wavefunction using the path integral for the region after the two slits. We
again utilize (111). Notice that the initial wavefunction $\psi(x_{0},0)$ in
that equation starts at time zero, so we will need to adjust the time at the
double slit wall so that it is set to zero. We can do this by substituting the
variable $t$ in the Kernel (111) with $t-t_{M}$. This can be done because time
has the unique property that any point can be arbitrarily set to $t=0$, and by
doing this swap we effectively set the time at the middle wall equal to zero
(starts at $t_{M}$ and evolves from there). So the integral we need to
evaluate will have the following form:

$\bra{x}\ket{\psi(t)}=N\int
dx_{M}\exp[\frac{im(x(t)-x_{M})^{2}}{2\hbar(t-t_{M})}]\psi(x_{M},t_{M})=$
$=N\int
dx_{M}\exp[\frac{im(x(t)-x_{M})^{2}}{2\hbar(t-t_{M})}]\exp[-\frac{1}{2}\frac{mx_{M}^{2}}{m\sigma_{x}^{2}+i\hbar
t_{M}}][C_{1}(x_{M})+C_{2}(x_{M})]$ (116)

where all constants, as is the custom, have been absorbed into $N$.

This expression above cannot be evaluated explicitly. However, because of the
properties of the step functions ([14]), it can be rewritten as

$\bra{x}\ket{\psi(t)}=N\Big{(}\int_{x_{A}-\delta}^{x_{A}+\delta}dx_{M}\exp[\frac{im(x(t)-x_{M})^{2}}{2\hbar(t-t_{M})}-\frac{1}{2}\frac{mx_{M}^{2}}{m\sigma_{x}^{2}+i\hbar
t_{M}}]+$
$+\int_{-x_{A}-\delta}^{-x_{A}+\delta}dx_{M}\exp[\frac{im(x(t)-x_{M})^{2}}{2\hbar(t-t_{M})}-\frac{1}{2}\frac{mx_{M}^{2}}{m\sigma_{x}^{2}+i\hbar
t_{M}}]\Big{)}$ (117)

You can consider each of the two integrals to represent an independent
wavefunction, so that you can rewrite the whole expression as

$\bra{x}\ket{\psi(t)}=N[\psi_{1}(x)+\psi_{2}(x)]$

You can think of $\psi_{1}$ as being the wavefunction coming out of
$\Delta_{1}$ and $\psi_{2}$ as the wavefunction associated with $\Delta_{2}$.

The equation (117) shows us that the phases of the two wavefunctions
$\psi_{1}$ and $\psi_{2}$ are out of phase. You can see this because the
integration is over two different intervals, so the two phases will turn out
different. This is significant, because wavefunctions out of phase is
characteristic of interference. When two wavefunctions with different phases
overlap, they form the characteristic interference pattern ([20]; [14]).

If we want the probability of finding a particle at the rightmost wall (the
interference wall), we need to know the wavefunction $\bra{x}\ket{\psi(t)}$ at
the rightmost wall. If it takes time $t_{R}$ to reach this wall, that
wavefunction can be simply represented as
$\bra{x}\ket{\psi(t_{R})}=\psi(x,t_{R})$ (equivalent to taking (117) and
plugging in $t=t_{R}$). Then, according to the postulates of section 2.5, we
just have to apply the Born Rule to find the probability density as a function
of $x$:

$\rho(x,t=t_{R})=\psi(x,t_{R})^{*}\psi(x,t_{R})=\bra{\psi(t_{R})}\ket{x}\bra{x}\ket{\psi(t_{R})}=$
$=\psi_{1}(x,t_{R})^{*}\psi_{1}(x,t_{R})+\psi_{2}(x,t_{R})^{*}\psi_{2}(x,t_{R})+\psi_{1}(x,t_{R})^{*}\psi_{2}(x,t_{R})+\psi_{2}(x,t_{R})^{*}\psi_{1}(x,t_{R})$

The last two terms are not identically zero ([14]). If we plot $\rho$ vs $x$,
we will get the typical interference pattern we expected, as seen in Figure
2b.

Figure 2: a. Left: Graph of KvN probability distribution. b. Right: Graph of
QM probability distribution. Adapted from [13]. Figure 3: a. Left: Example
graph of a single variable gaussian curve. b. Right: Example graph of a double
variable gaussian curve. Adapted from [24].

This, of course, comes as no surprise. We are still dealing with quantum
theory. Following the same steps, though, will we get the same results with
KvNM? We will follow a similar series of steps to evaluate classical KvN
theory.

