1\hat{1}+\alpha_x\hat{x}+\alpha_y\hat{y} + ... +
\alpha_{xyz}\at{xyz},
\end{equation}
where the $\alpha_\chi$ are scalars, that is, real numbers.  Because
the scalars are taken to be real numbers, the Pauli
algebra is a real algebra.


We\marginpar{\em Complex algebras versus real algebras} can also
suppose that the scalars are complex numbers; we then call the
algebra a ``complex algebra'', or an
``algebra over the complex numbers''.  In the first chapter we
showed that we could think of the pseudoscalar
element of the Pauli algebra, $\at{xyz}$, as the imaginary unit of
the complex numbers. In doing this we were making a bit of a
confusion of the notation because we would be writing $i\hat{1} =
i$, and confusing a scalar with the unit operator times that scalar.
In order to back away a bit from this confusion, we will extend our
hats over the imaginary unit where appropriate. That is, when $i$ is
geometric we will write $\hat{i}$ instead of $i$,
and $\at{ixy}$ instead of $i\at{xy}$.


Many people working in Geometric algebra try to replace all
imaginary numbers from their calculations
other than the geometric ones.  This makes sense from a physical
perspective. However, in solving equations in mathematics, imaginary
numbers can be very useful, and we can never know for sure that the
objects we discuss in physics are mathematical conveniences that
might have imaginary numbers present, or physical objects that must
be built from real numbers only.  For example, imaginary numbers can
appear naturally in Fourier transforms. Our
notation will allow these sorts of calculations to go forward
without more confusion than is inevitable.  In such a situation, it
is possible to have two imaginary units, both
squaring to $-1$ and both commuting with the entire algebra.


In the Pauli algebra, the element $\at{xyz}$ could
be thought of as an imaginary unit because it
squares to $-1$ and commutes with all elements in the algebra. Under
what conditions can an element commute with the
entire algebra?


Let $\at{\chi}$ be a unit multivector that
we wish to test to see if it commutes with the whole algebra.  Since
the algebra is made from products of vectors, it is necessary and
sufficient that $\at{\chi}$ commute with all the
vectors. Suppose that $\at{\chi}$ is a product of $m$ vectors and
the algebra is generated by $N$ vectors.


If $\at{\chi}$ is not a scalar then $1<m$ and we can pick a vector
$\hat{v}$ that is part of $\hat{\chi}$. In commuting $\hat{v}$ with
$\hat{\chi}$, we will perform $m$ commutations.  In $m-1$ of these,
the vectors being commuted will be different from $\hat{v}$ and a
minus sign will appear because Clifford vectors
anticommute. In $1$ of these, we will be
commuting $\hat{v}$ with itself and no minus sign will appear.
Therefore for $\hat{\chi}$ to commute with $\hat{v}$,
we must have that $m-1$ is even.


If $\hat{\chi}$ is not the pseudoscalar, that
is, if it is not the product of all the vectors, then $m<N$ and we
can pick a vector $\hat{v}$ that is not in the product that makes up
$\hat{\chi}$.  In this case, our commutation will be performed with
$m$ anticommutations and for $\hat{v}$ to commute
with $\hat{\chi}$ we must have that $m$ is even.  But $m$ and $m-1$
cannot both be even. Therefore any element of a Clifford algebra
that commutes with everything in the algebra must be either a scalar
or a pseudoscalar.


The scalars always commute with the entire algebra.  But the
pseudoscalar may or may not. In commuting a
vector $v$ with $\at{\chi}$, there will be $m=N-1$ anticommutations
and one commutation.  Therefore the pseudoscalar
will commute with the algebra if the algebra has an odd number of
basis vectors. And if there is an even number of basis vectors, then
the pseudoscalar anticommutes with the basis
vectors (and commutes with even products of basis
vectors).


To have a geometric imaginary unit,
the unit pseudoscalar must, in addition to
commuting with the rest of the algebra, also square to $-1$.  This
will happen in some Clifford algebras and not in others depending on
the signature and the number of basis vectors. First consider the
Clifford algebras of positive signature. It can be
shown that the unit pseudoscalar in such a
Clifford algebra will have positive signature if and only if the
number of basis vectors is equivalent to $0$ or $1$ modulo $4$.
Example: the Pauli algebra has $3$ basis vectors. This is not
equivalent to $0$ or $1$ modulo $4$, so the unit pseudoscalar of the
Pauli algebra squares to $-1$: that is, $\at{xyz}^2$ $= -1$.


If a Clifford algebra has an odd number of basis vectors with
negative signature, then the signature of the
pseudoscalar of that Clifford algebra will be
negated with respect to the pseudoscalar of a Clifford algebra of
purely positive signature. Example: the Dirac algebra has $4$ basis
vectors.  Four is equivalent to $0$ modulo $4$ so the unit
pseudoscalar of the purely positive Clifford algebra with four
vectors will have a positive signature.  But the Dirac algebra has
an odd number (i.e. one: $\hat{t}\;$) of basis vectors that have
negative signature.  So the pseudoscalar of the Dirac algebra
instead squares to $-1$: that is, $\at{xyzt}^2$
$= -1$.


The same rule that tells whether the unit pseudoscalar squares to
$+1$ or $-1$ tells whether any unit multivector
of a Clifford algebra squares to $\pm 1$. After a
certain amount of use, the reader will find that the rules for
determining whether the square of a unit multivector is $+1$ or $-1$
become intuitive and automatic. Until that time, photocopying
Table~(\ref{tbl:MVSquares}) and keeping it handy may assist.


% Table~(\ref{tbl:MVSquares})
\begin{table}[tbp]
\caption{\label{tbl:MVSquares}Signs of squares of Clifford algebra
MVs with $n_+$ positive vectors and $n_-$ negative vectors.} \center
\framebox[0.9\textwidth][c]{ $
\begin{array}{cc|cccccccc|} %
& & & &&n_+&& & &\\
& &0&1&2&3&4&5&6&7\\ \hline
&0&+&+&-&-&+&+&-&-\\
&1&-&+&+&-&-&+&+&-\\
&2&-&-&+&+&-&-&+&+\\
&3&+&-&-&+&+&-&-&+\\ n_-
&4&+&+&-&-&+&+&-&-\\
&5&-&+&+&-&-&+&+&-\\
&6&-&-&+&+&-&-&+&+\\
&7&+&-&-&+&+&-&-&+\\ \hline
\end{array} $}
\end{table}


Two Clifford unit multivectors will either commute or anticommute.
To find out which, you can anticommute the vectors they are made of
but there is an easier way.  First of all, in determining whether
they commute or anticommute we can
start from the assumption that their vectors are ordered in any
particular order, so long as we are consistent. Let's begin with the
two multivectors $a$ and $b$ written as follows:
\begin{equation}\label{eq:DetermineCommutation}
\begin{array}{rcl}
a &=& (a_1 a_2 ... a_j)(s_1 s_2 ... s_l),\\
b &=& (b_1 b_2 ... b_k)(s_1 s_2 ... s_l),
\end{array}
\end{equation}
where the $\{s_n\}$ are $l$ shared vectors, the $\{a_n\}$ are $j$
vectors that only $a$ has, and the $\{b_n\}$ are $k$ vectors that
only $b$ has. Then compute:
\begin{equation}\label{eq:CalcCommuteAB}
\begin{array}{rcl}
ab &=& (a_1 a_2 ... a_j)(s_1 s_2 ... s_l)(b_1 b_2 ... b_k)(s_1
s_2 ... s_l),\\
&=&(-1)^{kl}(a_1 a_2 ... a_j)(b_1 b_2 ... b_k)(s_1 s_2 ... s_l)(s_1
s_2 ... s_l),\\
&=&(-1)^{kl+jk}(b_1 b_2 ... b_k)(a_1 a_2 ... a_j)(s_1 s_2 ...
s_l)(s_1 s_2 ... s_l),\\
&=&(-1)^{kl+jk+jl}(b_1 b_2 ... b_k)(s_1 s_2 ... s_l)(a_1 a_2 ...
a_j)(s_1 s_2 ... s_l),\\
&=&(-1)^{kl+jk+jl}\;ba.
\end{array}
\end{equation}
Therefore\marginpar{\em Multivectors commutation rule.} the two
multivectors commute if $kl+jk+jl$ is even and anticommute if it is
odd.  So the rule is this:
Count the number of vectors that the two unit
multivectors share, and the numbers that each has unique. You have
three numbers. If more of them are odd then even, then the two unit
multivectors anticommute.
If more of them are even then odd, then the two unit multivectors
commute. Some examples:
\begin{equation}\label{eq:ExampleCliffordCommutation}
\begin{array}{cc|ccc|c}
\textrm{A}&\textrm{B}&j&k&l&\textrm{Commute?}\\ \hline
\at{xyzt}&\at{x}&3&0&1&\textrm{no}\\ %411
\at{xyz}&\at{xyt}&1&1&2&\textrm{no}\\ %332
\at{xy}&\at{yz}&1&1&1&\textrm{no}\\ %221
\at{xy}&\at{zt}&2&2&0&\textrm{yes}\\ %220
\at{xyz}&\hat{x}&2&0&1&\textrm{yes} \\ %311
\at{xzt}&\hat{y}&3&1&0&\textrm{no} \\ %310
\hat{x}&\hat{y}&1&1&0&\textrm{no}\\ %110
\hline
\end{array}
\end{equation}




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ThreeMatrixPIineq})
\section{\label{sec:TDiracAlgebra}  The Dirac Algebra}




The choice of signature for the basis
vectors of a Clifford algebra is not completely
arbitrary.\cite{Pezzaglia_9704048}  For this book we will follow the
lead of the Pauli algebra which has positive signature $(+++)$, and
give the spatial basis vectors of the Dirac algebra a positive
signature. Thus our unit vectors square as follows:
\begin{equation}\label{eq:DiracSquares}
\begin{array}{rclcrcl}
\hat{t}^2 &=& -1,&\;\;\;&\hat{x}^2 &=& +1,\\
\hat{y}^2 &=& +1,&\;\;\;&\hat{z}^2 &=& +1.
\end{array}
\end{equation}
This is a less common choice of signature.


As with the Pauli algebra, any product of Dirac
algebra vectors can be reduced to $\pm$ a product that includes each
vector at most once.  This gives $16$ Dirac ``unit
multivectors'', twice as many as the Pauli
unit multivectors. We will order the Dirac unit multivectors as
follows:\marginpar{\em Dirac
algebra unit multivectors.}
\begin{equation}\label{eq:DiracMVs}
\begin{array}{llll}
\hat{1},&\hat{x},&\hat{y},&\hat{z},\\
\hat{t},&\at{xt},&\at{yt},&\at{zt},\\
\at{xyz},&\at{yz},&\at{xz},&\at{xy},\\
\at{xyzt},\;\;\;&\at{yzt},\;\;\;& %
\at{xzt},\;\;\;&\at{xyt}.
\end{array}
\end{equation}
Note that the above has been arranged so that the columns have
orientation $n$, $x$, $y$, and $z$, respectively.  The
orientation we
called ``neutral'', Eq.~(\ref{eq:Orientation}), in the Dirac algebra
now has time vectors, but these are present in the other
orientations and we will still call it the neutral orientation.



With our choice of signature, $10$ of the Dirac
unit multivectors square to $+1$ and $6$
square to $-1$:
\begin{equation}\label{eq:DiracBilinSqrs}
\begin{array}{rcl}
\{\hat{1}, \hat{x}, \hat{y}, \hat{z}, \at{xt}, \at{yt}, \at{zt},
\at{yzt}, \at{xzt}, \at{xyt}\} &&\textrm{+1},\\
\{\at{yz}, \at{xz}, \at{xy}, \hat{t}, \at{xyz}, \at{xyzt}\}
&&\textrm{-1}.
\end{array}
\end{equation}
If we had chosen the opposite signature, we would
have only $6$ elements square to $+1$ and it would be harder to make
roots of unity in a real Dirac algebra.


The Dirac algebra is normally treated as a
complex algebra. This means that the Dirac algebra consists of sums
of complex scalar multiples of the $16$ Dirac unit
multivectors. For example,
\begin{equation}\label{eq:ExampleComplexDiracElement}
(1+3i)\hat{1} + (2+7i)\at{xyz} -3\at{zt}
\end{equation}
is in the complex Dirac algebra.  We
can also consider the real Dirac algebra,
it would consist of sums of real scalar multiples of Dirac unit
multivectors. An example of an element in the real Dirac algebra is:
\begin{equation}\label{eq:ExampleRealDelement}
2\hat{1} - 5\hat{x}+3.7\hat{y}+22\at{xyzt}
\end{equation}
For the elementary particles we have need only of the complex Dirac
algebra.  Our discussion of the real Dirac algebra is only for
pedagogical purposes. The real and complex Dirac algebras can be
faithfully represented by real and complex $4\times 4$ matrices,
respectively.


The Dirac equation is normally written with
spinors. In our geometric language, we write it as:
\begin{equation}\label{eq:DiracSpinorEqn}
\begin{array}{rcl}
0 &=& (\hat{x}\partial_x+\hat{y}\partial_y
+\hat{z}\partial_z+\hat{t}\partial_t+m)|\psi\ra,\\
&=&(\nabla + m)\;|\psi\ra,
\end{array}
\end{equation}
where we have introduced $\nabla$ as the geometric
gradient, the differential
part of the Dirac operator. We translate this into density operator
form by replacing the ket $|\psi\ra$ with the product of
the corresponding density operator with the vacuum
operator: $\rho_\psi\;\rho_0$.  The resulting equation
is:
\begin{equation}\label{eq:DiracMatrixEqn}
(\nabla + m) \rho_\psi\;\rho_0.
\end{equation}
In the above equation, as in the first chapter, $\rho_0$ is a
constant primitive idempotent, and is
quite arbitrary. Since $\rho_0$ is constant, it does not interact
with the partial derivatives. Since $\rho_0$ is arbitrary, we can
rewrite the above with several different values for $\rho_0$.


Suppose $\rho_0$, $\rho_0'$, $\rho_0''$ and $\rho_0'''$ are a
complete set of primitive idempotents of
the Dirac algebra so that $\hat{1} =
\rho_0+\rho_0'+\rho_0''+\rho_0'''$. Then we can write four different
Dirac equations (with four different $\rho_\psi$:
\begin{equation}\label{eq:Diracfourmatrixeqn}
\begin{array}{rcl}
0 &=& (\nabla + m)\rho_\psi\;\rho_0,\\
0 &=& (\nabla + m)\rho_\psi'\;\rho_0',\\
0 &=& (\nabla + m)\rho_\psi''\;\rho_0'',\\
0 &=& (\nabla + m)\rho_\psi'''\;\rho_0''',\\
\end{array}
\end{equation}
Define $\Psi$ as follows:
\begin{equation}\label{eq:PsiDefn}
\Psi = \psi\rho_0 + \psi'\rho_0' + \psi''\rho_0'' +
\psi'''\rho_0'''.
\end{equation}
Because a complete set of primitive idempotents annihilate one
another, we can pick out the $\psi$s from $\Psi$ by right
multiplying by the appropriate primitive
idempotent:
\begin{equation}\label{eq:psiFromPsi}
\begin{array}{rcl}
\psi &=& \Psi\;\rho,\\
\psi' &=& \Psi\;\rho',\\
\psi'' &=& \Psi\;\rho'',\\
\psi''' &=& \Psi\;\rho'''.
\end{array}
\end{equation}
This means that we can add up the four equations of
Eq.~(\ref{eq:Diracfourmatrixeqn}) into a single equation for
$\Psi$:
\begin{equation}\label{eq:RhoDiracEqn}
0 = (\nabla + m)\Psi.
\end{equation}
Now the above looks a lot like the usual spinor Dirac
equation, but it has four times as many degrees of freedom.  It
consists of four non interacting Dirac equations.  We will call this
the ``generalized Dirac'' equation, and where the context
is clear, may leave off the word ``generalized''. Since the operator
version of a wave function has four times as many degrees of
freedom, we will use capital letters to designate the operator
version as in $\Psi$, and small letters to designate the spinor
version as in $\psi$.


In particle physics, the Dirac
equation is used to compute the
propagators of the various spin$-1/2$
fermions. We can think of the four propagators
written as above as applying to four different (i.e.
``distinguishable'') spin$-1/2$ particles.
For example, we could assign $\psi$ to the electron
(and positron), and then assign $\psi'$, $\psi''$, and $\psi'''$ to
the three colors of the up quark.


If there were only four elementary fermions, we could
consider our task of unifying the elementary particles as well under
way, but in fact there are instead $24$ of them, that is, the
electron, the neutrino, three up
quarks, three down quarks, and then these eight
particles in two more generations. Thus our
equation is $6$ times too small. While the four spin$-1/2$ particles
contained in $\Psi$ cannot be easily assigned to four
elementary particles, we will retain this idea and will later apply
it to a preon model of the elementary
fermions. But the fact that we can assign multiple
fermions to a single generalized Dirac equation\footnote{The problem
of defining the Dirac equation in the spinor formalism
in geometric form is far more complicated than the density operator
formalism shown here. See
Baylis\cite{BaylisCommentOnSquareSpinorsDiracEqn_0202060} for an
enlightening discussion of various methods of geometrically defining
the Dirac equation for spinors.} suggests that we should take a look
at geometric transformations of the generalized Dirac equation, and
see if these can be related to the internal
symmetries of elementary particles.


Suppose that $\iota{\chi}$ is a unit multivector
from the Dirac algebra that happens to square to
$\hat{1}$.  This could be one of the commuting roots of unity that
we could use to define a complete set of primitive
idempotents. We can use $\iota_\chi$ to
define a transformation on the Dirac algebra as
follows:
\begin{equation}\label{eq:ChiModsDirac}
M \to \iota_{\chi} M \iota_{\chi}.
\end{equation}
The above transformation preserves $0$ and $\hat{1}$, and if $M$ and
$N$ are in the Dirac algebra, then their
product and sum are transformed in a manner that preserves
multiplication and addition:
\begin{equation}\label{eq:ChiTransformIsOkay}
\begin{array}{rcl}
0 &\to& \iota_\chi 0 \iota_\chi = 0,\\
\hat{1} &\to& \iota_\chi \hat{1}\iota_\chi = \iota_\chi^2 =
\hat{1},\\
M+N &\to& \iota_\chi(M+N)\iota_\chi = (\iota_\chi M\iota_\chi) +
(\iota_\chi N\iota_\chi),\\
MN &\to& \iota_\chi (MN) \iota_\chi = (\iota_\chi M\iota_\chi)
(\iota_\chi M\iota_\chi).\end{array}
\end{equation}
We can also apply this transformation to $\nabla$,
and since multiplication and addition are preserved, we can
transform a generalized Dirac equation
into a new generalized Dirac
equation as follows:
\begin{equation}\label{eq:TransformedDiracEqn}
\begin{array}{rcl}
   &0 =& (\nabla + m)\Psi,\\
\to&0 =& (\iota_\chi\nabla\iota_\chi + \iota_\chi
m\iota_\chi)\iota_\chi\Psi,
\end{array}
\end{equation}
where we have multiplied on the right by $\iota_\chi$.


Suppose that $\Psi$ is an eigenstate of
$\iota_\chi$. This could come about, for example, if $\Psi$ were a
primitive idempotent (which we associate
with the elementary particles) and $\iota_\chi$ were one of the
commuting roots of unity that define that
primitive idempotent. Then
$\iota_\chi\;\Psi_\pm$ $=\pm \Psi_\pm$, with the sign depending on
the eigenvalue. This overall sign can be canceled
from Eq.~(\ref{eq:TransformedDiracEqn}):
\begin{equation}\label{eq:TransformedPsiPM}
\begin{array}{rcl}
0 &=& (\iota_\chi\nabla\iota_\chi + \iota_\chi m\iota_\chi)\Psi_\pm,\\
&=&(\iota_\chi\nabla\iota_\chi \pm m)\Psi_\pm,\\
&=&((\iota_\chi\hat{x}\iota_\chi)\partial_x+
(\iota_\chi\hat{y}\iota_\chi)\partial_y+
(\iota_\chi\hat{z}\iota_\chi)\partial_z+
(\iota_\chi\hat{t}\iota_\chi)\partial_t\pm m)\Psi_\pm.
\end{array}
\end{equation}
Now $\Psi_\pm$ is a function of $(x,y,z,t)$.  And the
partial derivatives all have specific values at each position in
spacetime that depend on the details of how the wave function
depends on position and time (and not on how the wave function is
defined in the Clifford algebra).  Therefore we can relate
symmetries of the wave function to geometric properties of the
primitive idempotent $\Psi_\pm$.


As a first example, let's put $\iota_\chi =
\at{it}$, and find out what sort of properties a
wave function might possess, if it is a
eigenvector of this operator:
\begin{equation}\label{eq:ItSymmetry}
\at{it} \Psi_{\pm\at{it}} = \pm \Psi_{\pm\at{it}},
\end{equation}
and in addition satisfies the Dirac wave equation:
\begin{equation}\label{eq:DiracIt}
0 = (\nabla+m) \Psi_{\pm\at{it}}.
\end{equation}
Transform the above equation with $\at{it}$ to
obtain:
\begin{equation}\label{eq:DiracWithParity}
\begin{array}{rcl}
0&=&\at{it}(\nabla + m)\at{it}\at{it}\Psi_{\pm\at{it}}(x,y,z,t)\at{it},\\
&=& (\at{it}\hat{x}\at{it}\partial_x
+\at{it}\hat{y}\at{it}\partial_y +\at{it}\hat{z}\at{it}\partial_z
+\at{it}\hat{t}\at{it}\partial_t+\at{it}m\at{it})
\at{it}\Psi_{\pm\at{it}}(x,y,z,t)\at{it},\\
&=& (-\hat{x}\partial_x -\hat{y}\partial_y -\hat{z}\partial_z
+\hat{t}\partial_t+m)(\pm\Psi_{\pm\at{it}}(x,y,z,t))\at{it}.
\end{array}
\end{equation}
The trailing $\at{it}$ can be removed by multiplying on the right by
$\at{it}$.  This removes all the $\at{it}$ from the
above.  To get the differential operator back into
$\nabla$ form, make a substitution to the
coordinates:
\begin{equation}\label{eq:ParityTransform}
\begin{array}{ccc}
x \to -x,\;\;\;\;&y \to -y,\;\;\;\;&z \to -z.
\end{array}
\end{equation}
This will negate the partial derivatives with respect to $x$, $y$,
and $z$.  The result is:
\begin{equation}\label{eq:ParityXformedDirac}
0 = (\nabla + m) (\pm \Psi_{\pm\at{it}}(-x,-y,-z,t)).
\end{equation}
Comparing with Eq.~(\ref{eq:DiracIt}), we see that this will happen
automatically if
\begin{equation}\label{eq:ParityDefn}
\Psi_{\pm\at{it}}(x,y,z,t) = \pm \Psi_{\pm\at{it}}(-x,-y,-z,t),
\end{equation}
which\marginpar{\em Parity as an eigenvalue relation} is just the
definition of even ($+$) and odd ($-$) parity in a
wave function. Therefore, we define the geometric (or ``internal'')
parity operator $P$ as:
\begin{equation}\label{eq:ParityOpIsiT}
P = \at{it}.
\end{equation}
Note that in deriving the above we needed to compute how
$\at{it}$ transformed the unit vectors only.  In
the spinor version of this sort of thing one must also
figure out how $P$ changes the spinors, and since the spinors carry
arbitrary phases the process is more
complicated and less definitive.


There is a subtle distinction between the way we have defined the
parity operator $P$ in
Eq.~(\ref{eq:ParityOpIsiT}) and how this is done in most of the
literature on particle physics. We have
defined the parity operator as an operator that lives in the same
Clifford algebra as everything else.  The more usual practice is to
define the parity operator as something which acts on the state,
$\psi(x,y,z,t)$ and converts it into a new state $\psi'(x,y,z,t) =
\pm \psi(-x,-y,-z,t)$, with a factor of $\pm 1$ included as a
consequence of the ``internal parity'' of the associated particle.
In this book, since our wave functions $\Psi$, contain more than one
particle, the parity operator is an operator no
different from the rest of the unit multivectors.




We now move on to analyze the transformations of the other two
unoriented Dirac unit multivectors,
$\at{ixyz}$ and $\at{ixyzt}$. As before, we assume
eigenvectors of these operators:
\begin{equation}\label{eq:PsiEValEqns}
\begin{array}{rcl}
\at{ixyz}\Psi_{\pm\at{ixyz}} &=& \pm \Psi_{\pm\at{ixyz}},\\
\at{ixyzt}\Psi_{\pm\at{ixyzt}} &=& \pm \Psi_{\pm\at{ixyzt}}.
\end{array}
\end{equation}
The unoriented Dirac unit multivectors
transform the basis vectors as follows:
\begin{equation}\label{eq:ActionOtherNeutrals}
\begin{array}{c|ccc|}
\hat{\alpha}& \at{it}\hat{\alpha}\at{it}
&\at{ixyz}\hat{\alpha}\at{ixyz} &
\at{ixyzt}\hat{\alpha}\at{ixyzt}\\ \hline %
\hat{x} &-\hat{x} & +\hat{x}& -\hat{x}\\
\hat{y} &-\hat{y} & +\hat{y}& -\hat{y}\\
\hat{z} &-\hat{z} & +\hat{z}& -\hat{z}\\
\hat{t} &+\hat{t} & -\hat{t}& -\hat{t}\\ \hline
\end{array}.
\end{equation}
Applying the same analysis as with $\at{it}$, we find that the time
reversal operator $T$ is:
\begin{equation}\label{eq:TimeReversal}
T = \at{ixyz}.
\end{equation}
A\marginpar{\em Time reversal } wave function
$\Psi_{\pm\at{ixyz}}$ that is an
eigenstate of $T$ with eigenvalue $\pm 1$ and
satisfies the Dirac equation, is either negated or not upon the
action of time reversal according to the eigenvalue:
\begin{equation}\label{eq:TimeReversEqns}
\begin{array}{rcl}
\at{ixyz}\Psi_{\pm\at{ixyz}} &=& \pm \Psi_{\pm\at{ixyz}},\\
\Psi_{\pm\at{ixyz}}(x,y,z,t) &=& \pm \Psi_{\pm\at{ixyz}}(x,y,z,-t).
\end{array}
\end{equation}
Again, note that our definition of time reversal is different from
the usual.  And as before, since our states are operators, we need
only define time reversal as a transformation of operators, not
spinors. For an elementary discussion of the usual time
reversal operator when applied to operators as
well as fermions and bosons, see \cite[$\sec$ 4.4]{Sakurai}.


The remaining unoriented unit multivector, $\at{ixyzt}$ anticommutes
with all of the vectors and consequently is associated with the
combination of parity and time reversal.
We stress that this is a state operator that does not
commute with our $P$ or $T$ rather than an
operation that is performed on wave\marginpar{\em PT as an operator
on internal states.} functions:
\begin{equation}\label{eq:PTdefined}
\at{PT} = \at{ixyzt}.
\end{equation}
If $\Psi_{\pm\at{ixyzt}}$ is an eigenstate of
$\at{PT}$, and a solution of the generalized Dirac
equation, then it satisfies the combined
parity and time reversal
equation:
\begin{equation}\label{eq:PTOnPsi}
\Psi_{\pm\at{ixyzt}}(x,y,z,t) = \pm
\Psi_{\pm\at{ixyzt}}(-x,-y,-z,-t).
\end{equation}
This is also the same as the helicity operator we
will discuss later.

The other unit multivectors are oriented and
correspond to various possible spacetime symmetries
of a solution to the Dirac equation. For
example, $\at{zt}$ anticommutes with $\hat{z}$ and $\hat{t}$ and
commutes with the rest of the vectors, and therefore can be
associated with wave functions that are even or odd with respect to
the combination of time reversal and mirror
reflection in the $z$
direction. Some of these symmetries are very important but are best
understood in the context of exponential functions of multivectors
and we will discuss them later.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ParityOpIsiT})
\section{\label{sec:MatrixReps}  Matrix Representations}



The usual way of defining a matrix representation
of a Clifford algebra is to
specify the matrices corresponding to the basis vectors of the
Clifford algebra. One multiplies these together to get the matrices
for the various unit multivectors, if
desired.  This technique doesn't give much of a hint on how one
finds the matrix representation to begin with.


Since the density operator formalism associates quantum states with
the primitive idempotents, a more
natural way of defining a matrix
representation is through the primitive
idempotents.  We will now illustrate the
technique by finding a representation for the Dirac
algebra in $4\times 4$ matrices.

As with the $3\times 3$ matrices, it is easy to define the diagonal
$4\times 4$ primitive idempotent matrices.  There are four such
matrices.  We will call them $\rho_{\pm\pm}$ as follows:
\begin{equation}\label{eq:FourPIs}
\begin{array}{cc}
\rho_1 = \rho_{--} = \begin{array}({cccc}) %
1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{array} &
\rho_2 = \rho_{-+} = \begin{array}({cccc}) %
0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array} \\
&\\
\rho_3 = \rho_{+-} = \begin{array}({cccc}) %
0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0\end{array} &
\rho_4 = \rho_{++} = \begin{array}({cccc}) %
0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{array}
\end{array}
\end{equation}
We will associate the above four primitive idempotent matrices with
four primitive idempotent elements of
the Dirac algebra. These are the
diagonalized states.  But this choice is not
quite enough to define the representation.  We
need to choose one further primitive idempotent which we will call
$\rho_D$ where ``$D$'' stands for
``democratic''. It is the only idempotent matrix
that has all elements equal:
\begin{equation}\label{eq:DemocraticPI}
\rho_D = \frac{1}{4}\begin{array}({cccc})
1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{array}
\end{equation}


We've specified five $4x4$ primitive idempotent
matrices. Together, these matrices define a matrix
representation in the following way. Consider
products of the form:
\begin{equation}\label{eq:ProdsPIfours}
U_{mn} = 4 \rho_m\;\rho_D\;\rho_n,
\end{equation}
where $\rho_m$ and $\rho_n$ are independently any of
the four diagonal primitive idempotent matrices.  Then the matrix
$U_{mn}$ has all its elements zero except for the position $(m,n)$
where it has a one. For\marginpar{\em Fully specifying a matrix
representation requires the diagonalized states plus the democratic
state.} example:
\begin{equation}\label{eq:Example1mnDefn}
\begin{array}{rcl}
U_{24} &=& 4\rho_2\;\rho_D\;\rho_4,\\
&=&4\begin{array}({cccc})
0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array} \frac{1}{4}
\begin{array}({cccc})1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{array}
\begin{array}({cccc})
0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{array}\\
&&\\
&=&\begin{array}({cccc})
0&0&0&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{array}.
\end{array}
\end{equation}
Therefore if we specify these five primitive
idempotents, then the representation is
defined in that we can give a geometric interpretation of a $4\times
4$ matrix.


Of course we cannot pick any arbitrary set of $5$ primitive
idempotents and use them to define a
representation of the Dirac algebra.  The five primitive idempotent
Clifford algebra elements we choose have to match the algebra of the
corresponding $4\times 4$ primitive idempotent matrix elements of
Eq.~(\ref{eq:FourPIs}) and Eq.~(\ref{eq:DemocraticPI}).


For matching the algebras, we do not need to verify that addition
matches.  The Clifford algebra and the matrix algebra are both
linear spaces over their basis elements and addition will be
automatically equivalent.  It is in the multiplication that there
may be differences.


If it were not for the democratic primitive idempotent,
$\rho_D$, our task would be very simple.  The
diagonal primitive idempotents multiply very simply.
Any two different elements annihilate; they
multiply to zero. So we need only analyze how the democratic
primitive idempotent interacts with the others.


In the Pauli algebra, the diagonal primitive
idempotents correspond to the $\vec{z}$
and $-\vec{z}$ directions. The democratic primitive idempotent of
the $2\times 2$ matrices corresponds to
the Clifford algebra primitive idempotent $0.5(1+\sigma_x)$.  That
is:
\begin{equation}\label{eq:PauliDemocrat}
\frac{1}{2}\begin{array}({cc})1&1\\1&1\end{array} =
\frac{1}{2}(1+\sigma_x).
\end{equation}
There is a clue here.  The Pauli diagonal
primitive idempotents are perpendicular to the Pauli democratic
primitive idempotent. What we need is a
generalization of this relation to more complicated Clifford
algebras.


In the Pauli algebra, the fact that the two
vectors, $\vec{z}$ and $\vec{x}$, are perpendicular, can be written
in the geometric density operator language as:
\begin{equation}\label{eq:PerpPaulis}
P(z\to x) = 0.5 \rho_z = \rho_z\;\rho_x\;\rho_z.
\end{equation}
We need to understand how this comes about so that we can pick out
the structural elements of $\rho_x$ that cause it to be allowable as
a democratic primitive idempotent
when $\rho_z$ and $\bar{\rho_z}$ are chosen as the
diagonal primitive idempotents. Accordingly, rewrite
the above with $\rho_x$ written out as a sum of unit
multivectors:
\begin{equation}\label{eq:PerpPaulisWO}
\begin{array}{rcl}
P(z\to x) &=& \rho_z\;0.5(\hat{1}+\hat{x})\;\rho_z,\\
&=& 0.5 \rho_z\;\hat{1}\;\rho_z + 0.5\rho_z\;\hat{x}\;\rho_z.
\end{array}
\end{equation}
In the above form, the transition probability
has been written as two expectation
values with respect to the quantum state
$\rho_z$. Rewriting these in expectation value form we have:
\begin{equation}\label{eq:PerpPauliXV}
\begin{array}{rcl}
P(z\to x) &=& \la 0.5\hat{1} \ra_z + \la 0.5\hat{x}\ra_z,\\
&=&0.5 \rho_z + 0.
\end{array}
\end{equation}
Thus the transition probability is the expectation value of the
scalar part of $\rho_x$; the vector part of $\rho_x$ has an
expectation value of zero.


Now let us return to the $4\times 4$ problem and compute the
transition probability for going from
a diagonal primitive idempotent to the democratic primitive
idempotent.  For
example:
\begin{equation}\label{eq:TransDiagnDemoc}
\begin{array}{rcl}
P(\rho_2\to \rho_D) &=&\rho_2\;\rho_D\;\rho_2,\\
&=&\begin{array}({cccc})0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array}
\frac{1}{4}
\begin{array}({cccc})1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{array}
\begin{array}({cccc})0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array},\\
&=&\frac{1}{4}\rho_2.
\end{array}
\end{equation}
Therefore we require that the primitive idempotent from the Clifford
algebra that we choose to be represented by $\rho_D$ have an
expectation value of $1/4$ with respect to any of the diagonal
primitive idempotents of the Clifford algebra.  This is different
from the value $1/2$ that was obtained in the case of the Pauli
algebra, because the primitive idempotents of
the Dirac algebra have scalar values of $1/4$
instead of $1/2$.


Putting all this together, we have a procedure for designing a
representation of the Dirac algebra.  First choose two commuting
roots of unity.  These define four commuting
primitive idempotents.  Assign
them in any order to the four diagonal primitive
idempotents $\rho_1$, $\rho_2$, $\rho_3$, and $\rho_4$. Second, find
the expectation values for the unit multivectors with respect to
$\rho_1$, $\rho_2$, $\rho_3$ and $\rho_4$.  Some of these
expectation values will be zero.  Find a
primitive idempotent that can be written from these zero expectation
unit multivectors and assign it to $\rho_D$.  The result will be a
representation of the Dirac algebra in $4\times
4$ matrices.


We can make one more improvement to the procedure.  The process
``find the expectation values for the unit multivectors with respect
to $\rho_1$, $\rho_2$, $\rho_3$, and $\rho_4$'' can be time
consuming. There is a short cut for this calculation.  Write
$\rho_{\pm\pm}$ as a product as follows:
\begin{equation}\label{eq:ExpandRho}
\rho_{\pm\pm} = 0.25(1\pm\iota_1)(1\pm\iota_2),
\end{equation}
where $\iota_1$ and $\iota_2$ are commuting roots of unity.  Suppose
that $\iota_1$ and $\iota_2$ are Dirac unit multivectors.  Let
$\chi$ be the Dirac unit multivector for
which we wish to compute the expectation value. The computation is
as follows:
\begin{equation}\label{eq:ExpandExpVal}
\begin{array}{rcl}
\la \chi\ra_{\pm\pm} &=& \rho_{\pm\pm}\;\chi\;\rho_{\pm\pm},\\
&=&0.25(1\pm\iota_1)(1\pm\iota_2)\;\chi\;
0.25(1\pm\iota_1)(1\pm\iota_2).
\end{array}
\end{equation}
To compute the above, let us attempt to bring $\chi$ to the front of
the product of idempotents, that is, to factor
$\chi$ out. If $\chi$ commutes with $\iota_1$ and
$\iota_2$, then we can factor it out and the two primitive
idempotents multiply together. But if $\chi$
anticommutes with either or both of $\iota_1$ and
$\iota_2$, then in factoring out $\chi$ we will change the sign of
either or both of $\iota_1$ and $\iota_2$. That is, we will change
$(1\pm\iota_1)$ to $(1\mp\iota_1)$ or the same for $\iota_2$.  But
there are two copies of $(1\pm\iota_1)$ in the above, and if one of
them is changed to $(1\mp\iota_1)$ then the two of them will
annihilate and the expectation value will be zero.


Therefore, the unit multivectors that we can
use to make the democratic primitive
idempotent consist exactly of those unit
multivectors that
anticommute with either of $\iota_1$ or
$\iota_2$. Another way of saying the same thing is that we can use
any unit multivector that anticommutes with either of $\iota_1$ or
$\iota_2$. Since $\iota_1$ and $\iota_2$ form a set of commuting
roots of unity, it is clear that we cannot use $\iota_1$, $\iota_2$,
or their product, $\iota_1\iota_2$. Beyond that, if there were
another (non scalar) unit multivector $\chi$ that commuted with both
$\iota_1$ and $\iota_2$, we could use either $\chi$ or $i\chi$,
whichever squares to $+1$, to extend the set of commuting roots of
unity from having $2$ elements to a set of size $3$.  Clearly the
Dirac algebra can only have two commuting roots of unity,
so the simple rule for the choice of
democratic primitive idempotent is that we choose
a unit multivector that is not included in
the set $\{\iota_1, \iota_2, \iota_1\iota_2\}$.


In choosing the democratic primitive
idempotent there are a few more subtle
points. If we use $\iota_3$ and $\iota_4$ as the commuting roots of
unity that define the democratic primitive idempotent, then we have
four choices for the primitive idempotent according as we choose the
signs in:
\begin{equation}\label{eq:DemocratPIsigns}
\rho_D = 0.25(1\pm \iota_3)(1\pm \iota_4).
\end{equation}
This means that for any given choice of the commuting roots of
unity we have four different possible
democratic primitive idempotents. Suppose that we take the positive
signs. Then the democratic primitive
idempotent becomes:
\begin{equation}\label{eq:PlusDemoPIchoice}
\begin{array}{rcl}
\rho_D &=& 0.25(1+\iota_3)(1+\iota_4),\\
&=&0.25(1 + \iota_3 + \iota_4 + \iota_3\iota_4),
\end{array}
\end{equation}
so the democratic primitive idempotent actually has three non scalar
portions, not two.  And we must have that the expectation values of
each of these is zero.  This means that some pairs of commuting
roots of unity for the democratic primitive
idempotent, that would seem to be
compatible with the pair of commuting roots of unity chosen for the
diagonal roots of unity are actually not compatible.
We have to check not only the roots of unity,
but also their products for compatibility.


