INTRODUCTION TO KOOPMAN-VON NEUMANN MECHANICS

AN HONORS THESIS

SUBMITTED ON THE FIFTH DAY OF DECEMBER, 2021

TO THE DEPARTMENT OF PHYSICS AND ENGINEERING PHYSICS

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS

OF THE HONORS PROGRAM

OF NEWCOMB-TULANE COLLEGE

TULANE UNIVERSITY

FOR THE DEGREE OF

BACHELOR OF SCIENCES

WITH HONORS IN PHYSICS

BY

Daniel W. Piasecki

APPROVED:

Denys I. Bondar, Ph.D.

Director of Thesis

Ryan T. Glasser, Ph.D.

Second Reader

Frank J. Tipler, Ph.D

Third Reader

###### Abstract

This work is to consolidate current literature on Koopman-von Neumann (KvN)
Mechanics into a simple and easy to understand text. KvN Mechanics is a branch
of Classical Mechanics that has been recast into the mathematical language of
Quantum Mechanics. KvN Mechanics utilizes a Hilbert phase space with operators
to calculate the expectation values of observables of interest (expectation
values such as position, momentum, etc.) It is an important tool in
statistical physics and guide to illuminate the mysterious relationship
between Quantum and Classical Physics. A lot of important applications of KvN
Mechanics have been developed in the last decades.

## Acknowledgements

Thanksgiving is approaching at the time I’m writing this, so there are some
people who will get some messages of gratitude from me a few days early.

First, I would like to thank the Tulane Summer Research Program and the Tulane
Honors Program. This project would not have been possible without their
support.

Many thanks to the readers of this thesis: Dr. Ryan Glasser, Dr. Frank Tipler,
and especially my advisor for this project, Dr. Denys Bondar. I am grateful
for his advice on this paper and guidance for how I can take what I’ve learned
from it and apply it to the rest of my career, both in terms of new knowledge
about the subject matter and skills I’ve been practicing while working on
this. His openness and directness in advising me made a big difference, and
I’ve learned a lot from working on this project. Also, 80 pages is a lot to
read, so I am grateful for the time all three of these professors have
dedicated to reading my work.

******************************************************************

I am also very appreciative to my friends and family for their encouragement,
prayers, and patience with me. This thesis was a huge commitment for me, and
they were respectful of my time when I was busy and curious and uplifting
during my few moments of downtime. One of them also helped look over some of
it and did some proofreading, which I also am thankful for. Thanks goes
especially to my sweetheart Clover Robichaud for her continual advice,
patience, and support while the project has been ongoing. Also, thanks goes to
my good friend Kushal Singh for helping me prepare for the Thesis Defense
itself.

Finally, I am grateful to all other physicists and aspiring physicists who
read this; if you are one of these people, you are exploring a little known
topic that can make a big difference, and I appreciate that you are likely
branching out from the standard topics you are already taught in class. I am
hopeful that my work contributed to their knowledge of Koopman–von Neumann
Mechanics and that this will help them expand the field of Physics. This is
currently a little known branch of Physics, and the goal of this project was
to make this information more accessible to others unfamiliar with the topic,
both in Physics and otherwise. If anyone could take what I did and have that
help him or her in contributing to Physics, then I will know that I have made
a difference.

###### Contents

  1. 1 Introduction
  2. 2 General Review of Quantum Mechanics
    1. 2.1 Fundamentals of Hilbert Space
    2. 2.2 Operators and Commutators
    3. 2.3 Expectation Values and the Eigenvalue Problem of Quantum Mechanics
    4. 2.4 Review of Dirac Delta Functional
    5. 2.5 Summary of Quantum Mechanics and its Postulates
    6. 2.6 The Density Operator
  3. 3 Introduction to KvN Mechanics
  4. 4 Path Integral Formulation of Quantum and Classical Mechanics
    1. 4.1 Feynman Path Integral of Quantum Mechanics
    2. 4.2 Koopman-von Neumann Classical Path Integral
  5. 5 Classical and Quantum Comparison
    1. 5.1 The Phase in Koopman-von Neumann and Quantum Mechanics
    2. 5.2 Uncertainty Principle in Koopman-von Neumann and Quantum Mechanics
    3. 5.3 Measurement in Koopman-von Neumann and Quantum Mechanics
    4. 5.4 The Double Slit Experiment
  6. 6 Miscellaneous KvN Topics of Interest
    1. 6.1 Operational Dynamic Modeling
    2. 6.2 Wigner Quasiprobability Distribution
    3. 6.3 Time Dependent Harmonic Oscillator
    4. 6.4 Aharonov–Bohm Effect
  7. 7 Other Research
  8. A Useful Mathematics for Hilbert Space Physics
  9. B The Principle of Causality and Stone’s Theorem
  10. C Exponated Operators and the Propogator

###### List of Figures

  1. 1 This is the set up of the double slit experiment. Middle wall is at $y=y_{M}$ and rightmost wall is at $y=y_{R}$. Adapted from [13]
  2. 2 a. Left: Graph of KvN probability distribution. b. Right: Graph of QM probability distribution. Adapted from [13].
  3. 3 a. Left: Example graph of a single variable gaussian curve. b. Right: Example graph of a double variable gaussian curve. Adapted from [24].
  4. 4 Aharanov-Bohm Experiment set up. Reproduced from [1].

##  1 Introduction

Quantum Mechanics (QM) is known for its unique mathematical structure unlike
that traditionally found in other branches of Physics. In QM, one has
operators acting on the wavefunction in complex Hilbert Space in order to
compute what the expectation values of observables seen in the laboratory are
([20]). Classical Mechanics (CM), on the other hand, is based on continuous,
real valued functions of spatial and time dependent variables. One merely has
to plug in the values of the variables at a particular point, and one knows
the value of the function of interest. QM makes probabilistic predictions of
the quantum particle in question, whereas CM makes deterministic predictions
of all classical behavior.

The large differences in mathematical form has historically made it difficult
to compare QM to CM ([14]). There have been many attempts to make QM appear
more mathematically similar to CM ([14]). For instance, the Born and Madelung
approach to QM has a modified Hamilton-Jacobi equation as its basis ([14]).
But there have been few attempts to make CM more similar in form to QM. This
changed in the 1930’s, however, when a form of statistical CM was devised by
Bernard Koopman and John von Neumann that had the same operatorial
mathematical structure as QM ([10]; [22]). Koopman-von Neumann Mechanics
(KvNM), as it later became known, was subsequently used by mathematicians to
formulate ergodic theory, while physicists mostly forgot about the existence
of this field. Currently, there is a large renewal of interest by the physics
community in this relatively obscure field of study.

In this Thesis I intend to articulate many significant contributions of KvNM
to physics in a straightforward and easy to understand fashion. I am writing
this text in such a way that even an undergraduate with limited training in
Physics could understand. I will point out significant topics of research in
KvNM and how they are changing physics as we know it. By making this topic
more accessible to a variety of readers, I hope that many will find it of
interest and continue to build upon this novel and fascinating avenue of
study.

KvNM is statistical CM cast in the language of complex wavefunctions in
Hilbert Space. Although wavefunctions in Hilbert Space are thought to be tools
of QM, they can be used to exactly model the statistical properties of purely
classical systems. Classical wavefunctions have been used to study a wide
range of physical phenomena. For example, they have been used in studying
chaotic systems in quantum and classical theory ([23]), been used to develop a
model of stochasticity from determinism ([15]), been used to study the
interplay of connected quantum and classical systems ([3]), been used to study
free fall and Fischer Information ([19]), been used to reproduce equations
governing spin 1/2 relativistic particles ([6]), among many other areas of
research.

Of great interest, a field like KvNM might be able to decipher the true
meaning of QM, which has been hotly debated for over a century.111Despite a
century of progress, QM has no consensus interpretation of its equations.
‘Shut up and calculate’ rings as true today as it did a century ago. Feynman
gives us a glimmer of hope for the situation, however, as it always takes
several generations to understand a great, mysterious truth: “We have always
had a great deal of difficulty understanding the world view that quantum
mechanics represents. At least I do, because I’m an old enough man that I
haven’t got to the point that this stuff is obvious to me… You know how it
always is, every new idea, it takes a generation or two until it becomes
obvious that there’s no real problem.” ([8]) Perhaps we’re getting closer to
an understanding and perhaps a budding field like KvNM can hold clues to the
answer.

##  2 General Review of Quantum Mechanics

What follows is a brief review of Quantum Theory. QM depends on operations on
the vector $\ket{\psi}$ in Hilbert Space. The wavefunction $\ket{\psi}$ is a
complete description of the state of the system in QM ([20]). It contains
probabilistic information of the system’s behavior.

It has a number of convenient properties, such as the fact that it makes
switching between one set of bases and another is a rather straightforward
task. Hilbert Space useful for all sorts of mathematical procedures. It is,
however, not the typical tool of CM.

###  2.1 Fundamentals of Hilbert Space

To understand QM, we must first be comfortable with the concept of a Hilbert
Space. A Hilbert Space is an infinite dimensional vector space with a well
defined inner product. One can conceive of the Hilbert Space as a coordinate
system with an infinite number of perpendicular axes as a crude model. For QM,
the Hilbert Space we use not only contains real numbers, but has also been
extended to the complex numbers. Although there are Hilbert Spaces that only
contain real numbers, we will not need to concern ourselves with those when
working with QM or KvNM.

To see why Hilbert Spaces are so useful in QM, imagine a generic function
$f(x)$ which is continuous on its domain ([20]). Since any continuous function
can be represented by an infinite number of points, we can represent any
continuous function in Hilbert Space with state vectors. Each point along a
generic function $f(x)$ can be represented by a vector in Hilbert Space.
Because there is an infinite number of points along the function $f(x)$ (even
in a finite segment), we should not be surprised it will take an infinite
number of vectors to represent $f(x)$.

This collection of vectors can be represented in an orthogonal basis (i.e.,
each vector is perpendicular to any other vector in the collection or set, as
there are an infinite number of perpendicular axes in Hilbert Space.) A vector
can be represented with the following bra-ket notation (original to Dirac):

$\vec{v}=\begin{pmatrix}v_{1}\\\ v_{2}\\\
v_{3}\end{pmatrix}\Longrightarrow\ket{v}=\begin{pmatrix}v_{1}\\\ v_{2}\\\
v_{3}\end{pmatrix},\bra{v}=\begin{pmatrix}v_{1}^{*}&v_{2}^{*}&v_{3}^{*}.\end{pmatrix}$

where $\ket{v}$ is the ket vector and $\bra{v}$ is the bra vector ([20]). The
asterisk (*) represents the complex conjugate. Vectors like $\vec{v}$ are
ambiguous, because it is not often clear if to write them as columns (as
depicted above) or rows of numbers. However, the Dirac bra-ket notation
overcomes the ambiguity of typical vector notation in physics, by having that
the ket is always equivalent to a column and the bra is equivalent to a row of
values. Hilbert Space utilizes bra-ket notation in its infinite dimensional
mathematics. A traditional Hilbert Space ket might have more than three
entries therefore; it might be represented with an infinite number of entries.

With this in mind, the continuous function $f(x)$ would be represented in
Hilbert Space as:

$f(x)\Longrightarrow\ket{f(x)}=\begin{pmatrix}f(x_{1})\\\ f(x_{2})\\\
f(x_{3})\\\ \vdots\\\ f(x_{n})\\\ \vdots\end{pmatrix}.$ (1)

You can think about this representation of $f(x)$ as the value of each point
along $f(x)$, i.e. $f(x_{i})$, having a unique independent basis vector in
Hilbert Space multiplying it. So according to the rules of Linear Algebra you
may expand $\ket{f(x)}$, for example:

$\ket{f(x)}=\begin{pmatrix}1\\\ 0\\\ 0\\\ \vdots\\\ 0\\\
\vdots\end{pmatrix}f(x_{1})+\begin{pmatrix}0\\\ 1\\\ 0\\\ \vdots\\\ 0\\\
\vdots\end{pmatrix}f(x_{2})+\begin{pmatrix}0\\\ 0\\\ 1\\\ \vdots\\\ 0\\\
\vdots\end{pmatrix}f(x_{3})+...+\begin{pmatrix}0\\\ 0\\\ 0\\\ \vdots\\\ 1\\\
\vdots\end{pmatrix}f(x_{n})+...$

In the above expansion, notice that the column vectors form a unique set of
linearly independent basis vectors.

The way to switch between bras and kets in a complex Hilbert Space is to take
the transpose of the vector and complex conjugate of each of the values, like
so:

$\ket{v}=\begin{pmatrix}v_{1}\\\ v_{2}\\\ v_{3}\\\ \vdots\\\
\end{pmatrix}\Longleftrightarrow\bra{v}=\begin{pmatrix}v_{1}^{*}&v_{2}^{*}&v_{3}^{*}&\ldots\end{pmatrix}$

The asterisks represents the complex conjugate. This operation will be
represented by $\dagger$, so that $\ket{v}^{\dagger}=\bra{v}$ and vice versa.
Just like any ordinary vector, bra-ket vectors can be added component wise and
multiplied together:

$\begin{pmatrix}v_{1}\\\ v_{2}\\\ v_{3}\\\ \vdots\\\
\end{pmatrix}+\begin{pmatrix}w_{1}\\\ w_{2}\\\ w_{3}\\\ \vdots\\\
\end{pmatrix}=\ket{v}+\ket{w}=\begin{pmatrix}v_{1}+w_{1}\\\ v_{2}+w_{2}\\\
v_{3}+w_{3}\\\ \vdots\\\ \end{pmatrix}=\ket{v+w},$ (2)

$\begin{pmatrix}w_{1}^{*}&w_{2}^{*}&w_{3}^{*}&\ldots\end{pmatrix}\begin{pmatrix}v_{1}\\\
v_{2}\\\ v_{3}\\\ \vdots\\\
\end{pmatrix}=v_{1}w_{1}^{*}+v_{2}w_{2}^{*}+v_{3}w_{3}^{*}+\ldots=\innerproduct{w}{v},$

where $\innerproduct{w}{v}$ is a generalized “dot product” or inner product
for the vector space. Two vectors are said to be orthogonal if their inner
product is 0. So in order for two vectors $\ket{v}$ and $\ket{w}$ in Hilbert
Space to be orthogonal, we must have $\innerproduct{w}{v}=0$. Also, like for
any common vector, you can rescale a vector by a complex constant $c$ to
stretch or shrink its magnitude: $\ket{v^{\prime}}=c\ket{v}$.

The vectors additionally obey the following properties:

$(\innerproduct{w}{v})^{\dagger}=\innerproduct{v}{w},$
$\ket{c_{1}v+c_{2}w}^{\dagger}=(c_{1}\ket{v}+c_{2}\ket{w})^{\dagger}=(c_{1}\ket{v})^{\dagger}+(c_{2}\ket{w})^{\dagger}=c_{1}^{*}\bra{v}+c_{2}^{*}\bra{w}.$

Since continuous functions like $f(x)$ can be represented in Hilbert Space, we
can recast continuous functions of QM into the Hilbert Space formalism. We
will shortly see why this is useful to do.

The wavefunction $\psi$ is the backbone of the quantum mechanical description
of reality. The wavefunction $\psi$ is an equation of state, i.e., it provides
us with a description of the state of the system ([20]). 222One way you can
think of an ‘equation of state’ is, for instance, the ideal gas law, $PV=NkT$.
This equation will tell you what the state of the system, i.e., what the
pressure $P$, volume $V$, number of gas particles $N$, and temperature $T$,
for an ideal gas is. The wavefunction got its name from the fact that its form
resembles that of a wave equation. In configuration space, the wavefunction is
often written as:

$\psi(q_{1},q_{2},...,q_{n},t)=R(q_{1},q_{2},...,q_{n},t)e^{iS(q_{1},q_{2},...,q_{n},t)/\hbar},$
(3)

where $q_{1},q_{2},...,q_{n}$ are positions in configuration space and $t$ is
time. $S$ represents the Action and $R$ is the amplitude of the wave equation.
In light of the above discussion and since $\psi$ is a continuous function, we
can represent $\psi$ as $\ket{\psi(t)}$ in complex Hilbert Space.
Schrödinger’s equation, for example, becomes:

$i\hbar\frac{\partial\psi}{\partial
t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial
q^{2}}+V(q)\psi\Longrightarrow i\hbar\frac{\partial}{\partial
t}\ket{\psi(t)}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial
q^{2}}\ket{\psi(t)}+V(q)\ket{\psi(t)}.$ (4)

The benefits of using Hilbert Space instead of just utilizing $\psi$ is
because Hilbert Space provides a rich mathematical, abstract structure where
various manipulations are straightforward and easy to do ([20]; [4]). We can
very easily, for instance, change our basis of interest from one to another.
We can ‘deconstruct’ functions and put them in various representations. This
is not a flexibility given by simply using $\psi$. As we will see, changing
representations is a fundamental ingredient to QM, so the Hilbert Space form
is absolutely necessary.

###  2.2 Operators and Commutators

An operator is a mathematical object that “acts on” a function, vector, or
matrix to produce a new function (the derivative), a new vector, or a new
matrix. For example, the derivative $\frac{d}{dt}$ acting on a function $f(t)$
is an operator, as is a matrix $M$ acting on a vector $\ket{v}$. The operator
produces a new mathematical entity, such as the $\frac{df(t)}{dt}$ from $f(t)$
or a new vector $\ket{v^{\prime}}$ from the original vector
($\ket{v^{\prime}}=M\ket{v}$). All operators will be represented with
triangular “hats”, for example, $\hat{A}$.

Two important types of operators are Hermitian operators and anti-Hermitian
operators, which obey the following rules respectively:

$\hat{A}^{\dagger}=\hat{A},$ $\hat{A}^{\dagger}=-\hat{A},$

where the $\dagger$ symbol is the transpose with complex conjugate as before.
The operators we most use in QM are Hermitian operators. Hermitian operators
are incredibly useful to utilize, because if we take the $\dagger$ of an
expression with a Hermitian operator, the form of the Hermitian operator is
left unchanged. For example, for two vectors $\ket{v}$ and $\ket{w}$ and
Hermitian operator $\hat{H}$:

$(\bra{w}\hat{H}\ket{v})^{\dagger}=\bra{v}\hat{H}^{\dagger}\ket{w}=\bra{v}\hat{H}\ket{w}.$

An observable in QM is a property of a system you would like to measure or
observe. Examples of observables in QM would be the position of a particle,
its momentum, its spin, etc. Observables in QM are given by Hermitian
operators $\hat{A}$ acting on the state vector $\ket{\psi(t)}$, i.e.,
$\hat{A}\ket{\psi(t)}$. The state vector $\ket{\psi(t)}$ is a vector in
Hilbert Space that contains complete dynamical information about the system in
question.

Hermitian operators relate directly to properties you measure inside the
laboratory. For any generic operator $\hat{A}$, there exist special types of
ket vectors which when acted on by an operator $\hat{A}$ they equal themselves
rescaled by a constant. In other words, they obey the mathematical form of

$\hat{A}\ket{\psi(t)}=A\ket{\psi(t)},$ (5)

where ‘$A$’ is always a real number ([20]).

