% ****** Start of file DMAA.tex ******
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% Copyright (c) 2003-2004 Carl Brannen
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% Poetry:
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% 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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\begin{document}
\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}}
\addtocontents{toc}{\protect\contentsline {chapter}{\protect\chapternumberline {0} Fundamentals}{1}}
%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 High
\qbezier(107,44)(107,44)(183,38) %
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% Slit Scattering Low
\qbezier(107,36)(107,36)(183,30) %
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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&
\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})
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\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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\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