A more elegant way of defining whether the sets $\{\iota_1,
\iota_2\}$ and $\{\iota_3,\iota_4\}$ are compatible for defining the
diagonal and democratic primitive
idempotents is to look at the groups
generated by these sets under multiplication. We will designate the
group generated by multiplication as
``$G(\iota_1,\iota_2)$'', so the
groups we are interested in here have the following elements:
\begin{equation}\label{eq:GroupsGenByIotas}
\begin{array}{rcl}
G(\iota_1,\iota_2) &=& \{\hat{1},\iota_1,\iota_2,\iota_1\iota_2\},\\
G(\iota_3,\iota_4) &=& \{\hat{1},\iota_3,\iota_4,\iota_3\iota_4\}.
\end{array}
\end{equation}
Then $\{\iota_1,\iota_2\}$ is compatible with $\{\iota_3,\iota_4\}$
if and only if
\begin{equation}\label{eq:IotaCompatibDefn}
G(\iota_1,\iota_2) \cap G(\iota_3,\iota_4) = \{ \hat{1} \},
\end{equation}
that is, they are compatible if and only if the only element that
their generated groups share is the scalar.






% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ThreeMatrixPIineq})
\section{\label{sec:GammaMatrices}  Dirac's Gamma Matrices}



The derivation of this process has been somewhat complicated and
probably confusing as well. To solidify the concepts, let us use
this technique to work out some
representations of the Dirac
algebra in $4\times 4$ matrices.
The Dirac algebra is generally treated as a complex
algebra, however, as an illustration, let us find a real
representation of the real Dirac algebra.
This restricts us to using only the positive signature unit
multivectors listed in
Eq.~(\ref{eq:DiracBilinSqrs}):
\begin{equation}\label{eq:DiracPosBilin}
\begin{array}{ccccc}
\hat{1},&\hat{x},&\hat{y},&\hat{z},&\at{xt},\\
\at{yt},&\at{zt}^2,&\at{yzt},&\at{xzt},&\at{xyt}.
\end{array}
\end{equation}
Of course we cannot use $\hat{1}$.  The remaining $9$ unit
multivectors commute or anticommute
 with each other according to the ``$+$'' and
``$-$'' signs of the following table:
\begin{equation}\label{eq:DiracPosComTable}
\begin{array}{c|ccc|ccc|ccc|}
&\hat{x}&\hat{y}&\hat{z}&\at{xt}&\at{yt}&\at{zt}&\at{yzt}&\at{xzt}&\at{xyt}\\
\hline %
\hat{x}  &+&-&-&-&+&+&-&+&+\\
\hat{y}  &-&+&-&+&-&+&+&-&+\\
\hat{z}  &-&-&+&+&+&-&+&+&-\\ \hline %
\at{xt}  &-&+&+&+&-&-&-&+&+\\
\at{yt}  &+&-&+&-&+&-&+&-&+\\
\at{zt}  &+&+&-&-&-&+&+&+&-\\ \hline %
\at{yzt} &-&+&+&-&+&+&+&-&-\\
\at{xzt} &+&-&+&+&-&+&-&+&-\\
\at{xyt} &+&+&-&+&+&-&-&-&+\\ \hline %
\end{array}
\end{equation}.
If we were making a representation for some
particular purpose, we would choose two of these unit
multivectors so that our representation
would be diagonal for something we were particularly
interested in. For example, if we were associating spin$-1/2$ with
the $\at{iyz}$, $-\at{ixz}$, and $\at{ixy}$ elements we might want
our representation to have $\at{zt}$ diagonal, as is traditional.
The first two elements that commute in the above
table are $\hat{x}$ and $\at{yt}$.  We will choose these for the
commuting roots of unity that define the diagonal
primitive idempotents. We can assign the four primitive
idempotents to $\rho_n$ in any arbitrary
order. There are $4!=24$ choices.  A natural order is:
\begin{equation}\label{eq:DiagPIsExA}
\begin{array}{rcrcl}
\rho_{--} &=& \rho_1 &=& 0.25(1-\hat{x})(1-\at{yt}),\\
\rho_{-+} &=& \rho_2 &=& 0.25(1-\hat{x})(1+\at{yt}),\\
\rho_{+-} &=& \rho_3 &=& 0.25(1+\hat{x})(1-\at{yt}),\\
\rho_{++} &=& \rho_4 &=& 0.25(1+\hat{x})(1+\at{yt}).\\
\end{array}
\end{equation}
In defining the four diagonal primitive
idempotents we've used three unit
multivectors; $\hat{x}$, $\at{yt}$ and also
their product, $\at{xyt}$.


For the democratic primitive
idempotent we can use any two of the
remaining six positive signature unit
multivectors:
\begin{equation}\label{eq:DiracPosComTable2}
\begin{array}{c|cc|cc|cc|}
&\hat{y}&\hat{z}&\at{xt}&\at{zt}&\at{yzt}&\at{xzt}\\
\hline %
\hat{y}  &+&-&+&+&+&-\\
\hat{z}  &-&+&+&-&+&+\\ \hline %
\at{xt}  &+&+&+&-&-&+\\
\at{zt}  &+&-&-&+&+&+\\ \hline %
\at{yzt} &+&+&-&+&+&-\\
\at{xzt} &+&+&+&+&-&+\\ \hline
\end{array}.
\end{equation}
There are nine pairs of commuting roots of unity:
\begin{equation}\label{eq:DiracPosPairsOne}
\begin{array}{ccccc}
\{\hat{y},\at{xt}\},&    \{\hat{y},\at{zt}\},&
\{\hat{y},\at{yzt}\},&   \{\hat{z},\at{xt}\},&
\{\hat{z},\at{yzt}\},\\
\{\hat{z},\at{xzt}\},&  \{\at{xt},\at{xzt}\},&
\{\at{zt},\at{yzt}\},& \{\at{zt},\at{xzt}\}.&
\end{array}
\end{equation}
Some of these can't be used because the groups they
generate include non scalar elements that are in
$G(\iota_1,\iota_2) =$ $G(\hat{x},\at{yt}) =$
$\{\hat{1},\hat{x},\at{yt},\at{xyt}\}$. Since $\hat{y}\;\at{xt} =
-\at{xyt}$ and $\at{xyt}\in G(\hat{x},\att{yt})$, we cannot use the
pair $\{\hat{y},\at{xt}\}$. Since $\hat{z}\;\at{yzt} = -\at{yt}$,
and $\at{yt} \in G(\hat{x},\at{yt})$, we cannot use the pair
$\{\hat{z},\at{yzt}\}$. Finally, $\at{zt}\;\at{xzt} = \hat{x}$, so
we cannot use the pair $\{\at{zt},\at{xzt}\}$.  This leaves six
viable pairs for creating the democratic primitive
idempotent:
\begin{equation}\label{eq:DiracPosPairsTwo}
\begin{array}{ccc}
\{\hat{y},\at{zt}\},& \{\hat{y},\at{yzt}\},&   \{\hat{z},\at{xt}\},\\
\{\hat{z},\at{xzt}\},&  \{\hat{xt},\at{xzt}\},&
\{\hat{zt},\at{yzt}\}.
\end{array}
\end{equation}
We will choose the first of these pairs, $\{\hat{y},\at{zt}\}$. We
now have to choose which of the four possible primitive
idempotents to use, $\rho_D =$
$0.25(1\pm \hat{y})(1\pm\at{zt})$. We will take the $++$ case.  This
defines our representation of the Dirac
algebra:
\begin{equation}\label{eq:DiracPosRepDefined}
\begin{array}{rcl}
\rho_1 &=& 0.25(1-\hat{x})(1-\at{yt}),\\
\rho_2 &=& 0.25(1-\hat{x})(1+\at{yt}),\\
\rho_3 &=& 0.25(1+\hat{x})(1-\at{yt}),\\
\rho_4 &=& 0.25(1+\hat{x})(1+\at{yt}).\\
\rho_D &=& 0.25(1+\hat{y})(1+\at{zt}).
\end{array}
\end{equation}
After deciding that we would have $\hat{x}$ and $\at{yt}$ diagonal,
we had $4!=24$ choices for the order of $\rho_1$, $\rho_2$, $\rho_3$
and $\rho_4$.  Then we had six pairs of commuting roots of
unity available for the
democratic primitive idempotent.
Finally we had four sign choices for the democratic
primitive idempotent. There are therefore $24\times 6 \times 4$ $=
24^2 = 576$ representations of the Dirac
algebra that diagonalize
$\hat{x}$ and $\at{yt}$ (and use only unit multivectors).


Our representation,
Eq.~(\ref{eq:DiracPosRepDefined}), is written in the geometric
language and is quite elegant.  To convert it into matrix form, one
could use a fairly large amount of algebra but there is a simple
trick.  First, since we have defined the diagonal
primitive idempotents, we can quickly
find the solution for this part of the matrix.  To simplify our
notation for the off diagonal
elements, let
$U_{nm}$ designate the matrix which is all zero except for a $1$ in
position $(n,m)$ and rewrite Eq.~(\ref{eq:DiracPosRepDefined}) as:
\begin{equation}\label{eq:DiracPosUsualOne}
\begin{array}{rcl}
U_{11} &=& 0.25(\hat{1}-\hat{x}-\at{yt}+\at{xyt}),\\
U_{22} &=& 0.25(\hat{1}-\hat{x}+\at{yt}-\at{xyt}),\\
U_{33} &=& 0.25(\hat{1}+\hat{x}-\at{yt}-\at{xyt}),\\
U_{44} &=& 0.25(\hat{1}+\hat{x}+\at{yt}+\at{xyt}).
\end{array}
\end{equation}
The above amounts to four equations in four unknowns.  But they are
easier to solve then one might expect:
\begin{equation}\label{eq:DiracPosUsualTwo}
\begin{array}{rcl}
+U_{11}+U_{22}+U_{33}+U_{44} &=& \hat{1},\\
-U_{11}-U_{22}+U_{33}+U_{44} &=& \hat{x},,\\\
-U_{11}+U_{22}-U_{33}+U_{44} &=& \at{yt},,\\\
+U_{11}-U_{22}-U_{33}+U_{44} &=& \at{xyt},
\end{array}
\end{equation}
The same principle will be applied to the off diagonal
elements.


The off diagonal terms will have a similar form.
To find them, first note that we can write
\begin{equation}\label{eq:DiracPosUsualThree}
\begin{array}{rcl}
M_{mn} &=& 4\rho_m\; \rho_D\;\rho_n,\\
&=&\rho_m(\hat{1}+\hat{y}+\at{zt}+\at{yzt})\rho_n,\\
&=&\rho_m\;\rho_n +\rho_m(\hat{y}+\at{zt}+\at{yzt})\rho_n,\\
&=&0 + \rho_m(\hat{y}+\at{zt}+\at{yzt})\rho_n.
\end{array}
\end{equation}
The zero in the above follows from the fact that we are considering
off diagonal terms so $m \neq n$ and so $\rho_m$ will
annihilate $\rho_n$.  We will factor the three
inner unit multivectors,  $\hat{y}$,
$\at{zt}$, and $\at{yzt}$ to the left side of $\rho_m$. Since these
terms do not commute with $\rho_m$, doing this will change $\rho_m$.
The change will be that the signs of the commuting roots of
unity that make up $\rho_m$ will be changed if
that commuting root of unity does not commute with the unit
multivector we are factoring out. The result of these sign changes
will be that $\rho_m$ will be converted to some other $\rho_m'$.
Sometimes $\rho_m'$ will be equal to $\rho_n$ and these terms will
not annihilate.


This is easier done than described.  First we make a table showing
which of the unit multivectors we will
factor, commute with which of the roots
of unity of the diagonal
primitive idempotents:
\begin{equation}\label{eq:DiracPosCommRelations}
\begin{array}{c|cc|}
&\hat{x}&\at{yt}\\ \hline
\hat{y} &-&-\\
\at{zt} &+&-\\
\at{yzt}&-&+\\ \hline
\end{array}
\end{equation}
With the above, we can read off how $\rho_m$ will be altered by the
factoring out of our three terms:
\begin{equation}\label{eq:DiracPosFactorings}
\begin{array}{c|ccc|}
             &\hat{y}  &\at{zt}  &\at{yzt}\\ \hline
\rho_{--} \to&\rho_{++}&\rho_{-+}&\rho_{+-}\\
\rho_{-+} \to&\rho_{+-}&\rho_{--}&\rho_{++}\\
\rho_{+-} \to&\rho_{-+}&\rho_{++}&\rho_{--}\\
\rho_{++} \to&\rho_{--}&\rho_{+-}&\rho_{-+}\\ \hline
\end{array}.
\end{equation}
The above was obtained by negating signs in the $\rho_m$ on the left
according to the table of Eq.~(\ref{eq:DiracPosCommRelations}). It's
easier to use if we rewrite it as:
\begin{equation}\label{eq:DiracPosFactoring2}
\begin{array}{c|ccc|}
          &\hat{y}&\at{zt}&\at{yzt}\\ \hline
\rho_1 \to&\rho_4 &\rho_2 &\rho_3\\
\rho_2 \to&\rho_3 &\rho_1 &\rho_4\\
\rho_3 \to&\rho_2 &\rho_4 &\rho_1\\
\rho_4 \to&\rho_1 &\rho_3 &\rho_2\\ \hline
\end{array}.
\end{equation}
We can read off the terms that are not annihilated
in the off diagonal elements from the above.
The action of $\hat{y}$ is to convert $\rho_1 \to \rho_4$,
$\rho_2\to \rho_3$, $\rho_3\to\rho_2$ and $\rho_4\to\rho_1$.
Therefore the matrix elements that $\hat{y}$ will survive in are
$U_{14}$, $U_{23}$, $U_{32}$, and $U_{41}$:
\begin{equation}\label{eq:DiracPosY}
\begin{array}{rclcl}
U_{14} &=& \hat{y}\rho_4 &=& 0.25(+\hat{y}-\at{xy}+\hat{t}-\at{xt}),\\
U_{23} &=& \hat{y}\rho_3 &=& 0.25(+\hat{y}-\at{xy}-\hat{t}+\at{xt}),\\
U_{32} &=& \hat{y}\rho_2 &=& 0.25(+\hat{y}+\at{xy}+\hat{t}+\at{xt}),\\
U_{41} &=& \hat{y}\rho_1 &=& 0.25(+\hat{y}+\at{xy}-\hat{t}-\at{xt}).
\end{array}
\end{equation}
As with the diagonal terms, the above are four
equations in four unknowns that have an easy to find solution:
\begin{equation}\label{eq:DiracPosYSoln}
\begin{array}{rcl}
+U_{14}+U_{23}+U_{32}+U_{41} &=& \hat{y},\\
-U_{14}-U_{23}+U_{32}+U_{41} &=& \at{xy},,\\\
+U_{14}-U_{23}+U_{32}-U_{41} &=& \hat{t},\\
-U_{14}+U_{23}+U_{32}-U_{41} &=& \at{xt}..
\end{array}
\end{equation}
At this point in time, we've found the matrix
representation for $\hat{x}$ in
Eq.~(\ref{eq:DiracPosUsualTwo}), and the
representations for $\hat{y}$ and $\hat{t}$ in
Eq.~(\ref{eq:DiracPosYSoln}).  All that is left is $\hat{z}$, which
we can find by treating $\at{yzt}$ in the same manner as we just did
$\hat{y}$:
\begin{equation}\label{eq:DiracPosGetZ}
\begin{array}{rclcl}
U_{13}&=&\at{yzt}\rho_4&=&0.25(+\at{yzt}-\at{xyzt}+\hat{z}-\at{xz}),\\
U_{24}&=&\at{yzt}\rho_3&=&0.25(+\at{yzt}-\at{xyzt}-\hat{z}+\at{xz}),\\
U_{31}&=&\at{yzt}\rho_2&=&0.25(+\at{yzt}+\at{xyzt}+\hat{z}+\at{xz}),\\
U_{42}&=&\at{yzt}\rho_1&=&0.25(+\at{yzt}+\at{xyzt}-\hat{z}-\at{xz}).
\end{array}
\end{equation}
again four equations in four unknowns with the simple solution:
\begin{equation}\label{eq:DiracPosGotZ}
\begin{array}{rcl}
+U_{13}+U_{24}+U_{31}+U_{42} &=& \at{yzt},\\
-U_{13}-U_{24}+U_{31}+U_{42} &=& \at{xyztt},\\
+U_{13}-U_{24}+U_{31}-U_{42} &=& \hat{z},\\
-U_{13}+U_{24}+U_{31}-U_{42} &=& \at{xz}..
\end{array}
\end{equation}
We now have matrix representations for all
four basis vectors. Gathering them together:
\begin{equation}\label{eq:DiracPosGotAllgeo}
\begin{array}{rcl}
\hat{x} &=& -U_{11}-U_{22}+U_{33}+U_{44},\\
\hat{y} &=& +U_{14}+U_{23}+U_{32}+U_{41},\\
\hat{z} &=& +U_{13}-U_{24}+U_{31}-U_{42},\\
\hat{t} &=& +U_{14}-U_{23}+U_{32}-U_{41}. \\
\end{array}
\end{equation}
In the usual matrix notation:
\begin{equation}\label{eq:DiracPosGotAllMatrix}
\begin{array}{cc}
\hat{x}=\begin{array}({cccc})
-1&0&0&0\\0&-1&0&0\\0&0&+1&0\\0&0&0&+1\ennd{array},&
\hat{y}=\begin{array}({cccc})
0&0&0&+1\\0&0&+1&0\\0&+1&0&0\\+1&0&0&0\end{array},\\
&\\
\hat{z}=\begin{array}({cccc})
0&0&+1&0\\0&0&0&-1\\+1&0&0&0\\0&-1&0&0\end{array},&
\hat{t}=\begin{array}({cccc})
0&0&0&+1\\0&0&-1&0\\0&+1&0&0\\-1&0&0&0\end{array},
\end{array}
\end{equation}
As claimed, the above is a real representation of the Dirac
algebra.





% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TDGreps})
\section{\label{sec:TDGreps}  Traditional Gamma Representations}



Given a representation of Dirac's gamma
matrices, we can extract the geometric
description of it by reversing the above procedure. We need only
find which unit multivectors are
diagonal, and find the democratic
primitive idempotent.


A common definition of the gamma matrices\cite{PPATNM} is:
\begin{equation}\label{eq:CommonGammas}
\begin{array}{cc}
\hat{x}= \begin{array}({cccc})
0&0&0&+1\\0&0&+1&0\\0&-1&0&0\\-1&0&0&0\end{array} & %
\hat{y}=\begin{array}({cccc})
0&0&0&-i\\0&0&+i&0\\0&+i&0&0\\-i&0&0&0\end{array} \\
&\\
\hat{z}=\begin{array}({cccc})
0&0&+1&0\\0&0&0&-1\\-1&0&0&0\\0&+1&0&0\end{array} &
\hat{t}=\begin{array}({cccc})
+1&0&0&0\\0&+1&0&0\\0&0&-1&0\\0&0&0&-1\end{array} %
\end{array}
\end{equation}
Note that the above has signature the opposite of
the one that we are using.  One of the diagonal
roots of unity is given above, $\hat{t}$.


To get the other diagonal root of unity,
note that $\hat{x}$ and $\hat{y}$ multiply together to
give a diagonal matrix. But $\at{xy}$ squares to
$-1$, so the other diagonal root of unity is:
\begin{equation}\label{eq:CommonGOdrou}
\at{ixy} = \begin{array}({cccc})
+1&0&0&0\\0&-1&0&0\\0&0&+1&0\\0&0&0&-1\end{array}
\end{equation}
There is only one minor problem. The signs are reversed. To put
these into our (somewhat arbitrary) standard form that has
$\rho_{--}$ the diagonal primitive idempotent with $1$ in the top
left corner, we need to negate $\hat{t}$ and $\at{ixy}$.  Later we
will associate this root with spin, and the root
$\hat{t}$ with parity. Therefore the commuting roots
of unity are:
\begin{equation}\label{eq:CommonGcommroots}
\begin{array}{rcl}
\iota_1 &=& -\hat{t},\\
\iota_2 &=& -\at{ixy}.
\end{array}
\end{equation}
Thus the diagonal primitive idempotents are:
\begin{equation}\label{eq:CommonGdpis}
\rho_{\pm\pm} = 0.25(1\pm -\hat{t})(1\pm -\at{ixy}).
\end{equation}
The democratic primitive
idempotent takes a bit more work.


Computing $\at{xt} = \hat{x}\hat{t}$ gives:
\begin{equation}\label{eq:CommonGDpA}
\begin{array}{rcl}
\at{xt} &=& \begin{array}({cccc})
0&0&0&-1\\0&0&-1&0\\0&-1&0&0\\-1&0&0&0\end{array} \;\;\textrm{so}\\
&&\\
-\at{xt} &=& \begin{array}({cccc})
0&0&0&+1\\0&0&+1&0\\0&+1&0&0\\+1&0&0&0\end{array}.
\end{array}
\end{equation}
The same sort of inspiration suggests computing $\at{iyz}$ to
obtain:
\begin{equation}\label{eq:CommonGDpB}
\at{iyz} =
\begin{array}({cccc})
0&+1&0&0\\+1&0&0&0\\0&0&0&+1\\0&0&+1&0\end{array}
\end{equation}
These give $2$ out of the $3$ off diagonal sets
of elements that we need.  These $3$ are going to appear in the
geometric definition of the diagonal primitive
idempotent. The fourth term that appears
in the diagonal primitive idempotent is just the scalar.  Therefore,
we know that these two unit multivectors,
$-\at{xt}$ and $\at{iyz}$ are two of the commuting roots of
unity. In fact, they do commute
and square to $+1$. Therefore, the third term we need is simply
their product:
\begin{equation}\label{eq:CommonGDpC}
-\at{ixyzt} = \begin{array}({cccc})
0&0&+1&0\\0&0&0&+1\\+1&0&0&0\\0&+1&0&0\end{array}.
\end{equation}
Note that the above is commonly written as
$\gamma^5 = i\hat{t}\hat{x}\hat{y}\hat{z}$, the minus sign comes
about because we prefer to commute $\hat{t}$ around to the right.
The result is that we have:
\begin{equation}\label{eq:CommonGDpi}
\rho_D = 0.25(\hat{1}-\at{xt}+\at{iyz}-\at{ixyzt}).
\end{equation}
There are three commuting roots of unity
associated with the democratic primitive
idempotent, $\at{xt}$, $\at{iyz}$ and
$\at{ixyzt}$. Since we will associate $\at{ixy}$ with
spin in the $z-$ direction, we will associate $\at{iyz}$
with spin in the $x-$direction.  Note that spin in the $z$ and $x$
directions are incompatible
measurements, as is always the case between a
pair of commuting roots of unity when one is chosen from the
democratic primitive idempotent and the other from
the diagonal primitive idempotent.
The operator $\at{ixyzt}$ will be
associated with ``helicity'', and the operator
$\at{xt}$ will be associated with velocity (note,
velocity, \emph{not} momentum) in the $x$ direction.



The complete geometric description of the
representation is:
\begin{equation}\label{eq:CommonGdpis2}
\begin{array}{rclcl}
\rho_1 &=& \rho_{--} &=& 0.25(\hat{1} - -\hat{t})(\hat{1} - -\at{ixy}),\\
\rho_2 &=& \rho_{-+} &=& 0.25(\hat{1} - -\hat{t})(\hat{1} + -\at{ixy}),\\
\rho_3 &=& \rho_{+-} &=& 0.25(\hat{1} + -\hat{t})(\hat{1} - -\at{ixy}),\\
\rho_4 &=& \rho_{++} &=& 0.25(\hat{1} + -\hat{t})(\hat{1} + -\at{ixy}),\\
\rho_D &=&           && 0.25(\hat{1}-\at{xt})(\hat{1}+\at{iyz}).
\end{array}
\end{equation}
The group formed by the commuting roots of unity
for the diagonal primitive
idempotents is $G(-\hat{t},-\at{ixy}) =
$ $\{\hat{1},-\hat{t},-\at{ixy},\at{ixyt}\}$.  Note that, as we
required in our analysis of matrix representations
of Clifford algebras, this group has no non scalar
elements in common with the group formed by the commuting roots of
unity for the democratic primitive
idempotent,
$G(-\at{xt},\at{iyz})$. It is
interesting to note that to convert the above representation into a
representation for the signature $-+++$ we need
only modify one of the four commuting roots of unity,
that is, we replaced $-\hat{t}$ with $\at{it}$ or $-\at{it}$.



A slightly less common representation of the
Dirac algebra is the ``chiral representation'':
\begin{equation}\label{eq:ChiralGammas}
\begin{array}{cc}
\hat{x}= \begin{array}({cccc})
0&0&0&+1\\0&0&+1&0\\0&-1&0&0\\-1&0&0&0\end{array} & %
\hat{y}=\begin{array}({cccc})
0&0&0&-i\\0&0&+i&0\\0&+i&0&0\\-i&0&0&0\end{array} \\
&\\
\hat{z}=\begin{array}({cccc})
0&0&+1&0\\0&0&0&-1\\-1&0&0&0\\0&+1&0&0\end{array} &
\hat{t}=\begin{array}({cccc})
0&0&+1&0\\0&0&0&+1\\+1&0&0&0\\0&+1&0&0\end{array} %
\end{array}
\end{equation}
The above representation receives its name because the
helicity operator, $\gamma^5$
$=i\hat{t}\hat{x}\hat{y}\hat{z}$ $=-\at{ixyzt}$ 
is diagonalized:\footnote{One would suppose it should be called the
``helicity representation''.}
\begin{equation}\label{eq:ChiralGammaFive}
\gamma_5 = -\at{ixyzt} = \begin{array}({cccc})
-1&0&0&0\\0&-1&0&0\\0&0&+1&0\\0&0&0&+1\ennd{array}
\end{equation}
The $\hat{x}$, $\hat{y}$, and $\hat{z}$ representatives are
unchanged so the other diagonal operator is
$\at{ixy}$. As before, the signs are reversed and to put them into
our standard geometric form requires the roots of unity
as follows:
\begin{equation}\label{eq:ChiralGcommroots}
\begin{array}{rcl}
\iota_1 &=& -\at{ixyzt},\\
\iota_2 &=& -\at{ixy}.
\end{array}
\end{equation}
Thus the diagonal primitive idempotents are:
\begin{equation}\label{eq:CommonGdiagpis}
\rho_{\pm\pm} = 0.25(1\pm -\at{ixyzt})(1\pm -\at{ixy}).
\end{equation}
We continue on to the democratic primitive
idempotent.

As before, we look for the unit multivectors
that are off diagonal and constant.  An easy
catch is $\hat{t}$. And as before, $\at{iyz}$ is such:
\begin{equation}\label{eq:ChiralGDpB}
\at{iyz} =
\begin{array}({cccc})
0&+1&0&0\\+1&0&0&0\\0&0&0&+1\\0&0&+1&0\end{array}.
\end{equation}
The are commuting roots of unity, and as before, so is there
product:
\begin{equation}\label{eq:ChiralGDpC}
\at{iyzt} =
\begin{array}({cccc})
0&0&0&+1\\0&0&+1&0\\0&+1&0&0\\+1&0&0&0\end{array}.
\end{equation}
Thus we have the democratic primitive
idempotent of this
representation as:
\begin{equation}\label{eq:ChiralGDpi}
\rho_D = 0.25(\hat{1}+\hat{t}+\at{iyz}+\at{iyzt}).
\end{equation}
and the complete geometric description of the chiral
representation is:
\begin{equation}\label{eq:ChiralGdpis2}
\begin{array}{rclcl}
\rho_1 &=& \rho_{--} &=& 0.25(\hat{1} - -\at{ixyzt})(\hat{1} - -\at{ixy}),\\
\rho_2 &=& \rho_{-+} &=& 0.25(\hat{1} - -\at{ixyzt})(\hat{1} + -\at{ixy}),\\
\rho_3 &=& \rho_{+-} &=& 0.25(\hat{1} + -\at{ixyzt})(\hat{1} - -\at{ixy}),\\
\rho_4 &=& \rho_{++} &=& 0.25(\hat{1} + -\at{ixyzt})(\hat{1} + -\at{ixy}),\\
\rho_D &=&           && 0.25(\hat{1}+\hat{t})(\hat{1}+\at{iyz}).
\end{array}
\end{equation}
As usual, the groups formed by the commuting roots
of unity for the diagonal and democratic
primitive idempotents share only the scalar. Note that in comparison
with the usual representation, the above swaps the
parity $\at{it}$, and helicity
$\at{ixyzt}$ unit multivectors.  We now return to the usual
signature, $-+++$.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{\label{sec:Tricks} Algebra Tricks}

%\begin{quote}
\settowidth{\versewidth}{Now down the lines of tasselled pines the
yearning whispers wake--}
\begin{verse}[\versewidth]
%\em
His house is Gee \& Tellus' Sons,---so goes his jest with men---\\
He sold us Zeus knows what last year; he'll take us in again.\\
Disguised behind a livery-team, fur-coated, rubber-shod---\\
Yet Apis from the bull-pen lows---he knows his brother God!\\
$ $\\
Now down the lines of tasselled pines the yearning whispers wake---\\
Pithys of old thy love behold.  Come in for Hermes' sake!\\
How long since that so-Boston boot with reeling M\ae{n}ads ran?\\
\emph{Numen adest!} Let be the rest. Pipe and we pay, O Pan.
\end{verse}
%\end{quote}
$ $



\lettrine{I}{n order to maintain} a tight connection to the underlying
geometry, most of our calculations will be accomplished in the
geometric language.  Since quantum theory is usually taught by
choosing representations of symmetry groups or algebras, these
geometric techniques will be unfamiliar to many readers.  For
convenience, we've collected together some of the more common ones
in this chapter.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:ThreeMatrixPIineq})
\section{\label{sec:ExpAndTransforms} Exponentials and Transformations}


Define the exponential of a Clifford algebraic
element $M$ as usual:
\begin{equation}\label{eq:DefineExp}
\exp(M) = \hat{1} + M + M^2/2! + M^3/3! + ... .
\end{equation}
Since the exponent of a matrix is defined, and since one can
represent a Clifford algebra in matrix form, the above is well
defined.  We can check to see what happens when $M$ is various
things we've been working on.

Let $\iota$ be a root of unity, let $a$ be a
real number and compute $\exp(a\iota)$:
\begin{equation}\label{eq:ExpAiota}
\begin{array}{rcl}
e^{a\iota}
&=& \hat{1} + a\iota + a^2\iota^2/2! + a^3\iota^3/3! + ...,\\
&=& \hat{1} + a\iota + a^2\hat{1}/2! + a^3\iota/3! + ...,\\
&=& (1 + a^2/2! + ...)\hat{1} + (a +a^3/3! + ...)\iota,\\
&=&\cosh(a)\hat{1} + \sinh(a)\iota.
\end{array}
\end{equation}
Similarly, if $\chi_-$ squares to $-1$ and $a$ is a real number,
\begin{equation}\label{eq:ChiSqrsToM1}
e^{a\chi_-}=\cos(a)\hat{1}+\sin(a)\chi_-.
\end{equation}
If we limit ourselves to the subalgebra
generated by $\chi_-$, that is, elements of the form
\begin{equation}\label{eq:ChiGeneratedSA}
a_r\hat{1} + a_i\chi_-,
\end{equation}
where $a_r$ and $a_i$ are real numbers, we can treat $\chi_-$ as an
imaginary unit.  For example, given two
numbers of the above sort, their product:
\begin{equation}\label{eq:ProductComplexChi}
(a_r\hat{1}+a_i\chi_-)(b_r\hat{1}+b_i\chi_-) =
(a_rb_r-a_ib_i)\hat{1} + (a_ib_r+a_rb_i)\chi_-,
\end{equation}
acts just like the product of two complex numbers:
\begin{equation}\label{eq:ProductTwoComplx}
(a_r + ia_i)(b_r+ib_i) = (a_rb_r-a_ib_i) + i(a_rb_i+a_ib_r).
\end{equation}
This means that we can bring in the machinery of complex numbers in
making calculations.\footnote{This book is mostly concerned with
discrete symmetries of particles so we do not get into complex
analysis (i.e. things involving integration and differentiation)
here. However, it should be mentioned that David Hestenes'
works\cite{HestenesMvC} on the ``geometric calculus'' does this in a
way far more elegant, powerful, and geometrically natural than one
would suppose from the simple method shown here.}




In defining complete sets of primitive idempotents
we've been using complete sets of commuting roots of
unity. Our roots of unity have been quite
pedestrian, $\hat{x}$, $\at{yt}$, etc.  The Dirac
algebra includes much more general roots of
unity. For example, the following elements all square to $+1$:
\begin{equation}\label{eq:ExamplesRootsOfUnity}
\begin{array}{ccc}
0.6\hat{x}+0.8\hat{y},&& 2.6\hat{z}-2.4\hat{t},\\
0.6\at{yt}+0.8\hat{y},&\;\;\;\;& 2.6\hat{x}-2.4\at{yzt}.
\end{array}
\end{equation}
The elements in the right column are of ``mixed
signature'', that is, they are a sum over
unit multivectors of different signature. The
elements in the top row are made from vectors, but the elements in
the bottom row are multivectors.  In coming up with example
multivector roots of unity of the sort above, one must choose two
unit multivectors that imitate two unit
vectors, in that they must anticommute.


Among these more general roots of unity, one
can find sets of commuting roots, and from these one can define
complete sets of primitive idempotents.
Such complete sets can then be used to produce interesting
representations of the Dirac
algebra that you can use to pester\index{pester
students} your students or annoy the readers\index{annoy readers} of
your papers. It is a mathematical theorem of Clifford algebra that
any complete set of primitive idempotents
for a given Clifford algebra has the same
lattice, so this would seem to be an unnecessary
complication. However, we are associating the basis vectors with
coordinates for spacetime and this gives a physical interpretation
for such transformations.


If we replace $\hat{x}$ with $0.6\hat{x}+0.8\hat{y}$ (and modify
$\hat{y}$ accordingly), we recognize this transformation as
equivalent to a rotation around the $\hat{z}$ axis.
That is, modifying the coordinates by rotating the unit vectors
around the $z$ axis will have this effect on the vectors.  This is
the obvious physical interpretation of the upper left root of
unity shown in
Eq.~(\ref{eq:ExamplesRootsOfUnity}). Similarly, if we perform a
boost in the $-z$ direction to a coordinate system
moving with respect to the original one, we expect to see the $z$
and $t$ coordinates mixed in the manner seen in the upper right root
of unity of Eq.~(\ref{eq:ExamplesRootsOfUnity}).


We can produce transformations of the roots of unity (and therefore
the primitive idempotents, and so the
quantum states) by using
exponential functions as we will get to later in
this section. For the top row of Eq.~(\ref{eq:ExamplesRootsOfUnity})
these will physically correspond to rotations and
boosts. However, in doing this we will
also obtain transformations that move the usual roots of unity to
the mixed objects of the bottom row and these are more difficult to
give a physical interpretation. Nevertheless, the same mathematics
that produces the boosts and rotations also allow these, and we must
consider the possibility that they have a physical interpretation.


As a first example of an exponential
transformation, consider the unit multivector
$\at{xy}$, or rather, a real multiple of it, $a\at{xy}$. Since
$\at{xy}$ squares to $-1$, the exponential is:
\begin{equation}\label{eq:Axyexponential}
e^{\pm a\at{xy}} = \cos(a)\hat{1} \pm \sin(a)\at{xy}.
\end{equation}
Since $\at{xy}$ commutes with $\hat{z}$ and $\hat{t}$, so does
$\exp(a\at{xy})$.  But it anticommutes with $\hat{x}$ and $\hat{y}$.
Consider transformations on a Clifford algebra's elements of the
following form:
\begin{equation}\label{eq:TransformCA}
M \to e^{+a\at{xy}}\;M \;e^{-a\at{xy}},
\end{equation}
where $M$ is an element of the Clifford algebra. Note that the above
transformation is compatible with addition and multiplication on the
Clifford algebra. That is,
\begin{equation}\label{eq:TransformMA}
\begin{array}{rcccl}
M + N &\to& e^{-a\at{xy}}\;(M+N) \;e^{+a\at{xy}} &=&
e^{-a\at{xy}}\;M \;e^{+a\at{xy}} + e^{-a\at{xy}}\;N
\;e^{+a\at{xy}},\\
MN &\to& e^{-a\at{xy}}\;(MN) \;e^{+a\at{xy}} &=& e^{-a\at{xy}}\;M
\;e^{+a\at{xy}}\; e^{-a\at{xy}}\;N \;e^{+a\at{xy}},
\end{array}
\end{equation}
and that the transformation takes $0$ to $0$ and $\hat{1}$ to
$\hat{1}$.


The notions of primitive idempotents, commuting roots of unity, (and
everything else we discuss in the first three chapters of this book)
have been defined only through addition and multiplication, so the
transformation of Eq.~(\ref{eq:TransformCA}) will map complete sets
of primitive idempotents to complete sets of primitive idempotents,
and it will map complete sets of commuting roots of unity to
complete sets of commuting roots of unity.


Any element of the Clifford algebra can be described in terms of
multiplication and addition of basis vectors, so, since the
transformation is compatible with multiplication and addition, we
can fully understand its effect on a Clifford algebra by seeing what
it does to the basis vectors.  As usual, we compute the following by
again using the method of factoring the basis vectors
out to the left, but now, instead of having primitive
idempotents annihilate
against each other, we have two exponentials that
either cancel or multiply. The $\hat{x}$ vector is transformed by
rotating it towards $\hat{y}$:
\begin{equation}\label{eq:EffectOfAxyOnX}
\begin{array}{rcl}
\hat{x} &\to& (\cos(a)\hat{1}-\sin(a)\at{xy}) \hat{x}
(\cos(a)\hat{1}+\sin(a)\at{xy}),\\
&&=\hat{x}(\cos(a)\hat{1}+\sin(a)\at{xy})
(\cos(a)\hat{1}+\sin(a)\at{xy}),\\
&&=\hat{x}e^{2a\at{xy}},\\
&&=\hat{x}(\cos(2a)\hat{1}+\sin(2a)\at{xy}),\\
&&=\cos(2a)\hat{x} + \sin(2a)\hat{y}.
\end{array}
\end{equation}
Similarly, the $\hat{y}$ vector is transformed by
rotating it towards $-\hat{x}$:
\begin{equation}\label{eq:EffectOfAxyOnY}
\begin{array}{rcl}
\hat{y} &\to& (\cos(a)\hat{1}-\sin(a)\at{xy}) \hat{y}
(\cos(a)\hat{1}+\sin(a)\at{xy}),\\
&&=\hat{y}(\cos(2a)\hat{1}+\sin(2a)\at{xy}),\\
&&=\cos(2a)\hat{y} - \sin(2a)\hat{x}.
\end{array}
\end{equation}
Since $\hat{z}$ and $\hat{t}$ commute with $\at{xy}$, they are not
effected by the transformation.  Thus we see that this
transformation is a rotation around an axis through
the $\hat{z}$ direction that sends the $\hat{x}$ vector towards the
$\hat{y}$ vector.