Note that this special relationship is not true for just any vector
$\ket{\psi(t)}$. This is only true for specific vectors which we will call
eigenkets. The rescaling real number ‘$A$’ is known as the eigenvalue, and the
mathematical problem to identify for an operator $\hat{A}$ the correct
eigenkets and eigenvalues is called the eigenvalue problem([20])333The
function version of the eigenvalue problem is $\hat{A}\psi=A\psi$, where
$\psi$ is called an eigenfunction. To recover the Hilbert Space version,
remember the colloquial transformation $\psi\Longrightarrow\ket{\psi(t)}$. In
the literature, terms like eigenvectors and eigenfunctions are used
interchangeably, as they are technically the same thing, just one written in
an abstract Hilbert Space and one written without.. The eigenvalue problem is
central to QM because it directly relates to the observables measured in the
laboratory, as will be developed below.

Perhaps one of the most important properties of a Hermitian operator is that
its eigenvalues must necessarily always be real ([20]). Other types of
operators might give, for example, imaginary eigenvalues. Hermitian operators
will only give you real numbers, which are the types of values we would
measure inside the laboratory.

Two very important operators are the ones for the observables momentum and
position. In coordinate representation (as other representations are
possible), the momentum and coordinate operators are given respectively as
follows:

$\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial q},$ (6)

$\hat{q}=q.$ (7)

These operators are important, because as we shall see, they are related to
the average momentum and position of interest. 444For example, in standard
function form one can easily show that the $\hat{p}\psi=p\psi$ where $p$ is
the momentum expectation value measurable in the laboratory and $\hat{p}$ is
the above momentum operator. Earlier, we said that a common form of $\psi$ in
configuration space is $\psi=Re^{iS/\hbar}$, where $R$ is the amplitude of the
wavefunction and $S$ is the Action. Let’s assume a simple, one dimensional
case where $R=R(t)$ and $S=S(q,t)$. Then we would have:
$\hat{p}\psi=\frac{\hbar}{i}\frac{\partial}{\partial
q}Re^{iS/\hbar}=\frac{\hbar}{i}Re^{iS/\hbar}\frac{i}{\hbar}\frac{\partial
S}{\partial q}=\frac{\partial S}{\partial q}\psi.$ From CM, we know that
$p=\frac{\partial S}{\partial q}$, so the shoe fits. Please note that this
exercise is a bit of an oversimplification. Generally, $R=R(t)$ is not
considered a legitimate choice because it can lead to nonnormalizable
wavefunctions (see [21] for further discussion.) For the sake of a simple
demonstration of how operators acting on wavefunctions lead to eigenvalues
measured in the laboratory, I chose this oversimplified approach to highlight
connections between ideas .

Commutators describe relations between operators. The principle that two
numbers $A$ and $B$ obey $A\cdot B=B\cdot A$ is known as the principle of
commutativity. Operators do not need to, however, obey the principle of
commutativity. We might have, for example, that
$\hat{A}\hat{B}\neq\hat{B}\hat{A}$ for two operators $\hat{A}$ and $\hat{B}$.
This might be most explicitly clear when we think about the example of matrix
multiplication. The order matrices are multiplied in might give you different
end products. We say therefore that matrix multiplication is noncommutative.

In order to describe this property for operators, we use the idea of a
commutator. For two operators $\hat{A}$ and $\hat{B}$, the commutator is
defined as:

$[\hat{A},\hat{B}]\equiv\hat{A}\hat{B}-\hat{B}\hat{A}.$

If the commutator equals zero, the two operators commute:

$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}=0\Longrightarrow\hat{A}\hat{B}=\hat{B}\hat{A}.$

However, if the commutator is nonzero, then obviously the operators do not
commute.

The position and momentum operators are an example of noncommuting operators.
It can be demonstrated that:

$[\hat{q},\hat{p}]=\hat{q}\hat{p}-\hat{p}\hat{q}=q\frac{\hbar}{i}\frac{\partial}{\partial
q}-\frac{\hbar}{i}\frac{\partial}{\partial q}q.$
$[\hat{q},\hat{p}]\psi=q\frac{\hbar}{i}\frac{\partial\psi}{\partial
q}-\frac{\hbar}{i}\frac{\partial}{\partial
q}(q\psi)=q\frac{\hbar}{i}\frac{\partial\psi}{\partial
q}-\frac{\hbar}{i}\frac{\partial q}{\partial
q}\psi-q\frac{\hbar}{i}\frac{\partial\psi}{\partial q}=-\frac{\hbar}{i}\psi.$

Ergo, the commutator relation for $\hat{q}$ and $\hat{p}$ is

$[\hat{q},\hat{p}]\psi=-\frac{\hbar}{i}\psi=i\hbar\psi\Longrightarrow[\hat{q},\hat{p}]=i\hbar.$
(8)

This is the famous Heisenberg Canonical Commutator ([20]). In section 5.2 we
will derive the Heisenberg Uncertainty Principle from the Canonical
Commutator. The following are some important commutator relationships given
below with derivation for brevity:

$[\hat{A},\hat{B}]=-[\hat{B},\hat{A}].$ (9)

$[c\hat{A},\hat{B}]=[\hat{A},c\hat{B}]=c[\hat{A},\hat{B}].$ (10)

$[\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\hat{B}.$
(11)

$[\hat{A},\hat{B}\hat{C}]=\hat{B}[\hat{A},\hat{C}]+[\hat{A},\hat{B}]\hat{C}.$
(12)

$[\hat{A},\hat{B}+\hat{C}]=[\hat{A},\hat{B}]+[\hat{A},\hat{C}].$ (13)

###  2.3 Expectation Values and the Eigenvalue Problem of Quantum Mechanics

What exactly are the eigenvalues that we get out of the eigenvalue problem? I
have stated before that they are linked to measured observables, but in what
way exactly? Observables can be anything you would like to measure, such as
the momentum or position of a quantum particle.

Eigenvalues are going to be the numbers you measure directly inside the
laboratory. If you conduct an experiment on a quantum system, the eigenvalues
will be the exact numbers you record. Remember that for a Hermitian operator,
its eigenvalues will always be real numbers. This is critical, because any
number you measure in an experiment will be real instead of, say, complex.

Eigenvalues are connected to finding the expectation values of an experiment.
Imagine you conduct an experiment on a quantum system and you measure the
position or momentum. Instead of predicting individual momenta or position, QM
can predict the weighted mean value that you should expect. If you took many
measurements on identical systems, and took the weighted average of all those
measurements, the value you get out of this is the expectation value.

I will demonstrate how expectation values arise in QM in the following. I will
use both the Hilbert Space formulism and the function $\psi$ that some might
be more comfortable with. Both derivations will end up with identical
conclusions.

First, start with the eigenvalue expression $\hat{A}\psi=A\psi$ and multiply
both sides by the complex conjugate, like so:

$\psi^{*}\hat{A}\psi=\psi^{*}A\psi=A\psi^{*}\psi$

In Hilbert Space, the same can be achieved by multiplying the eigenvalue
expression $\hat{A}\ket{\psi(t)}=A\ket{\psi(t)}$ with a bra vector from the
left, so that the operator is ‘sandwiched’:

$\bra{\psi(t)}\hat{A}\ket{\psi(t)}=\bra{\psi(t)}A\ket{\psi(t)}$

For the typical function form, we are going to multiply both sides by the
‘operator’ $\int_{-\infty}^{\infty}dx$, in order to get:

$\int_{-\infty}^{\infty}\psi^{*}\hat{A}\psi
dx=\int_{-\infty}^{\infty}A\psi^{*}\psi dx$ (14)

We now introduce a fundamental postulate of QM, namely, the Born Rule. The
Born Rule states that $\rho=\psi^{*}\psi$ where $\rho$ represents the
probability density for continuous $\psi$. From probability theory, we know
that if a probability density is integrated over all possible states, it must
be equal to 1 (the renormalization condition):

$\int_{-\infty}^{\infty}\psi^{*}\psi dx=1$ (15)

In other words, the probability of something happening is certain (= 1) if all
possible outcomes occur. The equivalent renormalization condition in Hilbert
Space is that $\innerproduct{\psi(t)}{\psi(t)}=1$.555Both renormalization
conditions can be shown to be equivalent. From Appendix A, we have the
important completeness theorem, which says:  $I=\int dx\ket{x}\bra{x}$ Since
$I$ is just the identity operator (operator where $I\hat{A}=\hat{A}$ or
$I\ket{\psi}=\ket{\psi}$, just like multiplying something by the number ‘1’),
we can do the following operations:
$\innerproduct{\psi(t)}{\psi(t)}=\bra{\psi(t)}I\ket{\psi(t)}=\int\innerproduct{\psi(t)}{x}\innerproduct{x}{\psi(t)}dx$
Recall from before that we have said that
$\innerproduct{x}{\psi(t)}=\psi(x,t)$, so we get:
$\innerproduct{\psi(t)}{\psi(t)}=\int\psi^{*}\psi dx$

Equation (14) is defined to be the expectation value of the observable given
by operator $\hat{A}$, as can be seen from the similarity of both:

$\langle
f(x)\rangle=\int_{-\infty}^{\infty}f(x)\rho~{}dx\Longleftrightarrow\int_{-\infty}^{\infty}\psi^{*}\hat{A}\psi
dx=\int_{-\infty}^{\infty}A\psi^{*}\psi dx,$

where you can see that the eigenvalue $A$ takes on the role of $f(x)$ in the
above expression and $\psi^{*}\psi$ is the probability density. Please keep in
mind that there could be many different values you can measure for observable
$A$, so it makes sense it is equivalent to the function $f(x)$. Therefore the
expectation value of observable $A$ is given by

$\langle A\rangle=\int_{-\infty}^{\infty}\psi^{*}\hat{A}\psi
dx=\bra{\psi(t)}\hat{A}\ket{\psi(t)}.$ (16)

The expectation values measured in the laboratory happen to be the exact ones
predicted by the above equation. The expectation value expression was strictly
derived from the eigenvalue problem and the Born Rule.

###  2.4 Review of Dirac Delta Functional

The Dirac delta functional is important to QM for the renormalization of all
the basis vectors. For any Hilbert Space, one can find an orthogonal set of
bases to represent it. But we can do one better - one can always find an
orthonormal basis. An orthonormal basis is a basis where not only are all the
basis vectors orthogonal, but they are also normalized to unit length 1:

$\ket{A}_{normalized}=\frac{\ket{A}}{\sqrt{\innerproduct{A}{A}}}$ (17)

Any two generic vectors $\ket{A}$ and $\ket{A^{\prime}}$ in Hilbert Space are
orthonormal if and only if they satisfy

$\innerproduct{A}{A^{\prime}}=\delta_{AA^{\prime}}=\begin{cases}1\quad&\text{if}\,\ket{A}=\ket{A^{\prime}}\\\
0\quad&\text{if}\,\ket{A}\neq\ket{A^{\prime}}\\\ \end{cases}$

where $\delta_{AA^{\prime}}$ is the Kronecker delta. The Kronecker delta
functional is however only appropriate when dealing with strictly discrete
systems. Many problems in QM, however, can be either discrete or continuous.
We therefore need a functional that can encompass both discrete and continuous
systems to describe an orthonormal basis. This is done by using the Dirac
delta functional666Please note that the $A$ here is a generic placeholder
variable with arbitrary units. It could be a placeholder for position $q$,
momentum $p$, or some other measurable quantity. So you could, for example,
have $\innerproduct{q}{q^{\prime}}=\delta(q-q^{\prime})$ for position and
$\innerproduct{p}{p^{\prime}}=\delta(p-p^{\prime})$ for momentum, etc.:

$.\innerproduct{A}{A^{\prime}}=\delta(A-A^{\prime})$ (18)

The Dirac delta functional is a generalization of the Kronecker delta ([20]).
The Dirac delta functional obeys the following properties:

$\delta(A-A^{\prime})=0~{}\text{if}~{}A\neq A^{\prime}.$

$\int_{a}^{b}\delta(A-A^{\prime})~{}dA^{\prime}=1~{}\text{if}~{}a\leq A\leq
b.$

The functional is zero everywhere but at $A^{\prime}$ ([20]).

The Dirac delta functional should not be treated as a regular function. It has
its own unique properties and form, distinct from how functions behave. For
example, the area under a single point when $A=A^{\prime}$ should be zero (the
area under a point is zero). But for the Dirac delta functional, it is
precisely 1. This and other properties of this functional highlight the fact
that it should not by treated like the functions you have seen up to this
point. It is a functional, not a function. It is, however, fundamental to the
renormalization of vectors in Hilbert Space.

Other important properties of the functional we will utilize in the text of
this article include the following:

$\int\delta(A-A^{\prime})f(A^{\prime})dA^{\prime}=f(A).$ (19)

$\delta(A-A^{\prime})=\delta(A^{\prime}-A).$ (20)

$\frac{d}{dA}\delta(A-A^{\prime})=-\frac{d}{dA^{\prime}}\delta(A-A^{\prime}).$
(21)

Some useful relations for QM:

$\frac{d^{n}}{dA^{n}}\delta(A-A^{\prime})=\int\frac{d\omega}{2\pi}(-i\omega)^{n}e^{-i\omega(A-A^{\prime})}.$
(22)
$\int\frac{d^{n}}{dA^{n}}\delta(A-A^{\prime})f(A^{\prime})~{}dA^{\prime}=(-1)^{n}\frac{d^{n}f(A)}{dA^{n}}.$
(23)

$\delta(\alpha A)=\frac{\delta(A)}{|\alpha|},$ (24)

where $\alpha$ is a constant.

###  2.5 Summary of Quantum Mechanics and its Postulates

The postulates of QM can be reviewed as follows ([20]; [4]):

  1. 1.

The state of a quantum system (particle or collection of particles) is
represented by a vector $\ket{\psi}$ in complex Hilbert Space.

  2. 2.

For any observable $A$, there is an associated Hermitian operator $\hat{A}$
with eigenvalue problem $\hat{A}\ket{A}=A\ket{A}$ where $\ket{A}$ is the
specific eigenket and $A$ is the eigenvalue which is measured in the
laboratory. Examples of common observables are position of a particle $q$ and
momentum $p$. The variables position $q$ and momentum $p$ are represented by
Hermitian operators $\hat{q}$ and $\hat{p}$ which obey the following
properties: $\bra{q}\hat{q}\ket{q^{\prime}}=q\delta(q-q^{\prime})$ and
$\bra{q}\hat{p}\ket{q^{\prime}}=-i\hbar\frac{d}{dq}\delta(q-q^{\prime})$.

  3. 3.

Born Rule: For a continuous distribution, the probability density for a
particle is $\rho=\psi^{*}\psi$. Equivalently in Dirac Notation, it can be
written that the probability density of observing $A$ is
$\rho(A)\propto\innerproduct{\psi}{A}\innerproduct{A}{\psi}=\|\innerproduct{A}{\psi}\|^{2}$
where “$\propto$” is the proportionality sign. More precisely we could say
that the Born Rule dictates that

$\rho(A)=\frac{\innerproduct{\psi}{A}\innerproduct{A}{\psi}}{\innerproduct{\psi}{\psi}}$

As long as the above equation meets the renormalization condition
$\innerproduct{\psi}{\psi}=1$ (see equation (15) and associated footnote), the
Born Rule in its final form is often written as

$\rho(A)=\innerproduct{\psi}{A}\innerproduct{A}{\psi}$

For a discrete system, the Born Rule is written as:

$P(A)=\innerproduct{\psi}{A}\innerproduct{A}{\psi},$

assuming renormalization, where $P$ is the probability. For both the
continuous and discrete cases, the state of the system collapses from the
initial state $\ket{\psi(t)}$ to the final state $\ket{A}$ whenever a
measurement of the quantum system is undertaken. Under the Copenhagen
Interpretation of QM, the collapse of the wavefunction $\ket{\psi}$ to the
specific eigenket $\ket{A}$ is instantaneous and irreversible. If $A$ is
observed in experiment, the wavefunction collapses to the associated eigenket
$\ket{A}$.

  4. 4.

The state space of a composite system is the tensor product of the subsystem’s
state spaces

There are a couple other important things to say about the above overview of
QM. Firstly, although not included as a postulate, the expectation value
relationship (16) plays a critical role inside of QM. They are of fundamental
importance as we reviewed in section 2.3.

Not discussed thoroughly is the centrality of the Schrödinger equation to QM.
The state of the quantum system $\ket{\psi}$ obeys the Schrödinger equation:

$i\hbar\frac{\partial}{\partial
t}\ket{\psi}=\Big{[}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial
q^{2}}+V(q)\Big{]}\ket{\psi}=\hat{H}\ket{\psi}$ (25)

where $\hat{H}$ is the quantum Hamiltonian that bears great similarity to the
classical Hamiltonian777That is, a classical Hamiltonian where the potential
energy has no velocity or time dependence, which takes on the form
$H(q,p)=\frac{p^{2}}{2m}+V(q)$. Note that $p$ is momentum and $V(q)$ is a one
dimensional potential energy., except that the momentum and position are
replaced by the momentum and position operators of QM:

$H(q,p)=\frac{p^{2}}{2m}+V(q)\Longrightarrow\hat{H}(\hat{q},\hat{p})=\frac{\hat{p}^{2}}{2m}+V(\hat{q})=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial
q^{2}}+V(\hat{q})$

Of important note, the Hamiltonian operator which acts on the wavefunction in
equation (25) is Hermitian ([20]). Since it is Hermitian, its eigenvalues will
always be real instead of complex or imaginary, as required for numbers
measured in the laboratory.

There will be a slight adjustment of notation from the previous couple
sections. Although it was convenient to previously write
$\hat{A}\ket{\psi}=A\ket{\psi}$ as the eigenvalue problem, it is more precise
to write the eigenvalue problem as $\hat{A}\ket{A}=A\ket{A}$. As described
before, for an operator $\hat{A}$ there are specific eigenvalues and eigenkets
that satisfy the above equation. Although $\ket{\psi}$ might be used to
describe the specific eigenket in the context of the eigenvalue problem,
$\ket{\psi}$ is often the generic state vector of the quantum system, so this
might lead to some confusion. To highlight the fact that only specific
eigenkets $\ket{A}$ satisfy this relationship, $\hat{A}\ket{A}=A\ket{A}$ will
be used from now on.

Two very important eigenvalue problems are $\hat{q}\ket{q}=q\ket{q}$ for
position $q$ and $\hat{p}\ket{p}=p\ket{p}$ for momentum $p$. The expectation
values of the position and momentum measured in the laboratory will be the
eigenvalues $q$ and $p$. For any observable $A$, there is a relationship
$\hat{A}\ket{A}=A\ket{A}$ for which combined with the Born Rule axiom, you can
deduce the expectation value of $A$ (as described in section 2.3).

One of the initially confusing things about the Dirac notation is that the
symbols inside the bras and kets are just labels. So you might see vector bras
and kets with labels such as $\ket{1st}$, $\ket{2nd}$, $\ket{I}$, $\ket{II}$,
$\ket{\omega=2.5}$, etc. All of these have valid notation, as the labels
simply describe the associated vector, and that is all. Usually, the
eigenvalue is used to label the associated eigenket. Hence the eigenvalue
‘$A$’ being associated with eigenket $\ket{A}$ in the eigenvalue problem. It
is why in equation (2), we could make a statement such as
$\ket{v}+\ket{w}=\ket{v+w}$, because both $v$ and $w$ are just labels.