For the KvNM case, we will use a double gaussian function because we are
making considerations with both the position and momentum (KvNM is a phase
space theory). A double gaussian (a gaussian in x and y coordinates; see
Figure 3b) takes on the mathematical form:

$f(u,v)=\frac{1}{\sqrt{\pi{\sigma_{u}}{\sigma_{v}}}}e^{-(u-\mu_{u})^{2}/{2{\sigma_{u}}^{2}}}e^{-(v-\mu_{v})^{2}/{2{\sigma_{v}}^{2}}}$
(118)

We are interested in the interference pattern that lies along the $x$
direction. Taking the KvN wavefunction
$\psi(x,p_{x},t)=R(x,p_{x},t)e^{iG(x,p_{x},t)}$ (like in section 5.1) and
plugging in (118) for $R$ at $t=0$ we will have

$\psi(x,p_{x},t=0)=\frac{1}{\sqrt{\pi\sigma_{x}\sigma_{p}}}\exp[-\frac{x^{2}}{2{\sigma_{x}}^{2}}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}+iG(x,p_{x},t=0)]$
(119)

Please note here that $\sigma_{p}$ refers to the momentum in the $x$
direction. Following the Path Integral Formulism, in order to know the know
the wavefunction $\psi$ at some later time $t$ we will need to utilize the
Kernel. For this situation, the travelling classical particle is free of any
potential $V(x,y)$, so we will need the free particle wavefunction. As
$V(x,y)=0$, there will be no derivatives of the potential. We can take the
results (75) and (76) to get a straightforward and concise Kernel, which will
make the calculation we are about to do simple.

We can take (75) and rewrite it in the following way (dropping the constant
$2\pi$, as it can be normalized away, and replacing $q$ with the desired
variable $x$):

$\delta\Big{(}\frac{\Delta t}{m}(p_{j}-m\frac{x_{j+1}-x_{j}}{\Delta
t})\Big{)}=\delta\Big{(}\frac{p_{j}\Delta t}{m}-x_{j+1}+x_{j}\Big{)}$

In the continuum limit, we can just simply write this as:

$=\delta\Big{(}\frac{p_{0}t}{m}+x_{0}-x\Big{)}$

where $x_{0}$ is the initial position of the particle, $p_{0x}$ is the initial
momentum, and $m$ is the mass as before. This comes straight from elementary
kinematics, as since there is no potential, the derivative of the potential
must be 0, so there is no acceleration:

$F_{x}=-\frac{\partial V}{\partial x}=ma_{x}\Rightarrow a_{x}=0$

$x(t)=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2}\Rightarrow
x(t)=x_{0}+v_{0x}t=x_{0}+\frac{p_{0x}t}{m}$

Using the momentum relation $p_{x}=mv_{x}$ on the last step. In other words,
the Dirac delta functional is enforcing classical position evolution, $x(t)$.

The same can be done with (76) to get:

$\frac{1}{\Delta t}\delta\Big{(}\frac{p_{j+1}-p_{j}}{\Delta
t}+V^{\prime}(q_{j})\Big{)}=\delta\Big{(}p_{j+1}-p_{j}\Big{)}$

In the continuum limit remembering the direction is in $x$:

$=\delta\Big{(}p_{x}-p_{0x}\Big{)}$

This makes sense, considering that there are no forces present (because
potential is zero), and hence the momentum should be constant. This Dirac
delta functional enforces classical momentum conservation.

The Kernel $K$ can therefore be succinctly represented as

$K=\delta\Big{(}p_{x}-p_{0x}\Big{)}\delta\Big{(}x-\frac{p_{0x}t}{m}-x_{0}\Big{)}$
(120)

And plugging this Kernel into the original equation (68) shows us that indeed
it provides us with the correct result that we desire, forcing classical
evolution of the KvN wavefunction:

$\bra{x,p_{x}}\ket{\psi(t)}=\int
dx_{0}~{}dp_{0x}\delta\Big{(}p_{x}-p_{0x}\Big{)}\delta\Big{(}x-\frac{p_{0x}t}{m}-x_{0}\Big{)}\psi(x_{0},p_{0x},0)=$
$=\psi(x-\frac{p_{x}t}{m},p_{x},0)=\psi(x,p_{x},t),$

utilizing the property (19) . This statement checks out, as we know that
$\bra{x,p}\ket{\psi(t)}=\psi(x,p,t)$.