Similarly, the unit multivector $\at{zt}$ generates a
boost in the $\hat{z}$ direction.  The
exponential of $a\at{zt}$ gives a hyperbolic
function:
\begin{equation}\label{eq:ExpAzt}
e^{a\at{zt}} = \cosh(a)\hat{1} + \sinh(a)\at{zt}.
\end{equation}
This commutes with $\hat{x}$ and $\hat{y}$ so the associated
transformation:
\begin{equation}\label{eq:TransformAzt}
M \to e^{-a\at{zt}}\;M\;e^{+a\at{zt}},
\end{equation}
leaves $\hat{x}$ and $\hat{y}$ unchanged.  To find the effect on
$\hat{z}$ and $\hat{t}$, let us introduce a new way of factoring
that works with any analytic function of a Clifford
algebra element, and which will speed up this sort of calculation:
\begin{equation}\label{eq:FactorExponential}
\begin{array}{rcl}
f(+\at{zt})\;\hat{x} &=& \hat{x}\;f(+\at{zt}),\\
f(+\at{zt})\;\hat{y} &=& \hat{y}\;f(+\at{zt}),\\
f(+\at{zt})\;\hat{z} &=& \hat{z}\;f(-\at{zt}),\\
f(+\at{zt})\;\hat{t} &=& \hat{t}\;f(-\at{zt}).
\end{array}
\end{equation}
That is, so long as you can write the function $f$ as a series, the
odd terms will anticommute and the even terms
will commute. This will give the
factorization shown above. Thus the effect of the
transformation on $\hat{z}$:
\begin{equation}\label{eq:TransformAztOnZ}
\begin{array}{rcl}
\hat{z} &\to& e^{-a\at{zt}}\;\hat{z}\;e^{+a\at{zt}},\\
&&=\hat{z}\;e^{+2a\at{zt}} = \hat{z}(\cosh(2a)+\sinh(2a)\at{zt}),\\
&&=\cosh(2a)\hat{z} + \sinh(2a)\hat{t}.
\end{array}
\end{equation}
Similarly, the transformation on $\hat{t}$ is:
\begin{equation}\label{eq:TransformAztOnT}
\begin{array}{rcl}
\hat{t} &\to& \hat{t}\;e^{+2a\at{zt}} = \hat{t}(\cosh(2a)+\sinh(2a)\at{zt}),\\
&&=\cosh(2a)\hat{z} + \sinh(2a)\hat{t}.
\end{array}
\end{equation}
Let $v$ be a velocity between $0$ and $c$.  Define $a$ by
\begin{equation}\label{eq:TanhByBoostA}
a = 0.5\tanh^{-1}(v/c).
\end{equation}
Then $\tanh(2a) = v/c$.  From trigonometry, we have:
\begin{equation}\label{eq:SinhAndCosh}
\begin{array}{rcl}
\cosh(2a) &=& 1/\sqrt(1-v^2/c^2),\\
\sinh(2a) &=& (v/c)/\sqrt(1-v^2/c^2).
\end{array}
\end{equation}
Putting these into the transformations of $\hat{z}$ and $\hat{t}$,
we see that:
\begin{equation}\label{eq:TransformBoosts}
\begin{array}{rcl}
\hat{z} &\to& (\hat{z} + \hat{t}v/c)/\sqrt(1-v^2/c^2),\\
\hat{t} &\to& (\hat{t} + \hat{z}v/c)/\sqrt(1-v^2/c^2),
\end{array}
\end{equation}
which is the usual form for a boost in the $\hat{z}$
direction.


More generally, let $\vec{u} = (u_x,u_y,u_z)$ be a unit vector. Then
the bivector
\begin{equation}\label{eq:BivectorU}
U_R = u_x\at{yz} - u_y\at{xz} + u_z\at{xy}
\end{equation}
generates a rotation through the $\vec{u}$ axis.
That is, to rotate by an angle $\theta$ around the $\vec{u}$ axis,
one uses the transformation:
\begin{equation}\label{eq:TransformByU}
M \to e^{-\theta U/2}\;M\;e^{+\theta U/2}.
\end{equation}
Similarly, to boost in a direction $\vec{v} =
(v_x,v_y,v_z)$ one uses the bivector
\begin{equation}\label{eq:BivectorV}
V_B = v_x\at{xt} + v_y\at{yt} + v_z\at{zt}.
\end{equation}
In general, boosts and rotations do not
commute. However, let $\vec{v} = \vec{u}$, (or $=
-\vec{u}$) and compute the product of thee  two generators:
\begin{equation}\label{eq:BivectorProduct}
U_RU_B =(u_x\at{yz} - u_y\at{xz} + u_z\at{xy})(u_x\at{xt} +
u_y\at{yt} + u_z\at{zt}).
\end{equation}
The above consists of $9$ products.  The off diagonal products, such
as $u_x\at{yz}u_y\at{yt}$ and $u_y\at{yt}u_x\at{yz}$, come in pairs
and like this pair, their unit multivectors
anticommute and therefore the pairs all cancel.
The only terms left are the diagonal terms, and therefore:
\begin{equation}\label{eq:BivectorSquare}
\begin{array}{rcl}
U_RU_B &=& u_x^2\at{xyzt} + u_y^2\at{xyzt} + u_z^2\at{xyzt},\\
&=&\at{xyzt}.
\end{array}
\end{equation}
The above is $i$ times the helicity operator.


The rotation bivector $U_R$ of
Eq.~(\ref{eq:BivectorU}) squares to $-1$. To convert it to a root of
unity, we multiply it by $i$. On the other hand, the
boost bivector, $U_B$, is already a
root of unity. Since these two commute, they form a complete set of
commuting roots of unity as we have used to define primitive
idempotents.  The helicity operator is the third non scalar element
of the group generated by $U_R$ and $U_B$.  All accepted
experimental observations are apparently compatible with the
assumption that the laws of physics are symmetric under rotations
and boosts. It is therefore very
natural that helicity is so important to elementary
particles.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TransformCA})
\section{\label{sec:ContinuousSymmetries} Continuous Symmetries}


Our primary consideration in this book are the elementary particles,
and since we are associating the elementary particles with the
primitive idempotents, it is natural for
us to examine the exponential transformations
from the point of view of what they do to the
primitive idempotents.

Suppose that $\iota_\chi$ and $\iota_\chi'$ square to $+1$ and
commute so that they can be used as a complete set of commuting
roots of unity, which then define a complete set of primitive
idempotents.  Then these primitive idempotents are
eigenvectors of $\iota_\chi$ with
eigenvalues of $\pm 1$:
\begin{equation}\label{eq:IotaEValEqn}
\begin{array}{rcl}
\iota_\chi 0.25(\hat{1}\pm\iota_\chi)(\hat{1}+\kappa\iota_\chi') &=&
\pm
0.25(\hat{1}\pm\iota_\chi)(\hat{1}+\kappa\iota_\chi'),\\
\iota_\chi \rho_\pm &=& \pm \rho_\pm,
\end{array}
\end{equation}
where $\kappa$ is either $+1$ or $-1$, we don't care which. Then
$\rho_\pm$ commutes with $\iota_\chi$ and therefore
this primitive idempotent is left
unchanged by the exponential transformation
generated by $\iota_\chi$:
\begin{equation}\label{eq:ExpTranPIIotaChi}
\begin{array}{rcl}
\rho_\pm &\to& e^{-a\iota_\chi}\rho_\pm e^{+a\iota_\chi},\\
&&=\rho_\pm e^{-a\iota_\chi} e^{+a\iota_\chi},\\
&&= \rho_\pm.
\end{array}
\end{equation}
Therefore, the primitive idempotents that can be made from a given
root of unity are precisely the quantum
states that are the fixed points
of the exponential transformation
defined by that root of unity.


If $\chi$ squares to $-1$, it is not a root of unity.  In this case
we cannot make primitive idempotents out
of $\chi$ but we must instead use $i\iota_\chi = i\;\chi$.  The
above analysis will then apply. That is, the primitive idempotents
made from $\iota_\chi$ will be just those states that are unchanged
by the exponential transformation defined by $\chi$.


Before we examine the effect of the exponentials
of the remaining unit multivectors, we
should first discuss the classification of the Dirac unit
multivectors.  As with the Pauli unit multivectors, the Dirac unit
multivectors can be classified according to the number of vectors,
that is, by the blade.  There are $5$ blades:
\begin{equation}\label{eq:BladeListDirac}
\begin{array}{c}
\{\hat{1}\},\\
\{\hat{x},\hat{y},\hat{z},\hat{t}\},\\
\{\at{xy},\at{xz},\at{yz},\at{xt},\at{yt},\at{zt}\},\\
\{\at{yzt},\at{xzt},\at{xyt},\at{xyz}\},\\
\{\at{xyzt}\}.
\end{array}
\end{equation}
This classification makes sense from a Clifford algebra sense, but
not so much sense from a physics standpoint.  It is better to treat
$\hat{t}$ as different from the spatial vectors.  This gives eight
``time-blades'':
\begin{equation}\label{eq:BladeListTDirac}
\begin{array}{cc}
\{\hat{1}\}                , &\{\hat{t}\}\\
\{\hat{x},\hat{y},\hat{z}\}, &\{\at{xt},\at{yt},\at{zt}\},\\
\{\at{xy},\at{xz},\at{yz}\}, &\{\at{yzt},\at{xzt},\at{xyt}\},\\
\{\at{xyz}\}               , &\{\at{xyzt}\}.
\end{array}
\end{equation}


The scalar, $\hat{1}$ generates only the trivial exponential
transformation that doesn't move
anything around, so we won't discuss it further.  Of the remaining
seven, they each square to $+1$ or $-1$.  Following the mathematics
of the previous section, the unit multivectors
that square to $+1$ will generate
transformations that looks like boosts, that is, that uses
hyperbolic cosines and sines, while the unit
multivectors that square to $-1$ will generate transformations that
look like rotations.  But we can always multiply by
$i$ to convert unit multivectors from the $+1$ square type to the
$-1$ type or vice-versa so we need not concern ourselves much with
this.


A more concise way of describing the effects of the exponential
transformations is to describe
their Lie algebras instead of their Lie
groups. That is, we will look at the
infinitesimal boosts and
rotations instead of the boosts and rotations.
Practically, this means that instead of writing, for the exponential
transformation generated by $\at{xy}$ that $\hat{x} \to$
$\cos(2a)\hat{x} + \sin(2a)\hat{y}$, we will instead simply write
half the infinitesimal generator of the rotation, $+\hat{y}$.


We will list only the transformations for the unit
multivectors that are neutrally
oriented or are $z$ oriented. The remaining unit
multivectors are analogous to the $z$ oriented ones.  Then the
complete set of infinitesimal rotation
generators are:
\begin{equation}\label{eq:TableLieAlgebra}
\begin{array}{c|ccc|cccc|}
&\at{xy}&\at{zt}&\at{xyzt}&\hat{z}&\hat{t}&\at{xyz}&\at{xyt}\\
\hline %
\hat{x}&+\hat{y}&0&+\at{yzt}&+\at{xz}&+\at{xt}&0&0\\
\hat{y}&-\hat{x}&0&-\at{xzt}&+\at{yz}&+\at{yt}&0&0\\
\hat{z}&0&+\hat{t}&+\at{xyt}&0&+\at{zt}&0&+\at{xy} \\
\hat{t}&0&+\hat{z}&+\at{xyz}&-\at{zt}&0&-\at{xyzt}&0\\ \hline
\end{array}
\end{equation}
The first column gives the rotation around the $z$-axis.  The second
column gives the boost in the $z$ direction.  The third column is
helicity.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TransformCA})
\section{\label{sec:Velocity} Velocity}


In quantum mechanics, the operators for energy and
momentum arise as generators of infinitesimal
translations in time and position,
respectively. In defining this, there is no need to include any
geometric information as such, the same
translations apply to all sorts of particles of
whatever spin. As such, there is no geometric content of the
Clifford algebra sort involved in their definitions. In addition,
energy and momentum have units that involve mass, and
since the origin of mass is a primary topic of this book, were we to
begin by assuming energy and momentum we would be making an argument
that is arguably circular. Max Jammer\index{Jammer, Max} discusses
this problem at length in \cite{ConceptsOfMassICPAP_2000}.


On the other hand, position and velocity
 are concepts that do not require mass
to be defined. The concepts of position and time arise naturally
from our coordinate system, they are simply $(x,y,z)$ and $t$. And
it turns out that velocity is also a natural
geometric object in the Clifford algebra theory, which we now
discuss.  From here on, we will be working with the massless
Dirac equation instead of the massive
one.


Let $\Psi(x,y,z,t)$ be a plane wave that is moving
with velocity $\vec{v} = (v_x,v_y,v_z)$.  A plane
wave has the same value across the surface of a plane so that the
value of the wave can be completely determined for all positions and
time just by watching what values occur at a particular point over
all time.  The speed of the plane wave is $v =
\sqrt{v_x^2+v_y^2+v_z^2}$.  The part of this speed in the $x$
direction is $v_x/v$.  So the point $(x,0,0)$ gets its values
delayed by $x (v_x/v) /v$ from the value of the plane wave at
$(0,0,0)$.  Similarly for the point $(0,y,0)$ and $(0,0,z)$.
Accordingly, we can write:
\begin{equation}\label{eq:FixedPlaneWave}
\Psi(x,y,z,t) = \Psi(0,0,0,t- (xv_x + y/_y + zv_z)/v^2).
\end{equation}
We assume that $\Psi$ satisfies the generalized \emph{massless}
Dirac equation. This allows us to
compute as follows:
\begin{equation}\label{eq:ComputeVelocityPsi}
\begin{array}{rcl}
0 &=& (\nabla)\Psi(x,y,z,t),\\
&=&
(\hat{x}\partial_x+\hat{y}\partial_y+\hat{z}\partial_z+\hat{t}\partial_t)
\Psi(0,0,0,t- (xv_x + yv_y + zv_z)/v^2),\\
&=&(\hat{t} - (v_x\hat{x} + v_y\hat{y} +
v_z\hat{z})/v^2)\Psi_t(0,0,0,t).
\end{array}
\end{equation}
To put the above into a form more familiar from the point of view of
primitive idempotents, we multiply on the left by $-0.5 \hat{t}$. We
obtain:
\begin{equation}\label{eq:PrimitiveIdempotentVelocity}
0 = 0.5(\hat{1}-
(v_x\at{xt}+v_y\at{yt}+v_z\at{zt})/v^2)\Psi_t(0,0,0,t).
\end{equation}
Of course $\Psi(0,0,0,t) =$ constant solves the above.  But if we
want more interesting plane waves, that is, if
$\Psi_t$ is not everywhere zero, we need to have $\Psi_t$ multiply
against $0.5(\hat{1}-(\at{xt}v_x+\at{yt}v_y+\at{zt}v_z)/v^2)$ to
give zero.


Multiplying to give zero is what annihilating
idempotents are for. Idempotents are defined by roots of
unity, so we look for one in this problem. Note
that
\begin{equation}\label{eq:VelocityRootOfOne}
(\at{xt}v_x+\at{yt}v_y+\at{zt}v_z)^2 = (v_x^2+v_y^2+v_z^2)\hat{1},
\end{equation}
so $\iota_v = (\at{xt}v_x+\at{yt}v_y+\at{zt}v_z)/v^2$ is a root of
unity, if $v=1$, and defines annihilating
idempotents:
\begin{equation}\label{eq:AnnihilatingVelocity}
0.5(\hat{1} \pm \iota_v) = 0.5(\hat{1} \pm
(\at{xt}v_x+\at{yt}v_y+\at{zt}v_z)).
\end{equation}
Comparing with Eq.~(\ref{eq:PrimitiveIdempotentVelocity}) we see
that $\Psi_t$ will be annihilated if we put:
\begin{equation}\label{eq:AnnihilatedPsit}
\Psi_t = 0.5(\hat{1} +\at{xt}v_x+\at{yt}v_y+\at{zt}v_z)dK/dt,
\end{equation}
where $K(t)$ is an arbitrary\footnote{Okay, one that does not
annihilate $0.5(1+\iota_v)$.} Clifford algebra constant times any
scalar function of time. An example function for $K$ that has the
usual sinusoidal time dependence is $K(t) = (\at{xyz} + 3\hat{x} -
12)\sin(t)$.


What we have demonstrated here is that we can produce
velocity eigenstates of the massless
Dirac equation. In the previous
section we showed that the exponential
transformations produced from
$\at{xt}$, $\at{yt}$ or $\at{zt}$ corresponded to
boosts in the directions given by the orientation of
these unit multivectors. Now we see that the velocity eigenstates of
the massless Dirac equation are the states that are
eigenstates of boosts.



Furthermore, in examining the chiral
representation of the Dirac equation in Section (\ref{sec:TDGreps}),
we saw that the three commuting roots of unity
chosen to define the diagonal primitive idempotents
 of the
representation were the operator for
spin in the $z$ direction $\at{ixy}$, the
helicity operator $\at{ixyzt}$, and what we now see
to be the operator for velocity in the $z$
direction: $\at{zt}$.


That the Dirac equation produces natural velocity
eigenstates, with velocity eigenvalues
of $\pm 1$, has been known since the equation was first discovered.
The idea was explored in the ``zitterbewegung''
 theory of the electron. This
theory supposed that the electron exhibited very high speed motion
at speed $c$, and an approximately stationary electron is produced
only by the high frequency reversals of direction.  This theory has
been explored by David Hestenes\index{Hestenes, David} using the
geometric algebra in the context of the state vector
formulation of quantum
mechanics.\cite{HestenesZBW_FP20101990} This book will reopen this
idea, but from a density operator point of view, and with chiral
states.






% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TransformCA})
\section{\label{sec:HelicityaPT} Helicity and Proper Time}


This section is
rather speculative and is not entirely necessary, but is included
for completeness. The helicity operator, $\gamma^5 = -\at{ixyzt}$
has some interesting properties. It squares to $+1$ and it
anticommutes with the whole set of basis vectors:
\begin{equation}\label{eq:HelicityProps}
\begin{array}{rcl}
(-\at{ixyzt})^2 &=& 1,\\
\hat{x}\;(-\at{ixyzt}) &=& -(-\at{ixyzt})\;\hat{x},\\
\hat{y}\;(-\at{ixyzt}) &=& -(-\at{ixyzt})\;\hat{y},\\
\hat{z}\;(-\at{ixyzt}) &=& -(-\at{ixyzt})\;\hat{z},\\
\hat{t}\;(-\at{ixyzt}) &=& -(-\at{ixyzt})\;\hat{t}.
\end{array}
\end{equation}
These are exactly the attributes that we would require of a spatial
basis vector.  For example, compare the above with the following
equations where $-\at{ixyzt}$ is replaced by $\hat{z}$:
\begin{equation}\label{eq:HelicityComparedZ}
\begin{array}{rcl}
(\hat{z})^2 &=& 1,\\
\hat{x}\;(\hat{z}) &=& -(\hat{z})\;\hat{x},\\
\hat{y}\;(\hat{z}) &=& -(\hat{z})\;\hat{y},\\
\hat{t}\;(\hat{z}) &=& -(\hat{z})\;\hat{t}.
\end{array}
\end{equation}
We are therefore somewhat justified in thinking of $-\at{ixyzt}$ as
an additional spatial basis vector and write:
\begin{equation}\label{eq:IntroToS}
\hat{s} = -\at{ixyzt}.
\end{equation}
There is a problem; ``what direction should we associate with
$\hat{s}$?'',but ignoring that for the moment, the
above switch replaces our $4-$dimensional complex Clifford algebra
generated by the $4$ basis vectors
$\{\hat{x},\hat{y},\hat{z},\hat{t}\}$ with a $5-$dimensional real
Clifford algebra generated by the $5$ basis vectors
$\{\hat{x},\hat{y},\hat{z},$ $\hat{s},\hat{t}\}$.  In this section
we explore this version of the Dirac algebra.


In this new $5-$dimensional Clifford algebra, the imaginary unit
becomes a real geometric object:
\begin{equation}\label{eq:GeometricDiracI}
\begin{array}{rcl}
\hat{s} &=&-\at{ixyzt},\\
\hat{i}\hat{s}\hat{s} &=&-\hat{i}\hat{s}\at{ixyzt},\\
\hat{i} &=&+\hat{s}\;\at{xyzt},\\
&=& -\at{xyzst}.
\end{array}
\end{equation}
Thus the usual complex Dirac algebra can be
thought of as a real $5-$dimensional Clifford algebra.  But if we
are to do more than just think of $\hat{s}$as a
spatial vector, we need to associate some sort of spatial coordinate
with the basis vector.


The original reason Dirac designed the gamma matrices was to achieve
a relativistic quantum mechanics, so it is natural to return to
relativity for an inspiration on how to interpret $\hat{s}$.  The
signature of the Dirac algebra comes from the line
element, which we will choose to write as:
\begin{equation}\label{eq:ProperTimeDefined}
\begin{array}{rcl}
-ds^2 &=& -dt^2 + (dx^2+dy^2+dz^2),\\
&=&(\hat{t}dt)^2 + (\hat{x}dx)^2 + (\hat{y}dy)^2 + (\hat{z}dz)^2.
\end{array}
\end{equation}
The above is a bit ugly from a geometric point of view because the
right hand side is written as an operator, but the left hand side is
a scalar.  To treat $ds$ as if it were a differential associated
with a coordinate, we move it to the right hand side and bring in
the basis vector $\hat{s}$.The above line element
is transformed to
\begin{equation}\label{eq:NewLineElement}
0= (\hat{x}dx)^2 + (\hat{y}dy)^2 + (\hat{z}dz)^2 + (\hat{s}ds)^2 +
(\hat{t}dt)^2.
\end{equation}
The new line element describes a spacetime with
four spatial dimensions and one time dimension, but there is no
longer a scalar proper time on the left hand side.  This would be
appropriate for a theory that applied to massless particles that all
travel at the speed of light in the presence of a hidden dimension
with a circular coordinate $s$.


Similar ideas appear in Jose Almeida's\index{Almeida, Jose}
work\cite{AlmeidaMonogenicNullVectors_0601078}, which rewrites
gravitational dynamics,
electromagnetism, and quantum mechanics in a
common mathematical approach using Clifford / Geometric algebra by
supposing ``that all material particles follow null
geodesics of a $5$D space''.  He uses the same
signature, $-++++$ for the five dimensions as here.  The primary
difference is that he works in the state vector formalism instead of
the density operator formalism.  A less important difference is that
he does not choose to assume that $\hat{s}$ corresponds to a literal
hidden dimension.


It should be mentioned that if we were to choose the opposite
signature for a $5-$dimensional Clifford algebra, $+----$, there
would be no geometric imaginary unit.  Instead, while
$\at{xyzst}$ would still commute
with the algebra, it would square to $+1$ and we would be faced with
the predicament of having a commuting
pseudoscalar $1$ different from the usual one.











% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\chapter{\label{sec:Measurement} Measurement}

%\begin{quote}
\settowidth{\versewidth}{Strike hard who cares—--shoot straight who can—--}
\begin{verse}[\versewidth]
%\em
No proposition Euclid wrote,\\
\vin No formulae the text-books know,\\
Will turn the bullet from your coat,\\
\vin Or ward the tulwar's downward blow.\\
Strike hard who cares-—-shoot straight who can---\\
The odds are on the cheaper man. 
\end{verse}
%\end{quote}
$ $

\lettrine{W}{$\;$ithout measurement,} waves need never be associated
with particles. Measurement is the heart of quantum mechanics, the
difference between it and classical mechanics.

In the 1950s, Julian Schwinger wrote down an elegant but little
noticed foundation for quantum mechanics that he called the
``Measurement Algebra''.  In this chapter we introduce what is now
known as ``Schwinger's Measurement Algebra'', and apply the
primitive idempotent calculations of the previous chapters to the
problem of determining the structure of the natural elementary
particles of a Clifford algebra.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TDGreps})
\section{\label{sec:TSGE}  The Stern-Gerlach Experiment}


An experiment, simple in concept, that illustrates the difference
between classical and quantum effects is that originally carried out
by Otto Stern and Walter Gerlach in 1922.  An oven heats up atoms,
for example silver, until they form a gas.  In a vacuum, the silver
atoms travel in straight lines and can be formed into a beam by a
plate with a small hole in it.  The beam of atoms then travels
through the influence of a magnetic field to a screen.  See
Fig.~(\ref{fig:SternGerlachExp}).



% Fig.~(\ref{fig:SternGerlachExp})
\begin{figure}[!htp]
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\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,80)
% oven
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% Oven body
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\put( 20,50){\line( 1, 0){33}} %
% Oven heater wires
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\put( 50,10){\line( 0, 1){20}} %
% top curves
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\qbezier(44,48)(44,48)(48,52) %
\qbezier(48,52)(48,52)(50,50) %
% bottom curves
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\qbezier(42,28)(42,28)(38,32) %
\qbezier(45,28)(45,28)(41,32) %
\qbezier(48,28)(48,28)(44,32) %
% Plate
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\linethickness{2.0pt}
%\thicklines %
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\put(100,60){\line( 0,-1){17}} %
% Magnet
\put(130,70){Magnet} %
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% curves
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% straights
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\qbezier(145,50)(145,50)(145,43) %
\qbezier(142,43)(142,43)(145,43) %
\qbezier(158,50)(158,50)(158,43) %
\qbezier(155,50)(155,50)(155,43) %
\qbezier(158,43)(158,43)(155,43) %
% Screen
\put(185,70){Screen} %
\thicklines %
\put(200,15){\line(0,1){50}} %
% beam
\put(125,2){Beam} %
\thinlines %
% Scattering
\qbezier(53,40)(53,40)(83,45) % 10
\qbezier(53,40)(53,40)(83,35) %
\qbezier(53,40)(53,40)(81,50) % 20
\qbezier(53,40)(53,40)(81,30) %
\qbezier(53,40)(53,40)(79,55) % 30
\qbezier(53,40)(53,40)(79,25) %
\qbezier(53,40)(53,40)(76,59) % 40
\qbezier(53,40)(53,40)(76,21) %
\qbezier(53,40)(53,40)(72,63) % 50
\qbezier(53,40)(53,40)(72,17) %
% Straight
\qbezier(53,40)(53,40)(150,40) %
% Magnet Scattering
\qbezier(150,40)(150,40)(198,34) %
\qbezier(150,40)(150,40)(198,36) %
\qbezier(150,40)(150,40)(198,38) %
\qbezier(150,40)(150,40)(198,40) %
\qbezier(150,40)(150,40)(198,42) %
\qbezier(150,40)(150,40)(198,44) %
\qbezier(150,40)(150,40)(198,46) %
\put(120, 5){\vector(-1,4){8}} %
\end{picture}
} %
\caption{\label{fig:SternGerlachExp} The Stern-Gerlach experiment:
Atoms are heated to a gas in an oven.  Escaping atoms are formed
into a beam by a small hole in a plate. A magnet influences the
beam, which then forms a figure on a screen.}
\end{figure}


Classically, the atoms act like small magnets.  The reader is
familiar with how small magnets act in the presence of a magnetic
field because this is the principle on which the magnetic compass
operates.  The magnetic field induces a torsion on the compass
magnet, causing the compass to turn until it aligns itself with the
local magnetic field.  But there is no net attractive force on the
magnet.

On the other hand, the reader is also familiar with the fact that a
small magnet can feel not just torsion, but attraction and repulsion
when it is in the presence of the magnetic field created by another
small magnet.  The difference between this case and the compass is
that Earth's magnetic field is locally homogenous.  To arrange for a
small magnet to feel attraction and repulsion, one must arrange for
the magnetic field to be non homogenous.

So Stern and Gerlach ran the beam of atoms through a magnetic field
that was as non homogeneous as they could create.  They expected
that the various atoms would feel different forces depending on
their orientation and that this would spread out the pattern
deposited on the screen.

We can only imagine their surprise and consternation when the figure
turned out not to be simply an elongated spot, but instead split
into two distinct spots.  The usual explanation for an unexpected
result in an experiment such as this, is inattentiveness on the part
of a graduate student, but the effect was real.  An atom either went
up or down, it was as if there were only two possible orientations
for the atom's magnetic field, ``up'' or ``down''.

The tendency to go up or down in an inhomogeneous magnetic field
turned out to be preserved.  If one adjusted the screen to block
only the atoms in the ``down'' spot, leaving the ``up'' atoms as a
beam, and arranged for another inhomogeneous magnetic field for
these atoms to traverse, all of these atoms again took the ``up''
path.  Of course ``up'' and ``down'' are determined by the
orientation of the magnet, for example, the magnets might be
arranged so that their north poles point up and south poles down.


\SubSectionBreak

% Fig.~(\ref{fig:XandZSGOrientations})
\begin{figure}[!htp]
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\begin{picture}(220,132)
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% axes
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\put( 5,30){\vector(0,1){30}} % z
\put( 5,30){\vector(1,0){30}} % y
\put(16,  4){$x$} %
\put( 4, 62){$z$} %
\put(37, 29){$y$} %
% screen
\put(180, 40){\line(2,-3){20}} %
\put(180, 80){\line(2,-3){20}} %
\put(180, 90){\line(2,-3){20}} %
\put(180,130){\line(2,-3){20}} %
\put(180, 40){\line(0, 1){40}} %
\put(180, 90){\line(0, 1){40}} %
\put(200, 10){\line(0, 1){40}} %
\put(200, 60){\line(0, 1){40}} %
% Left beams
\qbezier(80,95)(80,95)(140,95) %
\qbezier(80,45)(80,45)(140,45) %
% Magnets
\thicklines %
\qbezier(137,100)(140,110)(143,100) % Top magnet
\qbezier(137,100)(137,100)(137, 97) %
\qbezier(143,100)(143,100)(143, 97) %
\put(136,112){$+z$}
\qbezier(140,40)(146, 37)(146,40) % Bottom magnet
\qbezier(140,40)(140, 40)(138,43) %
\qbezier(146,40)(146, 40)(144,43) %
\put(143,30){$+x$}
\thinlines %
% Split beams
\qbezier(140,95)(140,95)(190,105) %
\qbezier(140,95)(140,95)(190, 85) %
\qbezier(140,45)(140,45)(185, 52) %
\qbezier(140,45)(140,45)(195, 38) %
% spots
\put(190,105){\circle*{3}} %
\put(190, 85){\circle*{3}} %
\put(185, 52){\circle*{3}} %
\put(195, 38){\circle*{3}} %
\end{picture}
} %
\caption{\label{fig:XandZSGOrientations} Two Stern-Gerlach
experiments, one oriented in the $+z$ direction, the other oriented
in the $+x$ direction.}
\end{figure}


There are two orientations involved in the Stern-Gerlach experiment.
The experiment requires an inhomogeneous magnetic field, and the
orientation of this field determines how the beam is split.  This is
the obvious, macroscopic orientation.  The second orientation is the
orientation of the magnetic field of the atoms.  This orientation is
microscopic and our understanding of it is only indirect.

We can imagine arranging the magnetic field in orientations other
than up and down.  If the magnet is oriented horizontally instead of
vertically, the beam is deflected into two horizontal spots instead
of vertical, see Fig.~(\ref{fig:XandZSGOrientations}). This raises
the question of what happens when the Stern-Gerlach experiment is
performed on a beam of atoms that have previously been separated
with a different orientation.

It turns out that the results of this sort of experiment can be
accurately predicted using the projection operators of the Pauli
algebra.  The reason for this is that the two spots are evidence of
a spin-$1/2$ quantum state, just the states of the Pauli algebra.
When other atoms are sent through a Stern-Gerlach experiment, the
number of spots they produce can be different from two.  But the
elementary fermions are spin-$1/2$ particles and the case of two
spots is the one we will examine in great detail.


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TDGreps})
\section{\label{sec:FaBs}  Filters and Beams}



Operators as used in this book, have one input and one output. On
the other hand, the Stern-Gerlach experiment has one input and two
outputs.  To make the Stern-Gerlach experiment fit into our
mathematical apparatus, we have to rearrange it so that it has only
one output.  The way to do this is to imagine that after the beam is
split in two, only one of the beams is allowed to exit the
experiment.  The other is absorbed by a beam stop.  We will call
this type of apparatus a ``Stern-Gerlach Filter''.


We want to model the effect of sending the output beam of a
Stern-Gerlach filter directly into the input of another
Stern-Gerlach filter so we must ignore the details of the oven that
produced the atoms, the plate that defined the beam, and the screen
that absorbed the spots.  Instead, we will pay attention only to the
inhomogeneous magnetic field.  More, we will ignore the details of
the magnetic field and pay attention only to its orientation.


The inhomogeneous magnetic field of a Stern-Gerlach experiment is
arranged to cause the atoms to split in two beams going in slightly
different directions. Since a Stern-Gerlach filter keeps only one of
these two beams, we can define the orientation of the filter by the
direction that is kept.  For example, if a horizontal beam is split
by the experiment into two beams, one going slightly up and the
other going slightly down, and the filter keeps the beam going
slightly up, then the orientation we associate with this filter will
be ``up'', which we designate by a unit vector, $(0,0,1)$


Let $\vec{v} = (v_x,v_y,v_z)$ be a unit vector.  The projection
operator for spin in the $\vec{v}$ direction is:
\begin{equation}\label{eq:ProjOpSpinV}
\rho_v = 0.5(1 + v_x\hat{x} + v_y\hat{y} + v_z\hat{z}).
\end{equation}
This projection operator satisfies the idempotency relation:
\begin{equation}\label{eq:ProjOpSpinVI}
\rho_v\;\rho_v = \rho_v,
\end{equation}
and this matches the experimentally observed fact that a beam that
passes a Stern-Gerlach filters oriented in the $\vec{v}$ direction
will then pass another Stern-Gerlach filter oriented in that same
direction.

\SubSectionBreak


Given a Stern-Gerlach filter oriented in the direction $\vec{v}$, we
can imagine the complementary Stern-Gerlach filter that switches the
beam stop between the two split beams.  Instead of passing the up
beam and halting the down beam, the complementary Stern-Gerlach
filter does the reverse.  The projection operator associated with
this complementary Stern-Gerlach filter is the projection operator
associated with the vector in the opposite direction.


The Stern-Gerlach apparatus oriented in the $+z$ direction is
represented by the projection operator:
\begin{equation}\label{eq:ProjOpPzA}
\rho_{+z} = 0.5(1+\hat{z}),
\end{equation}
and so the complementary projection operator is
\begin{equation}\label{eq:ProjOpPzC}
\rho_{-z} = 0.5(1-\hat{z}).
\end{equation}
These two projection operators annihilate:
\begin{equation}\label{eq:ProjOpPzN}
\rho_{+z}\;\rho_{-z} =0,
\end{equation}
and this corresponds to the fact that two consecutive Stern-Gerlach
filters oriented in opposite directions will pass no beam -- they
act as a beam stop.  Since the associated projection operators
multiply to give zero, we associate zero with a beam stop.

Since we are representing consecutive experiments by multiplication,
the experiment that does nothing to the beam must be represented by
$\hat{1}$, as ``doing nothing'' must commute with all other possible
experiments.  On the other hand, the sum of two complementary
projection operators is unity.  For example:
\begin{equation}\label{eq:SumOfCompIsOne}
\rho_{+z} + \rho_{-z} = 0.5(1 + \hat{z}) + 0.5(1 - \hat{z}) =
\hat{1}.
\end{equation}
This meets up with the experimental fact that one can split a beam
with a beam splitter, recombine the two split beams, and the result
is a beam indistinguishable from the original.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:TDGreps})
\section{\label{sec:SBMs}  Statistical Beam Mixtures}


In the operator formalism, one represents quantum states as
operators.  With the Stern-Gerlach filters also represented by
operators, the mathematical language is unified and simplified.  But
the operators that represent Stern-Gerlach filters are projection
operators while quantum states are more general.

Let $\rho_{+z}$ and $\rho_{-z}$ be the projection operators that
represent Stern-Gerlach filters oriented in the $+z$ and $-z$
direction.  While we have been using these operators to represent
the Stern-Gerlach filters, we can also think of them as representing
the beam that results when an unpolarized beam is sent through their
respective Stern-Gerlach filter.  This is an important philosophical
point--we are using the same mathematical object to represent both
the beam and the filter that produces the beam.  But we can only do
this when the intensity of the beam is known, and this means that we
are assuming implicitly some unit intensity for the unpolarized beam
applied to the filter.

In recombining two beams, we earlier noted that $\rho_{+z} +
\rho_{-z} = \hat{1}$, that is, the act of splitting and recombining
is equivalent to leaving the beam unmodified.  Our use of the
notation here is very natural from the point of view of the
operators but it is at variance with the traditional way of doing
quantum statistical mechanics.  The difference amounts to a choice
of normalization.


\SubSectionBreak


In the standard formalism of quantum mechanics, the normalization of
quantum states is defined so that probabilities are easy to compute.
This book abandons this tradition and instead normalizes quantum
states so that their energies are defined naturally.  This is why we
can split and recombine beams and have their quantum states computed
by simple addition; In our formalism when one represents a beam by
an operator, the scalar part of the operator defines the beam
intensity in terms of particles per second or what have you.  When
an unpolarized beam is divided in two by a Stern-Gerlach experiment,
our intensity is divided as well.

To illustrate the standard method of describing a mixed beam, let us
consider making a beam that is partly spin up and partly spin down.
Since the standard method normalizes probabilities, we have to
choose two fractions that add to unity.  For example, we can have a
beam that is $p$ spin up and $q$ spin down where $p+q=1$, and
represent the combined beam, $\rho$ by:
\begin{equation}\label{eq:QuantumMixExamp}
\begin{array}{rcl}
\rho &=& p\rho_{+z} + q\rho_{-z},\\
&=&0.5p+0.5p\hat{z} + 0.5q-0.5q\hat{z},\\
&=& 0.5 +0.5(p-q)\hat{z},\\
&=&\left(\begin{array}{cc}p&0\\0&q\end{array}\right).
\end{array}
\end{equation}
The traditional quantum statistical entropy of a mixture is defined
by:
\begin{equation}\label{eq:TradQSE}
S = - k\;\tr(\rho\;\ln(\rho)),
\end{equation}
where $k$ is Boltzmann's constant.  For a diagonalized state, the
logarithm of the matrix is easy to compute and one finds:
\begin{equation}\label{eq:TradQSEp}
\begin{array}{rcl}
S_\rho &=& -k(p\ln(p) + q\ln(q)),\\
&=&-k\ln(p^pq^q).
\end{array}
\end{equation}
In the above, note that since $p$ and $q$ are between $0$ and $1$,
so are $p^p$ and $q^q$.  The entropy is therefore positive.  The
maximum value for the entropy occurs when $p=q=0.5$ and is
$k\ln(2)$.


\SubSectionBreak


In classical thermodynamics, whenever two separated systems are
brought into contact, their total entropy is expected to increase
or, at best, stay constant.  Therefore the quantum states that
maximize entropy are of particular interest.


Suppose that we have a complete set of $N$ mutually annihilating
primitive idempotents.  We can imagine having a beam made up of a
mixture of $p_1$ of the first state$\rho_1$, $p_2$, of the second
state $\rho_2$, and so on up to $p_N$ of the $N$th state$\rho_N$,
with $p_1 + p_2 + ... + p_N = 1$. Then the entropy is defined by
\begin{equation}\label{eq:EntropyNDefined}
S=-k\ln(p_1^{p_1}p_2^{p_2}...p_N^{p_N}).
\end{equation}
A small amount of calculus shows that this function is maximized
when $p_1 = p_2 = ... = p_N = 1/N$.