Finally, the collapse of the wavefunction describes one of the ongoing
mysteries of QM. It can be seen that the Schrödinger equation is fully
deterministic from equation (25), but the collapse of the wavefunction is (as
far as we know) indeterministic and random. There is no known formula to
describe how the wavefunction collapses. More will be said about wavefunction
collapse in section 5.3 of this article, in the context of QM and CM.

###  2.6 The Density Operator

Before we delve straight into KvNM, it is critical to briefly look at one more
tool that will be of use to us in this work. Instead of representing the state
of a system as a wavefunction vector in Hilbert Space $\ket{\psi}$, we can
instead use something called a density matrix for certain systems, represented
by $\hat{\rho}(t)$ (not to be confused with the previously used $\rho$ for
probability density.) The density matrix is defined by

$\hat{\rho}(t)=\sum_{j}p_{j}\ket{\psi_{j}(t)}\bra{\psi_{j}(t)},$ (26)

for the discrete case. The density matrix is used when you are not certain of
the configuration of your system at some point in time. You can represent it
as statistical ensemble of possible wavefunction configurations
$\ket{\psi_{j}(t)}$, each with probability of occurrence $p_{j}$. Note with
probability, we must have

$\sum_{j}p_{j}=1.$

We will see an example of the density matrix used in action later in section
5.3. Density matrices are good for describing the above described so-called
“mixed states” of a system (with the “pure states” being the individual
wavefunctions $\ket{\psi_{j}}$). If $j=1$ in (26) the density matrix
represents a pure state, but if it is greater than 1 it represents a mixed
state.

The probability of measuring $A$, or $P(A)$, is given by the density matrix
formulation as follows:

$P(A)=\Tr[\hat{\rho}(t)\ket{a}\bra{a}],$ (27)

where $\Tr[...]$ refers to the trace of the matrix. You can represent the
trace by

$\Tr[...]=\sum_{n}\bra{n}...\ket{n},$ (28)

where $\ket{n}$ are arbitrary basis of your choice. This is the density matrix
version of the Born Rule. It is easy to see that this is fully equivalent to
the Born Rule, for instance, when we plug (26) with $j=1$ into (27):

$P(A)=\bra{a}\ket{\psi_{1}}\bra{\psi_{1}}\ket{a}\bra{a}\ket{a}=|\bra{A}\ket{\psi_{1}(t)}|^{2},$
(29)

since $\bra{a}\ket{a}=1$ assuming orthonormality. The expectation value of
operator $\hat{A}$ is given by

$\langle A\rangle=\Tr[\hat{A}\hat{\rho}(t)],$ (30)

where $\hat{A}\hat{\rho}(t)$ form a matrix and $\Tr[...]$ refers to the trace
of the matrix. You can reduce (30) to (16) just by applying the trace formula
(28), easily demonstrated for the pure state $j=1$:

$\Tr[\hat{A}\hat{\rho}(t)]=\sum_{n}\bra{n}\hat{A}\ket{\psi_{1}(t)}\bra{\psi_{1}(t)}\ket{n}=\sum_{n}\bra{\psi_{1}(t)}\ket{n}\bra{n}\hat{A}\ket{\psi_{1}(t)}.$

Applying resolution of identity (Theorem II of Appendix A) we get the standard
expectation value formula:

$\Tr[\hat{A}\hat{\rho}(t)]=\bra{\psi_{1}(t)}\hat{A}\ket{\psi_{1}(t)}=\langle
A\rangle.$

Other useful properties of the density matrix:

  1. 1.

$\Tr[\hat{\rho}(t)]=1$, which is the normalization condition.

  2. 2.

$\Tr[\hat{\rho}(t)^{2}]=1$ for pure state.

  3. 3.

$\Tr[\hat{\rho}(t)^{2}]<1$ for mixed state.

  4. 4.

$\hat{\rho}(t)=\hat{\rho}(t)^{\dagger},$ in other words, $\hat{\rho}(t)$ is
Hermitian.

(Based on [16].) We will utilize the density operator in future sections.

##  3 Introduction to KvN Mechanics

Koopman-von Neumann Mechanics is unique because it is a form of classical
statistical mechanics that truly takes on a lot of the mathematical structure
of QM ([10]; [22]; [4]). It was invented in the 1930’s by Bernard Koopman and
John von Neumann. Subsequently forgotten for some time by most physicists, its
resurgence raises intriguing questions about the nature of QM to CM. Just like
with QM, we have a set of operators that act on vectors in Hilbert Space in
order to retrieve eigenvalues of observables ([10]; [22]; [4]). In order to
have a proper Hilbert Space, we first must define the inner product of that
Hilbert Space ([10]):

$\innerproduct{\psi}{\phi}=\int
dA\innerproduct{\psi}{A}\innerproduct{A}{\phi}=\int dA~{}\psi^{*}(A)\phi(A).$

CM always has distinct trajectories that particles travel along. In QM, we
have the Heisenberg Uncertainty Principle which states you cannot have both a
well defined position of a particle and traveling momentum (or velocity) at
the same time. Under the Copenhagen Interpretation, this is not due to some
principle of measurement, rather, this is a fundamental property of the fabric
of reality ([20]). Reality forbids quantum particles from having a precise
momentum and position simultaneously. In QM, trajectories could be viewed as
being “fuzzy” instead of well defined and localized like in CM. You can no
longer draw nice curves to describe how a particle travels in space under QM.

Because of this fact, both position and momentum would have a common set of
eigenstates in KvN Hilbert Space ([5]; [15]). The state vectors for KvN would
therefore have the following properties ([10]):

$\ket{A}=\ket{q,p}=\ket{q}\otimes\ket{p},$

$\hat{q}\ket{q,p}=q\ket{q,p},$

$\hat{p}\ket{q,p}=p\ket{q,p},$

$\innerproduct{q,p}{q^{\prime},p^{\prime}}=\delta(q-q^{\prime})\delta(p-p^{\prime}),$

$\int dp~{}dq\ket{q,p}\bra{q,p}=I.$ (31)

The observable $A$’s eigenket $\ket{A}$ is the KvN ket $\ket{q,p}$ because of
the common set of eigenstates between $q$ and $p$. Because it contains
information about $q$ and $p$, this is providing information in phase space
instead of the previously discussed configuration space ([10]). Previously for
configuration space we had $\innerproduct{q}{\psi(t)}=\psi(q,t)$,888Which can
be generalized into
$\innerproduct{q_{1},q_{2},q_{3},...,q_{n}}{\psi(t)}=\psi(q_{1},q_{2},q_{3},...,q_{n},t)$
for configuration space. whereas here we will have
$\innerproduct{q,p}{\psi(t)}=\psi(q,p,t)$ in phase space.

In QM, we have the Schrödinger equation (25) to dictate to us the time
evolution of the quantum particle described by $\ket{\psi}$. To estimate the
probability of an observable, we utilize
$\rho(A,t)=\psi^{*}(A,t)\psi(A,t)=|\innerproduct{A}{\psi(t)}|^{2}$ (Born
Rule). In CM, we have the following equation known as the Louiville Equation
to describe classical trajectories:

$\frac{\partial\rho}{\partial t}+\sum\Big{(}\frac{\partial H}{\partial
p_{j}}\frac{\partial\rho}{\partial q_{j}}-\frac{\partial H}{\partial
q_{j}}\frac{\partial\rho}{\partial p_{j}}\Big{)}=0$ (32)

The Louiville Equation is important because it tells you the probability
density $\rho$ of finding a particle in observable point $\omega=(q,p)$ in
phase space. In other words, it defines the probabilistic time evolution of an
ensemble of particles. We can define the Louiville operator to be ([14]):

$\hat{L}=\sum\Big{(}-i\frac{\partial H}{\partial
p_{j}}\frac{\partial}{\partial q_{j}}+i\frac{\partial H}{\partial
q_{j}}\frac{\partial}{\partial p_{j}}\Big{)}$ (33)

By doing so, we can rewrite equation (32) in operational terms:

$i\frac{\partial\rho}{\partial t}=\hat{L}\rho$ (34)

Notice each term in equation (33) has been multiplied by the imaginary unit
$i$. Just like in QM, we postulate a Born Rule for this probability density.
If $\rho(A,t)=\psi^{*}\psi$ (where $A$ is the observable), then we can easily
show that 999Demonstrated as follows. If $\rho=\psi^{*}\psi$, then for a
simple case with index $j=1$ we plug it into equation (34):
$i\frac{\partial\psi^{*}\psi}{\partial t}+i\frac{\partial H}{\partial
p}\frac{\partial\psi^{*}\psi}{\partial q}-i\frac{\partial H}{\partial
q}\frac{\partial\psi^{*}\psi}{\partial p}=0$
$i\frac{\partial\psi^{*}}{\partial t}\psi+i\frac{\partial\psi}{\partial
t}\psi^{*}+i\frac{\partial H}{\partial p}\frac{\partial\psi^{*}}{\partial
q}\psi+i\frac{\partial H}{\partial p}\frac{\partial\psi}{\partial
q}\psi^{*}-i\frac{\partial H}{\partial q}\frac{\partial\psi^{*}}{\partial
p}\psi-i\frac{\partial H}{\partial q}\frac{\partial\psi}{\partial
p}\psi^{*}=0$ $\Big{[}i\frac{\partial\psi^{*}}{\partial t}+i\frac{\partial
H}{\partial p}\frac{\partial\psi^{*}}{\partial q}-i\frac{\partial H}{\partial
q}\frac{\partial\psi^{*}}{\partial
p}\Big{]}\psi+\Big{[}i\frac{\partial\psi}{\partial t}+i\frac{\partial
H}{\partial p}\frac{\partial\psi}{\partial q}-i\frac{\partial H}{\partial
q}\frac{\partial\psi}{\partial p}\Big{]}\psi^{*}=0$ Since generally neither
$\psi$ nor $\psi^{*}$ are zero, this means the expressions in square brackets
must equal zero. The bracketed expressions are, of course, equation (35) and
its complex conjugate.

$i\frac{\partial\psi}{\partial t}=\hat{L}\psi$ (35)

This equation, of course, bears a striking resemblance to equation (25), the
Schrödinger equation in its operatorial form. The classical wavefunction
$\psi$ here is interpreted as a probability density amplitude, like under most
interpretations of QM ([20]). Just like with QM, KvN depends on the generic
state vector $\ket{\psi(t)}$ containing statistical information about the
system. Colloquially, just as before, $\psi\Longrightarrow\ket{\psi(t)}$.

An important property of KvN Mechanics is that the Louiville operator
$\hat{L}$ is Hermitian, just like the Hamiltonian operator of the Schrödinger
equation. The norm of $\innerproduct{\psi(t)}{\psi(t)}=\int
d\omega\psi^{*}(\omega)\psi(\omega)$ is therefore conserved, further
strengthening the Born’s Rule postulated for KvN Mechanics ([14]).

The Axioms of KvN Mechanics can be summarized as follows (based off [4]):

  1. 1.

The state of a system is represented by a vector $\ket{\psi}$ in a complex
Hilbert Space.

  2. 2.

For any observable $A$, there is an associated Hermitian operator $\hat{A}$
with eigenvalue problem $\hat{A}\ket{A}=A\ket{A}$ where $\ket{A}$ is the
specific eigenket and $A$ is the eigenvalue measured in the laboratory.

  3. 3.

Born Rule: The probability of measuring $A$ is given by
$P(A)=|\innerproduct{A}{\psi(t)}|^{2}$. The wavefunction $\ket{\psi}$
describing the system instantaneously collapses into the eigenket $\ket{A}$
associated with the observed eigenvalue $A$.

  4. 4.

The state space of a composite system is the tensor product of the subsystems’
state space

Notice, these are exactly the same postulates of QM in section 2.5, just with
anything “quantum” removed ([4]). The formula for the expectation value of the
system is still the same, since Axioms 2 and 3 remain unchanged, and therefore
the derivation in section 2.3 is the same. The very important expectation
value rule (16) therefore still applies in KvNM.

Given these four ‘universal’ postulates, one can derive both QM and KvNM from
the Ehrenfest Theorems. The Ehrenfest Theorems are as follows:

$\frac{d}{dt}\langle q\rangle=\frac{\langle p\rangle}{m},$ (36)
$\frac{d}{dt}\langle p\rangle=\langle-V^{\prime}(q)\rangle,$ (37)

and are usually considered strictly within the context of QM.

Under QM, the fabric of reality forbids the mutual existence of a well defined
position and momentum. Under CM, we can know the position and momentum to an
arbitrary precision. This difference is encoded with the commutator
relationship. In section 5.3, we will show that the Canonical Commutator of QM
(8) directly produces the Heisenberg Uncertainty Principle. We can also write
a commutator relationship for CM under the axioms of KvNM. We can say:

$[\hat{q},\hat{p}]=0$ (38)

I shall call this the “Classical Commutator”. This commutator relationship
would produce no uncertainty relationship between the classical position $q$
and momentum $p$. Ergo, position and momentum can be arbitrarily known.

For QM, one can show that the Canonical Commutator combined with the Ehrenfest
Theorem produces the traditional Schrödinger equation with quantum
Hamiltonian. We will demonstrate this as follows. According to the common set
of Axioms, we can say that the expectation value of observable $A$ is simply
$\bra{\psi(t)}\hat{A}\ket{\psi(t)}$ (16). Therefore we can easily say that

$\langle q\rangle=\bra{\psi(t)}\hat{q}\ket{\psi(t)}$

$\langle p\rangle=\bra{\psi(t)}\hat{p}\ket{\psi(t)}$

Then from the Ehrenfest Theorems one can carry out the following operations:

$\frac{d}{dt}\bra{\psi(t)}\hat{q}\ket{\psi(t)}=\frac{1}{m}\bra{\psi(t)}\hat{p}\ket{\psi(t)},$
$\frac{d}{dt}\bra{\psi(t)}\hat{p}\ket{\psi(t)}=\bra{\psi(t)}-V^{\prime}(\hat{q})\ket{\psi(t)}.$

We are using the Schrödinger picture of QM, so we assume time dependence
solely resides in $\ket{\psi(t)}$ and any operators are independend of time.
101010This is the opposite of the Heisenberg picture, which assumes no time
dependence in the wavefunction vector ket and puts all the time dependence
inside the operators! Using the chain rule of calculus, we write:

$\bra{\frac{d\psi(t)}{dt}}\hat{q}\ket{\psi(t)}+\bra{\psi(t)}\hat{q}\ket{\frac{d\psi(t)}{dt}}=\frac{1}{m}\bra{\psi(t)}\hat{p}\ket{\psi(t)},$
$\bra{\frac{d\psi(t)}{dt}}\hat{p}\ket{\psi(t)}+\bra{\psi(t)}\hat{p}\ket{\frac{d\psi(t)}{dt}}=\bra{\psi(t)}-V^{\prime}(\hat{q})\ket{\psi(t)}.$
(39)

Stone’s Theorem (see Appendix B) dictates that there exists a unique
Schrödinger-like equation $i\frac{\partial}{\partial
t}\ket{\psi}=\hat{G}\ket{\psi}$ where $\hat{G}$ is the Hermitian generator of
motion in operator form. We deduce there must be a $\hat{G}$ that governs the
evolution of CM, just as the Hamiltonian operator does for QM. The only thing
is to figure out what this generator $\hat{G}$ is.

For QM, we can derive what the generator of motion is by using the commutator
relationships. We can take Stone’s Theorem and then derive:

$\ket{\frac{d\psi(t)}{dt}}=\frac{\hat{H}}{i\hbar}\ket{\psi(t)}\iff\bra{\frac{d\psi(t)}{dt}}=-\bra{\psi(t)}\frac{\hat{H}}{i\hbar}$

Which means:

$-\bra{\psi(t)}\frac{\hat{H}}{i\hbar}\hat{q}\ket{\psi(t)}+\bra{\psi(t)}\hat{q}\frac{\hat{H}}{i\hbar}\ket{\psi(t)}=\frac{1}{m}\bra{\psi(t)}\hat{p}\ket{\psi(t)}$
$-\bra{\psi(t)}\frac{\hat{H}}{i\hbar}\hat{p}\ket{\psi(t)}+\bra{\psi(t)}\hat{p}\frac{\hat{H}}{i\hbar}\ket{\psi(t)}=\bra{\psi(t)}-V^{\prime}(\hat{q})\ket{\psi(t)}$

You can expand the vector ‘sandwich’ and rewrite it in terms of a commutator
to get:

$\bra{\psi(t)}-\hat{H}\hat{q}+\hat{q}\hat{H}\ket{\psi(t)}=-\bra{\psi(t)}[\hat{H},\hat{q}]\ket{\psi(t)}=\frac{i\hbar}{m}\bra{\psi(t)}\hat{p}\ket{\psi(t)}$
$\bra{\psi(t)}-\hat{H}\hat{p}+\hat{p}\hat{H}\ket{\psi(t)}=-\bra{\psi(t)}[\hat{H},\hat{p}]\ket{\psi(t)}=\bra{\psi(t)}-i\hbar
V^{\prime}(\hat{q})\ket{\psi(t)}$

Since the Ehrenfest Theorems are true regardless of the particular state of
the system $\ket{\psi(t)}$ (i.e., they are true for any state you might find),
we can remove the wavefunction vector from both sides to get:

$-[\hat{H},\hat{q}]=\frac{i\hbar}{m}\hat{p}$ $[\hat{H},\hat{p}]=i\hbar
V^{\prime}(\hat{q})$ (40)

We will make the assumption that $\hat{H}=H(\hat{q},\hat{p})$ and utilize
Theorem VI from Appendix A. By this Theorem, we have $\hat{q}=\hat{A}_{1}$,
$\hat{p}=\hat{A}_{2}$, and $H(\hat{q},\hat{p})=f(\hat{A}_{1},\hat{A}_{2})$.
Therefore:

$[\hat{H},\hat{q}]=[H(\hat{q},\hat{p}),\hat{q}]=\sum_{k=1}^{n=2}[\hat{A}_{k},\hat{q}]\frac{\partial
H(\hat{A}_{1},\hat{A}_{2})}{\partial\hat{A}_{k}}=[\hat{q},\hat{q}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{q}}+[\hat{p},\hat{q}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{p}}$

$[\hat{H},\hat{p}]=[H(\hat{q},\hat{p}),\hat{p}]=\sum_{k=1}^{n=2}[\hat{A}_{k},\hat{p}]\frac{\partial
H(\hat{A}_{1},\hat{A}_{2})}{\partial\hat{A}_{k}}=[\hat{q},\hat{p}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{q}}+[\hat{p},\hat{p}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{p}}$

A commutator with itself will always equal zero:

$[\hat{q},\hat{q}]=\hat{q}\hat{q}-\hat{q}\hat{q}=0$
$[\hat{p},\hat{p}]=\hat{p}\hat{p}-\hat{p}\hat{p}=0$

Ergo, the above expressions based in Theorem VI reduce to:

$[\hat{H},\hat{q}]=[\hat{p},\hat{q}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{p}}$

$[\hat{H},\hat{p}]=[\hat{q},\hat{p}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{q}}$

Plugging these expressions into (40) gives us:

$-[\hat{p},\hat{q}]\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{p}}=\frac{i\hbar}{m}\hat{p}$
$[\hat{q},\hat{p}]\frac{\partial H(\hat{q},\hat{p})}{\partial\hat{q}}=i\hbar
V^{\prime}(\hat{q})$ (41)