Since $\rho_1, ... \rho_N$ form a complete set of primitive
idempotents, they sum to unity.  Therefore, the quantum mixture that
maximizes entropy can be represented by the operator:
\begin{equation}\label{eq:MaxEntrN}
\rho_{\textrm{Max}} = (\rho_1 + \rho_2 + ... +\rho_N)/N = \hat{1}/N.
\end{equation}
Thus we see that the quantum mixture that has the maximum entropy
happens to be the mixture that is purely scalar.  Furthermore, for
small deviations from this maximum entropy mixture, the entropy is
quadratic in the non scalar amounts.  In a later chapter, when we
discuss the natural force between primitive idempotents, we will
revisit these observations.


\section{\label{sec:SMA}  Schwinger's Measurement Algebra}

In discussing the projection operator that picks out a quantum state
of type $a$, our notation has been to write $\rho_a$.  In our
notation, this describes both the fields of the Stern-Gerlach type
experiment that allows only that quantum state to pass, and and also
to describe the quantum state itself.  In Schwinger's
notation,\cite{QKAD} we would write this same object as $m(a)$. The
choice of notation is not significant, and to reduce confusion we
will stick with the density operator notation except for these few
paragraphs that will place the present effort in historical context.


In Schwinger's exposition, he quickly passes from the projection
operators $m(a)$ to symbols that represent Stern-Gerlach filters
that alter the beam.  For example, one could arrange for a filter
that accepts only spin oriented in the $+z$ direction, and releases
particles with spin oriented in the $+x$ direction.  Schwinger would
write such a filter with a more general notation as $m(+x,+z)$. In
this, Schwinger's notation exceeds what ours is capable of.

Schwinger measurement symbols of the type $m(a,b)$ incorporate a
disturbance to the quantum state.  We reject these for two reasons.
First, we want physical processes to occur when our quantum states
are disturbed.  Second, measurement symbols of this type
automatically imply an arbitrary choice of phase.

So long as the two quantum states $a$ and $b$ do not annihilate each
other, we can uniquely define the Schwinger measurement symbol
$m(a,b)$ as the product of the operators:
\begin{equation}\label{eq:SchMab}
m(a,b) = \rho_a\;\rho_b.
\end{equation}
But if the quantum states annihilate each other, the above
definition fails.  For example, $\rho_{+z} \rho_{-z} = 0$.


We could avoid the annihilation problem by choosing an intermediate
state.  For example, we could write:
\begin{equation}\label{eq:SchPrblm}
m(+z,-z) = \rho_{+z} \;\rho_{+x} \;\rho_{-z},
\end{equation}
but the result will be different depending on the choice of that
intermediate state.  For example,
\begin{equation}\label{eq:ExSchPrb}
\begin{array}{rcl}
\rho_{+z}\rho_{+x}\rho_{-z} &=&
+0.25(\hat{x}+\at{iy}),\;\;\;\textrm{but}\\
\rho_{+z}\rho_{-x}\rho_{-z} &=& -0.25(\hat{x}+\at{iy}).
\end{array}
\end{equation}
In fact we can get any arbitrary phase by judicious choice of the
intermediate state.  And furthermore, since spinors carry arbitrary
phases, each of these objects will convert a $-z$ ket to a $+z$ ket.


If the two states, $a$, and $b$ are chosen at random, it is highly
unlikely (i.e. a set of measure zero) that they will annihilate. Far
more likely is that $m(a,b)$ will be well defined by
$\rho_a\;\rho_b$.  From this one might suppose that one could use
the general measurement symbols with little likelihood of problem.
However, the general measurement symbols between annihilating states
are precisely the measurement symbols that the standard way of
formulating quantum mechanics finds most useful--they are the
raising and lowering operators.  Our inability to define their
phases in a unique way is equivalent to the fact that when one
defines Clebsch-Gordon coefficients one must choose arbitrary
complex phases.


The Schwinger notation can be made uniquely defined by assuming a
vacuum state $\rho_0$ that does not annihilate any quantum state.
And this is how Schwinger derives the usual state vector formalism
for quantum mechanics from his measurement algebra.  But this book
is devoted to the operator formalism, as opposed to the spinor
formalism and we will not assume what Schwinger refers to in
\cite{QKAD} as a ``fictitious'' vacuum state.  This ends our
historical digression, and we now return to the density operator,
$\rho_\chi$, notation.

\SubSectionBreak

Given that any physical quantity $A$ (for example, spin) assumes
only a finite number of distinct values, $a_1$, $a_2$, ... $a_n$,
the following rules apply:
\begin{equation}\label{eq:SMArules}
\begin{array}{rcl}
\rho_{aj}\;\rho_{aj} &=& \rho_{aj},\\
\rho_{aj}\;\rho_{ak} &=& 0, \;\;\;\;\textrm{if}\;j \ne k,\\
\sum_j \rho_{aj} = 1.
\end{array}
\end{equation}
The above are identical to the rules that govern a complete set of
primitive idempotents of a Clifford algebra so it is natural that we
will use Clifford algebra to analyze the elementary measurements of
Schwinger's measurement algebra.

So long as we restrict the physical quantity $A$ to be just spin we
have not gained anything above what we already know.  But we can
interpret the physical quantity $A$ as including more than just
spin.  $A$ could represent any other complete set of quantum states,
where ``complete'' depends on our context.


We've introduced primitive idempotents in a bottom up manner.  That
is, we begin with a Clifford algebra, define a complete set of
commuting roots of unity, and from that define a complete set of
primitive idempotents.  Schwinger's measurement algebra affords a
method of defining these objects from the top down.


For example, if we are concerned with solutions to the Dirac
equation for the electron, we could choose that $A$ includes
information about a state's spin and charge: $\{+ze, -ze,
+z\bar{e},-z\bar{e}\}$.  With the choice of these four states, the
primitive idempotents would be:
\begin{equation}\label{eq:PIsForDiracSpin}
\rho_{+ze}, \rho_{-ze}, \rho_{+z\bar{e}}, \rho_{-z\bar{e}}.
\end{equation}


In most quantum mechanics books, the operator for spin$-1/2$ in the
$\pm z$ direction squares to $1/4$, but we will instead define the
spin$-1/2$ operator as twice the usual definition so it will instead
square to $1$.  This makes $S_{+z}$ a ``root of unity'' that we can
use to define primitive idempotents, and we can define the
projection operator for spin$-1/2$ in the $\pm z$ directions as
\begin{equation}\label{eq:SpinProjOpInSz}
\rho_{\pm z} = 0.5(1 + S_{+z}).
\end{equation}
The above two equations can be solved to give $S_{+z}$ in terms of
$\rho_{\pm z}$:
\begin{equation}\label{eq:SpinProjOpRev}
S_{+z} = \rho_{+z} - \rho_{-z}.
\end{equation}
Thus we can derive the root of unity from the projection operators
of the Schwinger measurement algebra.


With the choice of primitive idempotents in
Eq.~(\ref{eq:PIsForDiracSpin}), we can define the projection
operator for spin$-1/2$ in the $\pm z$ direction as:
\begin{equation}\label{eq:SpinZinSMA}
\begin{array}{rcl}
\rho_{+z} &=& \rho_{+ze} + \rho_{+z\bar{e}},\\
\rho_{-z} &=& \rho_{-ze} + \rho_{-z\bar{e}}.
\end{array}
\end{equation}
Applying Eq.~(\ref{eq:SpinProjOpRev}), we derive:
\begin{equation}\label{eq:SpinFromSMA}
S_{+z} = (\rho_{+ze} + \rho_{+z\bar{e}}) - (\rho_{-ze} +
\rho_{-z\bar{e}}).
\end{equation}
The above spin operator, $S_{+z}$ is now defined entirely within the
Schwinger measurement algebra.  The rule for obtaining primitive
roots of unity from the Schwinger measurement algebra is to simply
add all the eigenstates with eigenvalue $+1$, and subtract the state
with eigenvalue $-1$.

Note that when an operator is defined as above, that is, as a sum
over a complete set of states each independently multiplied by $\pm
1$, the operator is automatically a root of unity.  When squared,
the cross terms will all be zero as different primitive idempotents
annihilate, and the squared terms will convert the $-$ signs to $+$.
Then this will be a sum over a complete set of primitive idempotents
and so is $\hat{1}$.


Thus we have ways of moving back and forth between a Schwinger
measurement algebraic description of a set of quantum states (i.e. a
list of sets of observed quantum numbers), and a description of a
set of quantum states that are primitive idempotents of a Clifford
algebra.  In later chapters we will apply this technology to the
problem of uniquely defining the elementary particles as composites
of primitive idempotents of a Clifford algebra.


\section{\label{sec:DOaSMA}  Clifford Algebra and the SMA}


As we add vectors to a Clifford algebra, we get larger numbers of
commuting roots of unity in our complete sets of commuting roots of
unity.  If there is only one vector, say $\hat{x}$, then there is
only one root of unity, $\hat{x}$.  In this case, there are only $2$
primitive idempotents, $0.5(1\pm \hat{x})$.

With two vectors, $\{\hat{x},\hat{y}\}$, we can still only define a
single commuting root of unity.  If we choose $\hat{x}$ the
remaining degrees of freedom in the Clifford algebra, namely
$\{\hat{y}, \at{xy}\}$, all anticommute with $\hat{x}$ and so we are
stuck with only one commuting root of unity.  In this situation, as
in the case with only one vector, there are only $2$ primitive
idempotents, $0.5(1\pm \hat{x})$.


The Pauli algebra has three vectors, $\{\hat{x}, \hat{y},
\hat{z}\}$.  If we choose $\hat{x}$ as a commuting root of unity,
the remaining degrees of freedom are $\{\hat{y},\hat{z},\at{yz},$
$\at{xy},\at{xz},\at{yz},\at{xyz}\}$.  Of these, $\at{yz}$ and
$\at{xyz}$ commute with $\hat{x}$, but neither of these squares to
$-1$.  Therefore the (real) Pauli algebra defines only two primitive
idempotents, $0.5(1\pm \hat{x})$.


On the other hand, if we define a Clifford algebra with the three
vectors $\{\hat{x}, \hat{y}, \hat{t}\}$, we can define two commuting
roots of unity, for example, $\hat{x}$ and $\at{yt}$, and we would
have four primitive idempotents, $0.25(1\pm \hat{x})(1\pm \at{yt})$.
So we see that the number of primitive idempotents that are
contained in a Clifford algebra can depend on its signature.


The Dirac algebra, with vectors
$\{\hat{x},\hat{y},\hat{z},\hat{t}\}$ also allows two commuting
roots of unity, for example, $\hat{x}$ and $\at{zt}$, and therefore
four primitive idempotents, $0.25(1\pm \hat{x})(1\pm \at{zt})$.

If we add another spatial vector, $\hat{s}$, to the Dirac algebra,
to get the ``proper time algebra'' as was discussed in
Sec.~(\ref{sec:HelicityaPT}), we can still only write two commuting
roots of unity, the same as with the Dirac algebra. To get three
commuting roots of unity we could instead add a second time vector
to the Dirac algebra, $\at{t'}$ with $(\at{t'})^2=-1$. Then three
commuting roots of unity would be $\{\hat{x},\at{yt},\at{zt'}\}$,
and we would have eight primitive idempotents.

Another way to get three commuting roots of unity would be to take
the proper time algebra to be complex rather than real.  Then we
could have three commuting roots of unity as
$\{\hat{x},\at{yt},\at{izs}\}$ again giving eight primitive
idempotents, that is,
$0.125(1\pm\hat{x})(1\pm\at{yt})(1\pm\at{izs})$.  If $\hat{s}$ is
treated as belonging to a compact or hidden dimension, then one can
consider the complex numbers as indicating position in the $s$
dimension.  But in doing this, one ends up with two natural
geometric imaginary units, $i$ and $\at{xyzst}$, both of which
square to $-1$ and commute with the geometry.

\SubSectionBreak

In any case, if we find $M$ commuting roots of unity, we find $2^M$
primitive idempotents.  Thinking of the $M$ commuting roots of unity
as operators, the $2^M$ primitive idempotents are eigenstates with
eigenvalues of $\pm 1$.  For example:
\begin{equation}\label{eq:ExampleEvEqnPI}
\hat{x}\;\;0.25(1-\hat{x})(1+\at{yt}) = -0.25(1-\hat{x})(1+\at{yt}),
\end{equation}
so $0.25(1-\hat{x})(1+\at{yt})$ is an eigenstate of $\hat{x}$ with
eigenvalue $-1$.

But the $M$ roots of unity commute, so a primitive idempotent
generated by these roots is also an eigenstate of any product of the
$M$ commuting roots of unity.  So rather than having the primitive
idempotents be eigenstates of $M$ commuting independent operators,
they are more generally eigenstates of $2^M$ commuting not
independent operators. This confusing but simple concept is best
illustrated with an example. The Dirac algebra is just a little too
simple to get the point across, so instead we will use the complex
proper time algebra.

Let our commuting roots of unity be $\{\hat{x}, \at{yt},\at{izs}\}$.
A particular primitive idempotent defined by these roots is:
\begin{equation}\label{eq:PIdefinpta}
\begin{array}{rcl}
\rho_{-+-} &=& 0.125(1-\hat{x})(1+\at{yt})(1-\at{izs}),\\
&=&0.125(\hat{1}-\hat{x}+\at{yt}-\at{izs}-\at{xyt}+\at{ixzs}-\at{iyzst}+\at{ixyzst}).
\end{array}
\end{equation}
The set of all products of the three commuting roots of unity is the
set with eight elements given by
$\{\hat{1},\hat{x},\at{yt},\at{xyt},$
$\at{izs},\at{ixzs},\at{iyzst},\at{ixyzst}\}$.  Note that these are
precisely the degrees of freedom present in the expansion of
$\rho_{-+-}$ above, Eq.~(\ref{eq:PIdefinpta}).


As each degree of freedom is represented in
Eq.~(\ref{eq:PIdefinpta}) with a plus or minus sign, so the
primitive idempotent $\rho_{-+-}$ is an eigenstate of that degree of
freedom with eigenvalue $+1$ or $-1$.  Using explicit
multiplication, the reader is invited to verify that $\rho_{-+-}$ is
an eigenvalue of these eight commuting roots of unity with
eigenvalues $\pm 1$ where the sign is given by the sign in the
expansion of $\rho_{-+-}$.

Using this principle, one can write down all the eigenvalues of all
eight primitive idempotents $\rho_{\pm \pm \pm}$:
\begin{equation}\label{eq:CommRootsAsOperators}
\begin{array}{r|cccccccc|}
&\hat{1}&\hat{x}&\at{yt}&\at{izs}&\at{xyt}&\at{ixzs}&\at{iyzst}&\at{ixyzst}\\
\hline %
\rho_{---}&+1&-1&-1&-1&+1&+1&+1&-1 \\
\rho_{--+}&+1&-1&-1&+1&+1&-1&-1&+1 \\
\rho_{-+-}&+1&-1&+1&-1&-1&+1&-1&+1 \\
\rho_{-++}&+1&-1&+1&+1&-1&-1&+1&-1 \\
\rho_{+--}&+1&+1&-1&-1&-1&-1&+1&+1 \\
\rho_{+-+}&+1&+1&-1&+1&-1&+1&-1&-1 \\
\rho_{++-}&+1&+1&+1&-1&+1&-1&-1&-1 \\
\rho_{+++}&+1&+1&+1&+1&+1&+1&+1&+1 \\ \hline
\end{array}
\end{equation}
In the above, it should be noted that the signs are somewhat
arbitrary in that $-\hat{x}$ makes just as good a root of unity as
$+\hat{x}$.  For this reason, not much can be made of the fact that
the last row has all $+1$ eigenvalues.

One of the $2^M$ commuting non independent roots of unity is
$\hat{1}$, and of course has trivial eigenvalues of $+1$ for all
primitive idempotents.  For the purpose of classifying quantum
states we can ignore this operator.  That leaves $2^M-1$ good
quantum numbers.  Each of the other commuting roots of unity has
equal numbers of eigenstates with $+1$ and $-1$ eigenvalues.
Therefore one has that the sum of the quantum numbers over all
states is zero, a fact that is reminiscent of the requirement for
anomaly cancelation in QFT, but is beyond the scope of this book.


\SubSectionBreak


Suppose that we have three commuting roots of unity, $\iota_1$,
$\iota_2$, and $\iota_3$.  The eight primitive idempotents are
$0.125(1\pm\iota_1)(1\pm\iota_2)(1\pm\iota_3)$.  Any of these eight
primitive idempotents has good eigenvalues with respect to any of
the eight operators in the Abelian group generated by the three
commuting roots of unity.  The table of eigenvalues will be
identical to Eq.~(\ref{eq:CommRootsAsOperators}) but we repeat it
here:
\begin{equation}\label{eq:CommIotasAsOperators}
\begin{array}{r|cccccccc|}
&\hat{1}&\iota_3&\iota_2&\iota_1&\iota_1\iota_2&\iota_1\iota_3&
\iota_2\iota_3&\iota_1\iota_2\iota_3\\
\hline %
\rho_{---}&+&-&-&-&+&+&+&- \\
\rho_{--+}&+&-&-&+&-&-&+&+ \\
\rho_{-+-}&+&-&+&-&-&+&-&+ \\
\rho_{-++}&+&-&+&+&+&-&-&- \\
\rho_{+--}&+&+&-&-&+&-&-&+ \\
\rho_{+-+}&+&+&-&+&-&+&-&- \\
\rho_{++-}&+&+&+&-&-&-&+&- \\
\rho_{+++}&+&+&+&+&+&+&+&+ \\ \hline
\end{array}
\end{equation}
written in terms of three generic commuting roots of unity and the
eight generic primitive idempotents they produce.


In standard quantum mechanics, different particles share the same
Dirac equation without any way of distinguishing between them except
in the notation of the physicist.  In the standard model, one must
anticommute identical fermion creation operators and the same with
identical annihilation operators.  This anticommutation enforces the
Pauli exclusion principle.  However, if two fermions are not
identical, one commutes their creation and annihilation operators
and there is no Pauli exclusion principle. In the standard
formulation, one keeps track of which fermions are identical or non
identical by notation.


In our formulation of quantum mechanics different primitive
idempotents are distinguished not merely by notation, but by having
different geometric values.  For this reason, we do not need to rely
on notation to keep track of whether our fermions are identical or
not.


\SubSectionBreak


Suppose that we have two distinct primitive idempotents in the same
complete set of primitive idempotents.  These two idempotents
annihilate, so their sum is also idempotent:
\begin{equation}\label{eq:SumOfPIsSqr}
\begin{array}{rcl}
(\rho_1 + \rho_2)^2 &=& (\rho_1)^2  + \rho_1\rho_2 + \rho_2\rho_1 +
(\rho_2)^2,\\
&=&(\rho_1)^2 + (\rho_2)^2,\\
&=& \rho_1 + \rho_2.
\end{array}
\end{equation}
The sum is not primitive so it is not an ``elementary particle'',
but it is an idempotent so we can think of it as two elementary
particles at the same point in space.


On the other hand, if we add two identical idempotents, their sum is
not an idempotent:
\begin{equation}\label{eq:SumOfIdenticalPIs}
(\rho_1 + \rho_1)^2 = 2(\rho_1 + \rho_1).
\end{equation}
We will interpret this as the primitive idempotent form of the Pauli
exclusion principle.  This book is devoted to discrete degrees of
freedom in the operator formalism of quantum mechanics. The author
plans to demonstrate the Pauli exclusion interpretation in a later
book devoted to spatial dependencies and continuous degrees of
freedom.


\SubSectionBreak


The set of all possible sums of distinct primitive idempotents in
our set generated by $\iota_1$, $\iota_2$, and $\iota_3$ includes
$2^8=256$ elements as each of the eight primitive idempotents can
either be included, or not, in the sum.  Ouch.

In Sec.~(\ref{sec:Velocity}) we saw that velocity is a natural
quantum number of a Clifford algebra.  The operator for velocity in
the $+z$ direction is $\at{zt}$.  Let us suppose that this is
$\iota_3$, one of the commuting roots of unity.  Now the eight
primitive idempotents are split into two groups of four each.  The
set  $\{\rho_{---},\rho_{--+},\rho_{-+-},\rho_{-++}\}$ are
eigenstates of $\at{zt}$ with eigenvalue $-1$ while
$\{\rho_{+--},\rho_{+-+},\rho_{++-},\rho_{+++}\}$ are eigenstates
with eigenvalue $+1$.  These two eigenvalues correspond to movement
in the $-z$ and $+z$ directions, and so composite particles made
from primitive idempotents selected from both these two groups
cannot share the same point in space for longer than an instant.


Furthermore, because we require physics to be equivalent under
rotations, the particles traveling in the $-z$ direction are
identical to the particles traveling in the $+z$ direction and we
need only study one of these sorts of particles to understand the
others. Therefore, in this section we will restrict our attention to
primitive idempotents that take a $+1$ eigenvalue with respect to
$\at{zt}$.  These are primitive idempotents that are traveling in
the same direction.  We will call a group of primitive idempotents
that travel together like this a ``snuark'', and they will be needed
to classify the elementary fermions.  Since there are only $4$
primitive idempotents with an eigenvalue for $\at{zt}$ of  $+1$,
there are only $2^4 = 16$ such snuarks and many of them are trivial.


\SubSectionBreak


In order to compare with the elementary particles of the standard
model, we need to compile quantum numbers for these sixteen
composite idempotents (snuarks). In the standard model of quantum
mechanics, quantum numbers are obtained as the eigenvalues of
operators.  We have followed this tradition with the primitive
idempotents, but if we continue the practice in the composite
idempotents we will end up with nonsensical results.


The difficulty arises from the fact that the standard model is built
from spinors.  If one has two distinct particle types and one has
spinors for each of them that are eigenstates of an operator $A$
with two eigenvalues, for example:
\begin{equation}\label{eq:StandModelCevals}
\begin{array}{rcl}
A|a1\ra &=& a_1|a1\ra,\\
A|a2\ra &=& a_2|a2\ra,\\
\end{array}
\end{equation}
then the combined state is also an eigenstate of $A$, but with an
eigenvalue given by the sum of the individual eigenvalues:
\begin{equation}\label{eq:StandModelCevalt}
\begin{array}{rcl}
A(|a1\ra|a2\ra) &=& (A|a1\ra)\;|a2\ra + |a1\ra\;(A|a2\ra),\\
&=&a_1 (|a1\ra|a2\ra) + a_2 (|a1\ra|a2\ra),\\
&=& (a_1+a_2)(|a1\ra|a2\ra).
\end{array}
\end{equation}
Therefore, to match the results for spinors, we must add the
eigenvalues of the primitive idempotents that contribute to the
composite idempotent.


\SubSectionBreak


Earlier, in the expansion of the primitive idempotents, we noted
that the eigenvalue of a primitive idempotent with respect to an
operator was given by the sign of the term in the primitive
idempotent's expansion.  When we add primitive idempotents, these
values will add.  Therefore, we see that our technique for assigning
eigenvalues to snuarks can be defined more simply than by just
copying the spinor method.  The eigenvalue of a snuark is just the
coefficient of the operator in the expansion of the snuark (and we
multiply by $2^M$ where $M$ is the number of commuting roots of
unity so that the eigenvalues of a primitive idempotent are $\pm
1$).


This concept is important and is less confusing than it may appear.
Consider the snuark $\rho = \rho_{+--} + \rho_{+-+} +\rho_{+++}$.
What is its quantum number for $\iota_1$?  There are two ways to
compute. First,
\begin{equation}\label{eq:QNsForRhoExamp}
\begin{array}{rcl}
\iota_1 \rho_{+--} &=& -\rho_{+--},\\
\iota_1 \rho_{+-+} &=& +\rho_{+-+},\\
\iota_1 \rho_{+++} &=& +\rho_{+++},
\end{array}
\end{equation}
so there are two primitive idempotents with quantum number $+1$, and
one with quantum number $-1$.  Therefore the answer is $-1+1+1 =
+1$.  But note that the snuark is \emph{not} an eigenstate of
$\iota_1$.  Instead, $\iota_1\;\rho$ gives $-\rho_{+--}
+\rho_{+-+}+\rho_{+++}$.


As a second way of finding the eigenvalue of $\rho$ with respect to
$\iota_1$, we can expand $\rho_{+--}$, $\rho_{+-+}$, and
$\rho_{+++}$ and add them together:
\begin{equation}\label{eq:QnForRhosEx2}
\begin{array}{rcl}
\rho&=& \rho_{+--} + \rho_{+-+} + \rho_{+++},\\
&=&\;\;0.125(1+\iota_3-\iota_2-\iota_1+\iota_2\iota_1-\iota_3\iota_1
-\iota_3\iota_2+\iota_3\iota_2\iota_1)\\
& &  + 0.125(1+\iota_3-\iota_2+\iota_1-\iota_2\iota_1+\iota_3\iota_1
-\iota_3\iota_2-\iota_3\iota_2\iota_1)\\
& &  + 0.125(1+\iota_3+\iota_2+\iota_1+\iota_2\iota_1+\iota_3\iota_1
+\iota_3\iota_2+\iota_3\iota_2\iota_1)\\
& =&  0.125(3+3\iota_3-\iota_2+\iota_1+\iota_2\iota_1+\iota_3\iota_1
-\iota_3\iota_2+\iota_3\iota_2\iota_1).
\end{array}
\end{equation}
Since the coefficient of $\iota_1$ in the sum is $+1$, (after
multiplying by $2^3 = 8$ to get rid of the common factor of $0.125$)
this is the eigenvalue.  The other eigenvalues can be read off the
last line.  For example, the eigenvalue of $\iota_3$ is $+3$, and
the eigenvalue of $\iota_3\iota_1$ is $+1$.  Note that while the
quantum numbers of a primitive idempotent are multiplicative, the
quantum numbers of a snuark are not.  Instead, since $\iota_3$ has
the same eigenvalue $+1$, for these four primitive idempotents, the
eigenvalue of $\iota_1$ is the same as the eigenvalue for
$\iota_3\iota_1$.


\SubSectionBreak


In combining the $4$ primitive idempotents $\rho_{+--}$,
$\rho_{+-+}$, $\rho_{++-}$ and $\rho_{+++}$ into sixteen possible
snuarks, only some of them are of interest to us. The snuarks follow
the binomial theorem in that there is $1$ with no primitive
idempotents, $4$ with exactly one, $6$ with exactly two, $4$ with
three, and $1$ with all four.  The one with no primitive idempotent
is simply zero and has no interpretation.  The four with single
primitive idempotents are simply the primitive idempotents
themselves and we already know these well.  When all four primitive
idempotents are added together the result is $\hat{1}$ which will
take eigenvalues of $+1$ with respect to all operators, and so is
not of interest. The four with exactly three primitive idempotents
will carry eigenvalues equal to the negative of the eigenvalues of
the missing primitive idempotent.  This could be equally
accomplished by using primitive idempotents and negating the
operators so these four are also not of interest.

The only snuarks of interest are the ones that have two primitive
idempotents.  Their eigenvalues are as follows:
\begin{equation}\label{eq:QNsOfCompVelEstates}
\begin{array}{r|cccc|ccc|}
&\rho_{+++}&\rho_{++-}&\rho_{+-+}&\rho_{+--}&\iota_2&\iota_1&\iota_{21}
\\ \hline %
\rho_{1100}&1&1& & &+2& 0& 0\\
\rho_{1010}&1& &1& & 0&+2& 0\\
\rho_{1001}&1& & &1& 0& 0&-2\\
\rho_{0011}& & &1&1&-2& 0& 0\\
\rho_{0101}& &1& &1& 0&-2& 0\\
\rho_{0110}& &1&1& & 0& 0&+2\\
\hline %
\end{array}
\end{equation}
In the above, we use, $\rho_{abcd}$ to designate which of the four
primitive idempotents,
$(\rho_{+++},\rho_{++-},\rho_{+-+},\rho_{+--})$ are included in the
snuark.  In a later chapter we will use these to model the
elementary fermions.


\section{\label{sec:GSGPlots}  Generalized Stern-Gerlach Plots}


The original Stern-Gerlach experiment split a beam of atoms in two.
This we modeled with the two primitive idempotents built from a
single root of unity.  More complicated Clifford algebras allow
larger numbers of commuting roots of unity which define larger
numbers of primitive idempotents.  In this section we discuss
generalized Stern-Gerlach experiments with these more general
particle beams.



% Fig.~(\ref{fig:SternGerlachFour})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,130)
\thinlines
% Beam
\put(20,60){\line(1,0){65}} % in
\qbezier(110,60)(110,60)(172,76) % TL
\qbezier(110,60)(110,60)(188,64) % TR
\qbezier(110,60)(110,60)(172,56) % BL
\qbezier(110,60)(110,60)(188,44) % BR
% Spots
\put(172,76){\circle*{3}} %
\put(188,64){\circle*{3}} %
\put(172,56){\circle*{3}} %
\put(188,44){\circle*{3}} %
% Screen
\put(156,37){\line(0,1){79}} %
\put(204, 1){\line(0,1){79}} %
\put(156, 37){\line(4,-3){48}} %
\put(156,116){\line(4,-3){48}} %
% Axes
\put(180,30){\vector(0, 1){60}} % T
\put(164,72){\vector(4,-3){32}} % R
% Axis labels
\put(192,40){$\iota_1$} %
\put(182,85){$\iota_2$} %
% Box
\thicklines %
\put( 79,54){\line(4,-3){12}} %
\put( 79,75){\line(4,-3){12}} %
\put( 99,75){\line(4,-3){12}} %
\put( 91,45){\line(1, 0){20}} %
\put( 91,66){\line(1, 0){20}} %
\put( 79,75){\line(1, 0){20}} %
\put( 79,54){\line(0, 1){21}} %
\put( 91,45){\line(0, 1){21}} %
\put(111,45){\line(0, 1){21}} %
\end{picture}
} %
\caption{\label{fig:SternGerlachFour} Stern-Gerlach experiment that
splits a beam of particles according to two commuting quantum
numbers, $\iota_1$ and $\iota_2$.}
\end{figure}


With two commuting roots of unity, we have four primitive
idempotents.  There are then four primitive Stern-Gerlach filters,
and the associated Stern-Gerlach experiment will have four outgoing
beams instead of just two.  See Fig.~(\ref{fig:SternGerlachFour}).
The choice of the operators $\iota_1$ and $\iota_2$ is arbitrary. We
could have chosen $-\iota_2$ and $\iota_1\iota_2$ and the diagram
would look exactly the same.  This is because the eigenvalues of the
four primitive idempotents are symmetric with respect to negation
and swapping any two operators:
\begin{equation}\label{eq:SymmetryOfEigenvalues}
\begin{array}{c|ccc|}
&\iota_2&\iota_1&\iota_2\iota_1\\ \hline %
\rho_{+--}&-&-&+ \\
\rho_{+-+}&-&+&- \\
\rho_{++-}&+&-&- \\
\rho_{+++}&+&+&+ \\ \hline
\end{array}
\end{equation}
That is, no matter which $2$ of the operators are chosen out of
$\{\pm\iota_2,\pm\iota_1,\pm\iota_2\iota_1\}$, there will be one
primitive idempotent which has eigenvalues of $(-1,-1)$, one which
has eigenvalues of $(+1,+1)$, and two with mixed eigenvalues of
$(-1,+1)$ and $(+1,-1)$.  No matter which operators are chosen for
the Stern-Gerlach apparatus, the result is the same, a square.


In these Stern-Gerlach experiments, the screen is $2$-dimensional,
so there is only room for two operators to be explored on it.  This
will not prevent our imagination from exploring Stern-Gerlach
experiments with three operators.  Instead of defining the beam
splitting by sending the beams to a screen, we will instead draw the
beams as spots in $3$-dimensions, with the eigenvalues giving the
coordinates of each spot.  If we have three eigenvalues that are all
$\pm 1$, the spots will fall on the corners of a cube.


As our first $3$-dimensional Stern-Gerlach experiment, we will draw
one with the three non independent operators $\iota_1$, $\iota_2$,
and $\iota_2\iota_1$.  Of course these define only four states, so
we will only have four spots instead of the eight that one might
expect from three operators.  The resulting $3$-dimensional
Stern-Gerlach drawing is shown as Fig.~(\ref{fig:SGthreeD})

% Fig.~(\ref{fig:SGthreeD})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,130)
\thinlines
% cube
\put(110, 50){\line(0,1){60}} %
\put(130, 10){\line(0,1){60}} %
\put(190, 10){\line(0,1){60}} %
\put(130, 10){\line(1,0){60}} %
\put(130, 70){\line(1,0){60}} %
\put(110,110){\line(1,0){60}} %
\put(110, 50){\line(1,-2){20}} %
\put(110,110){\line(1,-2){20}} %
\put(170,110){\line(1,-2){20}} %
% Spots
\put(110, 50){\circle*{4}} %
\put(130, 70){\circle*{4}} %
\put(170,110){\circle*{4}} %
\put(190, 10){\circle*{4}} %
% Quantum numbers
\put( 65, 50){$(-,+,-)$} %
\put(133, 75){$(+,+,+)$} %
\put(172,110){$(+,-,-)$} %
\put(180,  2){$(-,-,+)$} %
% Axes
\put(40,40){\vector( 0,1){30}} %
\put(40,40){\vector(-1,0){30}} %
\put(40,40){\vector(1,-2){10}} %
% Axis labels
\put(38, 72){$\iota_1$} %
\put( 2, 38){$\iota_2$} %
\put(43, 15){$\iota_2\iota_1$} %
\put( 5,110){$(\iota_1,\;\;\iota_2,\;\;\iota_2\iota_1)$}
\end{picture}
} %
\caption{\label{fig:SGthreeD} Stern-Gerlach experiment with the
three non independent operators, $\iota_1$, $\iota_2$, and
$\iota_2\iota_1$.  Spots are labeled with their eigenvalues, $\pm
1$, in the order $(\iota_1,\iota_2,\iota_2\iota_1)$. The cube is
drawn only for help in visualization.}
\end{figure}


\SubSectionBreak


When we assume that the particle beam is traveling in the same
direction we end up eliminating one of the commuting roots of unity
from use in describing the particle type of the beam.  This halves
the number of primitive idempotents that we have available for
modeling particles.  With $3$ commuting roots of unity, we get only
$2^2=4$ different particles.  But if these are all particles, we can
suppose that corresponding to each there is an antiparticle that can
be modeled by the particle traveling backwards in time.


So long as we distinguish between particles and antiparticles, this
gives us back our $2^3=8$ different quantum states.  In this case,
we can use $\iota_1$, $\;\iota_2$, and $\iota_3$ as independent
operators and our quantum states will occupy the corners of a cube.
We can think of $\iota_3$ as an operator that distinguishes between
particles (with eigenvalues of $+1$) and antiparticles (with
eigenvalues of $-1$).  If we choose $\iota_3$ as one of the three
operators that define the Stern-Gerlach beam plot, we will have that
the particles make up the four corners of the cube surrounding one
side, and the other four corners make up the antiparticles.


It is important to realize that the distribution of the primitive
idempotent quantum states on the corners of the cube is very
arbitrary in that it depends entirely on our choice of three
independent operators.  Rather than choosing $\iota_1$, $\;\iota_2$,
and $\iota_3$, we can choose any other set of independent operators
and we can choose them in any order.  For example, $\iota_1\iota_2$,
$\;\iota_2\iota_3$ and $-\iota_2$ are three independent commuting
roots of unity (i.e. operators) and will define a beam pattern for a
Stern-Gerlach experiment.  See Fig.~(\ref{fig:MixedCube}).

% Fig.~(\ref{fig:MixedCube})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,130)
\thinlines
% cube
\put(110, 50){\line(0,1){60}} %
\put(130, 10){\line(0,1){60}} %
\put(190, 10){\line(0,1){60}} %
\put(130, 10){\line(1,0){60}} %
\put(130, 70){\line(1,0){60}} %
\put(110,110){\line(1,0){60}} %
\put(110, 50){\line(1,-2){20}} %
\put(110,110){\line(1,-2){20}} %
\put(170,110){\line(1,-2){20}} %
% Spots
\put(110, 50){\circle*{4}} %
\put(110,110){\circle*{4}} %
\put(130, 10){\circle{4}} %
\put(130, 70){\circle{4}} %
\put(170, 50){\circle{4}} %
\put(170,110){\circle{4}} %
\put(190, 10){\circle*{4}} %
\put(190, 70){\circle*{4}} %
% Quantum numbers
\put( 65, 50){$(+,+,-)$} %
\put( 65,110){$(+,+,+)$} %
\put(113,  1){$(-,-,+)$} %
\put(133, 75){$(-,-,-)$} %
\put(172,110){$(-,+,+)$} %
\put(148, 40){$(-,+,-)$} %
\put(190,  1){$(+,-,+)$} %
\put(190, 75){$(+,-,-)$} %
% Axes
\put(40,30){\vector( 0,1){30}} %
\put(40,30){\vector(-1,0){30}} %
\put(40,30){\vector(1,-2){10}} %
% Axis labels
\put(38, 62){$\iota_1\iota_2$} %
\put( 2, 33){$\iota_2\iota_3$} %
\put(43,  5){$-\iota_2$} %
\put( 5, 90){$(\iota_3,\;\;\iota_2,\;\;\iota_1)$}
\end{picture}
} %
\caption{\label{fig:MixedCube} Stern-Gerlach plot with the three
operators $(\iota_1\iota_2,\;\iota_2\iota_3,\;-\iota_2)$.  The
labels for each corner are the $\iota_1$, $\iota_2$ and $\iota_3$
quantum numbers in order $(\iota_3,\;\iota_2,\;\iota_1)$.  The
antiparticles have $\iota_3=-1$ and are drawn as hollow circles.}
\end{figure}


\SubSectionBreak


As a final example of generalized Stern-Gerlach plots, we will plot
the six snuarks whose eigenvalues are shown in
Eq.~(\ref{eq:QNsOfCompVelEstates}).  Since the eigenvalues of
$\iota_1$ and $\iota_2$ are $0$ or $\pm 2$, we will divide these
eigenvalues by $2$ so that they fit more elegantly in cube form.


It is traditional to have antiparticles carry negative quantum
numbers.  To accomplish this, rather than plot the $\iota_1$ and
$\iota_2$ quantum numbers, we will instead plot $\iota_3\iota_1$ and
$\iota_3\iota_2$.  Since $\iota_3$ will be $-1$ for the
antiparticles, this will multiply the antiparticle quantum numbers
by $-1$.  With the antiparticles, we will have a total of $12$
quantum states to plot. The quantum numbers for the six composite
particles are:
\begin{equation}\label{eq:QNsOfCompVelEstates2}
\begin{array}{r|cccc|ccc|}
&\rho_{+++}&\rho_{++-}&\rho_{+-+}&\rho_{+--}&\iota_3&\iota_3\iota_2/2&\iota_3\iota_1/2
\\ \hline %
\rho_{1100}&1&1& & &+1&+1& 0\\
\rho_{1010}&1& &1& &+1& 0&+1\\
\rho_{1001}&1& & &1&+1& 0& 0\\
\rho_{0011}& & &1&1&+1&-1& 0\\
\rho_{0101}& &1& &1&+1& 0&-1\\
\rho_{0110}& &1&1& &+1& 0& 0\\
\hline %
\end{array}
\end{equation}
and the quantum numbers for the antiparticles are the negatives of
the above.  Note that two of the snuarks, $\rho_{1001}$ and
$\rho_{0110}$ appear degenerate, but if we also plotted
$\iota_2\iota_1$ (perhaps by adding a $4$th dimension to the plot)
the degeneracy would be broken.  The resulting $12$ particles in
$11$ spots are shown in Fig.~(\ref{fig:CompIdemCube}).  The
apparently degenerate particles will be eliminated by energy
considerations in the chapter on Force.


% Fig.~(\ref{fig:CompIdemCube})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,130)
\thinlines
% cube
%\put(110, 50){\line(0,1){60}} %
%\put(130, 10){\line(0,1){60}} %
%\put(190, 10){\line(0,1){60}} %
\put(130, 10){\line(1,0){60}} %
\put(130, 70){\line(1,0){60}} %
\put(110, 50){\line(1,0){60}} %
\put(110,110){\line(1,0){60}} %
\put(110, 50){\line(1,-2){20}} %
\put(170, 50){\line(1,-2){20}} %
\put(110,110){\line(1,-2){20}} %
\put(170,110){\line(1,-2){20}} %
% Coordinates:
% (110,50) (110,110)
% (130,10) (130, 70)
% (170,50) (170,110)
% (190,10) (190, 70)
% Snuarks
\put(120, 90){\circle*{4}} % 1010
\put(180, 90){\circle*{4}} % 0101
\put(140,110){\circle*{4}} % 0011
\put(160, 70){\circle*{4}} % 1100
\put(150, 90){\circle*{5}} % 1001+0110
% Anti Snuarks
\put(120, 30){\circle{4}} % /0101
\put(180, 30){\circle{4}} % /1010
\put(140, 50){\circle{4}} % /1100
\put(160, 10){\circle{4}} % /0011
\put(150, 30){\circle{5}} % /1001+0110
% Labels
\put( 95, 90){$\rho_{1010}$} % 1010
\put(185, 90){$\rho_{0101}$} % 0101
\put(126,116){$\rho_{0011}$} % 0011
\put(146, 63){$\rho_{1100}$} % 1100
\put(129, 82){$\rho_{1001},\rho_{0110}$} % 1001+0110
\put( 95, 30){$\bar{\rho}_{0101}$} % /0101
\put(185, 30){$\bar{\rho}_{1010}$} % /1010
\put(126, 56){$\bar{\rho}_{1100}$} % /1100
\put(146,  3){$\bar{\rho}_{0011}$} % /0011
\put(129, 22){$\bar{\rho}_{1001},\bar{\rho}_{0110}$} % /1001+0110
% Axes
\put(40,30){\vector( 0,1){30}} %
\put(40,30){\vector(-1,0){30}} %
\put(40,30){\vector(1,-2){10}} %
% Axis labels
\put(38, 63){$\iota_3$} %
\put( 2, 35){$\iota_3\iota_1$} %
\put(43,  3){$\iota_3\iota_2$} %
%\put( 5, 95){$\rho_{\iota_3\iota_2\iota_1}$}
\end{picture}
} %
\caption{\label{fig:CompIdemCube} Stern-Gerlach plot of $12$ snuarks
and their antiparticles.  See Eq.~(\ref{eq:QNsOfCompVelEstates2})
for the quantum numbers.  The antiparticles have $\iota_3=-1$ and
are drawn as hollow circles.}
\end{figure}





% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{\label{sec:Force} Force}

%\begin{quote}
\settowidth{\versewidth}{Yet sleek with oil, a Force was hid}
\begin{verse}[\versewidth]
%\em
Yet sleek with oil, a Force was hid\\
\vin Making mock of all they did\\
Ready at the opening hour\\
To yield up to Prometheus\\
The secular and well-drilled Power\\
The Gods secreted thus.
\end{verse}
%\end{quote}


\lettrine{W}{$\;$e need to} define a force that will bind together
primitive idempotents in a natural way.  Since the elementary
fermions are point-like particles, the force will have to create
bound states that act point-like.  To arrange this, we will define
the force as a potential energy that defines the total energy of a
set of primitive idempotents that are at the same point in space.






\section{\label{sec:PESG}  Potential Energy, Some Guesses}


The presence of three generations of elementary particles naturally
leads one to suppose that the higher generations are some sort of
excitation of the first generation.  This would explain the
hierarchy between generations.  However, experiments measuring the
total cross section for $e + \bar{e} \to \;\textrm{hadrons}$ show
that if there are four generations of elementary particles, the mass
of the $4$th neutrino has to be on the order of $10$ GeV$/c^2$,
which is unnaturally heavy.\cite[$\sec
11.2$]{PovhParticlesAndNuclei} This result argues that there should
only be three generations, no more and no less.


A natural way of obtaining exactly three generations of relatively
low mass particles is to suppose that, to first order, they are a
degenerate ground state and the masses we measure just come from a
second order splitting of the degeneracy.  This is the assumption we
will make.


Assuming that the elementary fermions are degenerate removes some of
the difficulty of assigning a potential energy as we need only
correctly guess the potential energy in the neighborhood of the
degeneracy.  There are several different forms for the potential
energy that come to mind and we will discuss them in this section.


\SubSectionBreak


In analyzing the Stern-Gerlach experiments of the previous chapter,
we saw that a primitive idempotent represented three things at once.
It represents an elementary particle.  It projects out the portion
of the incoming beam that consists of the elementary particle.  And
it represents the field configuration in the Stern-Gerlach filter
that separates that elementary particle from all others.  We can
treat this last observation as a clue to the form of the potential
energy.

Suppose that our universe of particles has only two particles in it,
a positive charged one and a negatively charged one, where
``charge'' we leave nebulously defined. Only one Stern-Gerlach
experiment exists, one that separates positively charged particles
from negatively charged ones.  The two particles are represented by
two primitive idempotents:
\begin{equation}\label{eq:SimpleChargedPIs}
\begin{array}{rcl}
\rho_+ &=& 0.5(1 + \hat{c}),\\
\rho_- &=& 0.5(1 - \hat{c}),
\end{array}
\end{equation}
where $\hat{c}$ is the root of unity, or operator, defining the
charge.

There is only one bound state, that consisting of a positively
charged particle and a negatively charged particle.  We need a
potential energy that will bind two oppositely charged particles
together.  As a function, potential energy needs to be real valued
and to give zero only when the two states are opposite in charge.

In the spinor language, a natural choice for potential energy that
meets these requirements is the probability of transition between
the states.  That is, since the two primitive idempotents
annihilate, there will be a zero probability of transition between
them.  We will write this guess as $V_1$:
\begin{equation}\label{eq:V1Defn}
V_1(A,B) = \la A|B\ra \la B|A\ra = |\la A|B\ra|^2.
\end{equation}
n $|A\ra$ is multiplied by an arbitrary phase, $\la A|$ is
multiplied by its complex conjugate so the above does not depend on
the arbitrary spinor phase.  We can rewrite this in operator form by
using the trace function:
\begin{equation}\label{eq:V1DefnB}
\begin{array}{rcl}
V_1(\rho_A,\rho_B) &=& \tr(\rho_A\;\rho_B).\\
&=& \tr( \begin{array}({cc})a_{11}&a_{12}\\a_{21}&a_{22}\end{array}
\begin{array}({cc})b_{11}&b_{12}\\b_{21}&b_{22}\end{array}),\\
&=& a_{11}b_{11}+a_{12}b_{21} + a_{21}b_{12} + a_{22}b_{22}.
\end{array}
\end{equation}


The above potential will work well for a more complicated algebra
than one that just allows two oppositely charged particles.  We can
assume that $|A\ra$ and $|B\ra$ are in the Pauli algebra.  In the
usual quantum mechanics, the Pauli exclusion principle would prevent
$A$ and $B$ from occupying other than opposite spin states.  Here we
instead allow them to occupy arbitrary states and obtain the Pauli
exclusion principle as a result of presuming that the known
particles are all ground states of this potential energy.


That the energy increases when $A$ and $B$ are not in opposite spin
states fits in well with our intuition.  When $A$ and $B$ are
distinct particles the Pauli exclusion principle does not prevent
them from occupying the same spin state and in such cases, one
typically finds that the energy is slightly lower when the two
particles are oppositely arranged.  An example of this effect is
atomic hydrogen.

To simplify Eq.~(\ref{eq:V1DefnB}), of course we put it into
geometric form.  To do this, recall that one converts a $2\times 2$
complex matrix into geometric form as follows:
\begin{equation}\label{eq:MatrixToGeo}
\begin{array}{rcl}
a_1 &=& \;(a_{11} + a_{22})/2,\\
a_x &=& \;(a_{12} + a_{21})/2,\\
a_y &=&  i(a_{12} - a_{21})/2,\\
a_z &=& \;(a_{11} - a_{22})/2.
\end{array}
\end{equation}
The inverse relations are then
\begin{equation}\label{eq:GeoToMatrix}
\begin{array}{rcl}
a_{11} &=& a_1 + a_z,\\
a_{12} &=& a_x +ia_y,\\
a_{21} &=& a_x -ia_y,\\
a_{22} &=& a_1 - a_z.
\end{array}
\end{equation}
Note that in order for our density matrices $\rho_A$ and $\rho_B$
must be Hermitian, and therefore $a_1$, $a_x$, $a_y$, and $a_z$ are
all real. Substituting these relations into Eq.~(\ref{eq:V1DefnB})
we obtain:
\begin{equation}\label{eq:GeoV1}
\begin{array}{rcl}
V(A,B) &=& (a_1+a_z)(b_1+b_z) + (a_x+ia_y)(b_x+ib_y)\\
&&+(a_x-ia_y)(b_x-ib_y) + (a_1-a_z)(b_1-b_z),\\
&=&2(a_1b_1 + a_xb_x + a_yb_y + a_zb_z).
\end{array}
\end{equation}
This is exactly what we would obtain if we treated the degrees of
freedom of $\rho_A$ and $\rho_B$ as if they were real, Euclidean
$4$-vectors and defined the potential energy as the dot product.
Thus we see that a potential energy that we defined as a $4$th order
function in spinors becomes a tractable $2$nd order function in
geometric density operators.


\SubSectionBreak


In choosing the potential energy to be the transition probability
between the states we found a reasonable potential energy for two
states, but what we need is a potential energy that can bind
together an arbitrary number of primitive idempotents.  To get this
sort of object, we can consider summing together the states to be
bound, and define the potential energy as a function of the sum.

As a first step at putting the potential energy in the form of a
summation of primitive idempotents, let us note that $\rho_+ +
\rho_-$ $= 1$.  Therefore we might suppose that bound states of
primitive idempotents must always sum to unity.  Then we can define
the potential as being the square of the difference between the sum
and unity.
\begin{equation}\label{eq:V2Try1}
V_2(\rho_A,\rho_B) = \left|\rho_A + \rho_B - \hat{1}\right|^2.
\end{equation}
To define $|\;\;|^2$, we simply take the sum of squared magnitudes
of the elements of the matrix.  This gives:
\begin{equation}\label{eq:V2Try2}
\begin{array}{rcl}
V_2(\rho_A,\rho_B) &=& \left|
\begin{array}({cc})a_{11}&a_{12}\\a_{21}&a_{22}\end{array} +
\begin{array}({cc})b_{11}&b_{12}\\b_{21}&b_{22}\end{array} -
\begin{array}({cc})1&0\\0&1\end{array}\right|^2,\\
&=&|a_{11}+b_{11}-1|^2 + |a_{12}+b_{12}|^2 \\
&&+ |a_{21}+b_{21}|^2 + |a_{22}+b_{22}-1|^2.
\end{array}
\end{equation}

To simplify the above, we again use Eq.~(\ref{eq:GeoToMatrix}) to
put it into geometric form.  Remembering that $a_1$, $a_x$, ...
$b_z$ are all real we have:
\begin{equation}\label{eq:GeomV2}
\begin{array}{rcl}
V_2(\rho_A,\rho_B) &=& |a_1+a_z+b_1+b_z-1|^2+|a_x+ia_y+b_x+ib_y|^2\\
&&+ |a_x-ia_y+b_x-ib_y|^2 +|a_1-a_z+b_1-b_z-1|^2,\\
&=&(a_1+b_1+a_z+b_z-1)^2 + 2(a_x+b_x)^2 +2(a_y+b_y)^2\\
&&+(a_1+b_1-a_z-b_z-1)^2,\\
&=&2(a_1+b_1)^2 + 2(a_x+b_x)^2 + 2(a_y+b_y)^2 +
2(a_z+b_z)^2\\
&&-4(a_1+b_1)+2,\\
&=&2(a_1^2+a_x^2+a_y^2+a_z^2) +2(b_1^2+b_x^2+b_y^2 +b_z^2)\\
&&+4(a_1b_1+a_xb_x+a_yb_y+a_zb_z) -4(a_1+b_1)+2.
\end{array}
\end{equation}
For a normalized primitive idempotent, $a_1=0.5$, and
$a_x^2+a_y^2+a_z^2 =0.25$, so the above can be considerably
simplified:
\begin{equation}\label{eq:GeomV2B}
\begin{array}{rcl}
V_2(\rho_A,\rho_B) &=& 2(0.25+0.25) +2(0.25+0.25)\\
&&+4(a_1b_1+a_xb_x+a_yb_y+a_zb_z) -4(0.5+0.5)+2,\\
&=&4(a_1b_1+a_xb_x+a_yb_y+a_zb_z).
\end{array}
\end{equation}
Comparing with Eq.~(\ref{eq:GeoV1}), we see that $V_2(A,B) =
2V_1(A,B)$.  In other words, these two very different approaches to
the potential energy gave identical potential energies, except for a
scaling factor.


\SubSectionBreak


Another way of defining the potential energy is to suppose that
there is some sort of statistical mechanical basis underlying
quantum mechanics, and use the entropy function defined in
Eq.~(\ref{eq:TradQSE}).  We suppose that the potential energy is
minimized when the entropy is maximized so we need a minus sign to
get a guess at an energy function:
\begin{equation}\label{eq:TradQSE2}
V_3(\rho_A,\rho_B) =  k\;\tr((\rho_A+\rho_B)\;\ln(\rho_A+\rho_B))
\;\;\;\;\textrm{1st guess}.
\end{equation}
The entropy is written under the assumption that $\rho$ is
normalized in the usual quantum mechanical way, that is, that
$\tr(\rho) = 1$.  For the Pauli algebra, this means that
$\rho=0.5\hat{1}$. Following the hint provided by the Stern-Gerlach
operators, we need a potential energy $V_3(\rho_A+\rho_B)$ that is
minimized when $\rho_A + \rho_B = \hat{1}$.

In addition, in terms of geometry, the trace, when applied to the
matrix representation of a Clifford algebra element, picks out the
scalar part.  Therefore, to convert the entropy formula into a
potential energy we must write:
\begin{equation}\label{eq:V3DefnA}
V_3(\rho_A,\rho_B) = V_3(\rho_A+\rho_B) =  k\; \la
0.5(\rho_A+\rho_B)\ln(0.5(\rho_A+\rho_B))\ra_0,
\end{equation}
where $\la\;\;\;\;\ra_0$ is the notation for the picking out of the
scalar part. Since the entropy is maximized when $\rho =
0.5\hat{1}$, the above will be minimized when $\rho_A+\rho_B =
\hat{1}$.


Since we are assuming that the potential energy is of the order of
the Planck mass, we need only worry about small perturbations around
the minimum.  As before, $\rho_A = 0.5\hat{1} + a_x\hat{x} +
a_y\hat{y} +a_z\hat{z}$ and similarly for $\rho_B$.  ``Small
perturbations'' means that $a_x+b_x$ is very small, and similarly
for $y$ and $z$ so:
\begin{equation}\label{eq:PerturbEntropy}
\rho_A+\rho_B = \hat{1} + (a_x+b_x)\hat{x} + (a_y+b_y)\hat{y} +
(a_z+b_z)\hat{z},
\end{equation}
where all the coefficients but the first are very small.  We write:
$\rho_A + \rho_B = 1 + \delta$.  Putting this into
Eq.~(\ref{eq:V3DefnA}) and keeping up to second order terms gives:
\begin{equation}\label{eq:V3DefnB}
\begin{array}{rcl}
V_3(\rho_A,\rho_B) &=&  k\; \la (\hat{1}+0.5\delta)\ln(0.5(\hat{1}+
0.5\delta))\ra_0,\\
&=&k\;\la (1+0.5\delta)(\ln(0.5) + \ln(1+0.5\delta))\ra_0,\\
&=&k\;\la (\ln(0.5) + \ln(1+0.5\delta)) + 0.5\delta(\ln(0.5) +
\ln(1+0.5\delta))\ra_0,\\
&=&k\;\la (\ln(0.5) + 0.5\delta-0.5\delta^2 + 0.5\delta\ln(0.5) +
0.25\delta^2)\ra_0,\\
&=&k\;\la (\ln(0.5) +0.5\delta(1+\ln(0.5))-0.25\delta^2).
\end{array}
\end{equation}
Note that the $\la\;\;\;\ra_0$ keeps only the scalar part and
$\delta$ is purely vector.  Therefore the above reduces to:
\begin{equation}\label{eq:V3DefnC}
\begin{array}{rcl}
V_3(\rho_A,\rho_B) &=& k\;(\ln(0.5)-0.25\delta^2),\\
&=&k\;(\ln(0.5) -0.25((a_x+b_x)^2 + (a_y+b_y)^2 + (a_z+b_z)^2)),\\
&=&k\;(\ln(0.5) -0.25(a_x^2 + a_y^2 + a_z^2 + b_x^2 + b_y^2 +
b_z^2)\\
&&+0.5(a_xb_x+a_yb_y+a_zb_z)),\\
&=&k\;(\ln(0.5) -0.125 - 0.5(a_xb_x+a_yb_y+a_zb_z)),\\
&=&k\;(\ln(0.5) - 0.5(a_1b_1+a_xb_x+a_yb_y+a_zb_z)).
\end{array}
\end{equation}
The $0.5\ln(0.5)$ is just an additive constant.  The rest of the
potential energy is simply a multiple of the same potential energy
we obtained for $V_1$ and $V_2$.


\section{\label{sec:GPE}  Geometric Potential Energy}


The transition probability $V_1$ is difficult to generalize to $n$
primitive idempotents.  Entropy, $V_3$ has the disadvantage of
having a logarithm.  And $V_2$ is written for two particles only.
But with these as clues, we can make a fourth guess:
\begin{equation}\label{eq:VflatDefn}
V_4(\rho_A,\rho_B, ...\rho_\chi) = (a_x+b_x + ... +\chi_x)^2 +
(a_y+b_y+ ... +\chi_y)^2 + (a_z+b_z+...+\chi_z)^2.
\end{equation}
That is, we sum the primitive idempotents, and calculate a squared
magnitude that includes everything but the scalar part.  When the
primitive idempotents sum to a scalar, this gives a potential energy
of zero which is the minimum.


For a Clifford algebra larger than the Pauli algebra, we can extend
the sum on the right to include the rest of the non scalar degrees
of freedom.  But in the above we are assigning different relative
weights to the scalar and vector components, should we also assign
differing weights to the bivector components?


If we were not assuming that the elementary particles are states
with zero energy, the natural interpretation of their different
energies would be that they give their respective masses.  So to
break the degeneracy, we can rewrite $V_4$ so that the scalar part
of the sum does contribute to the potential energy.  This will still
keep the particles as near degenerate in energy, but will allow us
to model the masses of the elementary particles.


\SubSectionBreak


Writing the potential energy with different weights for the scalar
and non scalar parts raises the question of how we should assign
weights for all the various non scalar parts.  The most complicated
Clifford algebra we are considering is $C(4,1)$, the complex algebra
generated by $\hat{x}$, $\hat{y}$, $\hat{z}$, $\hat{s}$ and
$\hat{t}$ where only the last has negative signature.  This Clifford
algebra has $32$ complex degrees of freedom or $64$ real degrees of
freedom.


Assuming that we give the potential energy full freedom to vary over
the $32$ complex degrees of freedom, we can write $V$ using $32$
possibly different weights, $v_\chi$:
\begin{equation}\label{eq:VariableV}
\begin{array}{rcl}
&&V(a_1\hat{1} + a_x\hat{x} + a_y\hat{y} + ... +
a_{xyzst}\at{xyzst}),\\
&=&V(a_1\hat{1}) + V(a_x\hat{x}) + V(a_y\hat{y}) + ... +
V(a_{xyzst}\at{xyzst}),\\
&=&v_1|a_1|^2 + v_x|a_x|^2 + v_y|a_y|^2 + ... +
v_{xyzst}|a_{xyzst}|^2.
\end{array}
\end{equation}
We need to choose the $32$ real values, $v_\chi$, a task which
appears formidable.


\SubSectionBreak


The various degrees of freedom for a Clifford algebra are related by
multiplication.  For example, $\at{xy} = \hat{x}\hat{y}$. It is
therefore natural to write the potential energy function as a
multiplicative function.  For example:
\begin{equation}\label{eq:MultPhi}
v_{xyzst} = v_x\;v_y\;v_z\;v_s\;v_t.
\end{equation}
This means that instead of defining $32$ real values, we need only
define $5$, and of those, it is natural to choose the spatial
potentials as:
\begin{equation}\label{eq:PhiSpatialChoices}
v_x=v_y=v_z=v_s.
\end{equation}
The scalar potential for $\hat{1}$ has to be the multiplicative
identity:
\begin{equation}\label{eq:PhiMultIdent}
V(\hat{1}) = v_1 = 1.
\end{equation}
Then we can write the potential $V$ completely in terms of $v_s$ and
$v_t$.  An example potential energy calculation:
\begin{equation}\label{eq:PhiExample}
\begin{array}{rcl}
V(\rho) &=& V(0.25(\hat{1} + \at{zt})(\hat{1}-\hat{s})),\\
&=& V(0.25(\hat{1} + \at{zt}-\hat{s}+\at{zst})),\\
&=&|0.25|^2 +v_sv_t|0.25|^2 +v_s|0.25|^2 +v_s^2v_t|0.25|^2,\\
&=&0.0625(1+v_sv_t)(1+v_s).
\end{array}
\end{equation}
Thus we can write potential energies as polynomials in $v_s$ and
$v_t$ that factor the same way as primitive idempotents factor into
idempotents.


The leptons are observed as free particles but the primitive
idempotents described in this book do not.  In order for this to be
a result of energy considerations, we need to have the energies of a
typical scalar be much smaller than the energy of a typical
primitive idempotent.  For example,
\begin{equation}\label{eq:EnergyComparison}
\begin{array}{rcl}
V(\hat{1}) &<<& V(0.25(\hat{1} +
\at{zt})(\hat{1}-\hat{s})),\;\;\;\textrm{or}\\
1 &<<& 0.625(1+v_sv_t)(1+v_s).
\end{array}
\end{equation}
The easy way to achieve this is to assume that $v_s$ and $v_t$ are
much greater than $1$, but we do not need to assume that these are
the same.


\SubSectionBreak


The potential energy function defines a map from any element of the
Clifford algebra to the non negative reals.  In particular, it is
multiplicative on a substantial number of the commuting roots of
unity.  For example:
\begin{equation}\label{eq:MultSubROU}
\begin{array}{rcl}
V(\at{ixyzt}) &=& v_xv_yv_zv_t,\\
&=&V(\at{ixy})\;V(\at{zt}).
\end{array}
\end{equation}
As we saw in the chapter on Primitive Idempotents, a set of
commuting roots of unity form an Abelian group.  Without having
defined the potential, we had no way of preferring one way of
generating the group from another.  For example, one could write an
idempotent in either of these equivalent ways:
\begin{equation}\label{eq:EquivIdems}
\begin{array}{rcl}
0.25(1+\at{ixy})(1+\at{zt}),\\
0.25(1+\at{ixy})(1+\at{ixyzt}),\\
\end{array}
\end{equation}
The equivalence boils down to the fact that the Abelian group of
commuting roots of unity, $\{1,\at{ixy},\at{zt},\at{ixyzt}\}$ can be
generated by either $\{\at{ixy}, \at{zt}\}$ or $\{\at{ixy},
\at{ixyzt}\}$.


Let us work on understanding the potential energy structure of the
commuting roots of unity of the most complicated Clifford algebra we
have considered, $C(4,1)$.  Since $\at{ixyzst}$ squares to $+1$ and
commutes with everything, it must be included. It remains to choose
two independent roots of unity.  Here, ``independent'' means with
respect to $\at{ixyzst}$.  For example, $\at{ixy}$ and $\at{zst}$
are not independent since they multiply to give $\at{ixyzst}$.
Because multiplication by $\at{ixyzst}$ will have the effect of
changing the number of vectors in a root of unity from $q$ to $5-q$,
without loss of generality we can suppose that the two roots of
unity we add to $\at{ixyzst}$ will have $2$ or fewer vectors.


Since two vectors anticommute, we cannot add two roots of unity that
are both made from one vector such as $\hat{x}$ or $\hat{y}$.  And
if the two roots we add both have two vectors, for example
$\at{ixy}$ and $\at{zt}$, then the only way they can commute is if
they share no vectors.  In that case, their product will have four
distinct vectors, and the product of that with $\at{ixyzst}$ will
have only one vector, in this case $\hat{s}$.  So we need only
consider the cases where we add a vector and a bivector. For these
to commute, they can share no vectors.


Keeping track only of the only time vector, $\hat{t}$ and the four
space vectors in no particular order, $\hat{a}$, $\hat{b}$,
$\hat{c}$, and $\hat{d}$, the only possible choices for a complete
set of independent commuting roots of unity are the following:
\begin{equation}\label{eq:TableOfCRoUZ}
\begin{array}{c}
\{\at{it}, \;\at{iab}, \;\at{icd}\},\\
\{\hat{a}, \;\at{ibc}, \;\at{dt}\}.
\end{array}
\end{equation}
That is, either $\at{it}$ is used as a root of unity, or it is in
one of the generating roots.  Since we will be wanting our primitive
idempotents to be eigenstates of velocity, such as $\at{zt}$, we
take the second case.


Writing out the full set of $8$ commuting roots of unity, we can
read off the potential energies of the operators:
\begin{equation}\label{eq:TableOfPotsOfRoU}
\begin{array}{cccccccc|}
\hat{1}&\at{iab}&\at{icd}&\at{abcd}&\at{it}&\at{iabt}&\at{icdt}&\at{ixyzst}\\
\hline %
1&v_s^2&v_s^2&v_s^4&v_t&v_tv_s^2&v_tv_s^2&v_tv_s^4
\end{array}
\end{equation}
If $v_s < v_t$, then the smallest potential energy is $v_tv_s$.
Otherwise, it is, $v_s^2$.  But in any case, the largest
contribution has to be from the $\at{ixyzst}$ term.


\section{\label{sec:SnuarksAsBoundStates}  Snuarks as Bound States}


The fact that $\at{ixyzst}$ is the largest contribution to the
energy of a primitive idempotent suggests that they will be deeply
bound together in order to cancel this contribution.  In order to
describe the ways that this can happen, let's specialize to
primitive idempotents traveling in the $+z$ direction.

In order to travel in the $+z$ direction, the primitive idempotents
must carry a factor of $(\hat{1}+\at{zt})$ as was described in
Sec.~(\ref{sec:Velocity}).  In order for their sum to cancel the
$\at{ixyzst}$ term, we can add two primitive idempotents, one with a
factor of $(\hat{1}+\at{ixyzst})$, the other with a factor of
$(\hat{1}-\at{ixyzst})$.  There remains one primitive idempotent to
specify, it is generated by a commuting root of unity that must
commute with $\at{zt}$.  Of course anything commutes with
$\at{ixyzst}$.


The stuff that commutes with $\at{zt}$ is generated by the four
elements $\{\hat{s},$ $\hat{x},$ $\hat{y},$ $\at{zt}\}$.  Somewhat
arbitrarily, we will assume that the third commuting root of unity
is $\hat{s}$.  We could, for example, have chosen $\at{ixyzt}$ or
$\at{ixy}$, but these have higher potential energy than $\hat{s}$.


We can suppose that the primitive idempotents bound by the
cancelation of $\at{ixyzst}$ can have either eigenvalue with respect
to $\hat{s}$.  The four primitive idempotents that we will consider
the binding potential on are therefore:
\begin{equation}\label{eq:FourPIsForSnarks}
\begin{array}{rcl}
\rho_{z--} &=& 0.125(1+\at{zt})(1-\hat{s})(1-\at{ixyzst}),\\
\rho_{z-+} &=& 0.125(1+\at{zt})(1-\hat{s})(1+\at{ixyzst}),\\
\rho_{z+-} &=& 0.125(1+\at{zt})(1+\hat{s})(1-\at{ixyzst}),\\
\rho_{z++} &=& 0.125(1+\at{zt})(1+\hat{s})(1+\at{ixyzst}),
\end{array}
\end{equation}
Multiplying these out we have:
\begin{equation}\label{eq:FourPIsForSnarksWO}
\begin{array}{rcl}
\rho_{z--}&=&0.125(1+\at{zt}-\hat{s}-\at{ixyzst}+\at{zst}+\at{ixys}-\at{ixyzt}-\at{ixy}),\\
\rho_{z-+}&=&0.125(1+\at{zt}-\hat{s}+\at{ixyzst}+\at{zst}-\at{ixys}+\at{ixyzt}+\at{ixy}),\\
\rho_{z+-}&=&0.125(1+\at{zt}+\hat{s}-\at{ixyzst}-\at{zst}+\at{ixys}+\at{ixyzt}+\at{ixy}),\\
\rho_{z++}&=&0.125(1+\at{zt}+\hat{s}+\at{ixyzst}-\at{zst}-\at{ixys}-\at{ixyzt}-\at{ixy}),
\end{array}
\end{equation}
The potential energy of the above four primitive idempotents are
identical, $(1+v_tv_s)(1+v_s)(1+v_s^2)/64$.  For large $v_s$ and
$v_t$, this will be dominated by the $v_tv_s^4$ term.  There are
four combinations whose sums cancel the $\at{ixyzst}$ term:
$\rho_{z--}+\rho_{z-+}$, $\rho_{z--}+\rho_{z++}$,
$\rho_{z+-}+\rho_{z-+}$, $\rho_{z+-}+\rho_{z++}$.  These are the
snuarks.  There are four of them for primitive idempotents traveling
in the $+z$ direction.



\SubSectionBreak


Because there are only three commuting roots of unity, these four
snuarks are all that we can define that cancel $\at{ixyzst}$.  We
will now continue with an analysis of the quantum numbers of the
snuarks.  To do this, we first explicitly write out the snuarks:
\begin{equation}\label{eq:SnuarksWO}
\begin{array}{rcl}
\rho_{z--}+\rho_{z-+}&=&0.25(1+\at{zt}-\hat{s}+\at{zst}),\\
\rho_{z+-}+\rho_{z++}&=&0.25(1+\at{zt}+\hat{s}-\at{zst}),\\
&&\\
\rho_{z--}+\rho_{z++}&=&0.25(1+\at{zt}-\at{ixyzt}-\at{ixy}),\\
\rho_{z+-}+\rho_{z-+}&=&0.25(1+\at{zt}+\at{ixyzt}+\at{ixy}).
\end{array}
\end{equation}
The four snuarks naturally group into two pairs, according to which
degrees of freedom they have not canceled out.


All four snuarks have a $\at{zt}$ term and are therefore eigenstates
of velocity.  The top two have a $\pm \hat{s}$ term and are
therefore eigenstates of $\hat{s}$ with eigenvalues of $\pm 1$.  But
the bottom two have no $\hat{s}$ terms;  but since they are made
from primitive idempotents that are eigenstates of $\hat{s}$, as
composite states, they are still eigenstates of $\hat{s}$ but have
eigenvalues of zero.


The primitive idempotents of the snuarks were made from three
commuting roots of unity and therefore have eight quantum numbers.
These are inherited by the snuarks as follows:
\begin{equation}\label{eq:QNsOfSnuarks}
\begin{array}{c|cccccccc|}
&1&\at{zt}&\at{ixys}&\at{ixyzst}&\hat{s}&\at{zst}&\at{ixyzt}&\at{ixy}\\ \hline %
\rho_{z--}+\rho_{z-+}&2&2&0&0&-2&+2& 0& 0\\
\rho_{z+-}+\rho_{z++}&2&2&0&0&+2&-2& 0& 0\\
\rho_{z--}+\rho_{z++}&2&2&0&0& 0& 0&-2&-2\\
\rho_{z+-}+\rho_{z-+}&2&2&0&0& 0& 0&+2&+2\\ \hline
\end{array}
\end{equation}
The first four columns have identical, and therefore uninteresting
quantum numbers.  The remaining four columns are each has the same
structure.  Two of the snuarks take zero quantum numbers, the other
two take $\pm 2$.  Later we will interpret this pattern as defining
the structure of the leptons, that is, a doublet and two singlets.
The four snuarks will be interpreted, after a rotation by the
Weinberg angle, as the left and right handed electron and neutrino.


Note that if instead of using $C(4,1)$ we were using the Dirac
algebra, we would have exhausted our commuting roots of unity at
$\at{zt}$ and $\at{ixyzt}$.  Instead of having a list of four
primitive idempotents as in Eq.~(\ref{eq:FourPIsForSnarksWO}), we
would have a list of only two, and these two would combine in only
one way.  Thus instead of having four snuarks, we would have only
one.  This is not enough snuarks to produce the standard model
fermions.  In particular, it would not reproduce the structure of
doublet and two singlets that characterizes weak isospin.


\SubSectionBreak


The snuarks defined above are composed of $2$ primitive idempotents
each and eliminate the $\at{ixyzst}$ degree of freedom.  One can
eliminate yet more higher degrees of freedom by adding together all
four primitive idempotents:
\begin{equation}\label{eq:DMsnuark}
\begin{array}{rcl}
\rho_{zDM} &=& \rho_{z--} + \rho_{z-+} + \rho_{z+-} + \rho_{z++},\\
&=&0.5(1+\at{zt}).
\end{array}
\end{equation}
This eliminates everything but the $\at{zt}$ non scalar degree of
freedom. We will associate this snuark with dark matter, hence the
abbreviation $\rho_{DM}$.


The electrons and neutrinos couple to the photon, $W^\pm$ and $Z^0$.
We will associate these interactions with the degrees of freedom and
quantum numbers of the four regular matter snuarks shown in
Eq.~(\ref{eq:SnuarksWO}) and Eq.~(\ref{eq:QNsOfSnuarks}).  These
degrees of freedom are absent in $\rho_{DM}$, and the quantum
numbers of $\rho_{DM}$ for these operators are zero.  Therefore
matter built from $\rho_{DM}$ will not participate in electroweak
interactions.  This is compatible with the characteristics of dark
matter postulated by the astronomers.


This book is primarily about the standard model and therefore
standard matter.  We will not discuss dark matter and $\rho_{DM}$
further in this book.  The topic is attractive as a research area.
For example, one might suppose that big bang produced large numbers
of primitive idempotents and that these primitive idempotents united
together to produce standard snuarks and dark matter snuarks.

Now it is known that standard matter does not decay at an
appreciable rate into dark matter.  Therefore one might suppose that
after the standard snuarks have condensed into standard matter, the
ratio of standard snuark to dark matter snuark will be frozen, and
therefore the present ratio of standard matter to dark matter will
be determined by statistical mechanics and the following reversible
reactions:
\begin{equation}\label{eq:DMtoSM}
\begin{array}{rcl}
(\rho_{z--}+\rho_{z-+}) + (\rho_{z+-}+\rho_{z++}) & <->&
\rho_{DM},\\
(\rho_{z--}+\rho_{z++}) + (\rho_{z+-}+\rho_{z-+}) & <->& \rho_{DM}.
\end{array}
\end{equation}
That is, the density of the four species of standard matter might be
related to the density of dark matter through the usual chemical
potential.  From this point on, when we refer to ``snuark'', the
reader can assume that we mean standard matter snuarks unless we
specify otherwise.


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:QNsOfCompVelEstates2})
\section{\label{sec:BST}  Binding Snuarks Together}


To bind together snuarks into elementary fermions, we need to cancel
off the remaining non scalar terms in Eq.~(\ref{eq:SnuarksWO}).  We
saw how the dark matter snuark $\rho_{DM}$ canceled these degrees of
freedom the easy way.  Now let us consider how to do it the hard
way.  We will consider the first two snuarks, that is, the
$(\rho_{z--}+\rho_{z-+})$ and $(\rho_{z+-}+\rho_{z++})$.  For
convenience, we repeat their values here:
\begin{equation}\label{eq:SnuarksWOtopTwo}
\begin{array}{rcl}
\rho_{z--}+\rho_{z-+}&=&0.25(1+\at{zt}-\hat{s}+\at{zst}),\\
\rho_{z+-}+\rho_{z++}&=&0.25(1+\at{zt}+\hat{s}-\at{zst}).
\end{array}
\end{equation}
The other pair of snuarks can be treated similarly.


Since the snuarks contain a $+\at{zt}$ degree of freedom, to cancel
this we need to add a snuark that will contain a $-\at{zt}$ degree
of freedom.  Such a snuark will travel in the $-z$ direction instead
of the $+z$ direction.  This is good because we will eventually need
to construct stationary matter, and having components that move in
opposite directions is necessary.


While a primitive idempotent contains $8$ degrees of freedom, it is
generated by only three choices of quantum numbers.  This says that
the degrees of freedom of a primitive idempotent are not
independent.  When we change one quantum number, in this case,
$\at{zt}$, we change four degrees of freedom.  The oriented degrees
of freedom change sign, the unoriented ones do not.

Similarly, changing the sign of $\at{zt}$ changes the sign of two of
the degrees of freedom of the snuarks.  We will follow the
crystallographer's convention and use $\bar{z}$ to designate a
snuark oriented in the $-z$ direction.  Then the two (of interest)
snuarks with $-1$ quantum numbers for $\at{zt}$ are:
\begin{equation}\label{eq:SnuarksWOtopTBar}
\begin{array}{rcl}
\rho_{\bar{z}--}+\rho_{\bar{z}-+}&=&0.25(1-\at{zt}-\hat{s}-\at{zst}),\\
\rho_{\bar{z}+-}+\rho_{\bar{z}++}&=&0.25(1-\at{zt}+\hat{s}+\at{zst}).
\end{array}
\end{equation}
We have four snuarks (i.e. the two in Eq.~(\ref{eq:SnuarksWOtopTwo})
and the two in Eq.~(\ref{eq:SnuarksWOtopTBar}) ) that we want to
combine into a scalar.  Let's suppose that we will have $A$, $B$,
$C$, and $D$ portions of each.  This sets up a matrix equation.
Ignoring the scalar component, the requirement that the non scalar
degrees of freedom add to zero gives us:
\begin{equation}\label{eq:MatrixSnuarkAdding}
\begin{array}({cccc})+1&+1&-1&-1\\-1&+1&-1&+1\\+1&-1&-1&+1\end{array}
\begin{array}({c})A\\B\\C\\D\end{array} =
\begin{array}({c})0\\0\\0\end{array}
\end{equation}
This set of equations has only one solution, $A=B=C=D=1$.  Thus we
must have all four snuarks contribute in order to get a scalar.


If we suppose that the two snuarks that travel in the $+z$ direction
are present at the same time, then these two snuarks will simply add
to $\rho_{DM}$ and we will be making dark matter.  Therefore, to
avoid the dark matter solution, we have to assume that these snuarks
are not present at the same time. In other words, we will have to
include these two as a linear superposition.