From the Canonical Commutator we know that $[\hat{q},\hat{p}]=i\hbar$.111111We
will also use the fact that $[\hat{q},\hat{p}]=-[\hat{p},\hat{q}]$. It’s not
hard to prove, just takes a couple lines of algebra:
$[\hat{q},\hat{p}]=\hat{q}\hat{p}-\hat{p}\hat{q}=-(-\hat{q}\hat{p}+\hat{p}\hat{q})=-(\hat{p}\hat{q}-\hat{q}\hat{p})=-[\hat{p},\hat{q}]$
This step is absolutely critical. It is in fact the step that is going to
distinguish QM from classical KvNM - the usage of the Canonical Commutator. We
are going to plug in the Canonical Commutator to the above expressions, so
that we will get:

$\frac{\partial H(\hat{q},\hat{p})}{\partial\hat{p}}=\frac{\hat{p}}{m}$
$\frac{\partial
H(\hat{q},\hat{p})}{\partial\hat{q}}=\frac{dV(\hat{q})}{d\hat{q}}$

Theorem VII of Appendix A will be useful, as we can now translate this
expression with operators into more familiar analytical functions in order to
do operations with derivations. We will have the following analytical
functions as a result of Theorem VII:

$\frac{\partial H(q,p)}{\partial p}=\frac{p}{m}$ $\frac{\partial
H(q,p)}{\partial q}=\frac{dV(q)}{dq}$

Some basic calculus reveals:

$\int\frac{\partial H(q,p)}{\partial
p}dp=H(q,p)=\int\frac{p}{m}dp=\frac{p^{2}}{2m}+C(q)$

where $C(q)$ is some function solely in terms of $q$ so that it disappears
when the partial with respect to $p$ is taken. Taking the partial derivative
with respect to $q$ of the above expression reveals:

$\frac{\partial H(q,p)}{\partial q}=\frac{dC(q)}{dq}=\frac{dV(q)}{dq}$

Ergo, $C(q)=V(q)$ so that the final solution is
$H(\hat{q},\hat{p})=\frac{\hat{p}^{2}}{2m}+V(\hat{q})$, which happens to be
equivalent to the quantum Hamiltonian. Therefore, plugging this solution of
$\hat{H}$ into the above version of Stone’s Theorem we reclaim the Schrödinger
equation(25):

$i\hbar\frac{\partial}{\partial
t}\ket{\psi}=\Big{[}\frac{\hat{p}^{2}}{2m}+V(\hat{q})\Big{]}\ket{\psi}$

The same series of steps can be carried out to derive KvNM. The only
difference in the derivation is that we will assume the Classical Commutator
instead of the Canonical Commutator. Once again we start with the Ehrenfest
Theorems and follow a similar series of steps. Starting from (39), we will
utilize Stone’s Theorem as $i\frac{\partial}{\partial
t}\ket{\psi}=\hat{K}\ket{\psi}$ to get:

$-\bra{\psi(t)}\frac{\hat{K}}{i}\hat{q}\ket{\psi(t)}+\bra{\psi(t)}\hat{q}\frac{\hat{K}}{i}\ket{\psi(t)}=\frac{1}{m}\bra{\psi(t)}\hat{p}\ket{\psi(t)}$
$-\bra{\psi(t)}\frac{\hat{K}}{i}\hat{p}\ket{\psi(t)}+\bra{\psi(t)}\hat{p}\frac{\hat{K}}{i}\ket{\psi(t)}=\bra{\psi(t)}-V^{\prime}(\hat{q})\ket{\psi(t)}$

Following a similar series of steps, we can derive that:

$-[\hat{K},\hat{q}]=\frac{i}{m}\hat{p}$
$[\hat{K},\hat{p}]=iV^{\prime}(\hat{q})$ (42)

Now we run into a problem. If we assume $\hat{K}=K(\hat{q},\hat{p})$ as
before, then we will derive once again (41). However, now we will utilize the
Classical Commutator instead of the Canonical Commutator, since we want well
defined classical trajectories. By imposing the Classical Commutator
$[\hat{q},\hat{p}]=0$, we run into a contradiction, namely:

$-[\hat{p},\hat{q}]\frac{\partial
K(\hat{q},\hat{p})}{\partial\hat{p}}=(0)\frac{\partial
K(\hat{q},\hat{p})}{\partial\hat{p}}=0=\frac{i}{m}\hat{p}\Longrightarrow p=0$
$[\hat{q},\hat{p}]\frac{\partial
K(\hat{q},\hat{p})}{\partial\hat{q}}=(0)\frac{\partial
K(\hat{q},\hat{p})}{\partial\hat{q}}=0=iV^{\prime}(\hat{q})\Longrightarrow\frac{dV}{dq}=0$

Obviously, this cannot be true for all systems, because many classical systems
definitely have a nonzero momentum. This contradiction between theory and
observation suggests that we must approach this from a different angle. One
such angle is to propose two new operators $\hat{\theta}$ and $\hat{\lambda}$
([4]). These operators obey the eigenvalue problems,
$\hat{\theta}\ket{\theta}=\theta\ket{\theta}$ and
$\hat{\lambda}\ket{\lambda}=\lambda\ket{\lambda}$, by postulate 2 of the
‘universal axioms’, but are not tied to any direct physically observable
features of reality (as noted in [18]).121212Even though these operators are
not directly physically observable, their existence has been seen through
indirect means. These operators are the famous Bopp Operators of the phase-
space formulation of QM ([5]). We will explore the Wigner Distribution in
section 6. They serve as an analogue to Lagrange Multipliers ([15]), serving
to constrain the realm of possible behavior to a very specific subset.

For KvNM, we will postulate the following behavior for these operators:

$[\hat{q},\hat{\theta}]=[\hat{p},\hat{\lambda}]=i$
$[\hat{q}_{c},\hat{p}_{c}]=[\hat{q},\hat{\lambda}]=[\hat{p},\hat{\theta}]=[\hat{\theta},\hat{\lambda}]=0$
(43)

where $i$ is the imaginary unit. Equations (43) make up what is known as the
Koopman-von Neumann Algebra. If we do this, then we can assume
$\hat{K}=K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})$. By Theorem VI of
Appendix A again, we will get:

$[K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda}),\hat{q}]=\sum_{k=1}^{n=4}[\hat{A}_{k},\hat{q}]\frac{\partial
K(\hat{A}_{1},\hat{A}_{2},\hat{A}_{3},\hat{A}_{4})}{\partial\hat{A}_{k}}=[\hat{q},\hat{q}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{q}}+[\hat{p},\hat{q}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{p}}$

$+[\hat{\theta},\hat{q}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\theta}}+[\hat{\lambda},\hat{q}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\lambda}}$

$[K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda}),\hat{p}]=\sum_{k=1}^{n=4}[\hat{A}_{k},\hat{p}]\frac{\partial
K(\hat{A}_{1},\hat{A}_{2},\hat{A}_{3},\hat{A}_{4})}{\partial\hat{A}_{k}}=[\hat{q},\hat{p}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{q}}+[\hat{p},\hat{p}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{p}}$

$+[\hat{\theta},\hat{p}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\theta}}+[\hat{\lambda},\hat{p}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\lambda}}$

Three terms from each expression will disappear because of the KvN Algebra
above and the Classical Commutator. The equations of motion will end up being:

$[\hat{\theta},\hat{q}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\theta}}=-\frac{i}{m}\hat{p}$
$[\hat{\lambda},\hat{p}]\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\lambda}}=iV^{\prime}(\hat{q})$
(44)

Once we substitute in $[\hat{q},\hat{\theta}]=[\hat{p},\hat{\lambda}]=i$ we
will simply have:

$\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\theta}}=\frac{\hat{p}}{m}$
$\frac{\partial
K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})}{\partial\hat{\lambda}}=-V^{\prime}(\hat{q})$

Utilizing Theorem VII again, we get the differential equations:

$\frac{\partial K(q,p,\theta,\lambda)}{\partial\theta}=\frac{p}{m}$
$\frac{\partial K(q,p,\theta,\lambda)}{\partial\lambda}=-V^{\prime}(q)$

Which we proceed to solve:

$\int\frac{\partial
K(q,p,\theta,\lambda)}{\partial\theta}d\theta=K(q,p,\theta,\lambda)=\int\frac{p}{m}d\theta=\frac{p\theta}{m}+C(q,p,\lambda)$
$\frac{\partial}{\partial\lambda}K(q,p,\theta,\lambda)=\frac{\partial}{\partial\lambda}(\frac{p\theta}{m}+C(q,p,\lambda))=\frac{\partial
C(q,p,\lambda)}{\partial\lambda}=-\frac{V(q)}{dq}$ $\int\frac{\partial
C(q,p,\lambda)}{\partial\lambda}d\lambda=C(q,p,\lambda)=-\int\frac{V(q)}{dq}d\lambda=-\frac{V(q)}{dq}\lambda+C(q,p)$

We can therefore derive

$\hat{K}=K(\hat{q},\hat{p},\hat{\theta},\hat{\lambda})=\frac{\hat{p}\hat{\theta}}{m}-V^{\prime}(\hat{q})\hat{\lambda}+C(\hat{q},\hat{p})$
(45)

which is the important Koopman Generator of KvNM. The Koopman Generator will
become very important when we discuss the Path Integral Formulation of QM and
CM (section 4).

We can show that the Koopman Generator gives us the Louiville Equations again
and produces the KvN classical wavefunction. To show this is quite
straightforward. Since the Born Rule for KvNM postulates that
$\rho=|\innerproduct{q,p}{\psi(t)}|^{2}$, let us try to use the Koopman
Generator to find an expression that governs the evolution of the classical
probability density.

First, refer to (31). From these basic properties, one can easily show the
following:

$\bra{q,p}\hat{q}\ket{\psi(t)}=[\bra{q,p}\hat{q}]\ket{\psi(t)}=\bra{q,p}q\ket{\psi(t)}=q\innerproduct{q,p}{\psi(t)}$
(46)

$\bra{q,p}\hat{p}\ket{\psi(t)}=[\bra{q,p}\hat{p}]\ket{\psi(t)}=\bra{q,p}p\ket{\psi(t)}=p\innerproduct{q,p}{\psi(t)}$
(47)

Using Theorem IV from Appendix A, we can demonstrate that the following are
true. Since $[\hat{q},\hat{\theta}]=i$, we can set $\kappa=1$,
$\ket{A}=\ket{q,p}$, and $B=\theta$ for the parameters in Theorem IV. We can
say that the following is true then:

$\bra{q,p}\hat{\theta}\ket{\psi(t)}=-i\frac{\partial}{\partial
q}\innerproduct{q,p}{\psi(t)}$ (48)

Notice that although $A=(q,p)$, we do not write a partial derivative with
respect to $p$, and that is because $[\hat{p},\hat{\theta}]=0$ per the Koopman
Algebra.

The same can be done with $[\hat{p},\hat{\lambda}]=i$. Using Theorem IV with
parameters $\kappa=1$, $\ket{A}=\ket{q,p}$, and $B=\lambda$, we get:

$\bra{q,p}\hat{\lambda}\ket{\psi(t)}=-i\frac{\partial}{\partial
p}\innerproduct{q,p}{\psi(t)}$ (49)

Again, there is no partial with respect to $q$ here, since
$[\hat{q},\hat{\lambda}]=0$ per the Koopman Algebra.

Plugging the Koopman Generator into Stone’s Theorem gives us:

$i\frac{\partial}{\partial
t}\ket{\psi(t)}=\hat{K}\ket{\psi(t)}=\Big{[}\frac{\hat{p}\hat{\theta}}{m}-V^{\prime}(\hat{q})\hat{\lambda}+C(\hat{q},\hat{p})\Big{]}\ket{\psi(t)}$

And now we will ‘sandwich’ it from the left hand side with the vector bra
$\bra{q,p}$ and work out the algebra:

$\bra{q,p}i\frac{\partial}{\partial
t}\ket{\psi(t)}=\bra{q,p}\Big{[}\frac{\hat{p}\hat{\theta}}{m}-V^{\prime}(\hat{q})\hat{\lambda}+C(\hat{q},\hat{p})\Big{]}\ket{\psi(t)}$

$i\frac{\partial}{\partial
t}\innerproduct{q,p}{\psi(t)}=\bra{q,p}\frac{\hat{p}\hat{\theta}}{m}\ket{\psi(t)}-\bra{q,p}V^{\prime}(\hat{q})\hat{\lambda}\ket{\psi(t)}+\bra{q,p}C(\hat{q},\hat{p})\ket{\psi(t)}$

Utilizing Theorems I of Appendix A131313This Theorem tells us that
$V^{\prime}(\hat{q})\ket{q,p}=V^{\prime}(q)\ket{q,p}$ and
$C(\hat{q},\hat{p})\ket{q,p}=C(q,p)\ket{q,p}$and equations (46),(47),(48) and
(49), we can derive:

$i\frac{\partial}{\partial
t}\innerproduct{q,p}{\psi(t)}=\frac{p}{m}\bra{q,p}\hat{\theta}\ket{\psi(t)}-V^{\prime}(q)\bra{q,p}\hat{\lambda}\ket{\psi(t)}+C(q,p)\innerproduct{q,p}{\psi(t)}$

$i\frac{\partial}{\partial
t}\innerproduct{q,p}{\psi(t)}=-i\frac{p}{m}\frac{\partial}{\partial
q}\innerproduct{q,p}{\psi(t)}+iV^{\prime}(q)\frac{\partial}{\partial
p}\innerproduct{q,p}{\psi(t)}+C(q,p)\innerproduct{q,p}{\psi(t)}$ (50)

We can derive the time evolution equation for the probability density by
utilizing equation (50) above. This will be done in this fashion:

$i\frac{\partial}{\partial
t}|\innerproduct{q,p}{\psi(t)}|^{2}=i\frac{\partial}{\partial
t}[\innerproduct{\psi(t)}{q,p}\innerproduct{q,p}{\psi(t)}]=$
$=i\frac{\partial}{\partial
t}[\innerproduct{\psi(t)}{q,p}]\innerproduct{q,p}{\psi(t)}+i\innerproduct{\psi(t)}{q,p}\frac{\partial}{\partial
t}[\innerproduct{q,p}{\psi(t)}]$

Into this equation plug in (50) and its complex conjugate:

$i\frac{\partial}{\partial
t}|\innerproduct{q,p}{\psi(t)}|^{2}=\Big{[}-i\frac{p}{m}\frac{\partial}{\partial
q}\innerproduct{\psi(t)}{q,p}+iV^{\prime}(q)\frac{\partial}{\partial
p}\innerproduct{\psi(t)}{q,p}-C(q,p)\innerproduct{\psi(t)}{q,p}\Big{]}\innerproduct{q,p}{\psi(t)}$
$+\innerproduct{\psi(t)}{q,p}\Big{[}-i\frac{p}{m}\frac{\partial}{\partial
q}\innerproduct{q,p}{\psi(t)}+iV^{\prime}(q)\frac{\partial}{\partial
p}\innerproduct{q,p}{\psi(t)}+C(q,p)\innerproduct{q,p}{\psi(t)}\Big{]}=$
$=-i\frac{p}{m}\innerproduct{q,p}{\psi(t)}\frac{\partial}{\partial
q}\innerproduct{\psi(t)}{q,p}+iV^{\prime}(q)\innerproduct{q,p}{\psi(t)}\frac{\partial}{\partial
p}\innerproduct{\psi(t)}{q,p}+$
$-\innerproduct{q,p}{\psi(t)}C(q,p)\innerproduct{\psi(t)}{q,p}-i\innerproduct{\psi(t)}{q,p}\frac{p}{m}\frac{\partial}{\partial
q}\innerproduct{q,p}{\psi(t)}$
$+i\innerproduct{\psi(t)}{q,p}V^{\prime}(q)\frac{\partial}{\partial
p}\innerproduct{q,p}{\psi(t)}+\innerproduct{\psi(t)}{q,p}C(q,p)\innerproduct{q,p}{\psi(t)}=$
$=-i\frac{p}{m}\innerproduct{q,p}{\psi(t)}\frac{\partial}{\partial
q}\innerproduct{\psi(t)}{q,p}-i\innerproduct{\psi(t)}{q,p}\frac{p}{m}\frac{\partial}{\partial
q}\innerproduct{q,p}{\psi(t)}$
$+iV^{\prime}(q)\innerproduct{q,p}{\psi(t)}\frac{\partial}{\partial
p}\innerproduct{\psi(t)}{q,p}+i\innerproduct{\psi(t)}{q,p}V^{\prime}(q)\frac{\partial}{\partial
p}\innerproduct{q,p}{\psi(t)}$

Recognizing the involvement of chain rule applied to the wavefunction:

$i\frac{\partial}{\partial
t}|\innerproduct{q,p}{\psi(t)}|^{2}=-i\frac{p}{m}\frac{\partial}{\partial
q}|\innerproduct{q,p}{\psi(t)}|^{2}+iV^{\prime}(q)\frac{\partial}{\partial
p}|\innerproduct{q,p}{\psi(t)}|^{2}$

Since $|\innerproduct{q,p}{\psi(t)}|^{2}=\rho$, the probability density, we
can see right away that this equation is fully equivalent to the Louiville
equation (32)141414Recall from Classical Mechanics that  $F=-\frac{\partial
H}{\partial q}=-V^{\prime}(q)$ $\frac{dq}{dt}=\frac{\partial H}{\partial
p}=\frac{p}{m}$ , and that $\innerproduct{q,p}{\psi(t)}=\psi(q,p,t)$ is the
previously identified KvN classical wavefunction in phase space.

In summary, KvNM is a classical theory of physics that utilizes complex
Hilbert Spaces, just like in QM. Like QM, it is probabilistic in nature, so it
can only make probabilistic predictions of classical systems. Like in QM, you
have a wavefunction that completely describes the state of the system (35).
This wavefunction is in phase space, and KvNM is a phase space theory. It
operates on the same axioms as QM, but is a fully classical theory. It is not
an extension of classical theory, but fully equivalent to it, recast in a non-
traditional form. The mathematical conveniences of QM exist in KvNM. For
example, one reason QM has its odd Hilbert Space form is to conveniently
change between vector bases. One can do this in KvNM with equal ease. Just
like in QM, the expectation value of observable $A$ is given by $\langle
A\rangle=\bra{\psi(t)}\hat{A}\ket{\psi(t)}$.

The backbone of KvNM lies in the standard Hilbert Space behaviors (31), the
Koopman Generator (45), and the Koopman Algebra (43), which lets one derive
the Louiville Equation governing classical probabilities via usage of the KvNM
classical wavefunction. One can derive KvNM from the Ehrenfest Theorems by
assuming the Classical Commutator. The main difference between Classical and
Quantum Physics appears to lie in the choice of the commutator for position
$q$ and momentum $p$ ([4]).

In the pages that follow, we will demonstrate the many applications of KvNM.
Even though technically it is strictly equivalent to any classical theory, it
allows for a wide variety of applications and uses. KvNM is a very important
and convenient tool for statistical physics.

##  4 Path Integral Formulation of Quantum and Classical Mechanics

###  4.1 Feynman Path Integral of Quantum Mechanics

Feynman Path Integral is another formulation of QM that is completely
equivalent to the Schrödinger equation. It was invented by the famous
physicist Richard Feynman in 1948 and has applications many usages in QM, some
which you will see in this Thesis. Whereas the Schrödinger equation is more
analogous to Hamiltonian Dynamics (for instance, see section 5.1 for the
similarity to Hamilton-Jacobi Theory), the Path Integral is more analogous to
Lagrangian Dynamics ([20]).