To move further with this idea requires that we have a better
understanding of what the mass interaction is, and how a left handed
particle changes into a right handed one.  This we postpone until
the chapter on mass.  For now, let us ignore the issue of how
snuarks change themselves into different snuarks (and so change
direction and various quantum numbers) and instead review the
general idea of how a particle with mass is built from particles
that travel at speed $c$.


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:QNsOfCompVelEstates2})

\section{\label{sec:tFCD}  The Feynman Checkerboard}


Each snuark consists of two primitive idempotents that travel
together, in the same direction at speed $c$.  Traveling in pairs
allows the elimination of the largest contribution to potential
energy, the $\at{ixyzst}$ term.  But in order to get a massive
fermion that is able to remain stationary, we need to combine enough
snuarks that the rest of the non scalar terms can be canceled. These
cancelations will be obtained by adding together snuarks that travel
in different directions and have different quantum numbers.

This concept is not entirely alien to particle physics.  Richard
Feynman came upon a similar idea when he was attempting to
understand the nature of the Dirac equation. He was able to get the
one dimensional Dirac equation by assuming that the electron moved
back and forth, always at speed $c$, that it changed directions at
random, and that each time it changed directions its wave function
was multiplied by a complex constant.

The idea is known as the ``Feynman Checkerboard'', or sometimes
chessboard, in reference to how the paths look when drawn on a $1+1$
space-time diagram.  The idea was originally presented as a problem
in the book on the path integral by Feynman and Hibbs.  A search of
the internet finds quite a few articles on the subject.  The first I
found, not necessarily the best, are
\cite{FosterFeynmanCheckerboard_0310166},
\cite{SmithFeynmanCheckerboard_9503015},
\cite{KullFeynmanCheckerboard_0212053},
\cite{KauffmanFeynmanCheckerboard_9603202}. When one tries to
generalize the Feynman checkerboard model to $3+1$ dimensions, an
issue that arises is that the speed of the particle making the
motion wants to be $c\sqrt{3}$.  This problem is discussed in Peter
Plavchan's excellent term paper.\cite{PlavchanFeynCheckerBoard_2002}
For the moment we will ignore this issue.  We will return to it in a
later chapter.


\SubSectionBreak


A primary objective of this paper is to derive relationships between
the masses of the leptons.  Particle masses are the energy of the
stationary, or non moving, particle.  Therefore, we will be
concentrating on particles which have no net motion.  To generalize
to particles that are moving, one simply takes appropriate linear
superpositions.  This will be the subject of a book to be written by
the author after the completion of the present one.


The Feynman checkerboard consists of a particle moving left and
right in $1$-dimensional space.  In our concept of this, we will
replace the left going particle with three snuarks oriented in the
$+x$, $+y$, and $+z$ directions, and we will replace the right going
particle with three snuarks oriented in the $-x$, $-y$, and $-z$
directions.  At the points where the left going particle turns into
a right going particle, we will instead have three snuarks changing
form into three other snuarks.  This will require considerably more
machinery than that used in previous Feynman checkerboard models,
but our machinery is stronger than that used previously.


In changing from one snuark form to another, spin is conserved. This
gives us a hint on how to interpret the various operators as spin.
Returning to the snuark quantum numbers, note that in
Eq.~(\ref{eq:QNsOfSnuarks}) all four snuarks shared the same non
zero quantum number for $\at{zt}$.  We will interpret this as a
precursor to spin, a sort of ``pre-spin'' in the $+z$ direction. Pre
spin in the other three directions is therefore:
\begin{equation}\label{eq:PreSpin}
\begin{array}{rcl}
S_x &=& \at{xt},\\
S_y &=& \at{yt},\\
S_z &=& \at{zt}.
\end{array}
\end{equation}
These three operators multiply to produce:
\begin{equation}\label{eq:SxSySzEq}
\begin{array}{rcl}
S_xS_yS_z &=&\at{xt}\;\at{yt}\;\at{zt},\\
&=&\at{xyzt},
\end{array}
\end{equation}
which does square to $-1$ as expected for the imaginary unit of the
Pauli algebra.


In addition, since $-\at{ixyzt} = \hat{s}\;\at{ixyzst}$ and
$\hat{s}$ and $\at{ixyzst}$ are two of the commuting roots of unity,
then so is $-\at{ixyzt}$, and multiplying this by $i$ still leaves
an operator that commutes with the rest of the primitive idempotent.
Therefore $\at{xyzt}$ will act just like an imaginary unit.  One can
then verify that $S_xS_y = \at{xyzt}S_z$ and cyclic permutations.
This completes the proof that $\{\at{xt},\at{yt},\at{zt}\}$ can be
interpreted as a set of $SU(2)$ spin operators.  Note that our
choice of $\hat{s}$ was not required for all this to happen.  The
various other possible choices would have led to analogous results.


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:QNsOfCompVelEstates2})

\section{\label{sec:aMthM}  Adding Mass to the Massless}


The standard model is founded on symmetry, while this book is
founded on geometry.  The most important symmetry is energy, and the
standard model uses a Lagrangian or Hamiltonian density to specify
particle interactions.  The simplest way to add mass to a massless
quantum field theory is simply to add a term that annihilates a
right handed particle and creates a left handed particle, with the
coupling constant being the mass itself, in addition, an additional
term, the Hermitian conjugate, is assumed but is sometimes not
mentioned.  For example, with the electron, one has a mass term in
the Lagrangian as follows:
\begin{equation}\label{eq:StandModelMass}
m\bar{e}_Le_R + \;\;\textrm{h.c.}\;\; =  m\bar{e}_Le_R +
m\bar{e}_Re_L,
\end{equation}
where ``h.c.'' stands for ``Hermitian conjugate''.


But the standard model uses a gauge principle to define the forces
between the elementary fermions and this principle is incompatible
with the above simple method of giving mass.  Instead, one requires
a scalar Higgs field that couples to the left and right handed
fields.  The gauge principle, like energy, is also a symmetry, and
this is in keeping with the fact that the standard model is built on
principles of symmetry.  Unfortunately, some of the symmetries are
not exact, so the standard model assumes that the symmetry is
broken.

The method used to break symmetry requires the assumption that the
vacuum of the physical world has less symmetry than the particles
possess.  The result is a theory that is complicated, difficult, and
while it does explain many experiments, it creates large numbers of
constants that can only be determined by experiment.  In the
operator formalism, there is no vacuum and this method of modeling
the elementary particles is not available.

Since we must break with the standard model over the existence of
the vacuum, we cannot logically follow the standard model in how
mass is defined.  Thus the natural inclination is to abandon the
Higgs mechanism and pursue the simpler method of giving masses to
fields, that is, with terms in the field theory such as
Eq.~(\ref{eq:StandModelMass}).


\SubSectionBreak


It is the author's intention, after this book is complete, to write
a book devoted to momentum and energy from a geometric operator
point of view.  However, let us briefly poach on the territory of
that second book and derive the massive electron propagator from the
massless propagators and the above Feynman diagrams.  The
justification for taking this material out of turn is that it will
motivate a generalization that is necessary to explain the structure
of the elementary fermions.  We will also use the idea to produce
formulas relating the masses of the leptons.


In the standard model, the principle of symmetry is at the
foundation.  In rejecting this and instead assuming that geometry is
fundamental we are released from slavishly copying all the methods
of deduction used in the standard model, but at the same time, we
can take advantage of the well tested calculational methods of the
standard model.  Methods of calculation are agnostic with respect to
symmetry versus geometry.


The standard model begins with a free Lagrangian.  One then perturbs
this Lagrangian by adding interaction terms.  Calculations are
simplified by the use of Feynman diagrams.  In turning this method
of calculation on its head, we will instead suppose that it is the
Feynman diagrams themselves that are physical, and the Lagrangian
just a mathematical convenience.  This will free us from having to
assume broken symmetries when our mathematical convenience turns out
to not be perfectly symmetric.

The mass term of Eq.~(\ref{eq:StandModelMass}) has no Feynman
diagrams associated with it.  The mass term is part of the
unperturbed field theory which can be solved exactly.  However,
since we have cut loose from the anchor of symmetry, we can treat
the mass term as if it were a perturbation of a massless theory, and
write down the Feynman diagrams that it would imply.

The two mass terms produce two Feynman diagrams.  The first converts
a right handed particle to left handed, the second converts the
opposite way:
\begin{equation}\label{eq:FeynmanLRandRL}
\begin{array}{rcl}
\feyn{!{fA}R \vertexlabel_{-im} !{fuA}L} \;\;,\;\;\;\; %
\feyn{!{fA}L \vertexlabel_{-im} !{fuA}R}
\end{array}
\end{equation}
The above two Feynman diagrams change the propagators from left
handed to right handed.  If we had chosen a different signature,
etc., we would have to use a different coupling constant.  The $-i$
is purely mathematical as the above is not a geometric theory.


Let $K_{LL}$ be the propagator for an electron that begins and ends
its journey as left handed, $K_{RL}$ be one that begins right handed
and ends left handed, etc.  We need to derive these propagators from
the massless propagators and the above Feynman diagrams.
Fortunately, the above Feynman diagrams are supremely simple and can
be easily resummed. We begin with $K_{LL}$:
\begin{equation}\label{eq:FeynmanLL}
\begin{array}{rcl}
K_{LL} &=& \feyn{!{fuA}L\;+\;\; !{fuA}L ![llft]{fdA}R !{fuA}L
\;\;+\;\; !{fuA}L ![llft]{fdA}R !{fuA}L ![llft]{fdA}R !{fuA}L
\;\; ...}\\
\end{array}
\end{equation}
From now we will leave off the coupling constants; there is one at
each vertex and they are all $-im$. The above series of Feynman
diagrams sum to:
\begin{equation}\label{eq:FeynmanLLSum}
\begin{array}{rcl}
K_{LL} &=&\left(\frac{-1}{\pbar}\right) +
\left(\frac{-1}{\pbar}\right)\left(\frac{-im}{1}\right)
\left(\frac{-1}{\pbar}\right)\left(\frac{-im}{1}\right)
\left(\frac{-1}{\pbar}\right) + ...\\
&=& \left(\frac{-1}{\pbar}\right)\left( 1 +
\left(\frac{-im}{1}\right)\left(\frac{-1}{\pbar}\right)
\left(\frac{-im}{1}\right)\left(\frac{-1}{\pbar}\right) + ... \right)\\
&=& \frac{-1}{\pbar}\left(1 - \frac{m^2}{p^2} + \frac{m^4}{p^4} -
\frac{m^6}{p^6} + ...\right)\\
&=& \frac{-1}{\pbar}\left(\frac{p^2}{p^2+m^2}\right)\\
&=& \frac{-\pbar}{p^2 + m^2},
\end{array}
\end{equation}
Similarly for the series that begin and end with right handed
propagators: $K_{RR} = -\pbar/(p^2+m^2)$.


The Feynman diagrams that begin with left handed propagators and end
with right handed propagators sum as follows:
\begin{equation}\label{eq:FeynmanLR}
\begin{array}{rcl}
K_{LR} &=& \feyn{!{fuA}L ![llft]{fdA}R  \;\;+\;\; %
!{fuA}L ![llft]{fdA}R !{fuA}L ![llft]{fdA}R  \;\;+\;\; %
!{fuA}L ![llft]{fdA}R !{fuA}L ![llft]{fdA}R !{fuA}L ![llft]{fdA}R
\;\; ...}\\
&=& %
\left(\frac{-1}{\pbar}\right)\left(\frac{-im}{1}\right)
\left(\frac{-1}{\pbar}\right) +\\
&&+\left(\frac{-1}{\pbar}\right)\left(\frac{-im}{1}\right)
\left(\frac{-1}{\pbar}\right)\left(\frac{-im}{1}\right)
\left(\frac{-1}{\pbar}\right)\left(\frac{-im}{1}\right)
\left(\frac{-1}{\pbar}\right) + ...\\
&=& %
\left(\frac{-im}{p^2}\right)\left(1 - \frac{m^2}{p^2} +
\frac{m^4}{p^4} - \frac{m^6}{p^6} + ...\right)\\
&=&\frac{-im}{p^2+m^2},
\end{array}
\end{equation}
and similarly for the right to left handed propagators: $K_{RL} =
-im/(p^2+m^2)$.


In the standard model, the left and right handed electron are
combined into a single particle.  This single particle can be split
back into its left and right handed parts using projection
operators.  What we've found above are the projections of the
propagators in their left and right handed forms.  In addition,
we've computed only the electron propagators.  The positron terms
will be similar.

To assemble $K_{LL}$, $K_{LR}$, $K_{RL}$ and $K_{RR}$ into a single,
massive, propagator, it is natural to choose a representation.  Our
choice of representation will also be used to define the gamma
matrices that are implicit in $\pbar$ and so these choices will
interact.  But regardless of the representation, when we define the
electron propagator to be the propagator of a particle that has left
and right components and includes both particle and antiparticle, we
will end up with a composite (i.e. massive) propagator that includes
the sums we have found:
\begin{equation}\label{eq:Kmassive}
K = (-\pbar-im)/(p^2+m^2).
\end{equation}
The above can be discussed at length, with more care and detail, and
the result will be the same.  The point is that the massive Dirac
propagator can be derived from massless Dirac propagators and the
resumming of two simple Feynman diagrams.




\section{\label{sec:CompCheckerboard}  A Composite Checkerboard}


The previous two sections gave two similar methods of giving mass to
massless particles.  The Feynman checkerboard of
Sec.~(\ref{sec:tFCD}) worked only in $1+1$ dimension, but it was
elegant and simple.  The left going and right going particles would
be eigenstates of velocity which makes for a consistent and simple
geometrization using Clifford algebra.  The two Feynman diagrams of
Sec.~(\ref{sec:aMthM}) worked in $3+1$ dimensions and was elegant
and simple, but it assume massless Dirac propagators and these are
not simple geometric objects.


In this section, we combine the two ideas into one which takes
advantage of the best parts of each.  We will use the Feynman
checkerboard technique of the earlier section so that we can use
primitive idempotents to define the subparticles, and we will use
the Feynman diagram technique of the later section so that we will
turn the problem into one of a resummation.


For simplicity, we will replace the snuarks with primitive
idempotents and use $L$ and $R$ just as labels to arbitrarily
distinguish between left and right handed particles without assuming
any difference in quantum numbers. This will simplify the problem
considerably.  We will return to the physical problem of the quarks
and leptons in the next chapter.


In the first part of this chapter we showed how primitive
idempotents could be bound together as snuarks.  A snuark is an
eigenstate of velocity and this limits how well it can minimize its
energy.  To reduce its energy to below the order of the Planck mass,
snuarks must combine together in ways that involve velocities
traveling in different directions. This is exactly the topic of the
Feynman checkerboard model and the resummation of massless
propagators to massive.


\SubSectionBreak


Pairs of primitive idempotents come equipped with a natural
probability, the familiar:
\begin{equation}\label{eq:PairsOfPIProbs}
\begin{array}{rcl}
P(A \to B) &=& \la A|B\ra \la B|A\ra,\\
&=&\tr(\rho_A\;\rho_B).
\end{array}
\end{equation}
Unfortunately for the Feynman checkerboard model, the above gives
zero when $\rho_A$ and $\rho_B$ are oriented in opposite directions.


We could always assume that gravity is a force that violates the
above assumption.  But we would still only have the Dirac propagator
in $1+1$ dimensions.  Instead, we will assume that the above is
true, that the usual $0.5(1+\cos(\theta))$ probability rule for
spin-$1/2$ states oriented in directions separated by an angle
$\theta$, does apply.

In making this assumption, it becomes impossible for a primitive
idempotent to ``turn on a dime''.  A primitive idempotent oriented
in $+z$, that is, one that carries a $+\at{zt}$ quantum number,
cannot transform immediately into one with a $-\at{zt}$ quantum
number.  It can, however, go to $\pm\at{xt}$ or $\pm\at{yt}$.

Therefore, we will assume that the mass interaction forces the bound
states of primitive idempotents to involve all six orientations:
\begin{equation}\label{eq:SnuarkQNOrientedSix}
\begin{array}{r|ccc|}
&\at{xt}&\at{yt}&\at{zt}\\ \hline %
\rho_{     x }&+2&0&0\\
\rho_{     y }&0&+2&0\\
\rho_{     z }&0&0&+2\\
\rho_{\bar{x}}&-2&0&0\\
\rho_{\bar{y}}&0&-2&0\\
\rho_{\bar{z}}&0&0&-2\\ \hline
\end{array}
\end{equation}
In the above we have included only the orientation quantum number.
The other quantum numbers add to the confusion, but our analysis
will cover them as well (in later chapters).  But our analysis will
be precise for the dark matter snuarks, $\rho_{DM}$, which have no
other quantum numbers.


We can write down a table showing the probabilities for primitive
idempotent transformations from one orientation to another:
\begin{equation}\label{eq:SnuarkTransProbs}
\begin{array}{c|cccccc|}
&\rho_x&\rho_y&\rho_z&\rho_{\bar{x}}&\rho_{\bar{y}}&\rho_{\bar{z}}\\
\hline %
\rho_x&1&0.5&0.5&0&0.5&0.5\\
\rho_y&0.5&1&0.5&0.5&0&0.5\\
\rho_z&0.5&0.5&1&0.5&0.5&0\\
\rho_{\bar{x}}&0&0.5&0.5&1&0.5&0.5\\
\rho_{\bar{y}}&0.5&0&0.5&0.5&1&0.5\\
\rho_{\bar{z}}&0.5&0.5&0&0.5&0.5&1\\ \hline
\end{array}
\end{equation}
In the above, again, we have ignored other quantum numbers.



If the mass interaction changed only the orientation quantum number
of the primitive idempotents, the above table of probabilities would
apply.  We do not specify the mass interaction, Nature does, and it
does not leave the other quantum numbers unchanged.  Therefore, we
have to divide the list of orientations into at least two groups. We
will distinguish these with $L$ and $R$, and the transition
probabilities will be zero inside either one of these groups.


Arbitrarily putting the $L$ into the $+x$, $+y$, and $+z$
orientations, the probability table looks like:
\begin{equation}\label{eq:SnuarkTransProbProb}
\begin{array}{c|cccccc|}
&\rho_{xL}&\rho_{yL}&\rho_{zL}&\rho_{\bar{x}R}&\rho_{\bar{y}R}&\rho_{\bar{z}R}\\
\hline %
\rho_{xL}&0&0&0&0&0.5&0.5\\
\rho_{yL}&0&0&0&0.5&0&0.5\\
\rho_{zL}&0&0&0&0.5&0.5&0\\
\rho_{\bar{x}R}&0&0.5&0.5&0&0&0\\
\rho_{\bar{y}R}&0.5&0&0.5&0&0&0\\
\rho_{\bar{z}R}&0.5&0.5&0&0&0&0\\ \hline
\end{array}
\end{equation}
The above can be obtained by postulating that three left handed
primitive idempotents transform to right handed primitive
idempotents and vice versa. This fits into a sort of Feynman
checkerboard scheme, but with three particles transforming into
three particles at each ``$m$'' vertex.


\SubSectionBreak


% Fig.~(\ref{fig:LtoRthree})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(240,100)
\thicklines % Propagators
\put( 10,40){\vector( 1,1){35}} % First
\put( 20,30){\vector( 1,1){45}} %
\put( 30,20){\vector( 1,1){25}} %
\put( 80,20){\line(-1,1){25}} %
\put( 90,30){\line(-1,1){45}} %
\put(100,40){\line(-1,1){35}} %
\put(140,40){\vector( 1,1){45}} % Second
\put(150,30){\vector( 1,1){25}} %
\put(160,20){\vector( 1,1){35}} %
\put(210,20){\line(-1,1){35}} %
\put(220,30){\line(-1,1){25}} %
\put(230,40){\line(-1,1){45}} %
\put(120,60){\line( 0,1){20}} % Plus
\put(110,70){\line( 1,0){20}} %
\thinlines
% Labels
\put(  9,23){$L$} %
\put(139,23){$L$} %
\put( 94,21){$R$} %
\put(224,21){$R$} %
\put( 15,53){$x$} % x,y,z
\put( 25,43){$y$} %
\put( 35,33){$z$} %
\put(145,53){$x$} %
\put(155,43){$y$} %
\put(165,33){$z$} %
\put( 90,54){$\bar{z}$} % /x,y,z
\put( 80,44){$\bar{y}$} %
\put( 70,34){$\bar{x}$} %
\put(220,54){$\bar{z}$} %
\put(210,44){$\bar{y}$} %
\put(200,34){$\bar{x}$} %
\end{picture}
} %
\caption{\label{fig:LtoRthree} Feynman diagrams that contribute to a
left handed primitive idempotent (snuark) becoming right handed.}
\end{figure}


The above table of probabilities allows only two possible
transitions each way.  Looking at $L$ to $R$ transitions, $x$ can
either transform to $\bar{y}$ or $\bar{z}$.  If $x$ transforms to
$\bar{y}$, then $z$ can only go to $\bar{x}$ and so $y$ can only go
to $\bar{x}$.  Thus there are only two Feynman diagrams that
contribute to the $L$ to $R$ process as shown in
Fig.~(\ref{fig:LtoRthree}).


For familiarity, let us write the array of transition amplitudes in
the spinor form.  Given an initial state $(|xL\ra,|yL\ra,|zL\ra)$,
the interaction transforms it to a new state
$(|\bar{x}R\ra,|\bar{y}R\ra,|\bar{z}R\ra)$.  The amplitudes for this
are:
\begin{equation}\label{eq:AmplitudesLtoRspin}
\begin{array}{c|ccc|}
&xL&yL&zL\\ \hline %
\bar{x}R&0 & \la\bar{x}R|yL\ra & \la\bar{x}R|zL\ra \\
\bar{y}R&\la\bar{y}R|xL\ra & 0 & \la\bar{y}R|zL\ra \\
\bar{z}R&\la\bar{z}R|xL\ra & \la\bar{z}R|yL\ra & 0 \\ \hline %
\end{array}
\end{equation}
We can replace the transition amplitudes with products of primitive
idempotents:
\begin{equation}\label{eq:AmplitudesLtoRrho}
\begin{array}{c|ccc|}
&\rho_{xL}&\rho_{yL}&\rho_{zL}\\ \hline %
\rho_{\bar{x}R}&0 & \rho_{\bar{x}R}\;\rho_{yL} & \rho_{\bar{x}R}\;\rho_{zL} \\
\rho_{\bar{y}R}& \rho_{\bar{y}R}\;\rho_{xL}& 0 & \rho_{\bar{y}R}\;\rho_{zL} \\
\rho_{\bar{z}R}& \rho_{\bar{z}R}\;\rho_{xL}& \rho_{\bar{z}R}\;\rho_{yL} & 0 \\ \hline %
\end{array}
\end{equation}
A similar set of transition amplitudes apply to the transformation
from $R$ to $L$:
\begin{equation}\label{eq:AmplitudesRtoLrho}
\begin{array}{c|ccc|}
&\rho_{\bar{x}L}&\rho_{\bar{y}L}&\rho_{\bar{z}L}\\ \hline %
\rho_{xR}&0 & \rho_{xR}\;\rho_{\bar{y}L} & \rho_{xR}\;\rho_{\bar{z}L} \\
\rho_{yR}& \rho_{yR}\;\rho_{\bar{x}L}& 0 & \rho_{yR}\;\rho_{\bar{z}L} \\
\rho_{zR}& \rho_{zR}\;\rho_{\bar{x}L}& \rho_{zR}\;\rho_{\bar{y}L} & 0 \\ \hline %
\end{array}
\end{equation}


\SubSectionBreak


Since we have transition amplitudes for $L$ to $R$ and $R$ to $L$,
we can combine these together to produce transition for $L$ to $R$
to $L$.  The general form will be obvious if we work out a few of
these by example.  We will abbreviate the intermediate states by
$\rho_R$:
\begin{equation}\label{eq:LtoRtoLExamples}
\begin{array}{rcl}
\rho_{xL}\;\rho_R\;\rho_{xL} &=&
\rho_{xL}\;\rho_{\bar{y}R}\;\rho_{xL}
+ \rho_{xl}\;\rho_{\bar{z}R}\;\rho_{xL},\\
\rho_{xL}\;\rho_R\;\rho_{yL} &=& \rho_{xL}\;\rho_{\bar{z}R}\;\rho_{yL},\\
\rho_{xL}\;\rho_R\;\rho_{zL} &=&
\rho_{xL}\;\rho_{\bar{y}R}\;\rho_{zL}.
\end{array}
\end{equation}
The other intermediate states, for example, $\rho_{\bar{x}R}$, are
eliminated by annihilation.


Since, for example, $\rho_{\bar{y}R} =  (\rho_{\bar{y}R})^2$, we can
rewrite the above as:
\begin{equation}\label{eq:LtoRtoLExamplet}
\begin{array}{rcl}
\rho_{xL} \;\rho_R\; \rho_{xL} &=&
(\rho_{xl}\;\rho_{\bar{y}R})(\rho_{\bar{y}R}\;\rho_{xL}) +
(\rho_{xl}\;\rho_{\bar{z}R})(\rho_{\bar{z}R}\;\rho_{xL}),\\
\rho_{xL}\;\rho_R \rho_{yL} &=& (\rho_{xl}\;\rho_{\bar{z}R}) (\rho_{\bar{z}R}\;\rho_{yL}),\\
\rho_{xL}\;\rho_R \rho_{zL} &=&
(\rho_{xl}\;\rho_{\bar{y}R})(\rho_{\bar{y}R}\;\rho_{zL}).
\end{array}
\end{equation}
And this is just the top line in the matrix product:
\begin{equation}\label{eq:AmplitudesLtoRtoL}
\begin{array}{l}
\begin{array}({ccc})
0 & \rho_{xL}\;\rho_{\bar{y}R} & \rho_{xL}\;\rho_{\bar{z}R} \\
\rho_{yL}\;\rho_{\bar{x}R}& 0 & \rho_{yL}\;\rho_{\bar{z}R} \\
\rho_{zL}\;\rho_{\bar{x}R}& \rho_{zL}\;\rho_{\bar{y}R} & 0
\end{array}
\begin{array}({ccc})
0 & \rho_{\bar{x}R}\rho_{yL} & \rho_{\bar{x}R}\rho_{zL} \\
\rho_{\bar{y}R}\rho_{xL}& 0 & \rho_{\bar{y}R}\rho_{zL} \\
\rho_{\bar{z}R}\rho_{xL}& \rho_{\bar{z}R}\rho_{yL} & 0
\end{array}\\
$ $\\
=\begin{array}({ccc})
\rho_{x}\rho_{\bar{y}}\rho_{x}+\rho_{x}\rho_{\bar{z}}\rho_{x}&
\rho_{xL}\;\rho_{\bar{z}R}\;\rho_{yL}&\rho_{xL}\;\rho_{\bar{y}R}\;\rho_{zL}\\
\rho_{yL}\;\rho_{\bar{z}R}\;\rho_{xL}&\rho_{y}\rho_{\bar{x}}\rho_{y}+
\rho_{y}\rho_{\bar{z}}\rho_{y}&\rho_{yL}\;\rho_{\bar{x}R}\;\rho_{zL}\\
\rho_{zL}\;\rho_{\bar{y}R}\;\rho_{xL}&\rho_{zL}\;\rho_{\bar{x}R}\;\rho_{yL}&
\rho_{z}\rho_{\bar{x}}\rho_{z}+\rho_{z}\rho_{\bar{y}}\rho_{z}
\end{array}
\end{array}
\end{equation}
where we have left off the $L$ and $R$ labels on the diagonal to fit
to the page. Thus it is natural for us to use matrices to represent
the $L$ to $R$ and $R$ to $L$ processes.


It is important to note that this sort of representation is not a
general feature of Feynman diagrams.  It only works here because we
are assuming (by energy considerations) that all three of the
initial and final states are filled by exactly one particle each.


If we represent our states $\rho_{\chi L}$ and $\rho_{\bar{\chi}R}$
in the Pauli algebra as $0.5(1+\sigma_\chi)$ and
$0.5(1-\sigma_\chi)$, respectively, we can simplify the above matrix
product.  Recall that any product that begins and ends with
$0.5(1+\sigma_\chi)$ is a real multiple of $0.5(1+\sigma_\chi)$. For
example:
\begin{equation}\label{eq:RecallxChix}
\begin{array}{l}
\rho_x (a_1\hat{1} + a_x\hat{x} + a_y\hat{y} +
a_z\hat{z}) \rho_x,\\
=0.5(1+\hat{x})(a_1\hat{1} + a_x\hat{x} + a_y\hat{y} +
a_z\hat{z})0.5(1+\hat{x}),\\
=(a_1 + a_x)0.5(1+\hat{x}),\\
=(a_1 + a_x)\rho_x.
\end{array}
\end{equation}
Thus $\rho_x\rho_{\bar{y}}\rho_x = 0.5\rho_x$, and the diagonal
elements in the matrix product of Eq.~(\ref{eq:AmplitudesLtoRtoL})
reduce to give:
\begin{equation}\label{eq:ReducedLtoL}
\begin{array}({ccc})
\rho_{xL}&\rho_{xL}\;\rho_{\bar{z}R}\;\rho_{yL}&\rho_{xL}\;\rho_{\bar{y}R}\;\rho_{zL}\\
\rho_{yL}\;\rho_{\bar{z}R}\;\rho_{xL}&\rho_{yL}&\rho_{yL}\;\rho_{\bar{x}R}\;\rho_{zL}\\
\rho_{zL}\;\rho_{\bar{y}R}\;\rho_{xL}&\rho_{zL}\;\rho_{\bar{x}R}\;\rho_{yL}&\rho_{zL}
\end{array}
\end{equation}
Similarly, using the methods of Sec.~(\ref{sec:ProdsDensOps}), the
off diagonal terms reduce.  For example:
\begin{equation}\label{eq:ExampleReduceOffDiagLtoL}
\begin{array}{l}
\rho_{xL}\;\rho_{\bar{y}R}\;\rho_{zL}\\
=0.5(1+i)\;\rho_{xL}\;\rho_{zL},\\
=\sqrt{0.5}\;e^{+i\pi/4}\rho_{xL}\;\rho_{zL},
\end{array}
\end{equation}
where $i=\at{xyz}$ as before.  One obtains the angle $\pm i\pi/4$
for the various off diagonal terms according to which diagonal they
are in. Factoring out $\sqrt{0.5}$, the fully reduced product is:
\begin{equation}\label{eq:FullyReducedLtoL}
\sqrt{0.5}
\begin{array}({ccc})
\sqrt{2}\rho_{xL}&e^{-i\pi/4}\rho_{xL}\rho_{yL}&e^{+i\pi/4}\rho_{xL}\rho_{zL}\\
e^{+i\pi/4}\rho_{yL}\rho_{xL}&\sqrt{2}\rho_{yL}&e^{-i\pi/4}\rho_{yL}\rho_{zL}\\
e^{-i\pi/4}\rho_{zL}\rho_{xL}&e^{+i\pi/4}\rho_{zL}\rho_{yL}&\sqrt{2}\rho_{zL}
\end{array}
\end{equation}
The elements of the above matrix are Clifford algebra constants
rather than the usual complex numbers.


Consider general $3\times 3$ matrices whose elements are complex
multiples of the corresponding elements of the above matrix.  Such
matrices are more general than complex matrices, however, since the
Clifford algebraic constant is specified for each position in the
matrix, the number of degrees of freedom is the same as that of
complex $3 \times 3$ matrices.


The sum of two such matrices will still be of this form, so this
type of matrix is closed under addition.  And addition will be
analogous to addition of complex matrices.  While multiplication
will be different from that of complex matrices, the reader can
quickly verify (using the primitive idempotent product reduction
equations) that matrices of this type are closed under
multiplication.

What's more, matrices of this sort include a natural unit matrix:
\begin{equation}\label{eq:UnitLtoLMatrix}
\hat{1} = \begin{array}({ccc}) \rho_{xL}&0&0\\
0&\rho_{yL}&0\\ 0&0&\rho_{zL} \end{array}.
\end{equation}
therefore we have an algebra.  That is, we have a set of objects
which we can add and multiply, a zero (i.e. the matrix of all
zeros), and a unit.


This is very significant.  The arrays of amplitudes for left handed
particles becoming right handed and returning to left handed form an
algebra.  We therefore can think of these objects as quantum states
and compute their primitive idempotents using the machinery we
developed in Chapter~(\ref{sec:PrimitiveIdempotents}).  This means
that we can derive the structure of bound states of primitive
idempotents from the structure of the primitive idempotents using
the same mathematics. This is how we will derive the elementary
fermions in Chapter~(\ref{sec:TheZoo}). But first, let us examine
exactly what this sort of sleight of hand means.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:AmplitudesLtoRtoL})
% \SubSectionBreak

\section{\label{sec:BoundStatePrimitiveIdempotents}  Bound State Primitive Idempotents}


Our original motivation for using a density operator formalism was
that the formalism worked for the elementary fermions.  But now our
primitive idempotents are assumed to be subparticles and the
elementary fermions that make up the elementary fermions.  This
implies that there must exist a way for relating our composite
models of fermions back to an idempotent form.  In this section we
address two objectives.  First, we solve the primitive idempotent
problem for the bound states.  Second, we give an interpretation in
terms of forces.


We are proposing that matrices of primitive idempotents are how one
must model the natural bound states of primitive idempotents.  That
is, we require our bound state $\rho$ be of the form:
\begin{equation}\label{eq:BoundStateMatrix}
\rho_a = \begin{array}({ccc}) %
a_{xx}\rho_x&a_{xy}\rho_x\;\rho_y&a_{xz}\rho_x\;\rho_z\\
a_{yx}\rho_y\;\rho_x&a_{yy}\rho_y&a_{yz}\rho_y\;\rho_z\\
a_{zx}\rho_z\;\rho_x&a_{zy}\rho_z\;\rho_y&a_{zz}\rho_z
\end{array},
\end{equation}
where $a_{jk}$ are complex numbers.  Let $\rho_b$ be another such
matrix, with coefficients $b_{jk}$.  To abbreviate our notation, let
us designate these sorts of matrices by leaving off the $\rho_\chi$
terms inside the matrix.  But to remind us that these are not the
usual complex matrices, and cannot be multiplied as such, we will
add a hat to the matrix as in $(\;)_\rho$.  Later, when we are
dealing with more complicated problems, we can add a designation
outside the matrix to indicate the internal quantum numbers of the
primitive idempotents.


The product $\rho_a\;\rho_b$ is of the same form, and can be
computed as follows:
\begin{equation}\label{eq:BoundStateProducts}
\begin{array}{rcl}
\rho_a\rho_b &=&
\begin{array}({ccc}) %
a_{xx}&a_{xy}&a_{xz}\\
a_{yx}&a_{yy}&a_{yz}\\
a_{zx}&a_{zy}&a_{zz}
\end{array}_\rho
\begin{array}({ccc}) %
b_{xx}&b_{xy}&b_{xz}\\
b_{yx}&b_{yy}&b_{yz}\\
b_{zx}&b_{zy}&b_{zz}
\end{array}_\rho\\
$ $\\
&=&\begin{array}({ccc})
a_{xx}b_{xx}+0.5a_{xy}b_{yx}+0.5a_{xz}b_{zx}  & ... & ...\\
a_{yx}b_{xx}+a_{yy}b_{yx}+0.5(1+i)a_{yz}b_{zx}& ... & ...\\
a_{zx}b_{xx}+0.5(1-i)a_{zy}b_{yx}+a_{zz}b_{zx}& ... & ...
\end{array}_\rho .
\end{array}
\end{equation}
where ``$...$'' stands for six more terms similar to those shown,
but which do not fit on the page.  The factors of $0.5$ come from
reduction of products like $\rho_x\;\rho_y\;\rho_x$, and the factors
of $0.5(1\pm i)$ come from products like $\rho_x\;\rho_y\;\rho_z$.


The form of the product implies a way that we can convert our
$(\;)_\rho$ matrices back and forth into the usual complex matrices.
The conversion runs as follows:
\begin{equation}\label{eq:ConvertRhoToComplx}
\begin{array}{l}
\begin{array}({ccc})
a_{xx}&a_{xy}&a_{xz}\\
a_{yx}&a_{yy}&a_{yz}\\
a_{zx}&a_{zy}&a_{zz}
\end{array}_\rho\\
$ $\\
\to
\begin{array}({ccc})
a_{xx}&\sqrt{0.5}\;e^{+i\pi/12}\;a_{xy}&\sqrt{0.5}\;e^{-i\pi/12}\;a_{xz}\\
\sqrt{0.5}\;e^{-i\pi/12}\;a_{yx}&a_{yy}&\sqrt{0.5}\;e^{+i\pi/12}\;a_{yz}\\
\sqrt{0.5}\;e^{+i\pi/12}\;a_{zx}&\sqrt{0.5}\;e^{-i\pi/12}\;a_{zy}&a_{zz}
\end{array}
\end{array}
\end{equation}
This conversion is linear.  That is, if $(a)_\rho + (b)_\rho$
$=(a+b)_\rho$, then $(a)+(b)$ $=(a+b)$.  And it preserves
multiplication, that is if $(a)_\rho\;(b)_\rho$ $=(ab)_\rho$, then
$(a)(b) = (ab)$.  Furthermore, the conversion preserves the unit
matrix, and so we have a way of doing computations in $(\;)_\rho$
matrices using regular complex matrices.


In short, despite the odd form of our $(\;)_\rho$ matrices, they act
in every way like the usual complex matrices.  In
Sec.~(\ref{sec:PIsOf33}) we analyzed the structure of the primitive
idempotents of $3\times 3$ complex matrices and that analysis can
immediately be applied here.  By symmetry, we need to treat the
three direction $x$, $y$, and $z$ the same.  Therefore the natural
choice of primitive idempotents are the circulant.


\SubSectionBreak


We found the circulant $3\times 3$ matrices in eigenvalue form as
Eq.~(\ref{eq:EvaluesCircIdempotents}), and include them here,
written out in matrix form, for convenience:
\begin{equation}\label{eq:CirculantPIsOverAgain}
\begin{array}{rcl}
\rho_{1} &=&\frac{1}{3}\begin{array}({ccc}) %
1&1&1\\1&1&1\\1&1&1 \end{array},\\
\rho_{\nu} &=&\frac{1}{3}\begin{array}({ccc}) %
1&\nu&\nu^*\\\nu^*&1&\nu\\\nu&\nu^*&1 \end{array},\\
\rho_{\nu^*} &=&\frac{1}{3}\begin{array}({ccc}) %
1&\nu^*&\nu\\\nu&1&\nu^*\\\nu^*&\nu&1 \end{array}.
\end{array}
\end{equation}
We have labeled these primitive idempotents with the complex numbers
making up their $(1,2)$ position.  There are three primitive
idempotents. Eventually we will associate these three solutions with
the three generations of elementary particles, but to do this
realistically we will need snuarks instead of the primitive
idempotents we are discussing here.


To translate the $3\times 3$ matrix primitive idempotents into
composite primitive idempotent form, we simply use the reverse of
the conversion given in Eq.~(\ref{eq:ConvertRhoToComplx}).  That is,
we multiply the off diagonal elements by $\sqrt{2}e^{\pm i\pi/12}$.
The three bound state primitive idempotents are then:
\begin{equation}\label{eq:CirculantPIsBoundStates}
\begin{array}{rcl}
\rho_{1} &=&\frac{1}{3}\begin{array}({ccc}) %
1&\sqrt{2}e^{-i\pi/12}&\sqrt{2}e^{+i\pi/12}\\
\sqrt{2}e^{+i\pi/12}&1&\sqrt{2}e^{-i\pi/12}\\
\sqrt{2}e^{-i\pi/12}&\sqrt{2}e^{+i\pi/12}&1 \end{array}_\rho,\\
\rho_{\nu} &=&\frac{1}{3}\begin{array}({ccc}) %
1&\sqrt{2}e^{+7i\pi/12}&\sqrt{2}e^{-7i\pi/12}\\
\sqrt{2}e^{-7i\pi/12}&1&\sqrt{2}e^{+7i\pi/12}\\
\sqrt{2}e^{+7i\pi/12}&\sqrt{2}e^{-7i\pi/12}&1 \end{array}_\rho,\\
\rho_{\nu^*} &=&\frac{1}{3}\begin{array}({ccc}) %
1&\sqrt{2}e^{-3i\pi/4}&\sqrt{2}e^{+3i\pi/4}\\
\sqrt{2}e^{+3i\pi/4}&1&\sqrt{2}e^{-3i\pi/4}\\
\sqrt{2}e^{-3i\pi/4}&\sqrt{2}e^{+3i\pi/4}&1
\end{array}_\rho.
\end{array}
\end{equation}
In the above, the angles are given by $-\pi/12$, $2\pi/3-\pi/12 =
7\pi/12$, and $-2\pi/3-\pi/12 = -3\pi/4$.  That is, the rather
arbitrary looking angles, $-\pi/12$, $7\pi/12$ and $-3\pi/4$ are
obtained by subtracting $\pi/12$ from the three complex cube roots
of unity.  Thus we can solve the bound state primitive idempotent
problem.


\SubSectionBreak


We've been using matrices of products of primitive idempotents, as
in Eq.~(\ref{eq:BoundStateMatrix}) to represent bound states.  The
diagonal elements are complex multiples of primitive idempotents,
$\rho_{x}$, $\rho_y$, and $\rho_z$.  However, in
Eq.~(\ref{eq:CirculantPIsBoundStates}) we see that the only
``complex multiples'' of the diagonal elements are $1/3$.  Thus the
diagonal elements are treated equally.  This is exactly what we
expect if these three primitive idempotents are combined into a
bound state.


It is natural to expect that there should only be one bound state
for the three primitive idempotents; instead we've found three. But
the difference between these three solutions is not in the diagonal
elements, but instead in the off diagonal.


The off diagonal terms are products of two primitive idempotents,
for example $\rho_x\;\rho_y$. In the operator formalism, products of
two primitive idempotents are what we use to represent a process
that converts a particle of one type into another.


In the standard model, one expects that the conversion of a fermion
from one type to another is accompanied by the creation or
annihilation of a gauge boson.  Thus we can think of the off
diagonal elements as having something to do with a particle
interaction.


In replacing the Feynman checkerboard interaction with a matrix, we
enforced the requirement that the number of particles of each type
was conserved.  That is, we began with one $\rho_{xL}$, one
$\rho_{yL}$, and one $\rho_{zL}$, and these became exactly one
$\rho_{\bar{x}R}$, one $\rho_{\bar{y}R}$, and one $\rho_{\bar{z}R}$.
One normally associates a conservation like this with an exchange
force.


% Fig.~(\ref{fig:LtoRBosons})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
$
\Diagram{ %
  &  &  &  &fu&fd&  &  &  &  \\
  &  &  &fu&gd&  &fd&  &  &  \\
  &  &fu&  &gu&gd&  &fd&  &  \\
  &fu&  &fu&fd&fu&fd&  &fd&  \\
fu&  &fu&  &fu&fd&  &fd&  &fd\\
  &fu&  &fu&  &  &fd&  &fd&  \\
  &  &fu&  &  &  &  &fd&  &  }
$} %
\caption{\label{fig:LtoRBosons} The mass interaction as an exchange
of three ``gauge bosons''.}
\end{figure}


When the standard model deals with gauge bosons, it must define a
propagator for the exchange particle.  In our case, the three
particles are transformed at the same point in space.  Thus there
are no propagators for these bosons, as they do not propagate
anywhere.  We can draw gauge bosons as in
Fig.~(\ref{fig:LtoRBosons}).


This gives us an interpretation for the bound state primitive
idempotents.  The diagonal elements correspond to the ``valence
primitive idempotents'', and the off diagonal elements are the
``primitive idempotent sea''.  In this interpretation, the three
generations of fermions correspond to the same valence primitive
idempotents (and therefore have the same quantum numbers), but
correspond to different levels of excitation in the sea.








% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:QNsOfCompVelEstates2})
% \SubSectionBreak
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{\label{sec:TheZoo} The Zoo}

%\begin{quote}
\settowidth{\versewidth}{Which they would not give to a lunatic, and
the competition so strong!}
\begin{verse}[\versewidth]
%\em
An' I sign for four-pound-ten a month and save the money clear,\\
An' I am in charge of the lower deck, an' I never lose a steer;\\
An' I believe in Almighty God an' preach His Gospel here.\\
$ $\\
The skippers say I'm crazy, but I can prove 'em wrong,\\
For I am in charge of the lower deck with all that doth belong---\\
{\em Which they would not give to a lunatic, and the competition so strong!}\\
\end{verse}
%\end{quote}
$ $

\lettrine{T}{$\;$he previous chapter} showed how it is natural to
suppose that primitive idempotents combine so as to cancel their
strongest degrees of freedom, and the resulting ``snuarks''
naturally form into the same doublet and dual singlet form seen in
the elementary fermions.  Then we showed how to make idempotency
calculations with collections of primitive idempotents that are
bound together in a manner similar to how mass binds left and right
handed particles.  The result was that the bound states exhibited
three levels of excitation of the sea, and we claimed that these
could be thought of as the three generations of elementary fermions.


For simplicity, these two ideas, binding primitive idempotents into
snuarks and collecting primitive idempotents into bound states, were
applied separately. In this chapter, we apply them simultaneously,
and show how the result can be interpreted as the three generations
of elementary fermions.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:AmplitudesLtoRtoL})
% \SubSectionBreak

\section{\label{sec:TheMassInteraction}  The Mass Interaction}


In Chapter~(\ref{sec:Force}), we associated the mass interaction
with three primitive idempotents exchanging gauge bosons.  For
example three left handed primitive idempotents can be changed into
right handed primitive idempotents.  Our expectation is that the
left and right handed composite states will give a sum that has no
non scalar parts left.  The resulting purely scalar sum will define
the mass.

Looking at this from the point of view of potential energy, we
expect the left and right handed states to have non scalar parts
that cancel.  That is, the non scalar part of the left handed state
needs to be the negative of the non scalar part of the right handed
state.  But this implies that the gauge bosons have completely
negated the non scalar part of the left handed state.

Since the non scalar parts of a quantum state are the parts that
have Planck energies, we expect this process to leave a gauge boson
with energy around the Planck energy.  But this seems rather
unphysical.


\SubSectionBreak


To explain how the mass interaction can change the non scalar parts
of quantum states, we need to reexamine three facts from the
previous chapter: First, we found that a natural potential energy
needs to ignore, or almost ignore, the scalar part of a Clifford
algebra element. Second, we found that the natural mass interaction
needs to change the sign of Clifford algebra primitive idempotents
as in $0.5(1+\at{zt})$ to $0.5(1-\at{zt})$. Third, we found that
primitive idempotents will bind in groups of three, and when doing
this, there will be three different choices for the sea that binds
them together.

This book has been devoted to the discrete symmetries of elementary
particles or quantum states.  In so concentrating, we have ignored
the fact that elementary particles form interference patterns.  This
suggests that instead of thinking of our elementary particles as
idempotents such as $0.5(1+\at{zt})$, we should instead think of
them as having some sort of phase, for example, $e^{i\omega
t}0.5(1+\at{zt})$.  In other words, we should think of
$-0.5(1+\at{zt})$ as representing the same elementary particle as
$0.5(1+\at{zt})$.


Since the natural potential energy needs to almost ignore the scalar
part of a Clifford algebra primitive idempotent, this suggests that
we should represent the mass interaction as a negating of the scalar
part of a quantum object.  This will give something that is not
normalized, so we then negate the whole object.  We will call this
operation $M$.  For example:
\begin{equation}\label{eq:DefnOfMop}
\begin{array}{rcl}
M(0.5(1+\at{zt})) &=&- (0.5(-1+\at{zt})),\\
&=&+0.5(1-\at{zt}).
\end{array}
\end{equation}
In general, $M$ will negate all but the scalar part of an element.
In the above example, $M$ maps an idempotent to an idempotent, but
we are more interested in what $M$ does to primitive idempotents.


Note that our mass interaction, $M$, preserves our potential energy,
$V$. That is, since our potential energy is a sum of multiples of
squares, changing the signs of the vector part does not change the
potential energy:
\begin{equation}\label{eq:MassInteractionPreserves}
V(M(\rho)) = V(\rho).
\end{equation}
Any modification which consists of changing the signs of some set of
components of the Clifford algebra will preserve potential energy,
but the one we are using minimizes the change by assuming that the
change is to the scalar part only.


\SubSectionBreak


To see what $M$ does to primitive idempotents, let's choose
primitive roots of unity $\hat{s}$, $\at{zt}$, and $\at{ixyzst}$.  A
typical primitive idempotent looks like:
\begin{equation}\label{eq:typPIforexpA}
\begin{array}{rcl}
\rho_{z++} &=& 0.125(1+\at{zt})(1+\hat{s})(1+\at{ixyzst}),\\
&=&0.125(1+\at{zt}+\hat{s}+\at{ixyzst}-\at{zst}-\at{ixys}-\at{ixyzt}-\at{ixy}).
\end{array}
\end{equation}
Applying $M$ to this primitive idempotent negates all but the scalar
term giving:
\begin{equation}\label{eq:typMPIforexp}
\begin{array}{rcl}
M(\rho_{z++}) &=&
0.125(1-\at{zt}-\hat{s}-\at{ixyzst}+\at{zst}+\at{ixys}+\at{ixyzt}+\at{ixy}).
\end{array}
\end{equation}
The above is not a primitive idempotent.  The first four terms look
like the first four terms of $\rho_{\bar{z}--}$, but the next three
terms have the wrong signs:
\begin{equation}\label{eq:typPIforexpB}
\begin{array}{rcl}
\rho_{z--} &=& 0.125(1-\at{zt})(1-\hat{s})(1-\at{ixyzst}),\\
&=&0.125(1-\at{zt}-\hat{s}-\at{ixyzst}-\at{zst}-\at{ixys}-\at{ixyzt}+\at{ixy}),\\
&\neq& M(\rho_{z++}).
\end{array}
\end{equation}
Thus $M$ does not map primitive idempotents to primitive
idempotents.  The result is not a quantum state, it is instead a
mixture.


In quantum mechanics, when a state propagates to a mixture, a
measurement of the quantum numbers of the state gives a result
proportional to the transition probabilities between the mixed state
and the various states contributing to the mixture.  This is exactly
what we want in a mass interaction.  But instead of applying $M$ to
a primitive idempotent, we need to apply it to bound states of the
sort we found in the previous chapter.


The mass interaction defined in Eq.~(\ref{eq:typPIforexpB}) will mix
states completely.  We will associate the usual elementary particles
with mixtures that are preserved by this interaction.  Rather than
solve this problem immediately, let's take a look at how the
solution will look.


If the mixture were equally distributed over all the states, then
all the quantum numbers would be zero and the particle would be dark
matter.  For standard matter, the solutions of interest will be the
ones that are not evenly balanced.  We will now assume that the
states of interest are categorized by having one snuark represented
more than the others.  Thus, for the purposes of looking at quantum
numbers, we can look at the quantum numbers of the snuarks
themselves.



% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:AmplitudesLtoRtoL})
% \SubSectionBreak
\section{\label{sec:SnuarkAntis}  Snuark Antiparticles}


In the standard model, the quantum numbers of antiparticles are the
negatives of the quantum numbers of the particles.  This is somewhat
at odds with what we know about the quantum numbers of primitive
idempotents.  For example, a typical primitive idempotent is:
\begin{equation}\label{eq:QNsOfPIForAntiA}
\begin{array}{rcl}
\rho_{z-+} &=& 0.125(1+\at{zt})(1-\hat{s})(1+\at{ixyzst}),\\
&=&0.125(1+\at{zt}-\hat{s}-\at{ixyzst}+\at{zst}-\at{ixys}+\at{ixyzt}+\at{ixy}),
\end{array}
\end{equation}
and it's eigenvalues can be read off the signs of the second line,
that is, the seven (non trivial) quantum numbers are $+1$, $-1$,
$+1$, $+1$, $-1$, $+1$, and $-1$.

If we are to define the antiparticle as the particle that has the
negatives of all these quantum numbers, the particle will therefore
be:
\begin{equation}\label{eq:QNsOfPIForAntiB}
\bar{\rho_{z-+}} =
0.125(1-\at{zt}+\hat{s}-\at{ixyzst}-\at{zst}+\at{ixys}-\at{ixyzt}-\at{ixy}).
\end{equation}
But the above does not satisfy the idempotency equation
$\rho^2=\rho$ and is therefore not an idempotent, primitive or
otherwise.


The difficulty arises from the fact that quantum numbers for
primitive idempotents are inherently multiplicative.  As soon as our
Clifford algebra is complicated enough to include two commuting
roots of unity, the product of those two quantum numbers will not be
negated when the two quantum numbers are negated.  For the moment,
we will ignore this problem, which is similar to what happens with
the mass interaction.


\SubSectionBreak


In defining an antiparticle as having negated quantum numbers, if we
are to assume that they are primitive idempotents we have to define
which three of the seven possible quantum numbers are to be negated.
The set of seven commuting roots of unity for our example are:
\begin{equation}\label{eq:SevenCRoU}
\begin{array}{cccc}
\hat{s},\;\;\;&\at{ixy},\;\;\;&\at{zt},&\\
\at{zst},\;\;\;&\at{ixys},\;\;\;&\at{ixyzt},\;\;\;&\at{ixyzst}
\end{array}
\end{equation}
The quantum numbers of an observed particle are the quantum numbers
that have not been canceled in the binding together of the particle.
Therefore, it makes sense to choose the three of these seven that
have the least energy.  The potential energy of those in the top row
are $v_s$, $v_s^2$, and $v_sv_t$.  None of these are smaller than
the potential energies of those in the second row, $v_s^2v_t$,
$v_s^3$, $v_s^3v_t$, and $v_s^4v_t$.  Therefore the top row are the
commuting roots of unity with minimum potential energy.


Negating the $\at{zt}$ quantum number means that the direction of
travel of the antiparticle is the opposite of the particle.  This is
compatible with the idea that the antiparticles correspond to
particles traveling backwards in time.  Therefore, we will classify
the antiparticles that carry a $-1$ quantum number for $\at{zt}$ as
traveling in the same direction as the particles with a $+1$ quantum
number.


The question then arises whether a primitive idempotent and the
antiparticle of a primitive idempotent can combine into a snuark.
While we ignored this possibility in the previous section, we now
rectify this.


\SubSectionBreak


In contrast to the previous chapter, by taking into account
antiparticles there are now eight primitive idempotents, and we
label them by $(\at{zt},\hat{s},\at{ixy})$:
\begin{equation}\label{eq:QNsOf8PIs}
\begin{array}{c|ccc|cccc|}
&\at{zt}&\hat{s}&\at{ixy}&\at{zst}&\at{ixys}&\at{ixyzt}&\at{ixyzst}\\ \hline %
\rho_{\bar{z}--}&-&-&-&-&+&+&-\\
\rho_{\bar{z}-+}&-&-&+&-&-&-&+\\
\rho_{\bar{z}+-}&-&+&-&+&-&+&+\\
\rho_{\bar{z}++}&-&+&+&+&+&-&-\\
\rho_{z--}      &+&-&-&+&+&-&+\\
\rho_{z-+}      &+&-&+&+&-&+&-\\
\rho_{z+-}      &+&+&-&-&-&-&-\\
\rho_{z++}      &+&+&+&-&+&+&+\\ \hline
\end{array}
\end{equation}
In the above we have represented the quantum numbers of $\pm 1$ by
$\pm$.


There are four primitive idempotents with positive $\at{ixyzst}$
quantum numbers and four with negative.  This gives sixteen possible
snuarks that cancel $\at{ixyzst}$.  After $\at{ixyzst}$, the next
largest commuting root of unity is $\at{ixyzt}$.  Of the sixteen
snuarks, eight of these also have zero for $\at{ixyzt}$.  Leaving
off the two columns of zero for $\at{ixyzst}$ and $\at{ixyzt}$
quantum numbers, these eight snuarks are:
\begin{equation}\label{eq:QNsOf8Snuarks}
\begin{array}{c|c|cc|cc|}
&\hat{s}&\at{zt}&\at{zst}&\at{ixy}&\at{ixys}\\ \hline %
\rho_{z+-,z++}            &+&+&-&0&0\\
\rho_{\bar{z}+-,\bar{z}++}&+&-&+&0&0\\
\rho_{z--,z-+}            &-&+&+&0&0\\
\rho_{\bar{z}--,\bar{z}-+}&-&-&-&0&0\\ \hline
\rho_{z+-,\bar{z}+-}      &+&0&0&-&-\\
\rho_{z++,\bar{z}++}      &+&0&0&+&+\\
\rho_{z--,\bar{z}--}      &-&0&0&-&+\\
\rho_{z-+,\bar{z}-+}      &-&0&0&+&-\\ \hline
\end{array}
\end{equation}
In the above we represent quantum numbers of $\pm 2$ by $\pm$.


% Fig.~(\ref{fig:FirstGraphSnuarks})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,150)
\thinlines
% Snuark cube
\put( 90, 50){\line(0,1){60}} %
\put(170, 10){\line(0,1){60}} %
\put(210, 50){\line(0,1){60}} %
\put( 90, 50){\line(2,-1){80}} %
\put( 90,110){\line(2,-1){80}} %
\put(130,150){\line(2,-1){80}} %
\put( 90,110){\line(1,1){40}} %
\put(170, 70){\line(1,1){40}} %
\put(170, 10){\line(1,1){40}} %
% Regular cube
\put(110, 70){\line(0,1){60}} %
\put(130, 30){\line(0,1){60}} %
\put(190, 30){\line(0,1){60}} %
\put(130, 30){\line(1,0){60}} %
\put(130, 90){\line(1,0){60}} %
\put(110,130){\line(1,0){60}} %
\put(110, 70){\line(1,-2){20}} %
\put(110,130){\line(1,-2){20}} %
\put(170,130){\line(1,-2){20}} %
% Cube Spots
\put(110, 70){\circle*{3}} %
\put(110,130){\circle*{3}} %
\put(130, 30){\circle*{3}} %
\put(130, 90){\circle*{3}} %
%\put(170, 70){\circle*{3}} %
\put(170,130){\circle*{3}} %
\put(190, 30){\circle*{3}} %
\put(190, 90){\circle*{3}} %
% Snuark Spots
\put( 90, 50){\circle{6}} %
\put( 90,110){\circle{6}} %
%\put(130, 90){\circle{6}} %
\put(130,150){\circle{6}} %
\put(170, 10){\circle{6}} %
\put(170, 70){\circle{6}} %
\put(210, 50){\circle{6}} %
\put(210,110){\circle{6}} %
% Axes
\put(40,30){\vector( 0,1){30}} %
\put(40,30){\vector(-1,0){30}} %
\put(40,30){\vector(1,-2){10}} %
% Axis labels
\put(38, 62){$\hat{s}$} %
\put( 2, 33){$\at{zt}$} %
\put(50,  2){$\at{ixy}$} %
\put( 2, 90){$(\at{zt},\hat{s},\at{ixy})$} %
% Snuark Labels
\put( 65, 35){$\rho_{z--,z-+}$} %
\put( 45,115){$\rho_{z+-,z++}$} %
%\put(130, 90){$\rho_{z--,\bar{z}--}$} %
\put(137,150){$\rho_{z+-,\bar{z}+-}$} %
\put(177, 10){$\rho_{z-+,\bar{z}-+}$} %
\put(125, 62){$\rho_{z++,\bar{z}++}$} %
\put(210, 40){$\rho_{\bar{z}--,\bar{z}-+}$} %
\put(210,119){$\rho_{\bar{z}+-,\bar{z}++}$} %
\end{picture}
} %
\caption{\label{fig:FirstGraphSnuarks} Quantum number plot of the
eight snuarks with zero $\at{ixyzst}$ and $\at{ixyzt}$. The
primitive idempotents are marked with small solid circles and form a
cube aligned with the axes.}
\end{figure}


While there are five non zero quantum numbers in the snuarks of
Eq.~(\ref{eq:QNsOf8PIs}), there are only three that are independent.
We can graph them using $\hat{s}$, $\at{zt}$ and $\at{ixy}$. See
Fig.~(\ref{fig:FirstGraphSnuarks}), and compare with
Fig.~(\ref{fig:MixedCube}).  The primitive idempotents form a cube
with edges of length $2$.  The eight snuarks, by contrast, form the
corners of a block with two edges of length $2\sqrt{2}$ and one edge
of length $4$, at least in the above choice of quantum numbers.
Since quantum numbers can be scaled, the snuark quantum numbers
cannot be distinguished from a cube.





% Fig.~(\ref{fig:MixedCube}) and Eq.~(\ref{eq:AmplitudesLtoRtoL})
% \SubSectionBreak

\section{\label{sec:Quarks}  Quarks}

Fig.~(\ref{fig:FirstGraphSnuarks}) gives the quantum numbers for one
snuark.  Following the modified Feynman checkerboard model of the
previous chapter, we expect the chiral fermions to require three
snuarks each. In combining snuarks, it is natural to group three
snuarks of the same sort, but with different orientations. For
example, one could combine $\rho_{x+-,\bar{x}+-}$,
$\rho_{y+-,\bar{y}+-}$, and $\rho_{z+-,\bar{z}+-}$.  We will
associate the leptons with these sorts of combinations.


According to our modified Feynman checkerboard model, three snuarks
differing only by orientation will combine to produce a chiral
fermion that will be able to produce a particle whose potential
energy is purely scalar. What would be the effect of a substitution
of one of the snuarks? The result will be that the potential energy
of the combination will no longer be purely scalar.  Since we have a
model of the potential energy, we can minimize the change in
potential energy caused by the substitution of a snuark.


\SubSectionBreak


As an example, let suppose that our three snuarks are of type
$\rho_{*+-,*++}$ where $*$ represents $x$, $y$, and $z$, and that
this snuark contributes to a fermion which is purely scalar.  Then
we have
\begin{equation}\label{eq:LeptonTypeSum}
\rho_{x+-,x++} + \rho_{y+-,y++} + \rho_{z+-,z++} + \;\;\textrm{RH} =
\;\;\textrm{scalar},
\end{equation}
where ``RH'' stands for the same sum but for the right handed
particle.  (At the moment we ignore the difficulties in defining how
a mass interaction is mapped to the primitive idempotents and
snuarks.)


The snuarks have five nonzero quantum numbers, $\hat{s}$, $\at{zt}$,
$\at{ixy}$, $\at{zst}$ and $\at{ixys}$.  The potential energy of
these commuting roots of unity are $v_s$, $v_sv_t$, $v_s^2$,
$v_s^2v_t$, and $v_s^3$.  Of these, the two highest potential
energies are $v_s^2v_t$ and $v_s^3$, corresponding to $\at{zst}$ and
$\at{ixys}$.  Following the same energy principle we've been using
for combining primitive idempotents into snuarks, we assume that
when snuarks substitute for one another, these two degrees of
freedom must be conserved.


Referring to Eq.~(\ref{eq:QNsOf8Snuarks}), we see that requiring
that these quantum numbers not be changed in a substitution, groups
the snuarks into four groups of two:
\begin{equation}\label{eq:SnuarksBySubstitution}
\begin{array}{c|c|cc|cc|}
&\hat{s}&\at{zt}&\at{zst}&\at{ixy}&\at{ixys}\\ \hline %
\rho_{z+-,z++}            &+&+&-&0&0\\
\rho_{\bar{z}--,\bar{z}-+}&-&-&-&0&0\\ \hline %
\rho_{\bar{z}+-,\bar{z}++}&+&-&+&0&0\\
\rho_{z--,z-+}            &-&+&+&0&0\\ \hline %
\rho_{z+-,\bar{z}+-}      &+&0&0&-&-\\
\rho_{z-+,\bar{z}-+}      &-&0&0&+&-\\ \hline %
\rho_{z--,\bar{z}--}      &-&0&0&-&+\\
\rho_{z++,\bar{z}++}      &+&0&0&+&+\\ \hline %
\end{array}
\end{equation}
When a snuark is substituted, the quantum numbers of the resulting
quark will be intermediate between the quantum numbers of the
corresponding leptons.


Referring again to Fig.~(\ref{fig:FirstGraphSnuarks}), our
substitution rule allows snuarks to pair if they are on opposite
corners.  Rather than graph them in this manner, let us replace our
$\at{zt}$ and $\at{ixy}$ axes with $\at{zst}$ and $\at{ixys}$. That
is, instead of graphing $\hat{s}$ with the next two smallest
potential degrees of freedom, we will instead graph $\hat{s}$ with
the two largest potential degrees of freedom.


This sort of change to the choice of independent quantum numbers
used in plotting the particles is done for convenience only.  The
student should remember that there is no physical $3-$dimensional
space for these graphs.


By our energy principle, the large degrees of freedom that will be
conserved in substitution, and by graphing with these degrees of
freedom, the quarks will form columns between the leptons. See
Fig.~(\ref{fig:SecondGraphSnuarks}).


% Fig.~(\ref{fig:SecondGraphSnuarks})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,150)
\thinlines
% Snuark cube
\put( 90, 50){\line(0,1){60}} %
\put(170, 10){\line(0,1){60}} %
\put(210, 50){\line(0,1){60}} %
\put( 90, 50){\line(2,-1){80}} %
\put( 90,110){\line(2,-1){80}} %
\put(130,150){\line(2,-1){80}} %
\put( 90,110){\line(1,1){40}} %
\put(170, 70){\line(1,1){40}} %
\put(170, 10){\line(1,1){40}} %
% Regular cube
\put(110, 70){\line(0,1){60}} %
\put(130, 30){\line(0,1){60}} %
\put(190, 30){\line(0,1){60}} %
\put(130, 30){\line(1,0){60}} %
\put(130, 90){\line(1,0){60}} %
\put(110,130){\line(1,0){60}} %
\put(110, 70){\line(1,-2){20}} %
\put(110,130){\line(1,-2){20}} %
\put(170,130){\line(1,-2){20}} %
% Cube Spots
\put(110, 70){\circle*{3}} %
\put(110,130){\circle*{3}} %
\put(130, 30){\circle*{3}} %
\put(130, 90){\circle*{3}} %
%\put(170, 70){\circle*{3}} %
\put(170,130){\circle*{3}} %
\put(190, 30){\circle*{3}} %
\put(190, 90){\circle*{3}} %
% Lepton Spots
\put( 90, 50){\circle{6}} %
\put( 90,110){\circle{6}} %
%\put(130, 90){\circle{6}} %
\put(130,150){\circle{6}} %
\put(170, 10){\circle{6}} %
\put(170, 70){\circle{6}} %
\put(210, 50){\circle{6}} %
\put(210,110){\circle{6}} %
% Quark Spots
\put( 90, 70){\circle*{6}} %
\put( 90, 90){\circle*{6}} %
\put(170, 30){\circle*{6}} %
\put(170, 50){\circle*{6}} %
\put(210, 70){\circle*{6}} %
\put(210, 90){\circle*{6}} %
% Axes
\put(40,30){\vector( 0,1){30}} %
\put(40,30){\vector(-1,0){30}} %
\put(40,30){\vector(1,-2){10}} %
% Axis labels
\put(38, 62){$\hat{s}$} %
\put( 2, 33){$-\at{*st}$} %
\put(50,  2){$\at{i*xyzs}$} %
\put( 2, 90){$(\at{*t},\hat{s},\at{ixy})$} %
% Snuark Labels
\put( 65, 35){$\rho_{\bar{*}--,\bar{*}-+}$} %
\put( 45,115){$\rho_{*+-,*++}$} %
%\put(130, 90){$\rho_{*-+,\bar{*}-+}$} %
\put(137,150){$\rho_{*+-,\bar{*}+-}$} %
\put(177, 10){$\rho_{*--,\bar{*}--}$} %
\put(125, 62){$\rho_{*++,\bar{*}++}$} %
\put(210, 40){$\rho_{*--,*-+}$} %
\put(210,119){$\rho_{\bar{*}+-,\bar{*}++}$} %
\end{picture}
} %
\caption{\label{fig:SecondGraphSnuarks} Quantum number plot of the
eight leptons and sixteen quarks in snuark form relative to the
primitive idempotents.  Primitive idempotents are small filled
circles, leptons are large hollow circles and quarks ($\times 3$)
are large filled circles.}
\end{figure}


Each quark comes in three colors.  In our model, this corresponds to
the fact that there are three ways the substitution can take place.
In the example of substitutions of snuarks of type $\rho_{*+-,*++}$,
the alternatives are snuarks of type $\rho_{\bar{*}--,\bar{*}-+}$.
Composites of this type form the leftmost column of
Fig.~(\ref{fig:SecondGraphSnuarks}).  The two leptons and six
quarks, with their (non zero) quantum numbers, are:
\begin{equation}\label{eq:TwoLepsAndSixQs}
\begin{array}{c|cc|c|}
&\hat{s}&\at{zt}&\at{zst}\\ \hline %
\rho_{x+-,x++}\;\;\;\;\rho_{y+-,y++}\;\;\;\;\rho_{z+-,z++}  &+6&+6&-6\\ \hline %
\rho_{\bar{x}--,\bar{x}-+}\;\;\;\;\rho_{y+-,y++}\;\;\;\;\rho_{z+-,z++}  &+2&+2&-6\\
\rho_{x+-,x++}\;\;\;\;\rho_{\bar{y}--,\bar{y}-+}\;\;\;\;\rho_{z+-,z++}  &+2&+2&-6\\
\rho_{x+-,x++}\;\;\;\;\rho_{y+-,y++}\;\;\;\;\rho_{\bar{z}--,\bar{z}-+}  &+2&+2&-6\\ \hline %
\rho_{x+-,x++}\;\;\;\;\rho_{\bar{y}--,\bar{y}-+}\;\;\;\;
\rho_{\bar{z}--,\bar{z}-+}&-2&-2&-6\\
\rho_{\bar{x}--,\bar{x}-+}\;\;\;\;\rho_{y+-,y++}\;\;\;\;
\rho_{\bar{z}--,\bar{z}-+}&-2&-2&-6\\
\rho_{\bar{x}--,\bar{x}-+}\;\;\;\;\rho_{\bar{y}--,\bar{y}-+}\;\;\;\;
\rho_{z+-,z++}&-2&-2&-6\\ \hline %
\rho_{\bar{x}--,\bar{x}-+}\;\;\;\;\rho_{\bar{y}--,\bar{y}-+}\;\;\;\;
\rho_{\bar{z}--,\bar{z}-+}&-6&-6&-6\\ \hline %
\end{array}
\end{equation}
The leptons are the top and bottom particles.


\SubSectionBreak


There are a total of $8$ leptons and $24$ quarks whose quantum
numbers are shown in Fig.~(\ref{fig:SecondGraphSnuarks}).  By the
same method shown in
Section~(\ref{sec:BoundStatePrimitiveIdempotents}), each of these
$32$ composites comes in three choices of sea, and these are the
three generations.  For the remainder of this chapter we will ignore
the details of the generations.


The strongest force between the $32$ leptons and quarks arises from
their largest potential degree of freedom.  This force corresponds
to the ``strong'' force of the standard model.  This is the
uncanceled vector potential of the quarks. The leptons add to
scalars and so do not partake in the strong force.


Since the strong force is associated with a non scalar, its strength
is of the magnitude of the Planck energy, and therefore quarks will
bind together so strongly that ordinary energies will be unable to
separate them.


To determine how the strong force acts between quarks, we need to
compare the degrees of freedom of a snuark with the snuark it can
substitute for.  That is, we must compute the differences between
the pairs of snuarks of Eq.~(\ref{eq:SnuarksBySubstitution}):
\begin{equation}\label{eq:SnuarkDifferences}
\begin{array}{c|c|cc|cc|}
&\hat{s}&\at{zt}&\at{zst}&\at{ixy}&\at{ixys}\\ \hline %
\rho_{z--,z-+} - \rho_{\bar{z}+-,\bar{z}++}&-4&+4&0& 0&0\\
\rho_{z-+,\bar{z}-+} - \rho_{z+-,\bar{z}+-}&-4& 0&0&+4&0\\ \hline
\rho_{z+-,z++} - \rho_{\bar{z}--,\bar{z}-+}&+4&+4&0& 0&0\\
\rho_{z++,\bar{z}++} - \rho_{z--,\bar{z}--}&+4& 0&0&+4&0\\ \hline
\end{array},
\end{equation}
and similarly for $x$ and $y$.  The negatives of these are also
available.


We will define the color charges as red, green, and blue, and, in
reference to a lepton made from snuarks traveling in the $+x$, $+y$,
and $+z$ directions, we will associate red with a substitution of
the $+x$ snuark, green with a substitution of the $+y$ snuark, and
blue with a substitution of the $+z$ snuark.


To make an elementary fermion requires that we give the left handed
as well as the right handed part.  Therefore, to compute the
potential energy between a collection of quarks we will have to add
together the right and left handed portions. At this time we have
not yet fully explained the mass interaction. However, peeking
ahead, we will find that the mass interaction puts the snuarks (or
the equivalents to snuarks) into pairs, and we have conveniently
arranged for the snuark differences listed in
Eq.~(\ref{eq:SnuarkDifferences}) to be arranged in pairs.  That is,
they are listed in left handed / right handed pairs.


Eq.~(\ref{eq:SnuarkDifferences}) are written in terms of $+z$
snuarks, so they define the blue quantum numbers.  Since the left
handed state has to accompany the right handed state, we have to add
these together by pairs to define the blue quantum numbers.  The two
pairs give different sums, $B_1$ and $B_2$ as follows:
\begin{equation}\label{eq:QuarkDifferences}
\begin{array}{c|c|cc|cc|}
&\hat{s}&\at{zt}&\at{zst}&\at{ixy}&\at{ixys}\\ \hline %
B_1&-8&+4&0&+4&0\\
\bar{B}_1&-8&-4&0&-4&0\\
B_2&+8&+4&0&+4&0\\
\bar{B}_2&+8&-4&0&-4&0\\ \hline
\end{array}.
\end{equation}
For example, $B_1 = (\rho_{z--,z-+} - \rho_{\bar{z}+-,\bar{z}++})$
$+ (\rho_{z-+,\bar{z}-+} - \rho_{z+-,\bar{z}+-})$, while $B_2$ is
the sum of the last two lines of Eq.~(\ref{eq:SnuarkDifferences}).


In computing $B_1$ and $B_2$ as differences between snuarks we took
the sum over all degrees of freedom.  The $\hat{s}$ degree of
freedom has the lowest potential energy, $v_s$, and is small
compared to that of the strong force, $v_s^2v_t$ or $v_s^3$.  In the
standard model, the forces are divided according to their strength.
This suggests that we need to think of the blue quantum number as
the average of $B_1$ and $B_2$.


With this interpretation, we ignore the $\hat{s}$ degree of freedom.
The red, green, and blue quantum numbers are:
\begin{equation}\label{eq:RedGreenBlueA}
\begin{array}{c|cccccc|}
&\at{xt}&\at{iyz}&\at{yt}&-\at{ixz}&\at{zt}&\at{ixy}\\ \hline %
     R &+4&+4& 0& 0& 0& 0\\
     G & 0& 0&+4&+4& 0& 0\\
     B & 0& 0& 0& 0&+4&+4\\
\bar{R}&-4&-4& 0& 0& 0& 0\\
\bar{G}& 0& 0&-4&-4& 0& 0\\
\bar{B}& 0& 0& 0& 0&-4&-4\\
\end{array}. \;\;\textrm{(not quite right)}
\end{equation}
To get a scalar sum, we can add $R$ to $\bar{R}$, or the same with
green and blue.  Collections of colored particles that satisfy this
requirement are allowed in the standard model, but the standard
model also allows sums that have equal amounts of red, green, and
blue.


In Section~(\ref{sec:TheMassInteraction}), we made the assumption
that ``the states of interest are categorized by having one snuark
represented more than the others.''  In other words, we approximated
the solution by representing solutions that were predominantly
oriented in one direction by solutions that were oriented only in
that direction.  This approximation prevents our having colored
states add to a colorless state.

To allow colorless states to appear as a sum of the $R$, $G$ and $B$
states, we need to replace the $0$s of Eq.~(\ref{eq:RedGreenBlueA})
with $\pm 1$ and divide the $\pm 4$ values by two:
\begin{equation}\label{eq:RedGreenBlueB}
\begin{array}{c|cccccc|}
&\at{xt}&\at{iyz}&\at{yt}&-\at{ixz}&\at{zt}&\at{ixy}\\ \hline %
     R &+2&+2&-1&-1&-1&-1\\
     G &-1&-1&+2&+2&-1&-1\\
     B &-1&-1&-1&-1&+2&+2\\
\bar{R}&-2&-2&+1&+1&+1&+1\\
\bar{G}&+1&+1&-2&-2&+1&+1\\
\bar{B}&+1&+1&+1&+1&-2&-2\\
\end{array}.
\end{equation}
These quantum numbers will give the correct behavior for quarks.  We
will justify these assumptions later in the book.


% Fig.~(\ref{fig:MixedCube}) and Eq.~(\ref{eq:RedGreenBlue})
% \SubSectionBreak

\section{\label{sec:SpinorAndOperatorSymmetry}  Spinor and Operator Symmetry}


It is time to make contact with the standard model.  This book is
written from a geometric point of view.  Our objective is to
understand the elementary particles as geometric objects, and to
maintain as close a relationship to the physics of the particles as
possible.  Our method is a ``top down'' approach.  We begin with
geometry, that is, a Clifford algebra.  We make the assumption that
the elementary particles are primitive idempotents, and we then find
out as much as we can about primitive idempotents of the Clifford
algebra.


By contrast, the standard model is built from a ``bottom up''
procedure.  One looks to experiment to find things that are
conserved (for example, momentum, energy, electric charge), or what
is the same thing, a symmetry, and one incorporates these
experimental observations into a model of the elementary particles
that obey them.


For example, Einstein showed that experiments could be interpreted
as implying that the results of an experiment could never depend on
the absolute motion of the experiment.  The modeler then builds this
observation into the model of the elementary particles.  This is an
indirect way of understanding the physical world and it is subject
to several complications and limitations.


\SubSectionBreak


Newton developed a method of modeling physical situations that was
very direct and had clear and simple physical interpretations.
Particles were assumed to be connected to other particles by forces.
Forces cause acceleration of the particles according to the mass of
the particle:
\begin{equation}\label{eq:FemA}
F_x = m d^2 x/dt^2.
\end{equation}
Mass, $m$, is simply a constant of proportionality between
acceleration in a direction, $d^2x/dt^2$, and force in that
direction, $F_x$. Since force is arbitrarily defined, so is mass,
and one can always redefine them.