Under Lagrangian Dynamics in Classical Theory, you select two points, a
starting point and an endpoint point. In the allowed time interval, the
classical particle would take the path where the function $S$, called the
Action, would be minimized. From CM, the term $T-V$ is called the Lagrangian
(where $T$ is kinetic energy and $V$ is potential energy) and the Action is
defined as the time integral of the Lagrangian:

$S=\int_{0}^{t}dt(T-V)$ (51)

This function has to be minimized to define the classical path. The
minimization rule for classical particles - called the Least Action Principle
or Hamilton’s Principle - is often written as:

$\delta\int_{0}^{t}dt(T-V)=0$

Things are not so simple in Quantum Theory.

We begin with a simple derivation of the Feynman Path Integral to understand
the differences between QM and Lagrangian Theory. We start with the following
expression, which is simply a rewrite of the Schrödinger equation:

$\ket{\psi(t)}=e^{-\frac{i}{\hbar}t\hat{H}}\ket{\psi(0)}$ (52)

In Appendix C, one can see how this expression is completely equivalent to the
Schrödinger equation. One might also notice that the operator $\hat{H}$ is
exponated, but this is not unusual in QM (more information about exponated
operators and what they mean is found in the same Appendix). The term
$\exp{-\frac{i}{\hbar}t\hat{H}}$ is called the unitary time-evolution operator
and defines how the quantum particle will evolve in time from its starting
condition $\ket{\psi(0)}$ to its final condition $\ket{\psi(t)}$ at time $t$
(see Appendix B for more on unitary operators).

To begin, we insert a position $q$ bra on both sides of (52):

$\bra{q}\ket{\psi(t)}=\psi(q,t)=\bra{q}e^{-\frac{i}{\hbar}t\hat{H}}\ket{\psi(0)}$
(53)

And then insert resolution of identity (Appendix A Theorem II) for initial
position $q^{\prime}$:

$\bra{q}\ket{\psi(t)}=\psi(q,t)=\int
dq^{\prime}\bra{q}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q^{\prime}}\bra{q^{\prime}}\ket{\psi(0)}=\int
dq^{\prime}\bra{q}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q^{\prime}}\psi(q^{\prime},0)$
(54)

The term $\bra{q}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q^{\prime}}$ is called the
kernell of the Path Integral. The kernell represents the probability amplitude
while traveling from the starting point to the final point. Think of the
$\ket{q^{\prime}}=\ket{q_{0}}$ as the starting point in the Lagrangian
formulism and $\ket{q}=\ket{q_{f}}$ as the final prespecified point.

Let $\Delta t$ be a very small change in time, where $\Delta t=t_{j+1}-t_{j}$.
We will break the total allowed time $t$ into $N$ equal increments, so that we
have $\Delta t=t/N$. We can therefore write:

$\bra{q}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q^{\prime}}=\bra{q}e^{-\frac{i}{\hbar}(\Delta
t_{1}+\Delta t_{2}+...+\Delta
t_{N})\hat{H}}\ket{q^{\prime}}=\bra{q}e^{-\frac{i}{\hbar}\Delta
t_{1}\hat{H}-\frac{i}{\hbar}\Delta t_{2}\hat{H}+...-\frac{i}{\hbar}\Delta
t_{N}\hat{H}}\ket{q^{\prime}}$
$\bra{q}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q^{\prime}}=\bra{q}e^{-\frac{i}{\hbar}\Delta
t_{1}\hat{H}}e^{-\frac{i}{\hbar}\Delta t_{2}\hat{H}}\cdot\cdot\cdot
e^{-\frac{i}{\hbar}\Delta t_{N}\hat{H}}\ket{q^{\prime}}$

The subscripts on each of the $\Delta t$ are there just to make clear that we
are breaking the total time $t$ into a large number of $N$ increments. Since
the $\Delta t$ here are very small, we will drop the subscripts from here on
out. Just keep in mind there are now a total $N$ exponential terms inside the
above expression.

Using Theorem II of Appendix A we can insert between each exponential the
resolution of identity:

$\bra{q_{f}}e^{-\frac{i}{\hbar}\Delta t\hat{H}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\cdot\cdot\cdot e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\ket{q_{0}}=\bra{q_{f}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}Ie^{-\frac{i}{\hbar}\Delta t\hat{H}}I\cdot\cdot\cdot
Ie^{-\frac{i}{\hbar}\Delta t\hat{H}}\ket{q_{0}}=$

$=\int\int\cdot\cdot\cdot\int dq_{N-1}dq_{N-2}\cdot\cdot\cdot
dq_{1}\bra{q_{f}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\ket{q_{N-1}}\bra{q_{N-1}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\ket{q_{N-2}}\times$
$\times\bra{q_{N-2}}\cdot\cdot\cdot\ket{q_{1}}\bra{q_{1}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\ket{q_{0}}$ (55)

We will take one of the “bra-ket” sandwiches above and do some further
operations on it. We can represent each “sandwich” as
$\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta t\hat{H}}\ket{q_{j}}$ and plug in the
value of the Hamiltonian operator $\hat{H}$:

$\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\ket{q_{j}}=\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t[\frac{\hat{p}^{2}}{2m}+V(\hat{q})]}\ket{q_{j}}=\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\frac{\hat{p}^{2}}{2m}}e^{-\frac{i}{\hbar}\Delta tV(\hat{q})}\ket{q_{j}}$

From Theorem I of Appendix A we know that:

$e^{-\frac{i}{\hbar}\Delta tV(\hat{q})}\ket{q_{j}}=e^{-\frac{i}{\hbar}\Delta
tV({q_{j}})}\ket{q_{j}}$

Next we apply the completeness theorem with momentum $p$:

$\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\frac{\hat{p}^{2}}{2m}}Ie^{-\frac{i}{\hbar}\Delta
tV(\hat{q})}\ket{q_{j}}=\int dp\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\frac{\hat{p}^{2}}{2m}}\ket{p}\bra{p}e^{-\frac{i}{\hbar}\Delta
tV({q}_{j})}\ket{q_{j}}$

Again applying Theorem I we know:

$e^{-\frac{i}{\hbar}\Delta
t\frac{\hat{p}^{2}}{2m}}\ket{p}=e^{-\frac{i}{\hbar}\Delta
t\frac{{p}^{2}}{2m}}\ket{p}$

so that the whole thing becomes

$\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\frac{\hat{p}^{2}}{2m}}e^{-\frac{i}{\hbar}\Delta tV(\hat{q})}\ket{q_{j}}=\int
dp\bra{q_{j+1}}\ket{p}\bra{p}\ket{q_{j}}e^{-\frac{i}{\hbar}\Delta
t[\frac{p^{2}}{2m}+V({q_{j}})]}$

Since QM obeys the Canonical Commutator $[\hat{q},\hat{p}]=i\hbar$, we know
what $\bra{q_{j+1}}\ket{p}$ and $\bra{p}\ket{q_{j}}$ are from Theorem V of
Appendix A where we set $\kappa=\hbar$. Plugging these results into the above
equation we get:

$\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\frac{\hat{p}^{2}}{2m}}e^{-\frac{i}{\hbar}\Delta tV(\hat{q})}\ket{q_{j}}=\int
dp~{}e^{-\frac{i}{\hbar}\Delta
t[\frac{p^{2}}{2m}+V({q_{j}})]}\frac{1}{2\pi\hbar}e^{\frac{i}{\hbar}pq_{j+1}}e^{-\frac{i}{\hbar}pq_{j}}=$

$=\int dp~{}e^{-\frac{i}{\hbar}\Delta
t[\frac{p^{2}}{2m}+V({q_{j}})]}\frac{1}{2\pi\hbar}e^{\frac{i}{\hbar}p(q_{j+1}-q_{j})}$

This is known as a Gaussian Integral. From Theorem VIII of Appendix A, we know
how to compute Gaussian Integrals. We simply set
$a=\frac{i}{\hbar}\frac{\Delta t}{2m}$, $b=\frac{i}{\hbar}(q_{j+1}-q_{j})$,
and $c=-\frac{i}{\hbar}\Delta tV(q_{j})$ in this Theorem and conduct the
following steps:

$\frac{1}{2\pi\hbar}\int dp~{}e^{-\frac{i}{\hbar}\frac{\Delta
t}{2m}p^{2}+\frac{i}{\hbar}(q_{j+1}-q_{j})p-\frac{i}{\hbar}\Delta
tV({q_{j}})}=\frac{1}{2\pi\hbar}\sqrt{\frac{\pi}{a}}e^{\frac{b^{2}}{4a}+c}=$
$=\frac{1}{2\pi\hbar}\sqrt{\frac{\pi}{(\frac{i}{\hbar}\frac{\Delta
t}{2m})}}\exp[\frac{(\frac{i}{\hbar}(q_{j+1}-q_{j}))^{2}}{4(\frac{i}{\hbar}\frac{\Delta
t}{2m})}-\frac{i}{\hbar}\Delta tV(q_{j})]$

$\therefore\bra{q_{j+1}}e^{-\frac{i}{\hbar}\Delta
t\hat{H}}\ket{q_{j}}=\frac{1}{2\pi\hbar}\sqrt{\frac{2\pi m\hbar}{i\Delta
t}}\exp[\frac{im(q_{j+1}-q_{j})^{2}}{2\hbar\Delta t}-\frac{i}{\hbar}\Delta
tV(q_{j})]$ (56)

Plugging equation (56) into (55) we get the following for the total time $t$:

$\bra{q_{f}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q_{0}}=\Big{(}\frac{1}{2\pi\hbar}\Big{)}^{N}\Big{(}{\frac{2\pi
m\hbar}{i\Delta t}}\Big{)}^{\frac{N}{2}}\int\int\cdot\cdot\cdot\int
dq_{N-1}dq_{N-2}\cdot\cdot\cdot dq_{1}\times$
$\times\exp[\sum_{j=0}^{N-1}\Big{(}\frac{im(q_{j+1}-q_{j})^{2}}{2\hbar\Delta
t}-\frac{i}{\hbar}\Delta tV(q_{j})\Big{)}]$

This hideous looking expression can be notationally simplified a bit by
treating all the integrals with respect to $q_{j}$ as a product of integrals

$\int\int\cdot\cdot\cdot\int dq_{N-1}dq_{N-2}\cdot\cdot\cdot
dq_{1}=\prod_{j=0}^{N-1}\int dq_{j}$

where $\Pi$ just means product from $j=0,1,...,N-1$, similarly like $\Sigma$
means sum from $j=0,1,...,N-1$. Compactly written:

$\bra{q_{f}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q_{0}}=\Big{(}\frac{1}{2\pi\hbar}\Big{)}^{N}\Big{(}{\frac{2\pi
m\hbar}{i\Delta t}}\Big{)}^{\frac{N}{2}}\prod_{j=0}^{N-1}\int
dq_{j}\exp[\sum_{j=0}^{N-1}\Big{(}\frac{i}{\hbar}\frac{m(q_{j+1}-q_{j})^{2}}{2\Delta
t}-\frac{i}{\hbar}\Delta tV(q_{j})\Big{)}]$ (57)

Factoring out a factor of $\frac{i\Delta t}{\hbar}$ from the exponent:

$\bra{q_{f}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q_{0}}=\Big{(}\frac{1}{2\pi\hbar}\Big{)}^{N}\Big{(}{\frac{2\pi
m\hbar}{i\Delta t}}\Big{)}^{\frac{N}{2}}\prod_{j=0}^{N-1}\int
dq_{j}\exp[\sum_{j=0}^{N-1}\Big{(}\frac{i\Delta
t}{\hbar}\Big{(}\frac{m(q_{j+1}-q_{j})^{2}}{2\Delta
t^{2}}-V(q_{j})\Big{)}\Big{)}]$ (58)

Now, let us take the limit as $\Delta t\rightarrow 0$. Since $\Delta t=t/N$,
this is equivalent to taking $N\rightarrow\infty$.

$\lim_{\Delta t\rightarrow 0}\Big{[}\frac{q_{j+1}-q_{j}}{\Delta
t}\Big{]}^{2}=\Big{(}\frac{dq}{dt}\Big{)}^{2}$ (59)

This is simply the definition of a derivative. And:

$\lim_{N\rightarrow\infty}\sum_{j=0}^{N-1}=\int_{0}^{t}$ (60)

We end up with:

$\bra{q_{f}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q_{0}}=\Big{(}\frac{1}{2\pi\hbar}\Big{)}^{N}\Big{(}{\frac{2\pi
m\hbar}{i\Delta t}}\Big{)}^{\frac{N}{2}}\prod_{j=0}^{N-1}\int
dq_{j}\exp[\frac{i}{\hbar}\int_{0}^{t}dt\Big{(}\frac{m}{2}\Big{(}\frac{dq}{dt}\Big{)}^{2}-V(q)\Big{)}]$
(61)

Since $q$ is the position, $\frac{dq}{dt}$ is obviously the velocity. A quick
glance will reveal that the exponated term $\frac{m}{2}(\frac{dq}{dt})^{2}$ is
simply the Kinetic Energy $T$. Ergo, if you compare the exponent to (51), the
last term becomes $e^{iS/\hbar}$, which should be familiar. It is exactly the
same term that appears in equation (3), the wavefunction represented in radial
form.

We can write some of the messy, ugly terms into a nice format. We say that

$\lim_{N\rightarrow\infty}\Big{(}\frac{-i2\pi m}{\Delta
t}\Big{)}^{\frac{N}{2}}\prod_{j=0}^{N-1}\int dq_{j}=\int Dq(t)$

And therefore the entire thing can be written simply as:

$\bra{q_{f}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q_{0}}=\int
Dq(t)~{}\exp[\frac{i}{\hbar}\int_{0}^{t}dt\Big{(}\frac{m}{2}\Big{(}\frac{dq}{dt}\Big{)}^{2}-V(q)\Big{)}]$

$\bra{q_{f}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{q_{0}}=\int Dq(t)~{}e^{iS/\hbar}$
(62)

Equation (62) represents the final Kernel of the Feynman Path Integral. This
integral is fascinating because it says that for a fixed starting point
$\ket{q_{0}}$ and ending point $\ket{q_{f}}$, the quantum particle takes all
possible paths inbetween within the allotted time interval. This is like
saying that if you are throwing darts at a dart board, the flying dart takes
all possible paths from your hand to the bullseye. This shocking outcome
highlights another strange property of quantum particles. This goes hand in
hand with the earlier comments about quantum particles having “fuzzy”
trajectories.

###  4.2 Koopman-von Neumann Classical Path Integral

Just like a Path Integral can be formulated for Quantum Theory, one can also
formulate a path integral for classical theory. The path integral, however,
must enforce Newton’s Laws and particles must travel along well defined paths
instead of the “fuzzy” trajectories of QM. This classical path integral is
possible thanks to the KvN formulism ([14]; [15]).

First, in order to develop the KvN Path Integral, let us develop some initial
useful tools. These will be ingredients we will use in a future step in making
the Path Integral. Using Theorem V of Appendix A, you can compute what
expressions like $\bra{q,\lambda}\ket{q^{\prime},p}$ are. Let $A=p$,
$B=\lambda$, and $\kappa=1$. Utilizing the Koopman-von Neumann Algebra (43),
you can get:

$\bra{q,\lambda}\ket{q^{\prime},p}=\bra{q}\otimes\bra{\lambda}~{}\ket{q^{\prime}}\otimes\ket{p}=\innerproduct{q}{q^{\prime}}\innerproduct{\lambda}{p}=\frac{1}{\sqrt{2\pi}}\delta(q-q^{\prime})e^{-ip\lambda}$
(63)

Note that there is no need to evaluate $\innerproduct{q}{p}$ and
$\innerproduct{\lambda}{q^{\prime}}$ since $[\hat{q},\hat{p}]=0$ and
$[\hat{\lambda},\hat{q}]=0$. From (18), recall
$\innerproduct{q}{q^{\prime}}=\delta(q-q^{\prime})$. You can follow a similar
procedure to calculate the following two expressions:

$\bra{\theta,p}\ket{q,p^{\prime}}=\bra{\theta}\otimes\bra{p}~{}\ket{q}\otimes\ket{p^{\prime}}=\innerproduct{\theta}{q}\innerproduct{p}{p^{\prime}}=\frac{1}{\sqrt{2\pi}}\delta(p-p^{\prime})e^{-i\theta
q}$ (64)

$\bra{\theta,p}\ket{q,\lambda}=\bra{\theta}\otimes\bra{p}~{}\ket{q}\otimes\ket{\lambda}=\innerproduct{\theta}{q}\innerproduct{p}{\lambda}=\frac{1}{{2\pi}}e^{ip\lambda-i\theta
q}$ (65)

The three expressions above are critical ingredients which we will utilize
soon.

Now we can begin with the proper formulation of the Path Integral. Similar to
(52), we will utilize

$\ket{\psi(t)}=e^{-it\hat{K}}\ket{\psi(0)}$ (66)

with the Koopman Generator $\hat{K}$ (45) instead of the quantum Hamiltonian
$\hat{H}$.151515More information about exponated operators in Appendix C.
Since KvNM deals with phase space, we will attack this expression with a $q,p$
bra on the left to get the following:

$\bra{q,p}\ket{\psi(t)}=\psi(q,p,t)=\bra{q,p}e^{-it\hat{K}}\ket{\psi(0)}$ (67)

And like before, we will introduce the resolution of identity, this time with
respect to $q^{\prime},p^{\prime}$:

$\bra{q,p}\ket{\psi(t)}=\int
dq^{\prime}~{}dp^{\prime}\bra{q,p}e^{-it\hat{K}}\ket{q^{\prime},p^{\prime}}\bra{q^{\prime},p^{\prime}}\ket{\psi(0)}=$
$=\int
dq^{\prime}~{}dp^{\prime}\bra{q,p}e^{-\frac{i}{\hbar}t\hat{K}}\ket{q^{\prime},p^{\prime}}\psi(q^{\prime},p^{\prime},0)$
(68)

The KvN Kernel is
$\bra{q,p}e^{-\frac{i}{\hbar}t\hat{K}}\ket{q^{\prime},p^{\prime}}$ with
$\ket{q^{\prime},p^{\prime}}=\ket{q_{0},p_{0}}$ being the initial state of the
system in phase space and $\ket{q,p}=\ket{q_{f},p_{f}}$ being the final state
of the system. Again, let $\Delta t$ be a small change in time. As before in
the quantum case, it is a very small increment where we will have $\Delta
t=t/N$. We will get:

$\bra{q,p}e^{-it\hat{K}}\ket{q^{\prime},p^{\prime}}=\bra{q,p}e^{-i\Delta
t_{1}\hat{K}}e^{-i\Delta t_{2}\hat{K}}\cdot\cdot\cdot e^{-i\Delta
t_{N}\hat{K}}\ket{q^{\prime},p^{\prime}}$ (69)

The subscripts on each of the $\Delta t$ are there just to make clear that we
are breaking the total time $t$ into a large number of $N$ increments. Since
the $\Delta t$ here are very small, we will drop the subscripts from here on
out. Just keep in mind there are now a total $N$ exponential terms inside the
above expression.

Using Theorem II we can insert between each exponential a completeness
relationship, as before:

$\bra{q_{f},p_{f}}e^{-i\Delta t\hat{K}}e^{-i\Delta t\hat{K}}\cdot\cdot\cdot
e^{-i\Delta t\hat{K}}\ket{q_{0},p_{0}}=$ $=\int\cdot\cdot\cdot\int
dq_{N-1}dp_{N-1}\cdot\cdot\cdot dq_{1}dp_{1}\bra{q_{f},p_{f}}e^{-i\Delta
t\hat{K}}\ket{q_{N-1},p_{N-1}}\times$ $\times\bra{q_{N-1},p_{N-1}}e^{-i\Delta
t\hat{K}}\ket{q_{N-2},p_{N-2}}\bra{q_{N-2},p_{N-2}}\cdot\cdot\cdot\ket{q_{1},p_{1}}\bra{q_{1},p_{1}}e^{-i\Delta
t\hat{K}}\ket{q_{0},p_{0}}$ (70)

We will take one of the “bra-ket” sandwiches above and do some further
operations on it. We can represent each “sandwich” as
$\bra{q_{j+1},p_{j+1}}e^{-i\Delta t\hat{K}}\ket{q_{j},p_{j}}$ and plug in the
value of the Koopman Generator $\hat{K}$:

$\bra{q_{j+1},p_{j+1}}e^{-i\Delta
t\hat{K}}\ket{q_{j},p_{j}}=\bra{q_{j+1},p_{j+1}}\exp(-i\Delta
t\Big{[}\frac{\hat{p}\hat{\theta}}{m}-V^{\prime}(\hat{q})\hat{\lambda}\Big{]})\ket{q_{j},p_{j}}=$
$=\bra{q_{j+1},p_{j+1}}e^{-i\Delta t\frac{\hat{p}\hat{\theta}}{m}}e^{i\Delta
tV^{\prime}(\hat{q})\hat{\lambda}}\ket{q_{j},p_{j}}$

Please note, in keeping with [15], we have omitted the constant of integration
$C(\hat{q},\hat{p})$ from (45) in the following calculations. It will not
affect the dynamics of the experiment for classical systems ([4]; [15]). Next
we apply the completeness theorem twice with $\theta,p$ and $q,\lambda$:

$\bra{q_{j+1},p_{j+1}}e^{-i\Delta
t\hat{K}}\ket{q_{j},p_{j}}=\bra{q_{j+1},p_{j+1}}e^{-i\Delta
t\frac{\hat{p}\hat{\theta}}{m}}e^{i\Delta
tV^{\prime}(\hat{q})\hat{\lambda})}\ket{q_{j},p_{j}}=$
$=\bra{q_{j+1},p_{j+1}}e^{-i\Delta t\frac{\hat{p}\hat{\theta}}{m}}IIe^{i\Delta
tV^{\prime}(\hat{q})\hat{\lambda}}\ket{q_{j},p_{j}}=$ $=\iiiint
d\theta~{}dp~{}dq~{}d\lambda\bra{q_{j+1},p_{j+1}}e^{-i\Delta
t\frac{\hat{p}\hat{\theta}}{m}}\ket{\theta,p}\bra{\theta,p}\ket{q,\lambda}\bra{q,\lambda}e^{i\Delta
tV^{\prime}(\hat{q})\hat{\lambda}}\ket{q_{j},p_{j}}$ (71)

Note that by Theorem I, we have that

$\exp({-i\Delta t\frac{\hat{p}\hat{\theta}}{m}})\ket{\theta,p}=\exp({-i\Delta
t\frac{p\theta}{m}})\ket{\theta,p}$

And also that

$\bra{q,\lambda}\exp({i\Delta
tV^{\prime}(\hat{q})\hat{\lambda}})=\bra{q,\lambda}\exp({i\Delta
tV^{\prime}(q){\lambda}})$

Therefore, plugging into the expression (71) we get:

$\iiiint
d\theta~{}dp~{}dq~{}d\lambda\bra{q_{j+1},p_{j+1}}\ket{\theta,p}\bra{\theta,p}\ket{q,\lambda}\bra{q,\lambda}\ket{q_{j},p_{j}}e^{-i\Delta
t\frac{p\theta}{m}}e^{i\Delta tV^{\prime}(q)\lambda}$

For the expressions $\bra{q_{j+1},p_{j+1}}\ket{\theta,p}$,
$\bra{\theta,p}\ket{q,\lambda}$, and $\bra{q,\lambda}\ket{q_{j},p_{j}}$, we
substitute in our ingredients we derived earlier (63), (64), and (65):

$\iiiint
d\theta~{}dp~{}dq~{}d\lambda\frac{1}{\sqrt{2\pi}}\delta(p-p^{\prime})e^{i\theta
q_{j+1}}\frac{1}{{2\pi}}e^{ip_{j+1}\lambda-i\theta
q_{j}}\frac{1}{\sqrt{2\pi}}\delta(q-q^{\prime})e^{-ip_{j}\lambda}e^{-i\Delta
t\frac{p\theta}{m}}e^{i\Delta tV^{\prime}(q)\lambda}$

Remember that according to (19), the integrals of Dirac delta functionals will
reduce to the following, neater expression:

$=\iint d\theta~{}d\lambda\frac{1}{\sqrt{2\pi}}e^{i\theta
q_{j+1}}\frac{1}{{2\pi}}e^{ip_{j+1}\lambda-i\theta
q_{j}}\frac{1}{\sqrt{2\pi}}e^{-ip_{j}\lambda}e^{-i\Delta
t\frac{p\theta}{m}}e^{i\Delta tV^{\prime}(q)\lambda}$

And simplifying:

$=\frac{1}{(2\pi)^{2}}\iint d\theta~{}d\lambda~{}e^{i\theta q_{j+1}-i\theta
q_{j}-i\Delta t\frac{p\theta}{m}}e^{ip_{j+1}\lambda-ip_{j}\lambda+i\Delta
tV^{\prime}(q)\lambda}$

Factoring out the observables $\theta$ and $\lambda$ from each exponent we
get:

$\frac{1}{(2\pi)^{2}}\iint d\theta~{}d\lambda~{}\exp(i\theta\Delta
t\Big{[}\frac{q_{j+1}-q_{j}}{\Delta t}-\frac{p}{m}\Big{]})\exp(i\lambda\Delta
t\Big{[}\frac{p_{j+1}-p_{j}}{\Delta t}+V^{\prime}(q)\Big{]})$

Inserting this important expression into equation (70) we will get that:

$\bra{q_{f},p_{f}}e^{-it\hat{K}}\ket{q_{0},p_{0}}=$ $=\int\cdot\cdot\cdot\int
dq_{N-1}dp_{N-1}\cdot\cdot\cdot dq_{1}dp_{1}\bigg{[}\frac{1}{(2\pi)^{2}}\iint
d\theta_{N-1}~{}d\lambda_{N-1}~{}\exp\Big(i\theta_{N-1}\Delta
t\Big{[}\frac{q_{j+1}-q_{j}}{\Delta
t}-\frac{p_{j}}{m}\Big{]}\Big{missing})\times$
$\times\exp\Big(i\lambda_{N-1}\Delta t\Big{[}\frac{p_{j+1}-p_{j}}{\Delta
t}+V^{\prime}(q_{j})\Big{]}\Big{missing})\bigg{]}\cdot\cdot\cdot\bigg{[}\frac{1}{(2\pi)^{2}}\iint
d\theta_{1}~{}d\lambda_{1}~{}\exp\Big(i\theta_{1}\Delta
t\Big{[}\frac{q_{j+1}-q_{j}}{\Delta
t}-\frac{p_{j}}{m}\Big{]}\Big{missing})\times$
$\times\exp\Big(i\lambda_{1}\Delta t\Big{[}\frac{p_{j+1}-p_{j}}{\Delta
t}+V^{\prime}(q_{j})\Big{]}\Big{missing})\bigg{]}$

We can utilize some of the simplifying notation we used for Feynman’s Path
Integral:

$\bra{q_{f},p_{f}}e^{-it\hat{K}}\ket{q_{0},p_{0}}=\frac{1}{(2\pi)^{2N}}\iiiint\prod_{j=1}^{N-1}dq_{j}~{}dp_{j}~{}d\theta_{j}~{}d\lambda_{j}\times$
$\times\exp\Big(i\Delta
t\sum_{j=0}^{N-1}\Big{[}\theta_{j}\Big{(}\frac{q_{j+1}-q_{j}}{\Delta
t}-\frac{p_{j}}{m}\Big{)}\Big{]}\Big{missing})\exp\Big(i\Delta
t\sum_{j=0}^{N-1}\Big{[}\lambda_{j}\Big{(}\frac{p_{j+1}-p_{j}}{\Delta
t}+V^{\prime}(q_{j})\Big{)}\Big{]}\Big{missing})$ (72)

Just as before, let $\Delta t\rightarrow 0$ (equivalent to
$N\rightarrow\infty$) to get:

$\lim_{\Delta t\rightarrow 0}\frac{q_{j+1}-q_{j}}{\Delta t}=\frac{dq}{dt}$

$\lim_{\Delta t\rightarrow 0}\frac{p_{j+1}-p_{j}}{\Delta t}=\frac{dp}{dt}$

$\lim_{N\rightarrow\infty}\sum_{j=0}^{N-1}=\int_{0}^{t}$

One more piece of notation. Just as before, let us define that for a variable
$f$, that we will have:

$Df=\lim_{N\rightarrow\infty}\prod_{n}^{N}\frac{df_{n}}{\sqrt{2\pi}}$ (73)

And so the classical analog of the path integral can be written:

$\bra{q_{f},p_{f}}e^{-it\hat{K}}\ket{q_{0},p_{0}}=\iiiint
Dq~{}Dp~{}D\theta~{}D\lambda~{}\exp\bigg(i\int_{0}^{t}dt\Big{[}\lambda(t)\Big{(}\frac{dq}{dt}-\frac{p}{m}\Big{)}+\theta(t)\Big{(}\frac{dp}{dt}+V^{\prime}(q)\Big{)}\Big{]}\bigg{missing})$
(74)

What is the relevance of this? This encodes the equations of CM ([15]). The
path integral is a kind of redundancy that can be shown to enforce the laws of
CM. We can show that it leads to a Dirac delta functional that enforces
Newton’s Laws.

Recall that the Dirac delta functional $\delta(A-A^{\prime})$ is nonzero when
$A=A^{\prime}$ but zero if $A\neq A^{\prime}$. Let us use equation (72) above,
as it is written in a slightly easier to mathematically manipulate form
(following the footsteps of [15]). Let us look at the portion of the integral
with respect to $\theta_{j}$:

$\int d\theta_{j}\exp\Big(i\Delta
t\Big{[}\theta_{j}\Big{(}\frac{q_{j+1}-q_{j}}{\Delta
t}-\frac{p_{j}}{m}\Big{)}\Big{]}\Big{missing})=\int
d\theta_{j}\exp\Big(i\frac{\Delta
t}{m}\Big{[}\theta_{j}\Big{(}m\frac{q_{j+1}-q_{j}}{\Delta
t}-p_{j}\Big{)}\Big{]}\Big{missing})$

Next, we will utilize equation (22) for our convenience. We will take the 0th
derivative of the Dirac delta functional in the formula (which is the same as
taking no derivative at all):

$\frac{d^{0}}{dA^{0}}\delta(A-A^{\prime})=\delta(A-A^{\prime})=\int\frac{d\omega}{2\pi}(-i\omega)^{0}e^{-i\omega(A-A^{\prime})}=\int\frac{d\omega}{2\pi}e^{-i\omega(A-A^{\prime})}$

If you set $\theta=\omega/{2\pi}$, you can plug into the integral with respect
to $\theta_{j}$ to get:

$\int d\theta_{j}\exp\Big(i\frac{\Delta
t}{m}\Big{[}\theta_{j}\Big{(}m\frac{q_{j+1}-q_{j}}{\Delta
t}-p_{j}\Big{)}\Big{]}\Big{missing})=\int\frac{d\omega}{2\pi}\exp\Big(-i\frac{\Delta
t}{m}\Big{[}2\pi\omega\Big{(}p_{j}-m\frac{q_{j+1}-q_{j}}{\Delta
t}\Big{)}\Big{]}\Big{missing})=$
$=\int\frac{d\omega}{2\pi}\exp\Big(-i\omega\Big{[}2\pi\frac{\Delta
t}{m}\Big{(}p_{j}-m\frac{q_{j+1}-q_{j}}{\Delta t}\Big{)}\Big{]}\Big{missing})$

By comparing it to the Dirac delta expression above, we see right away that
$(A-A^{\prime})=\Big{[}2\pi\frac{\Delta
t}{m}(p_{j}-m\frac{q_{j+1}-q_{j}}{\Delta t})\Big{]}$, and so:

$\int d\theta_{j}\exp\Big(i\Delta
t\Big{[}\theta_{j}\Big{(}\frac{q_{j+1}-q_{j}}{\Delta
t}-\frac{p_{j}}{m}\Big{)}\Big{]}\Big{missing})=\delta\Big{(}\frac{\Delta
t}{2\pi m}(p_{j}-m\frac{q_{j+1}-q_{j}}{\Delta t})\Big{)}=$
$=2\pi\frac{m}{\Delta t}\delta\Big{(}p_{j}-m\frac{q_{j+1}-q_{j}}{\Delta
t}\Big{)}$ (75)

utilizing equation (24) for the last step. This reinforces the dynamical
equation $p_{j}-m\frac{q_{j+1}-q_{j}}{\Delta t}=0$, which is simply the
discretized version of the momentum relation, $p=mv$. If this delta functional
is plugged back into equation (74), then you can reduce the total number of
variables integrated over from four to just $q$ and $\lambda$ ([15]). The two
variable version of the Classical Path Integral ensures CM holds ([14]; [15]).

The above integration over $\theta$ can also be conducted over the variable
$\lambda$ instead to get

$\int d\lambda_{j}\exp\Big(i\Delta
t\Big{[}\lambda_{j}\Big{(}\frac{p_{j+1}-p_{j}}{\Delta
t}+V^{\prime}(q_{j})\Big{)}\Big{]}\Big{missing})=\int\frac{d\omega}{2\pi}\exp\Big(-i\omega\Big{[}\frac{\Delta
t}{2\pi}\Big{(}-\frac{p_{j+1}-p_{j}}{\Delta
t}-V^{\prime}(q_{j})\Big{)}\Big{]}\Big{missing})=$ $=\frac{2\pi}{\Delta
t}\delta\Big{(}\frac{p_{j+1}-p_{j}}{\Delta t}+V^{\prime}(q_{j})\Big{)}$ (76)

This reinforces the dynamical equation $\frac{p_{j+1}-p_{j}}{\Delta
t}+V^{\prime}(q_{j})=0$, which is simply the discrete version of Newton’s 2nd
Law of Motion, $\frac{dp}{dt}=-V^{\prime}(q)=F$.

All the above equations are elaborate formulations of the laws of CM. These
are different ways of enforcing the laws of CM hold true. We will go over some
practical usages of the Classical Path Integral in the pages that follow.

##  5 Classical and Quantum Comparison

###  5.1 The Phase in Koopman-von Neumann and Quantum Mechanics

In QM, the interaction of the quantum phase $S$ with the wavefunction
amplitude $R$ gives rise to quantum mechanical behavior ([14]). Recall that
one way of expressing the wavefunction $\psi(t)$ is (3). If we plug this
expression into the Schrödinger equation (4) we get two equations:

$\frac{\partial S}{\partial t}+\frac{(\gradient
S)^{2}}{2m}-\frac{\hbar^{2}}{2mR}\laplacian{R}+V(q)=0$ (77)

which is the classical Hamilton-Jacobi theory with quantum potential161616As
an interesting aside, one can trivially show that the quantum potential
disappears if one assumes the Classical Commutator. For instance, equation
(77) can be rewritten since we know that $(i\hbar)^{2}=-\hbar^{2}$:
$\frac{\partial S}{\partial t}+\frac{(\gradient
S)^{2}}{2m}+\frac{[\hat{q},\hat{p}]^{2}}{2mR}\laplacian{R}+V(q)=0$ If you
assume the Canonical Commutator, you retrieve the Schrödinger equation as is.
If you assume the Classical Commutator, you retrieve the completely classical
Hamilton-Jacobi equation. This result is trivial, but still interesting to
consider from the perspective of unity of ideas., and

$\frac{\partial R^{2}}{\partial t}+\gradient\cdot(R^{2}\frac{\gradient
S}{m})=0$ (78)

which is the continuity equation for density $R^{2}$. 171717Keep in mind that
if $\rho=\psi^{*}\psi$, then
$\rho=\psi^{*}\psi=Re^{-iS/\hbar}Re^{iS/\hbar}=R^{2}$. From the above
expression, you can see that the $S$ and $R$ are coupled together and cannot
be separated from each other in QM.

This is not the case in CM. From the KvNM equation (35), you can see if you
plug in $\psi(q,p,t)=R(q,p,t)e^{iG(q,p,t)}$ you will get complete decoupling
of the amplitude $R$ from the phase $G$. All one has to do is plug the phase
space wavefunction $\psi(q,p,t)$ into (35), like so:

$i\frac{\partial\psi}{\partial t}=\hat{L}\psi=i\frac{\partial}{\partial
t}R(q,p,t)e^{iG(q,p,t)}=\Big{[}-i\frac{p}{m}\frac{\partial}{\partial
q}+i\frac{dV}{dq}\frac{\partial}{\partial p}\Big{]}R(q,p,t)e^{iG(q,p,t)}$

This process begets:

$i\frac{\partial R}{\partial t}=\hat{L}R$ (79)

$i\frac{\partial G}{\partial t}=\hat{L}G$ (80)

The amplitude and the phase are completely seperate. Quantum mechanics has
therefore been described as the theory that tells how the phase interacts with
the amplitude ([14]).

If the phase and the amplitude evolve separately in KvNM, one may ask what is
the practical use of the KvNM formulism? Why the use of complex wavefunctions
if the exponated phase is completely independent of the amplitude? While it is
true that in the momentum position representation $\ket{q,p}$ of KvNM the
formulism can be redundant, one can always use the convenient properties of
Hilbert Spaces to change to other representations, which gives us new insights
into CM ([4]; [5]; [15], etc.) We will see examples of this change in basis
utilized in section 6. In order to have the most general form of CM, it is
important to write the classical wavefunction in this complex form ([14]).

###  5.2 Uncertainty Principle in Koopman-von Neumann and Quantum Mechanics

One of the hallmarks of QM is the existence of an uncertainty principle
between the position and momentum. The uncertainty principle tells you that
the more precisely you measure where a wavefunction is localized, the less
precisely you know the momentum distribution. This contrasts with CM, where
you can measure the values of both the momentum and position to an arbitrary
precision at the same time. This trade off is another re-articulation of the
strange property of “fuzzy” trajectories in QM and well-defined trajectories
in CM.

One can naturally derive the Heisenberg Uncertainty Principle from the
Canonical Commutator. Through the same series of steps, one can also
demonstrate the lack of an uncertainty principle between position and momentum
for the KvN classical case. The procedure is quite straightforward.