When one applies Newton's equations to a collection of $N$
particles, one obtains a set of $N$ differential equations.  To
predict the motion of the system, one inserts an initial state into
the differential equations and solves them.  Initial states were
position and velocity.  To find the final state, for example, the
state at time $T$, one takes the solution of the differential
equation at time $T$.  The differential equation also provided
solutions for the intermediate states.


In the $19$th century, physicists and mathematicians found that they
could rewrite Newton's laws as a variational principle.  Instead of
writing down $N$ differential equations, they could instead write
down a ``Lagrangian'', a real valued function of the positions and
velocities of the $N$ particles.  They next suppose that the motion
of the system maximizes (or minimizes) this function. Newton's
differential equations can then be derived by the Euler-Lagrange
equations.


The Lagrangian formalism had various advantages over Newton's
equations.  Being only a single function, it was simpler than the
full set of differential equations.  One could model constrained
systems without having to keep track of the constraint forces.  But
for our story, the most important advantage was that it was very
easy to incorporate symmetry into the Lagrangian function.


\SubSectionBreak


In the $1920$s, it was realized that since elementary particles can
have such small masses, the initial and final states of an
experiment can never be determined with perfect precision.  Thus the
model of an experiment would have to be statistical in nature.  To
achieve this, the positions and momentums (i.e. real functions of
time) of classical mechanics were replaced by position and momentum
wave functions (i.e. probability densities, or functions of space
and time).


But the probability densities of classical statistical mechanics
turned out to be inefficient at solving quantum problems.  While
they could be made to work (i.e. Bohmian mechanics), a much simpler
technique was to use complex valued functions of space and time,
that is, wave functions.  The method could be written in Lagrangian
form, and therefore models could be easily made that would
incorporate symmetries observed in experiment.


Einstein's relativity defines a symmetry, Lorentz invariance.  Since
linear equations are easier to solve than nonlinear ones, Dirac
searched for, and found, a linear equation that would satisfy
Lorentz invariance.  The Dirac equation was wildly successful at
modeling the electron (and lies at the heart of the Clifford algebra
used in this book).


With the success of the Dirac equation, physics was hooked.
Experimental results were examined for other symmetries, and as
these were discovered, they were incorporated into quantum
mechanics.  Most physicists believed that nature was, at heart, a
collection of symmetries.


\SubSectionBreak


If one postulates that a differential equation satisfies a simple
symmetry principle, then one can obtain some knowledge about the
solutions of the differential equation even if one does not know the
differential equation, cannot write it down, and cannot solve it
exactly. In this case, one typically ends up with some arbitrary
constants that depend on the details of the differential equations,
but one cannot solve the differential equations by symmetry alone.
And this is precisely the problem with the standard model today, too
many arbitrary constants.

On the other hand, if one postulates that a differential equation is
of a simple form, then one can solve the differential equation
directly.  Thus the assumption that equations of physics are simple
is far more powerful than the assumption that the equations of
physics have simple symmetries.

What's more, physicists found experimental situations where symmetry
was only approximate.  This should have been a clue that symmetry
was a blind alley, but since great advances had been made using
symmetry, physicists refused to abandon it.



When physicists attempt to extend the standard model, they typically
do this by postulating yet more symmetry.  The most important thing
for them is how this more general symmetry is broken to the observed
symmetry, $SU(3)\times SU(2) \times U(1)$. The particular
representations of that symmetry, that is, the elementary particles
themselves, are treated secondarily. It is the symmetry that the
usual approach to physics treats primarily.


Since we are deriving the standard model in the reverse direction,
from the bottom up, our method must instead concentrate on the
representatives of the symmetry.  We then show that these objects
possess the symmetry given them in the standard model. As a first
step, we've already seen in
Section~(\ref{sec:BoundStatePrimitiveIdempotents}) that our bound
states will appear in three varieties that we interpret as the
generations.


\SubSectionBreak


In Section~(\ref{sec:Quarks}), we showed that our bound states
appeared in the correct numbers to give $24$ quarks and $8$ leptons,
under the interpretation that the leptons were pure mixtures and the
quarks were not.  And we showed that the rule that quarks only
appear in colorless mixtures was plausible.  The leptons were
singlets for color, while the quarks were triplets. This singlet /
triplet structure reminds one of the $SU(3)$ structure of the
leptons and quarks, but the symmetry of our model is simpler than
$SU(3)$.


Our representation was based on a choice of three perpendicular
vectors in $3$ spatial dimensions.  We used the example of
$\hat{x}$, $\hat{y}$, and $\hat{z}$ as indicating those vectors. The
symmetry of this sort of choice is not $SU(3)$, but instead is only
$SO(3)$, the symmetry of proper rotations of $3$-dimensional real
space.  Both $SU(3)$ and $SO(3)$ are proper rotations, that is, they
both are connected components that include the identity rotation.


The quantum states of the standard model are represented by spinors,
while we are using density operators.  Spinors possess an arbitrary
complex phase, while density operators do not.  This accounts for
the difference in symmetry between $SU(3)$ and $SO(3)$.  That is,
$SU(3)$ preserves the length of $3$-dimensional complex vectors,
while $SO(3)$ preserves the length of $3$-dimensional real vectors.
Thus the symmetry of our quarks matches the $SU(3)$ symmetry of the
standard model quarks, after taking into account the difference
between spinors and density operators.


\SubSectionBreak


The $SU(2)$ symmetry of the $SU(3)\times SU(2)\times U(1)$ standard
model is also called ``weak isospin''.  The Pauli algebra also
corresponds to an $SU(3)$ symmetry, but it is distinct from weak
isospin in more than just application.

The Pauli algebra contains three generators, $\hat{x}$, $\hat{y}$
and $\hat{z}$.  Given any unit vector $(u_x,u_y,u_z)$, we can define
the operator $u_x\hat{x}+u_y\hat{y}+u_z\hat{z}$ and define quantum
states that are (spin-$1/2$) eigenstates of this operator.  And all
these different states can be produced in the lab.

By contrast, weak isospin defines the relationship between left
handed and right handed states.  The left handed states show up in
doublet representations while the right handed states are singlets.
If one were to imagine these states defined with a different axis
(as can be had with the Pauli algebra), we could have a state that
is mixed between electron and neutrino.  These sorts of things are
not observed in the lab.  In standard quantum mechanics they are
excluded by superselection rules.  Consequently, in showing that our
particles have an $SU(2)$ symmetry equivalent to that of the
standard model, we do not have to produce a full set of generators
for this $SU(2)$ symmetry.


\SubSectionBreak


Consider the Clifford algebra with only one dimension, say
$\hat{x'}$.  There are two eigenstates, $0.5(1+\hat{x'})$ and
$0.5(1-\hat{x'})$.  We can think of these two states as being an
$SU(2)$ doublet similar to the $SU(2)$ doublet of weak isospin.  We
can always increase our Clifford algebra by adding two more vectors,
$\hat{y'}$, and $\hat{z'}$, to obtain a true $SU(2)$ representation,
but in doing this our $0.5(1\pm\hat{x'})$ eigenstates will not
change.


In Section~(\ref{sec:SnuarksAsBoundStates}) we showed that it was
natural that two primitive idempotents that differ in their
$\at{ixyzst}$ quantum numbers, but have the same velocity, would
bind together.  The quantum numbers of the bound states,
Eq.~(\ref{eq:QNsOfSnuarks}), we repeat here for convenience:
\begin{equation}\label{eq:QNsOfSnuarksAgin}
\begin{array}{c|cccccccc|}
&1&\at{zt}&\at{ixys}&\at{ixyzst}&\hat{s}&\at{zst}&\at{ixyzt}&\at{ixy}\\ \hline %
\rho_{z--}+\rho_{z-+}&2&2&0&0&-2&+2& 0& 0\\
\rho_{z+-}+\rho_{z++}&2&2&0&0&+2&-2& 0& 0\\
\rho_{z--}+\rho_{z++}&2&2&0&0& 0& 0&-2&-2\\
\rho_{z+-}+\rho_{z-+}&2&2&0&0& 0& 0&+2&+2\\ \hline
\end{array}
\end{equation}
The four rightmost columns all show the same structure.  There are
two snuarks with quantum number $0$, and a pair with quantum number
$\pm 2$.  Any of these has the same structure as weak isospin.


The reader may note that these four quantum numbers are oriented.
They depend on the velocity direction that the snuark, but since any
given snuark does have a velocity, we do not have a choice for how
these four quantum numbers are oriented.  For example, if we suppose
$\at{zst}$ represents weak isospin, the first two states form the
doublet, and the second two states are singlets.

With this interpretation, it is satisfying to see that weak isospin
does have a full $SU(2)$ symmetry, that is, continuing our example,
the set $\{\at{xst}, \at{yst}, \at{zst}\}$ does form a set of
generators for an $SU(2)$.  (I.e., like the vector generators of a
Clifford algebra, each squares to $+1$ and they anticommute.) We can
thus interpret the superselection rule as arising from the fact that
weak isospin is only defined relative to the chiral states, which
are massless, travel at $c$, and have an orientation.


We have treated the standard model symmetries $SU(3)$ and $SU(2)$
differently in that we took advantage of the difference between
spinors and density operators to explain the difference between
$SU(3)$ and $SO(3)$, but we did not rely on this sort of argument
with $SU(2)$.  In analogy with the $SU(3)$ situation, we can reduce
$SU(2)$ to a real rotation that preserves real vectors of length
$2$, that is, $SO(2)$.  There is then only one generator, and the
representations include the ones observed in the snuarks.


Finally, in moving from spinors to density operators, the arbitrary
complex phase of the $U(1)$ symmetry of the standard model is
eliminated. Thus we see that the symmetry of the $32$ states
described in Section~(\ref{sec:Quarks}) matches that of the standard
model.  It remains, however, to check that the whole collection of
states have quantum numbers that are compatible overall with those
of the standard model.


% Fig.~(\ref{fig:MixedCube}) and Eq.~(\ref{eq:QNsOfSnuarks})
% \SubSectionBreak

\section{\label{sec:WHaIsospin}  Weak Hypercharge and Isospin}


It remains to compare the quantum numbers of one generation of the
standard model fermions with the quantum numbers of the quarks and
leptons made from snuarks.  The snuark version of quarks are built
from a combination of primitive idempotent particles and primitive
idempotent antiparticles, so in comparing quantum numbers we must
compare the full set of particles and antiparticles.


To make the comparison, let us graph the quarks and leptons
according to their weak isospin and weak hypercharge quantum
numbers.  The quantum numbers of the left handed antiparticles are
simply the negatives of the quantum numbers of the right handed
particles, and the same for the right handed antiparticles and left
handed particles.  See Table~(\ref{fig:WeakQNums}).

% Fig.~(\ref{fig:WeakQNums})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{ $
\begin{array}{c|cc|}
&       t_0&t_3 \\ \hline %
e_R   & -1 &  0 \\
e_L   &-1/2&-1/2\\
\nu_R &  0 &  0 \\
\nu_L &-1/2&+1/2\\
d_R   &-1/3&  0 \\
d_L   &+1/6&-1/2\\
u_R   &+2/3&  0 \\
u_L   &+1/6&+1/2\\ \hline
\end{array} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
\begin{array}{c|cc|}
&             t_0&t_3 \\ \hline %
\bar{e}_{L}   & +1 &  0 \\
\bar{e}_{R}   &+1/2&+1/2\\
\bar{\nu}_{L} &  0 &  0 \\
\bar{\nu}_{R} &+1/2&-1/2\\
\bar{d}_{L}   &+1/3&  0 \\
\bar{d}_{R}   &-1/6&+1/2\\
\bar{u}_{L}   &-2/3&  0 \\
\bar{u}_{R}   &-1/6&-1/2\\ \hline
\end{array}
$} %
\caption{\label{fig:WeakQNums} Weak hypercharge ($t_0$) and weak
isospin ($t_3$) quantum numbers for the first generation particles
and antiparticles.}
\end{figure}


Weak hypercharge and weak isospin are sufficient to distinguish all
the elementary fermions except for the right handed neutrino,
$\nu_R$ and the left handed antineutrino, $\bar{\nu_L}$.  In the
original ``standard model'', the neutrino was assumed massless and
these two states were not included.  The density operator model
automatically includes these states and so we have added them to the
standard model for comparison.


In extending the standard model to include neutrino masses, there
are several possibilities, including the usual Dirac mass term, or a
Majorana mass term.  More complicated ways of giving mass to the
neutrino involves the addition of ``sterile'' neutrinos.  In the
next chapter we will discuss this at great length, for now, we will
simply add these two neutrinos as if we were adding a Dirac mass
term.


Other quantum numbers for the fermions are sometimes given.  The
weak isospin quantum number shown is weak isospin in the $3$rd,
direction, $t_3$. Some authors add a quantum number $t$ to
distinguish the doublets, $\{e_L,\nu_L, d_L, u_L\}$, (which take
$t=1/2$) from the singlets, $\{e_R, \nu_R, d_R, u_R\}$ (which take
$t=0$).


The standard model particles are defined according to how they
interact with each other.  The electromagnetic coupling is called
the electric charge, $Q$, and is given by the sum of the weak
hypercharge and weak isospin:
\begin{equation}\label{eq:ElectricCharge}
Q = t_3 + t_0.
\end{equation}
An analogous coupling for the weak force is called \cite[Table
6.2]{QLAGF} the ``neutral charge'', $Q'$.  To obtain the neutral
charge from weak hypercharge and weak isospin one requires the
Weinberg angle $\theta_W$:
\begin{equation}\label{eq:WeakCharge}
Q' = t_3\cot(\theta_W) - t_0\tan(\theta_W),
\end{equation}
where $\sin^2(\theta_W)$ is approximately $1/4$.  The transformation
from weak hypercharge to weak isospin is a linear one, so we can
plot them together, see Fig.~(\ref{fig:WeakQNumPlot}).


% Fig.~(\ref{fig:WeakQNumPlot})
\begin{figure}[!htp]
\setlength{\unitlength}{1pt} %
\center \framebox[0.9\textwidth][c]{
\begin{picture}(220,150)
\thinlines
% Snuark cube
\put( 80, 40){\line(0,1){60}} %
\put(140, 10){\line(0,1){60}} %
\put(200, 40){\line(0,1){60}} %
\put(140, 10){\line(-2,1){60}} %
\put(140, 70){\line(-2,1){60}} %
\put(200,100){\line(-2,1){60}} %
\put( 80,100){\line(2,1){60}} %
\put(140, 70){\line(2,1){60}} %
\put(140, 10){\line(2,1){60}} %
% Lepton Spots
\put( 80, 40){\circle{6}} %
\put( 80,100){\circle{6}} %
\put(140, 10){\circle{6}} %
\put(140, 70){\circle{6}} %
\put(140,130){\circle{6}} %
\put(200, 40){\circle{6}} %
\put(200,100){\circle{6}} %
% Quark Spots
\put( 80, 60){\circle*{6}} %
\put( 80, 80){\circle*{6}} %
\put(140, 30){\circle*{6}} %
\put(140, 50){\circle*{6}} %
\put(140, 90){\circle*{6}} %
\put(140,110){\circle*{6}} %
\put(200, 60){\circle*{6}} %
\put(200, 80){\circle*{6}} %
% Weak Axes
\put(40,60){\vector( 0,1){45}} %
\put(40,60){\vector(-1,0){30}} %
% Weak labels
\put(38,112){$t_0$} %
\put( 2, 63){$t_3$} %
% Charge Axes
\put(40, 60){\vector(-2,3){30}} %
\put(40, 60){\vector(-2,-1){30}} %
% Charge labels
\put( 2, 92){$Q$} %
\put( 2, 35){$Q'$} %
% Particle Labels
\put( 62, 40){$\nu_L$} %
\put( 62, 60){$\bar{d}_{R}$} %
\put( 62, 80){$u_L$} %
\put( 62,100){$\bar{e}_{R}$} %
\put(147,  5){$e_R$} %
\put(147, 25){$\bar{u}_{L}$} %
\put(147, 45){$d_R$} %
\put(147, 65){$\bar{\nu}_{L}$} %
\put(147, 95){$\bar{d}_{L}$} %
\put(147,115){$u_R$} %
\put(147,135){$\bar{e}_{L}$} %
\put(207, 40){$e_L$} %
\put(207, 60){$\bar{u}_{R}$} %
\put(207, 80){$d_L$} %
\put(207,100){$\bar{\nu}_{R}$} %
\end{picture}
} %
\caption{\label{fig:WeakQNumPlot} Weak hypercharge, $t_0$, and weak
isospin, $t_3$, quantum numbers plotted for the first generation
standard model quantum states.  Leptons are hollow circles and
quarks ($\times 3$) are filled circles.  Electric charge, $Q$, and
neutral charge, $Q'$ also shown.}
\end{figure}


\SubSectionBreak


The similarity between Fig.~(\ref{fig:SecondGraphSnuarks}) and
Fig.~(\ref{fig:WeakQNumPlot}) show that the quantum numbers of bound
states of snuarks come in a $3$-dimensional pattern (four parallel
edges of a cube) that is easy to match to the observed quantum
numbers of the elementary fermions.


There are still two difficulties.  First, we did not solve the mass
interaction completely.  Instead, we approximated its solutions by
assuming that the solution states could be approximated by the
snuarks themselves. Second, there are a number of different ways of
associating the axes of the two figures.  In addition to the obvious
possibility of rotating the two cubes, one can also imagine that we
could choose different commuting roots of unity and thereby could
transform the snuark cube.

In solving these issues, we will finally obtain complete geometric
models for the elementary particles.  This will allow us to begin
computing attributes of the elementary particles from first
principles.  This will be the topic of the next chapter.  The first
attributes we will solve for will be the masses of the elementary
fermions, which can be thought of as the mass charges.


% Fig.~(\ref{fig:MixedCube}) and Eq.~(\ref{eq:QNsOfSnuarks})
% \SubSectionBreak

\section{\label{sec:Antiparticles}  Antiparticles and the Arrow of Time}


Following Feynman, we define antiparticles as particles that travel
backwards in time.  However, given a pair of states, one a particle,
the other an antiparticle, we have no way of determining which is
the one that travels backwards in time and which travels forwards. A
particle and its antiparticle form a related set, but we cannot
logically apply the label "antiparticle" to one of them rather than
the other.  If we were able to distinguish the true direction of
travel in time of a quantum state, forwards in time and therefore a
particle, versus backwards in time and therefore an antiparticle, we
would have a way of defining the ``arrow of time'' in quantum
mechanics.


But the standard model includes no such arrow of time.  Instead, we
define antiparticles by arbitrarily assuming that the states we see
in everyday matter are made up of particles, and their complementary
states are antiparticles.  In fact, there are bound states of
particles with antiparticles, for example, pions:
\begin{equation}\label{eq:PionAntiParticles}
\begin{array}{rcl}
\pi^+ &=& u\bar{d},\\
\pi^- &=& d\bar{u}.
\end{array}
\end{equation}
Clearly these mixed bound states are inherently neither particles
nor antiparticles.  We have the choice of defining them either way.
Looked at this way, particle / antiparticle works best as a relative
relationship between two states rather than a description that can
be absolutely applied to a single state.


\SubSectionBreak


Our primitive idempotents are eigenstates of velocity, and we
defined their antiparticles as particles that carried the same
quantum number for velocity, but travel in the opposite direction.
As such, the primitive idempotent model does have an arrow of time,
one that is consistent across all the primitive idempotents, if not
the elementary particles.


Of course in fitting the model to experiment we will have two
choices, and the arrow of time will melt away, but unlike the
standard model, in the density operator model we can naturally
extend the particle / antiparticle relationship to all the
elementary particles.




We therefore examine the particle / antiparticle content of the
snuark model of the elementary fermions.  Repeating
Eq.~(\ref{eq:QNsOf8Snuarks}), there are eight snuarks:
\begin{equation}\label{eq:QNsOf8SnuarksYetAgain}
\begin{array}{c|c|cc|cc|}
&\hat{s}&\at{zt}&\at{zst}&\at{ixy}&\at{ixys}\\ \hline %
\rho_{z+-,z++}            &+&+&-&0&0\\
\rho_{\bar{z}+-,\bar{z}++}&+&-&+&0&0\\
\rho_{z--,z-+}            &-&+&+&0&0\\
\rho_{\bar{z}--,\bar{z}-+}&-&-&-&0&0\\ \hline
\rho_{z+-,\bar{z}+-}      &+&0&0&-&-\\
\rho_{z++,\bar{z}++}      &+&0&0&+&+\\
\rho_{z--,\bar{z}--}      &-&0&0&-&+\\
\rho_{z-+,\bar{z}-+}      &-&0&0&+&-\\ \hline
\end{array}.
\end{equation}
Of these, two are pure antiparticle, $\rho_{\bar{z}--,\bar{z}-+}$
and $\rho_{\bar{z}+-,\bar{z}++}$, two are pure particle,
$\rho_{z+-,z++}$ and $\rho_{z--,z-+}$, and the remaining four are
mixed.


Returning to the snuark and quark quantum number plot
Fig.~(\ref{fig:SecondGraphSnuarks}), there are two lepton states
that are pure particle, two that are pure antiparticle, and the
remaining four are mixed.  The four mixed leptons are either the
weak isospin singlets or the doublets.  All the quarks are mixed.
Comparing with Fig.~(\ref{fig:WeakQNumPlot}), it appears that the
mass interaction converts pure snuarks to mixed snuarks.  But we
must again stress that these purported lepton states were made under
the assumption that the true leptons were snuarks, an assumption
that we know is incompatible with our assumed gravitational
interaction.


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:QNsOf8Snuarks})
% \SubSectionBreak

%sec:FiniteDensityOperators Density Operators
%sec:Geometry               Geometry
%sec:PrimitiveIdempotents   Primitive Idempotents
%sec:Representations        Representations
%sec:Tricks                 Algebra Tricks
%sec:Measurement            Measurement
%sec:Force                  Force
%sec:Zoo                    The Zoo
%sec:Mass                   Mass
%sec:CosmicHaze             Cosmic Haze

% Snuark quantum numbers revisited.
% Snuark antiparticles
% Weak hypercharge and weak isospin
%


%===================




% Fig.~(\ref{fig:LtoRBosons}) and Eq.~(\ref{eq:CirculantPIsBoundStates})
% \SubSectionBreak


% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of Chapter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{\label{sec:Mass} Mass}

%\begin{quote}
\settowidth{\versewidth}{Down and through the big fat marshes that
the virgin ore-bed stains,}
\begin{verse}[\versewidth]
%\em
Up along the hostile mountains, where the hair-poised snow-slide
shivers---\\
Down and through the big fat marshes that the virgin ore-bed stains,
\\
Till I heard the mile-wide mutterings of unimagined rivers \\
And beyond the nameless timber saw illimitable plains!
\end{verse}
%\end{quote}
$ $


\lettrine{W}{$\;$e have postulated} a fundamental force that binds
primitive idempotents.  We have shown that the elementary fermions
of the standard model have a structure similar to what you would
expect from a complete cancelation of the non scalar parts of this
force. We have guessed at a form for the mass interaction.  We have
speculated that the measured masses of the elementary particles is
given by the scalar remnant of the fundamental force after the
vector portion of that force is completely canceled.  We have
assumed that the observed leptons correspond to pure snuarks, and we
have seen that the resulting set of states are similar in nature to
the quarks and leptons of the standard model.  It remains to improve
this assumption and to calculate the observed elementary fermion
masses.



% Fig.~(\ref{fig:MixedCube}) and Eq.~(\ref{eq:QNsOfSnuarks})
% \SubSectionBreak

\section{\label{sec:StatisticalMixtures}  Statistical Mixtures}



Suppose that we have a set of $N$ probabilities $p_n$ (i.e. numbers
between $0$ and $1$) that add up to unity, and I have $N$ primitive
idempotents $\rho_n$, defined according to the same commuting roots
of unity.  Then we have:
\begin{equation}\label{eq:SumsOfProbsAndPIs}
\begin{array}{rcl}
p_1 + p_2 + ... + p_n &=& 1,\\
\rho_1 + \rho_2 + ... + \rho_n &=&\hat{1}.
\end{array}
\end{equation}
We define a statistical mixture as a sum over the primitive
idempotents:
\begin{equation}\label{eq:StatMixPIs}
\rho = p_1\rho_1 + p_2\rho_2 + ... + p_N\rho_N.
\end{equation}
The signs of the vector parts of the primitive idempotents depend on
the quantum numbers of the particular primitive idempotent.  Only
the scalar part always has the same sign.  For the $C(4,1)$ Clifford
algebra, there are $3$ commuting roots of unity and therefore the
scalar part is always $2^{-3} = 1/8$.  Since the probabilities add
up to one, the scalar part of a statistical mixture will also be
$1/8$.



In Section Eq.~(\ref{sec:TheMassInteraction}) we postulated that the
mass interaction, $M$, has the effect of negating the signs of all
the non scalar portions of a primitive idempotent:
\begin{equation}\label{eq:MassInteractSM}
\begin{array}{rcl}
M(A) &=& M(a_1\hat{1} + a_x\hat{x} + a_y\hat{y} + ...
+\at{ixyzst}),\\
&=&a_1\hat{1} - a_x\hat{x} - a_y\hat{y} - ... -\at{ixyzst}.
\end{array}
\end{equation}
We noted that $M$ has the disquieting effect of not preserving
primitive idempotents.  That is, if $\rho_\chi$ is a primitive
idempotent, $M(\rho_\chi)$ is not.  Instead, it is a more general
operator.


However, $M$ preserves the scalar part of a primitive idempotent:
\begin{equation}\label{eq:MpreservesScalar}
\la M(a_1\hat{1}+ a_x\hat{x}+...)\ra_0 = a_1 = %
\la (a_1\hat{1}+ a_x\hat{x}+...)\ra_0.
\end{equation}
and consequently we can hope that $M(\rho_\chi)$ can be interpreted
as a function that maps statistical mixtures to statistical
mixtures.


\SubSectionBreak


In defining the modification to the Feynman checkerboard, we assumed
that the transition amplitudes were given by products of the
primitive idempotents.  We then assembled the amplitudes into a $3
\times 3$ matrix of amplitudes and associated the bound states with
the matrices that were primitive idempotents in the algebra of all
such matrices.


To make the Feynman checkerboard generalization work with our mass
interaction, we need to replace its primitive idempotents with
statistical mixtures.  To see how this works, let's consider
products of two different statistical mixtures, $\rho$ and $\rho'$.
Each of these will have $N$ probabilities and primitive idempotents,
but following our earlier efforts in the Feynman checkerboard
matrices, we do not assume that they use the same sets of primitive
idempotents:
\begin{equation}\label{eq:StatMixMults}
\begin{array}{rcl}
\rho  &=& p_1\rho_1 + p_2\rho_2 + ... + p_N\rho_N,\\
\rho' &=& p_1'\rho_1' + p_2'\rho_2' + ... + p_N'\rho_N'.
\end{array}
\end{equation}
The contribution to the transition amplitude, $\rho_m\;\rho_n'$,
needs to be multiplied by the probability that both of these are the
actual primitive idempotent. We assume that the probabilities are
independent, so this probability is just the product of the
probabilities, $p_np_m'$. Summing up over all possible choices of
the two primitive idempotents, the statistical mixture of the
transition amplitude is simply:
\begin{equation}\label{eq:StatMixTAs}
\begin{array}{rcl}
\sum_m\sum_n(p_mp_n'\;\rho_m\;\rho_n') &=& (\sum_m p_m\rho_m)
(\sum_n p_n'\rho_n'),\\
&=& \rho\;\rho'.
\end{array}
\end{equation}
Thus the machinery for making computations with statistical mixtures
of primitive idempotents is similar to the machinery we've already
been using with the primitive idempotents themselves.


If $M(\rho_{z++})$ is a statistical mixture, then we can write it as
a sum over primitive idempotents.  To compute it as a sum over
primitive idempotents oriented in the $z$ direction, the
probabilities are computed as the scalar parts of
$8M(\rho_{z++})\;\rho_{\pm z\pm\pm}$.  The $8$ comes from the
conversion from the ``trace'' function used in matrices to the
``scalar part'' function used in geometry.  We first write out
$M(\rho_{z++})$:
\begin{equation}\label{eq:TransProbMzpp}
\begin{array}{rcl}
M(\rho_{z++}) &=&
M(0.125(1+\at{zt}+\hat{s}+\at{ixyzst}-\at{zst}-\at{ixys}-\at{ixyzt}-\at{ixy})),\\
&=&0.125(1-\at{zt}-\hat{s}-\at{ixyzst}+\at{zst}+\at{ixys}+\at{ixyzt}+\at{ixy}).
\end{array}
\end{equation}
Then, after a certain amount of calculation, the transition
probabilities are:
\begin{equation}\label{eq:ProbArrayZtoZ}
\begin{array}{llll}
p_{z++}=-0.75,\;\;&p_{z+-}=+0.25,\;\;&p_{z-+}=+0.25,\;\;&p_{z--}=+0.25,\\
p_{\bar{z}++}=+0.25,\;\;&p_{\bar{z}+-}=+0.25,\;\;&
p_{\bar{z}-+}=+0.25,\;\;&p_{\bar{z}--}=+0.25.
\end{array}
\end{equation}
Note that while the probabilities do add to unity, one of them, the
one corresponding to the probability of the mass interaction leaving
the primitive idempotent unchanged, is negative.  This indicates the
mass interaction does not map the primitive idempotent $\rho_{z++}$
to a statistical mixture, which should not be very surprising.  The
result is obviously general, none of the primitive idempotents
$\rho_{z\chi\chi}$ will be mapped to a statistical mixture.


We can also compute the transition probabilities for $M(\rho_{z++})$
to primitive idempotents oriented in the $x$ direction.  We obtain:
\begin{equation}\label{eq:ProbArrayZtox}
\begin{array}{llll}
p_{z++}=-0.25,\;\;&p_{z+-}=+0.25,\;\;&p_{z-+}=+0.25,\;\;&p_{z--}=+0.25,\\
p_{\bar{z}++}=-0.25,\;\;&p_{\bar{z}+-}=+0.25,\;\;&
p_{\bar{z}-+}=+0.25,\;\;&p_{\bar{z}--}=+0.25.
\end{array}
\end{equation}
Again, the transition probabilities sum to unity, but two of them
are negative.  The negative ones are the ones that leave the last
two quantum numbers unchanged.


\SubSectionBreak


We next look at the action of $M$ on snuarks.  Let us repeat the
calculation with $M(\rho_{z+-,z++})$.  Referring to
Eq.~(\ref{eq:SnuarksBySubstitution}) for the snuark values:
\begin{equation}\label{eq:TransProbMzpmzpp}
\begin{array}{rcl}
M(\rho_{z+-,z++}) &=& M(0.25(1+\at{zt}+\hat{s}-\at{zst}),\\
&=&0.25(1-\at{zt}-\hat{s}+\at{zst}).
\end{array}
\end{equation}
Since snuarks are combinations of two primitive idempotents, the
scaling factor is changed.  To see what it is, note that
$(\rho_{z+-,z++})^2$ $=\rho_{z+-,z++}$, so the transition
probability should be unity.  But the scalar part of
$\rho_{z+-,z++}$ is $0.25$, so the scaling factor needs to be $4$.
Accordingly, compute the scalar part of $4M(\rho_{z+-,z++})\;
\rho_{\chi,\chi}$:
\begin{equation}\label{eq:ProbArraySnztoSnz}
\begin{array}{llll}
p_{z+-,z++}=-0.5,\;\;&p_{\bar{z}--,\bar{z}-+}=+0.5,
\;\;&p_{\bar{z}+-,\bar{z}++}=+0.5,\;\;&p_{z--,z-+}=+0.5,\\
p_{z+-,\bar{z}+-}=+0.5,\;\;&p_{z-+,\bar{z}-+}=+0.5,\;\;&
p_{z--,\bar{z}--}=+0.5,\;\;&p_{z++,\bar{z}++}=+0.5.
\end{array}
\end{equation}
The probabilities sum two $2$ because the snuarks doubly represent
the number of degrees of freedom.  And again we have a negative
probability for the transition from number.  Also, since our snuarks
are degenerate, that is, since they share degrees of freedom, the
probabilities sum to more than $1$.


Snuarks with different orientation can share only two degrees of
freedom, $\hat{1}$ and $\hat{s}$.  Therefore, the transition
probabilities will all be $(1-1+0+0)/4 = 0$ or $(1+1+0+0)/4 = 0.5$,
depending on the eigenvalue of $\hat{s}$. Referring to
Eq.~(\ref{eq:SnuarksBySubstitution}), the transition probabilities
are:
\begin{equation}\label{eq:ProbArraySnztoSnx}
\begin{array}{llll}
p_{x+-,x++}= 0,\;\;&p_{\bar{x}--,\bar{x}-+}=+0.5,
\;\;&p_{\bar{x}+-,\bar{x}++}=0,\;\;&p_{x--,x-+}=+0.5,\\
p_{x+-,\bar{x}+-}=0,\;\;&p_{x-+,\bar{x}-+}=+0.5,\;\;&
p_{x--,\bar{x}--}=+0.5,\;\;&p_{x++,\bar{x}++}=0.
\end{array}
\end{equation}
Note that none of the above are negative, and that $M$ will map a
snuark to a statistical mixture of four snuarks.  And that
statistical mixture is a little special; it is equal for all four
values.


\SubSectionBreak


We represent Stern-Gerlach filters with projection operators, and
the key attribute of Stern-Gerlach filters is that after they
measure an attribute of a particle they preserve that attribute.
This is the physical reason why it is natural for us to use
primitive idempotents to represent the elementary particles.


However, the idempotency equation, $\rho^2$ is never equal to $\rho$
for a statistical mixture that is not pure.  On the other hand, we
can imagine a more general Stern-Gerlach filter that would preserve
a statistical mixture.  As before, we will represent such a
Stern-Gerlach filter with the same mathematical object that
represents the statistical mixture, $\rho$ itself.


Given a complete set of $8$ primitive idempotents, the most complete
mixture one can make is to take $0.125$ of each:
\begin{equation}\label{eq:ExCompStatMix}
\rho_c = 0.125(\rho_1 + \rho_2 + ... + \rho_8) = 0.125 \hat{1}.
\end{equation}
Since this is a mixture, it is no longer idempotent, but it is
suspiciously close:
\begin{equation}\label{eq:NotIdempMix}
\rho_c^2 = 0.125 \rho_c.
\end{equation}
That is, $\rho_c$ is just an eighth of the unity operator, which is
idempotent, but not a primitive idempotent.  This suggests that a
natural extension of the idempotency equation to statistical
mixtures is to allow
\begin{equation}\label{eq:GenIdemEqn}
\rho^2 = k \rho,
\end{equation}
where $k$ is a real constant.  The value $0.125 \hat{1}$ is a pure
scalar and therefore has a very low potential energy.  This suggests
that we should consider the potential energies of statistical
mixtures.










\SubSectionBreak

\SubSectionBreak

\SubSectionBreak










% Fig.~(\ref{fig:LtoRBosons}) and Eq.~(\ref{eq:CirculantPIsBoundStates})
% \SubSectionBreak




In making calculations with primitive idempotents, one of the most
useful techniques we used was the fact that any product that begins
with a primitive idempotent and ends with another, will be a complex
multiple of the two primitive idempotents on the ends.  For example:
\begin{equation}\label{eq:PrimitiveIdempotentProds}
\rho_m\;X\;\rho_n' = x_{mn}\;\rho_m\;\rho_n',
\end{equation}
where $X$ is any operator, and $x_{mn}$ is a complex number.  From a
physical point of view, this corresponds to putting $X$ between two
Stern-Gerlach filters.  Because the filters allow only a single
state to exit, the stuff in the middle, $X$, can only have the
effect of decreasing the amplitude, or changing its phase.



% Fig.~(\ref{fig:LtoRBosons}) and Eq.~(\ref{eq:CirculantPIsBoundStates})
% \SubSectionBreak



\section{\label{sec:TheKoideRelation}  The Koide Relation}


In 1982, Yoshio Koide postulated
\cite{KoideOriginalMassRelation_1982} a relationship between the
masses of the charged leptons.




% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:RuvwDefn})
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% End of Chapter
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\chapter{\label{sec:CosmicHaze} Cosmic Haze}

%\begin{quote}
\settowidth{\versewidth}{And he yearned to the flare of Hell-gate
there as the light of his own hearth-stone.}
\begin{verse}[\versewidth]
%\em
The Spirit gripped him by the hair, and sun by sun they fell\\
Till they came to the belt of Naughty Stars that rim the mouth of Hell.\\
The first are red with pride and wrath, the next are white with pain,\\
But the third are black with clinkered sin that cannot burn again:\\
They may hold their path, they may leave their path, with never a soul to mark, \\
They may burn or freeze, but they must not cease in the Scorn of the Outer Dark. \\
The Wind that blows between the Worlds, it nipped him to the bone,\\
And he yearned to the flare of Hell-gate there as the light of his own hearth-stone.\\
\end{verse}
%\end{quote}
$ $

\lettrine{W}{$\;$hile snuarks and} anions are too high energy for
man to create them with current technology, nature has access to
higher energies and we can suppose that they were present at the big
bang, and that they may be let loose by the extreme conditions at
black holes.


\chapter{\label{sec:Conclusion} Conclusion}

%\begin{quote}
\settowidth{\versewidth}{The long bazar will praise, but thou---}
\begin{verse}[\versewidth]
%\em
Small mirth was in the making---now\\
I lift the cloth that cloaks the clay,\\
And, wearied, at thy feet I lay\\
My wares, ere I go forth to sell.\\
The long bazar will praise, but thou---\\
Heart of my heart---have I done well?
\end{verse}
%\end{quote}
$ $

\lettrine{P}{$\;$erhaps we've applied} density operator formalism with too much enthusiasm and have over reached a calculation or two. Apologies to the reader. Certainly we've added this chapter mostly to provide an opportunity to quote from yet one more Kipling poem.






% Fig.~(\ref{fig:FermCube}) and Eq.~(\ref{eq:DiracPosCommRelations})
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\backmatter

\bibliography{DMAA} % Produces the bibliography via BibTeX.
\bibliographystyle{unsrt}

\input{DMAA.ind}

\end{document}
% End of document DMAA.tex


When elementary particles emit or absorb gauge bosons, they are
changed, and this must be equivalent to changes to their primitive
idempotents.  Suppose that a primitive idempotent undergoes a
process that begins in an initial state $\rho_I$ and ends in a final
state $\rho_F$.  We can think of this as the replacement of a state
$\rho_I$ with a state $\rho_F$.  We now look at this process from
the point of view of potential energy.


Initially, the elementary particle is made up of some collection of
primitive idempotents that add up to a scalar potential.  Replacing
one of the primitive idempotents with a new primitive idempotent
would change the sum to something other than a scalar and that would
create a forbidden state.  Instead, some combination of primitive
idempotents must be simultaneously changed.


We suppose that the various forces differ in their strength
according to the number of primitive idempotents that must be
altered in order to effect the transition.  Gravity, being the
weakest, we can assume is a force that changes all the primitive
idempotents together.  The electroweak force will change the next
largest, and the strong force fewer still.


In the emission of a gauge boson, an elementary particle changes
form from one sort to another.  Since we are modeling potential
energies as sums of idempotents, the conversion of a primitive
idempotent from one sort to another acts like a replacement of one
primitive idempotent with another.  A reasonable guess is that the
initial and final primitive idempotents are in the same complete set
of primitive idempotents.  In this case, they can be related by