Firstly, from statistics we know that the standard deviation of some measured
quantity $A$ is given by:

$\sigma_{A}=\sqrt{\langle A^{2}\rangle-\langle A\rangle^{2}}$ (81)

And this will be a crucial ingredient in deriving the Heisenberg Uncertainty
Principle. From equation (16), we know that

$\langle A\rangle=\bra{\psi(t)}\hat{A}\ket{\psi(t)}$ $\langle
A^{2}\rangle=\bra{\psi(t)}\hat{A}^{2}\ket{\psi(t)}$

Treat $\langle A\rangle$ as a constant. Then we can see the following
important relationship:

$(\hat{A}-\langle A\rangle)^{2}=\hat{A}^{2}-2\langle A\rangle\hat{A}+\langle
A\rangle^{2},$
$\bra{\psi(t)}[\hat{A}^{2}-2\hat{A}\langle A\rangle+\langle
A\rangle^{2}]\ket{\psi(t)}=\bra{\psi(t)}\hat{A}^{2}\ket{\psi(t)}-\bra{\psi(t)}2\langle
A\rangle\hat{A}\ket{\psi(t)}+\bra{\psi(t)}\langle A\rangle^{2}\ket{\psi(t)}=$
$=\bra{\psi(t)}\hat{A}^{2}\ket{\psi(t)}-2\langle
A\rangle\bra{\psi(t)}\hat{A}\ket{\psi(t)}+\langle
A\rangle^{2}\innerproduct{\psi(t)}{\psi(t)}=$
$=\bra{\psi(t)}\hat{A}^{2}\ket{\psi(t)}-2\langle A\rangle^{2}+\langle
A\rangle^{2}=\langle A^{2}\rangle-\langle A\rangle^{2}.$

utilizing (16) and $\innerproduct{\psi(t)}{\psi(t)}=1$.

We have all the ingredients we need to derive the Heisenberg Uncertainty
Principle. First, let us compute the variance $\sigma^{2}$ for both the
position and momentum. Definitionally following (81) and our above result,
they would be

$\sigma_{q}^{2}=\langle q^{2}\rangle-\langle
q\rangle^{2}=\bra{\psi(t)}(\hat{q}-\langle q\rangle)^{2}\ket{\psi(t)}$
$\sigma_{p}^{2}=\langle p^{2}\rangle-\langle
p\rangle^{2}=\bra{\psi(t)}(\hat{p}-\langle p\rangle)^{2}\ket{\psi(t)}$

We can define a new vector to simplify our calculation. We can define
$\ket{Q}=(\hat{q}-\langle q\rangle)\ket{\psi(t)}$ and
$\ket{P}=(\hat{p}-\langle p\rangle)\ket{\psi(t)}$ in order to write the
variance of position and momentum simply as

$\sigma_{q}^{2}=\innerproduct{Q}{Q}$ $\sigma_{p}^{2}=\innerproduct{P}{P}$

Then we apply the Cauchy-Schwarz Inequality (Theorem III of Appendix A) which
tells us that:

$\sigma_{p}^{2}\sigma_{q}^{2}=\innerproduct{P}{P}\innerproduct{Q}{Q}\geq|\innerproduct{P}{Q}|^{2}$
(82)

We know that for any complex number $z=a+bi$, that

$|z|^{2}=a^{2}+b^{2}\geq b^{2}$

where we know from complex analysis that $b=\frac{z-z^{*}}{2i}$. Ergo,

$|z|^{2}=|\innerproduct{P}{Q}|^{2}\geq\Big{(}\frac{z-z^{*}}{2i}\Big{)}^{2}=\Big{(}\frac{\innerproduct{P}{Q}-\innerproduct{Q}{P}}{2i}\Big{)}^{2},$

where

$\innerproduct{P}{Q}=\bra{\psi(t)}(\hat{p}-\langle p\rangle)(\hat{q}-\langle
q\rangle)\ket{\psi(t)}=\bra{\psi(t)}[\hat{p}\hat{q}-\hat{q}\langle
p\rangle-\langle q\rangle\hat{p}+\langle q\rangle\langle
p\rangle]\ket{\psi(t)}=$
$=\bra{\psi(t)}\hat{p}\hat{q}\ket{\psi(t)}-\bra{\psi(t)}\hat{q}\langle
p\rangle\ket{\psi(t)}-\bra{\psi(t)}\langle
q\rangle\hat{p}\ket{\psi(t)}+\bra{\psi(t)}\langle q\rangle\langle
p\rangle]\ket{\psi(t)},$

and

$\innerproduct{Q}{P}=\bra{\psi(t)}\hat{q}\hat{p}\ket{\psi(t)}-\bra{\psi(t)}\hat{q}\langle
p\rangle\ket{\psi(t)}-\bra{\psi(t)}\langle
q\rangle\hat{p}\ket{\psi(t)}+\bra{\psi(t)}\langle q\rangle\langle
p\rangle]\ket{\psi(t)}.$

With the calculations above, we plug into (82) and derive:

$|\innerproduct{P}{Q}|^{2}\geq\Big{(}\frac{\innerproduct{P}{Q}-\innerproduct{Q}{P}}{2i}\Big{)}^{2}=\Big{(}\frac{\bra{\psi(t)}\hat{p}\hat{q}\ket{\psi(t)}-\bra{\psi(t)}\hat{q}\hat{p}\ket{\psi(t)}}{2i}\Big{)}^{2},$

$|\innerproduct{P}{Q}|^{2}\geq\Big{(}\frac{\bra{\psi(t)}[\hat{p},\hat{q}]\ket{\psi(t)}}{2i}\Big{)}^{2}.$
(83)

From this, it is simple to see that:

$\sigma_{p}\sigma_{q}\geq\frac{|\bra{\psi(t)}[\hat{p},\hat{q}]\ket{\psi(t)}|}{2}$
(84)

If we plug in the Canonical Commutator for $[\hat{p},\hat{q}]$ we derive the
Heisenberg Uncertainty Principle,

$\sigma_{p}\sigma_{q}\geq\hbar/2$ (85)

If we plug in the Classical Commutator for $[\hat{p},\hat{q}]=0$ there is no
uncertainty relation between the position and momentum, hence there is no
theoretical restriction to knowledge of precision of both at the same time.
Depending on your choice of the Commutator, you derive either Classical or
Quantum Physics ([4]).

This does not, however, mean there are no uncertainty relations for KvNM.
Because the Koopman Algebra (43) tells us that $[\hat{q},\hat{\theta}]=i$ and
$[\hat{p},\hat{\lambda}]=i$, we can develop uncertainty relationships for
$q$-$\theta$ and $p$-$\lambda$ by plugging the commutators into expressions
analogous to (84):

$\sigma_{p}\sigma_{\lambda}\geq\frac{|\bra{\psi(t)}[\hat{p},\hat{\lambda}]\ket{\psi(t)}|}{2}=\frac{1}{2}$
$\sigma_{q}\sigma_{\theta}\geq\frac{|\bra{\psi(t)}[\hat{q},\hat{\theta}]\ket{\psi(t)}|}{2}=\frac{1}{2}$

These would be the ‘classical’ uncertainty relationships. Since neither
$\theta$ nor $\lambda$ are directly observable ([18]), these uncertainty
relations will not impact our observations of classical systems. For KvNM,
many “quantum” phenomena are tucked into the not directly observable operators
$\hat{\theta}$ and $\hat{\lambda}$.

###  5.3 Measurement in Koopman-von Neumann and Quantum Mechanics

QM is known to have the strange property that it makes probabilistic
predictions instead of deterministic ones. One strange feature related to this
fact that tucked into the previously mentioned postulates of QM you have a
phenomenon termed the collapse of the wavefunction. Any wavefunction can be
expanded across a basis using the completeness theorem (Theorem II of Appendix
A):

$\ket{\psi}=I\ket{\psi},$

where $I$ is the identity operator (analogous to the number ‘1’.) Inserting
the discrete181818Note that in (86) I used the discrete case instead of the
continuous completeness identity. This is simply due to the ease of
highlighting certain concepts. The continuous completeness identity would do
the same thing as the discrete case, but might obscure for some readers
results we will discuss. Remember that the integral is just a continuous
summation of infinitely small things. completeness relation we get:

$\ket{\psi}=\sum_{j}\ket{j}\bra{j}\ket{\psi}=\ket{A}\bra{A}\ket{\psi}+\ket{B}\bra{B}\ket{\psi}+\ket{C}\bra{C}\ket{\psi}+\cdot\cdot\cdot,$
(86)

where the collection of all $\ket{j}$ vectors ($j=A,B,C,...$) are your basis
vectors according to the principles of Linear Algebra and $\bra{j}\ket{\psi}$
are scalar values or functions (recalling that $\bra{j}\ket{\psi}=\psi(j)$).
This is simply expanding the state vector $\ket{\psi}$ along a particular set
of basis vectors $\ket{j}$. $A$, $B$, $C$, etc., are different eigenvalues
associated with each ‘eigenbasis’, for example, $\hat{A}\ket{A}=A\ket{A}$.

If you measure $B$ then the initial wavefunction $\ket{\psi}$ instantaneously
collapses the entire quantum system to state $\ket{B}$. How the state of the
system $\ket{\psi}$ instantaneously changes to $\ket{B}$ when $B$ is measured
is still a modern day quantum mystery. As described by Postulate 3 in section
2, the probability for observing $B$ is $P(B)=|\innerproduct{B}{\psi}|^{2}$
for discrete systems and the probability density is
$\rho(B)=|\innerproduct{B}{\psi}|^{2}$ in continuous systems.

The strange phenomenon of collapse is preserved in KvNM. You have a KvN
classical wavefunction that again can be expanded over a particular basis of
your choice

$\ket{\psi}=\ket{A}\bra{A}\ket{\psi}+\ket{B}\bra{B}\ket{\psi}+\ket{C}\bra{C}\ket{\psi}+\cdot\cdot\cdot,$

and that when a particular eigenvalue is observed, the system collapses to its
associated eigenbasis. The same Born Rule applies. You have the same postulate
for collapse from the set of ‘universal axioms’ .

The common behavior of the wavefunction collapsing during measurement in QM
and KvNM is mystifying, and perhaps one day further light can be shed on it.
Maybe it is a common feature of probabilistic models built in Hilbert Space,
or perhaps there will be some other explanation developed. As far as we know,
the collapse is instant, irreversible, and no causal process exists to explain
it.

There is an important difference though in the collapse of the wavefunction
during measurement for the KvNM versus QM cases. The wavefunction will not
collapse for non-selective measurements in the KvN case but will in the QM
case ([9]; [14]). Non-selective measurements are measurements conducted on the
system where the results are not ‘read out loud’, meaning the system would be
hypothetically collapsed but we do not ‘record’, remember, or consider which
eigenbasis it collapses to. We remain completely ignorant of whether the
system state $\ket{\psi}$ hypothetically collapses into $\ket{A}$, $\ket{B}$,
$\ket{C}$, or some other state. This difference in collapse for KvN and QM
wavefunctions will be demonstrated (following the work of [9]; [14]).

First, let us consider the QM case. Consider a Hamiltonian $\hat{H}$ which
obeys the following eigenvalue relationships:

$\hat{H}\ket{+}=\hbar\omega\ket{+},$ (87)

$\hat{H}\ket{-}=-\hbar\omega\ket{-},$ (88)

where $\ket{+}$ and $\ket{-}$ are the eigenvectors and $\hbar\omega$ and
$-\hbar\omega$ are the energy eigenvalues of the operator $\hat{H}$,
respectively. 191919This should bring to mind, for instance, that the energy
of a photon is given by $E=\hbar\omega$, where $\omega$ is the angular
frequency. Let us consider an observable with operator $\hat{\Omega}$, which
has the eigenvectors $\ket{a}$ and $\ket{b}$ defined by:

$\ket{a}=\frac{1}{\sqrt{2}}\Big{[}\ket{+}+\ket{-}\Big{]},$ (89)

$\ket{b}=\frac{1}{\sqrt{2}}\Big{[}\ket{+}-\ket{-}\Big{]}.$ (90)

For the sake of our analysis, we can choose the initial state of the system to
be the following superposition of states ([14]):

$\ket{\psi(t=0)}=\frac{1}{2}\ket{+}+\sqrt{\frac{3}{4}}\ket{-}$ (91)

We will utilize the Schrödinger equation in the form of (52) (see also
Appendix C) order to evolve the above wavefunction, $\ket{\psi(0)}$. We want
to evolve the system to $t=2\tau$ and see what the probability of measuring
$a$, that is $P(a)$, will be ([14]).

$\ket{\psi(t=2\tau)}=e^{-\frac{i}{\hbar}\hat{H}2\tau}\ket{\psi(0)}=\exp[-\frac{i}{\hbar}\hat{H}2\tau]\Big{[}\frac{1}{2}\ket{+}+\sqrt{\frac{3}{4}}\ket{-}\Big{]}.$
(92)

We will utilize the very useful Theorem I from Appendix A alongside (87) and
(88). This will give us:

$\ket{\psi(t=2\tau)}=\frac{1}{2}\exp[-\frac{i}{\hbar}\hat{H}2\tau]\ket{+}+\sqrt{\frac{3}{4}}\exp[-\frac{i}{\hbar}\hat{H}2\tau]\ket{-}=$
$=\frac{1}{2}\exp[-\frac{i}{\hbar}2\tau(\hbar\omega)]\ket{+}+\sqrt{\frac{3}{4}}\exp[-\frac{i}{\hbar}2\tau(-\hbar\omega)]\ket{-}=$
$=\frac{1}{2}\exp[-2i\tau\omega]\ket{+}+\sqrt{\frac{3}{4}}\exp[2i\tau\omega]\ket{-}.$
(93)

To find the probability of observing $a$ at $t=2\tau$ we will use the Born
Rule:

$P(a)=\bra{\psi(2\tau)}\ket{a}\bra{a}\ket{\psi(2\tau)}=|\bra{a}\ket{\psi(2\tau)}|^{2},$

where we can evaluate:

$\bra{a}\ket{\psi(2\tau)}=\frac{1}{2}\exp[-2i\tau\omega]\bra{a}\ket{+}+\sqrt{\frac{3}{4}}\exp[2i\tau\omega]\bra{a}\ket{-}=$
$=\frac{1}{2}\exp[-2i\tau\omega]\frac{1}{\sqrt{2}}\Big{[}\bra{+}+\bra{-}\Big{]}\ket{+}+\sqrt{\frac{3}{4}}\exp[2i\tau\omega]\frac{1}{\sqrt{2}}\Big{[}\bra{+}+\bra{-}\Big{]}\ket{-}.$

Using the orthonormality condition, we know that $\innerproduct{+}{-}=0$ while
$\innerproduct{+}{+}=\innerproduct{-}{-}=1$, so that the above expression
reduces to

$\bra{a}\ket{\psi(2\tau)}=\frac{1}{2\sqrt{2}}\exp[-2i\tau\omega]+\sqrt{\frac{3}{8}}\exp[2i\tau\omega].$

Therefore, taking
$(\bra{a}\ket{\psi(2\tau)})^{\dagger}=\bra{\psi(2\tau)}\ket{a}$ and plugging
into the Born Rule:

$P(a)=\bra{\psi(2\tau)}\ket{a}\bra{a}\ket{\psi(2\tau)}=\Big{(}\frac{1}{2\sqrt{2}}\exp[2i\tau\omega]+\sqrt{\frac{3}{8}}\exp[-2i\tau\omega]\Big{)}\times$
$\times\Big{(}\frac{1}{2\sqrt{2}}\exp[-2i\tau\omega]+\sqrt{\frac{3}{8}}\exp[2i\tau\omega]\Big{)}=$
$=\frac{1}{2\sqrt{2}}\exp[2i\tau\omega]\frac{1}{2\sqrt{2}}\exp[-2i\tau\omega]+\frac{1}{2\sqrt{2}}\exp[2i\tau\omega]\sqrt{\frac{3}{8}}\exp[2i\tau\omega]+$
$+\sqrt{\frac{3}{8}}\exp[-2i\tau\omega]\frac{1}{2\sqrt{2}}\exp[-2i\tau\omega]+\sqrt{\frac{3}{8}}\exp[-2i\tau\omega]\sqrt{\frac{3}{8}}\exp[2i\tau\omega]=$
$=\frac{1}{8}+\frac{\sqrt{2}}{4}\sqrt{\frac{3}{8}}\exp[4i\omega\tau]+\frac{\sqrt{2}}{4}\sqrt{\frac{3}{8}}\exp[-4i\omega\tau]+\frac{3}{8}.$
(94)

There are one important trigonometric identity we can make use of. It is:

$\cos(\theta)=(1/2)(e^{i\theta}+e^{-i\theta})$ (95)

Taking the identity (95) with (94), we can calculate that

$P(a)=\frac{1}{2}\Big{(}1+\sqrt{\frac{3}{4}}\cos(4\omega\tau)\Big{)}.$ (96)

The above usage of the wavefunction is for “pure states” (term defined in
section 2.6).

Now, instead of letting the quantum system evolve continuous from $t=0$ to
$t=2\tau$, let us introduce a non-selective measurement at $t=\tau$. How will
this affect the probability of measuring $a$ at $t=2\tau$? We will start with
the same $\ket{\psi(0)}$. We will, however, use the density operator formulism
to describe the system (see section 2.6) instead of strictly using only the
‘pure’ wavefunction. This is because we have here what is called a “mixed
state”. When we take the measurement at $t=\tau$, the original system given by
(91) collapses, but we do not know whether it collapses into $\ket{a}$ or
$\ket{b}$, as we do not ‘read out’ or record what the collapse leads to. We
can use (26) to write down:

$\hat{\rho}(\tau)=p_{a}(\tau)\ket{a}\bra{a}+p_{b}(\tau)\ket{b}\bra{b},$ (97)

where $p_{a}(\tau)$ and $p_{b}(\tau)$ are the probabilistic weights to the
‘statistical mixture’ of possible states $\ket{a}$ and $\ket{b}$ of the
system. The factor $p_{a}(\tau)$ refers to the probability that we observe $a$
at $t=\tau$ (and hence the system collapsed to $\ket{a}$) and the term
$p_{b}(\tau)$ refers to the probability that we observe $b$ at $t=\tau$ (and
hence the system collapsed to $\ket{b}$). Since we are ignorant of which way
the wavefunction collapses, the probabilities became weights of likelihood in
(97).

We calculate what the factors $p_{a}(\tau)$ and $p_{b}(\tau)$ are in a similar
fashion to calculating (96):

$p_{a}(\tau)=\bra{\psi(\tau)}\ket{a}\bra{a}\ket{\psi(\tau)}=\Big{(}\frac{\sqrt{2}}{4}\exp[-i\omega\tau]+\sqrt{\frac{3}{8}}\exp[i\omega\tau]\Big{)}\times$
$\times\Big{(}\frac{\sqrt{2}}{4}\exp[i\omega\tau]+\sqrt{\frac{3}{8}}\exp[-i\omega\tau]\Big{)}=$
$=\frac{1}{2}\Big{(}1+\sqrt{\frac{3}{4}}\Big{[}\frac{1}{2}\exp[-2i\omega\tau]+\frac{1}{2}\exp[2i\omega\tau]\Big{]}\Big{)}=\frac{1}{2}\Big{(}1+\sqrt{\frac{3}{4}}\cos(2\omega\tau)\Big{)}$
(98)
$p_{b}(\tau)=\bra{\psi(\tau)}\ket{b}\bra{b}\ket{\psi(\tau)}=\Big{(}\frac{\sqrt{2}}{4}\exp[-i\omega\tau]-\sqrt{\frac{3}{8}}\exp[i\omega\tau]\Big{)}\times$
$\times\Big{(}\frac{\sqrt{2}}{4}\exp[i\omega\tau]-\sqrt{\frac{3}{8}}\exp[-i\omega\tau]\Big{)}=$
$=\frac{1}{2}\Big{(}1-\sqrt{\frac{3}{4}}\Big{[}\frac{1}{2}\exp[-2i\omega\tau]+\frac{1}{2}\exp[2i\omega\tau]\Big{]}\Big{)}=\frac{1}{2}\Big{(}1-\sqrt{\frac{3}{4}}\cos(2\omega\tau)\Big{)}$
(99)

We will use (52) again as before to evolve the system then from (97) at
$t=\tau$ to $t=2\tau$, in the following fashion:

$\ket{\psi_{a}(2\tau)}=e^{-\frac{i}{\hbar}\hat{H}(2\tau-\tau)}\ket{a}=\exp[-\frac{i}{\hbar}\hat{H}\tau]\frac{1}{\sqrt{2}}\Big{[}\ket{+}+\ket{-}\Big{]},$
(100)

$\ket{\psi_{b}(2\tau)}=e^{-\frac{i}{\hbar}\hat{H}(2\tau-\tau)}\ket{b}=\exp[-\frac{i}{\hbar}\hat{H}\tau]\frac{1}{\sqrt{2}}\Big{[}\ket{+}-\ket{-}\Big{]},$
(101)

utilizing the expressions (89) and (90) above.

As before, utilizing the eigenvalue expressions for $\hat{H}$ (87) and (88),
and Theorem I we get:

$\ket{\psi_{a}(2\tau)}=\frac{1}{\sqrt{2}}\exp[-\frac{i}{\hbar}\tau(\hbar\omega)]\ket{+}+\frac{1}{\sqrt{2}}\exp[-\frac{i}{\hbar}\tau(-\hbar\omega)]\ket{-},$
(102)

$\ket{\psi_{b}(2\tau)}=\frac{1}{\sqrt{2}}\exp[-\frac{i}{\hbar}\tau(\hbar\omega)]\ket{+}-\frac{1}{\sqrt{2}}\exp[-\frac{i}{\hbar}\tau(-\hbar\omega)]\ket{-}.$
(103)

The density operator therefore at $t=2\tau$ can be represented as the
following using the above expressions:

$\hat{\rho}(2\tau)=p_{a}(\tau)\ket{\psi_{a}(2\tau)}\bra{\psi_{a}(2\tau)}+p_{b}(\tau)\ket{\psi_{b}(2\tau)}\bra{\psi_{b}(2\tau)}$
(104)

And we can use this to evaluate the probability of measuring $a$ by using
(27):

$P(a)=\Tr[\hat{\rho}(2\tau)\ket{a}\bra{a}]=\bra{a}\hat{\rho}(2\tau)\ket{a}\bra{a}\ket{a}+\bra{b}\hat{\rho}(2\tau)\ket{a}\bra{a}\ket{b}.$

Remember that $\bra{a}\ket{b}=0$ and $\bra{a}\ket{a}=1$, so:

$P(a)=\bra{a}\hat{\rho}(2\tau)\ket{a}.$

Plugging in (104) we get:

$P(a)=p_{a}(\tau)\bra{a}\ket{\psi_{a}(2\tau)}\bra{\psi_{a}(2\tau)}\ket{a}+p_{b}(\tau)\bra{a}\ket{\psi_{b}(2\tau)}\bra{\psi_{b}(2\tau)}\ket{a}.$
(105)

And now we can substitute the expressions (102), (103), (98), and (99) to get
after some algebra our final expression:

$P(a)=\frac{1}{2}\Big{(}1+\sqrt{\frac{3}{4}}\cos^{2}(2\omega\tau)\Big{)}.$
(106)

Obviously, comparing (96) with (106) shows that they are two different
equations. This shows us that in the quantum case making a non-selective
measurement will change the outcome of a later measurement. A non-selective
measurement will modify the probability of outcome at a later point in time
$t$.

So far, what we looked at is an abstract example to demonstrate non-selective
measurements in QM. Let us look at a more concrete example in QM, and then
figure out what the implications are for KvNM (following [14]). Starting off,
let us measure the probability of a quantum particle’s position from $t=0$ to
$t=\tau$.

This will be given by (26) except we are no longer dealing with a discrete
case. Since position is continuous, we replace the summation sign with an
integral (as an integral is simply continuous summation of infinitely small
things). We start with a pure state with wavefunction $\ket{\psi(0)}$ at
$t=0$. The probability density $\rho$ (not to be confused with the density
matrix) of observing $q$ at time $t=\tau$ will be given by

$\rho(q,\tau)=\Tr[\hat{\rho}(\tau)\ket{q}\bra{q}]=|\psi(q,\tau)|^{2},$ (107)

where $\hat{\rho}(\tau)=\ket{\psi(\tau)}\bra{\psi(\tau)}$ (pure state density
matrix).

Let us say that we make a non-selective measurement immediately after $t=0$,
which we will represent as time $t=0_{+}$. The pure state density matrix gets
replaced with a mixed density matrix. We utilize a continuous version of (26)
to write the density matrix

$\hat{\rho}(0_{+})=\int dq_{0}|\psi(q_{0},0)|^{2}\ket{q_{0}}\bra{q_{0}}.$

To represent a state of the system immediately before $t=0$, we can write time
$t=0_{-}$ and say that:

$\hat{\rho}(0_{-})=\ket{\psi(0_{-})}\bra{\psi(0_{-})}$

The system goes from $\hat{\rho}(0_{-})$ to $\hat{\rho}(0_{+})$ as a non-
selective measurement is made at $t=0$.

In QM, the measurement right before and the measurement right after will
observe the same thing ([14]). You can convince yourself of this by computing
the following:

$\rho(q,0_{-})=\Tr[\hat{\rho}(0_{-})\ket{q}\bra{q}]=\psi^{*}(q,0)\psi(q,0)$
$\rho(q,0_{+})=\frac{\Tr[\hat{\rho}(0_{+})\ket{q}\bra{q}]}{\Tr[\hat{\rho}(0_{+})]}=\psi^{*}(q,0)\psi(q,0)$

Remember that the denominator, $\Tr[\hat{\rho}(0_{+})]$, just represents the
normalization condition (described in section 2.6). We can evolve
$\psi^{*}(q,0)\psi(q,0)$ to see that in the long term the non-selective
measurement does affect the quantum system, just like in the above worked out
example ([14]). The probability density will be the same instantaneously right
before and right after measurement, but will alter in the long run in QM.

Under KvNM, however, we can observe some differences arise. If we start with a
pure state again

$\hat{\rho}(0_{-})=\ket{\psi(0_{-})}\bra{\psi(0_{-})},$

and conduct no measurements, we will have the KvN wavefunction evolve to

$\rho(q,p,\tau)=\Tr[\hat{\rho}(\tau)\ket{q,p}\bra{q,p}]=|\psi(q,p,\tau)|^{2},$
(108)

similarly to the quantum case (107). Also like the quantum case, in KvNM
measuring the probability density right before and after will not alter the
distribution:

$\frac{\Tr[\hat{\rho}(0_{-})\ket{q,p}\bra{q,p}]}{\Tr[\hat{\rho}(0_{-})]}=\frac{\Tr[\hat{\rho}(0_{+})\ket{q,p}\bra{q,p}]}{\Tr[\hat{\rho}(0_{+})]}$

However, a non-selective measurement can be shown to leave the KvN
wavefunction unaltered, whereas a quantum wavefunction will be changed due to
the non-selective measurement .

$\rho(0_{+})=\int
dq_{0}dp_{0}|\psi(q_{0},p_{0},0)|^{2}\ket{q_{0},p_{0}}\bra{q_{0},p_{0}}.$

The above probability density, taken right after the measurement, can be shown
to naturally evolve into (108), as if no measurement has taken place ([14]).

There is one final similarity worth mentioning between QM and KvNM. Under QM,
when a measurement is taken for position $q$, it affects the outcome of the
distribution of $p$. Perhaps, this is not surprising, considering the
Heisenberg Uncertainty Principle. Under KvNM, a measurement taken for $(q,p)$
will alter the statistical distribution of the variable $\lambda$ ([14]). This
should also not be surprising because, as we discussed in section 5.2, there
exists a classical uncertainty relationship between $p$ and $\lambda$. The
effects of QM are still present in KvNM therefore, if we replace $q$ with
$(q,p)$ and $p$ with $\lambda$.

###  5.4 The Double Slit Experiment

The Double Slit Experiment is at the center of Quantum Theory. QM has photons,
electrons, neutrons, and other quanta going through two tiny slits and forming
an interference pattern on the other side. Even if we send one particle at a
time through the double slit, the particle build up on the other side will
reflect an interference pattern. Even if the particle does not seem to
interfere with any other particle, it still is part of an interference
pattern. This shows that the interference patterns are due to something more
fundamental in nature. 202020Interestingly, if you cover up on of the two
slits, you will recover particle travelling in straight trajectories without
any interference pattern, just as in CM. This makes the relationship between
CM and QM even more bizarre.

The wavefunction mathematically describes this strange behavior of elementary
particles, even though the exact interpretation of the wavefunction is not
known. Feynman believed that the strange behavior must lie in the complex
nature of the wavefunction ([14]). KvNM also contains complex wavefunctions in
Hilbert Space, so we might logically ask the obvious question: if KvNM has
complex wavefunctions, will it successfully preserve our knowledge of CM,
without interference patterns arising? Here we will test the outcome of
complex wavefunctions in classical phase space. This analysis is based on the
work of ([14]).

Figure 1: This is the set up of the double slit experiment. Middle wall is at
$y=y_{M}$ and rightmost wall is at $y=y_{R}$. Adapted from [13]

The double slit experiment is set up in Figure 1 above. The y axis is aligned
parallel to the direction of travel. The x axis will be the coordinate
parallel to the back wall where the interference is supposed to occur for the
quantum case. In the middle, at $y=y_{M}$ you have the wall with the two
slits. At $y=y_{F}$ you have the wall where interference occurs in the quantum
case.

Under KvNM, we should expect no interference build up at the backwall at
$y=y_{R}$. Under QM (and in reality), we want there to be an interference
build up at the backwall. We will start with analyzing the quantum case, and
how it works. We want a gaussian wave emerge from $y=0$ and travel right until
it hits the double slit wall. Some of the wavefunction will travel through the
two slits and eventually interfere at $y=y_{R}$ if the system behaves quantum.

A gaussian function has the following form:

$f(u)=\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-(u-\mu)^{2}/{2\sigma^{2}}}$ (109)

where $\mu$ is the average value $\mu=\langle u\rangle$ and $\sigma$ is the
standard deviation, given by $\sigma=\sqrt{\langle u^{2}\rangle-\langle
u\rangle^{2}}$.

Recalling the form of a quantum wavefunction (3), we can make it a gaussian
wavefunction by setting $R$ equal to (109) with a generic renormalization
constant $N$.212121In order to keep things nice and streamlined, the constant
in front of (109) will simply be absorbed into the Renormalization Constant.
Let $\mu=0$ in order to center the wavefunction of the double slit experiment.
The equation will have the form:

$\psi(x,t=0)=N\exp[-\frac{x^{2}}{2\sigma_{x}^{2}}]\exp[\frac{i}{\hbar}S]$

Note that for this wavefunction, $S=0$, however. This is obvious when you
consider the definition of the Action (51) for a free particle. Since we
decided that the time at the leftmost wall is $t=0$, the Action at the
leftmost wall would be:

$S=\int_{0}^{0}dtT=0$

Therefore, a better way to write the previous equation at $t=0$ is:

$\psi(x,t=0)=N\exp[-\frac{x^{2}}{2\sigma_{x}^{2}}]$ (110)

Following the Path Integral Formulism (54), in order to know the know the
wavefunction $\psi(x,t)$ at some later time $t$ we will need to utilize the
Kernel (57). The Kernel $\bra{x}e^{-\frac{i}{\hbar}t\hat{H}}\ket{x_{0}}$ will
evolve the system to the time of interest.

$\bra{x}\ket{\psi(t)}=\int
dx_{0}\bra{x}e^{-\frac{i}{\hbar}t\hat{H}}\ket{x_{0}}\psi(x_{0},0)$

We will utilize a very special trick in order to evaluate the above
expression. Since we are dealing with a free particle, there are no
potentials, or $V(x,y)=0$ (hence no forces or accelerations). Because of this,
we do not have to use the full, intimidating form of the Feynman Kernel (57).
In this semiclassical treatment there is no acceleration of the quantum
particle in the y direction, and so the Kernel can be written in a much
simpler form.

Instead of taking on the form (57), it will instead have the form (56). This
is because we can treat the final position of the wavefunction as $q_{j+1}$
and the initial position as $q_{j}$ if there is no acceleration. The term
$q_{j+1}-q_{j}$ simply becomes one large step instead of a miniscule step as
before. This is analogous to how in kinematics, when there is no acceleration,
the formula for instantaneous velocity is $v=\Delta x/\Delta t$, even though
in every other case the formula gives you average velocity (as it is a slope
formula). You can also think of this simplification as taking (57) and setting
$N=1$ (one large step due to no acceleration). Please note that this is not an
approximation, but this will give us an exact expression.

Our Kernel $K$ (not to be confused with the Koopman Generator) is therefore:

$K=\sqrt{\frac{m}{2\pi
i\hbar(t-t_{0})}}\exp[\frac{im(x(t)-x_{0})^{2}}{2\hbar(t-t_{0})}]$ (111)

And then we can write the following Path Integral:

$\bra{x}\ket{\psi(t)}=\int_{-\infty}^{\infty}dx_{0}\sqrt{\frac{m}{2\pi
i\hbar(t-t_{0})}}\exp[\frac{im(x(t)-x_{0})^{2}}{2\hbar(t-t_{0})}]N\exp[-\frac{x_{0}^{2}}{2\sigma_{x}^{2}}]$

We want to evaluate the path integral up until the point it reached the middle
wall with the double slits, at time $t_{M}$. Remembering that we have set the
initial time $t_{0}$ equal to 0 and treating $t$ and $x(t)$ as constants, we
can evaluate the integral. First, we will pull out all constants and absorb
them into $N$, the renormalization constant (recall the renormalization
condition described in section 2.3). Then we do the following algebraic
manipulations, in preparation to evaluating the integral itself:

$\exp[\frac{im(x(t)-x_{0})^{2}}{2\hbar
t}]\exp[-\frac{x_{0}^{2}}{2\sigma_{x}^{2}}]=\exp[\frac{im(x(t)^{2}-2x(t)x_{0}+{x_{0}}^{2})}{2\hbar
t}-\frac{x_{0}^{2}}{2\sigma_{x}^{2}}]=$ $=\exp[\Big{(}\frac{im}{2\hbar
t}-\frac{1}{2\sigma_{x}^{2}}\Big{)}x_{0}^{2}-\frac{imx(t)}{\hbar
t}x_{0}+\frac{imx(t)^{2}}{2\hbar t}]$

We can then use Theorem VIII of Appendix A to evaluate the integral. Comparing
the two, we can recognize the following coefficients of the Gaussian integral:

$a=-\Big{(}\frac{im}{2\hbar t}-\frac{1}{2\sigma_{x}^{2}}\Big{)}$
$b=-\frac{imx(t)}{\hbar t}$ $c=\frac{imx(t)^{2}}{2\hbar t}$

And so by the Theorem, the integral turns out to be:

$N\int_{-\infty}^{\infty}dx_{0}\exp[\Big{(}\frac{im}{2\hbar
t}-\frac{1}{2\sigma_{x}^{2}}\Big{)}x_{0}^{2}-\frac{imx(t)}{\hbar
t}x_{0}+\frac{imx(t)^{2}}{2\hbar t}]=$
$=N\sqrt{\frac{\pi}{-\Big{(}\frac{im}{2\hbar
t}-\frac{1}{2\sigma_{x}^{2}}\Big{)}}}\exp[\frac{(\frac{imx(t)}{\hbar
t})^{2}}{-4\Big{(}\frac{im}{2\hbar
t}-\frac{1}{2\sigma_{x}^{2}}\Big{)}}]\exp[\frac{imx(t)^{2}}{2\hbar t}]=$
$=N\sqrt{\frac{2\pi\hbar t\sigma_{x}^{2}}{-im\sigma_{x}^{2}+\hbar
t}}\exp[\frac{m^{2}x(t)^{2}\sigma_{x}^{2}}{2\hbar t(-im\sigma_{x}^{2}+\hbar
t)}]\exp[\frac{imx(t)^{2}}{2\hbar t}]$

$\therefore\bra{x}\ket{\psi(t)}=N\sqrt{\frac{2\pi\hbar
t\sigma_{x}^{2}}{-im\sigma_{x}^{2}+\hbar
t}}\exp[-\frac{1}{2}\frac{mx(t)^{2}}{m\sigma_{x}^{2}+i\hbar t}]$

The above wavefunction $\bra{x}\ket{\psi(t)}=\psi(x,t)$ is true for anytime
$0<t\leq t_{M}$, where $t_{M}$ is the time to reach the middle barrier with
the two slits. This is because the above wavefunction is that of a free
particle, and once it encounters a barrier (or potential), the particle is
obviously no longer free.

Next, since the expression $\psi(x,t_{M})$ gives you the quantum wavefunction
at the double slits, we have to take into consideration the portion of the
wavefunction that goes through the double slits. To describe the wavefunction
going through the double slit we will utilize step functions. The step
functions will closely model the fact that the wavefunction that hits the
middle wall and stops will not be important in creating any effects on the
other side of the wall. Only the wavefunction that goes through the two slits
will be important in creating the diffraction pattern on the other side ([20];
[14]). The step function, also called the $\theta$-Heaveside function, is
given by:

$H(x)=\left\\{\begin{array}[]{ll}1&\quad x~{}\textgreater~{}0\\\ 0&\quad x\leq
0\end{array}\right.$ (112)

Each slit has a width of $2\delta$. The first slit has width
$\Delta_{1}=(x_{A}-\delta,x_{A}+\delta)$ and the second slit has spans
$\Delta_{2}=(-x_{A}-\delta,-x_{A}+\delta)$ (following the notation of [14]).

Then, the wavefunction right after the two slits can be written as

$\psi(x,t_{M}+dt)=N\psi(x,t_{M})[C_{1}(x)+C_{2}(x)]=N\exp[-\frac{1}{2}\frac{mx^{2}}{m\sigma_{x}^{2}+i\hbar
t}][C_{1}(x)+C_{2}(x)]$ (113)

where the following functions can be defined:

$C_{1}(x)=H(x-x_{A}+\delta)-H(x-x_{A}-\delta)$ (114)
$C_{2}(x)=H(x+x_{A}+\delta)-H(x+x_{A}-\delta)$ (115)

This is done in order to filter out the effects of the wavefunction that
actually goes through the slits. We do not care for the portion of the
wavefunction that does not go through the double slit. They will produce no
effects. The above equations make certain that only the wavefunction that goes
through slits $\Delta_{1}$ and $\Delta_{2}$ will be studied. The $N$ in (113)
is a renormalization constant, so that $\int dx\psi^{*}\psi=1$ (the
renormalization condition described in section 2.3).

Using our Kernel for the free particle again, we will evaluate the