Returning to (119), we will evaluate the same integral utilizing the Kernel.
By equation (68), the evolution of the wavefunction from its original state
when $t=0$ is given by

$\bra{x,p_{x}}\ket{\psi(t)}=\frac{1}{\sqrt{\pi\sigma_{x}\sigma_{p}}}\int
dx_{0}~{}dp_{0x}\delta\Big{(}p_{x}-p_{0x}\Big{)}\delta\Big{(}x-\frac{p_{0x}t}{m}-x_{0}\Big{)}\times$
$\times\exp[-\frac{x_{0}^{2}}{2{\sigma_{x}}^{2}}-\frac{p_{0x}^{2}}{2{\sigma_{p}}^{2}}+iG(x_{0},p_{0x},0)]=$
$=\frac{1}{\sqrt{\pi\sigma_{x}\sigma_{p}}}\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}+iG\Big{(}x-\frac{p_{x}t}{m},p_{x}\Big{)}]$
(121)

Note that the above is strictly in terms of the position and momentum in the
$x$ direction. The above equation is true at least until it hits the middle
wall with the slits, because up until that point the wavefunction is that of a
free particle. Using geometric considerations, we can figure out the
wavefunction at the double slit wall. Since there is no acceleration in the
direction of travel (as $-\partial V/\partial y=0$), we know from kinematics:

$v_{y}=\frac{p_{0y}}{m}=\frac{\Delta y}{\Delta
t}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}$

Under the assumption that $y(0)=0$ we can write:

$\frac{y_{M}}{t}=\frac{p_{0y}}{m}\Rightarrow t_{M}=\frac{y_{M}m}{p_{0y}},$
(122)

where $t_{M}$ is the time to the double slit wall. We can substitute this
value into (121) to know what is the classical wavefunction at that point in
time.

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

$\psi(x,p_{x},t_{M}+dt)=N\psi(x,p_{x},t_{M})[C_{1}(x)+C_{2}(x)]$ (123)

Utilizing, as before, the functions (114) and (115) we had defined in the
quantum picture above. 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 (123) is a renormalization constant, similar to before, so
that $\int dx~{}dp_{x}\psi^{*}\psi=1$ (the renormalization condition for the
KvN wavefunction.)

Using our Kernel for the free particle again, we will evaluate the
wavefunction using the path integral for the region after the two slits. We
again utilize (68). Notice that the initial wavefunction $\psi(q_{0},p_{0},0)$
in that equation starts at time zero, so we will need to adjust the time at
the double slit wall so that it is set to zero. We can do this by substituting
the variable $t$ in the Kernel (120) with $t-t_{M}$. This can be done because
time has the unique property that any point can be arbitrarily set to $t=0$,
and by doing this swap we effectively set the time at the middle wall equal to
zero (starts at $t_{M}$ and evolves from there). So the integral we need to
evaluate has the following form:

$\bra{x,p_{x}}\ket{\psi(t-t_{M})}=\int
dx_{M}~{}dp_{Mx}\delta\Big{(}p_{x}-p_{Mx}\Big{)}\delta\Big{(}x-\frac{p_{Mx}(t-t_{M})}{m}-x_{M}\Big{)}\times$
$\times N\psi(x_{M},p_{Mx},t_{M})[C_{1}(x_{M})+C_{2}(x_{M})]$

Plugging everything in:

$\bra{x,p_{x}}\ket{\psi(t-t_{M})}=N\int
dx_{0}~{}dp_{0x}\delta\Big{(}p_{x}-p_{0x}\Big{)}\delta\Big{(}x-\frac{p_{0x}(t-t_{M})}{m}-x_{0}\Big{)}\times$
$\times\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x_{0}-\frac{p_{0x}t_{M}}{m}}\Big{)}^{2}-\frac{p_{0x}^{2}}{2{\sigma_{p}}^{2}}+iG\Big{(}x_{0}-\frac{p_{0x}t_{M}}{m},p_{0x}\Big{)}][C_{1}(x_{0})+C_{2}(x_{0})]$

And then using (19) to evaluate:

$\bra{x,p_{x}}\ket{\psi(t-t_{M})}=Nexp\Big{[}-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}(t-t_{M})}{m}-\frac{p_{x}t_{M}}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}+$
$+iG\Big{(}x-\frac{p_{x}(t-t_{M})}{m}-\frac{p_{x}t_{M}}{m},p_{x}\Big{)}\Big{]}\Big{[}C_{1}\Big{(}x-\frac{p_{x}(t-t_{M})}{m}\Big{)}+C_{2}\Big{(}x-\frac{p_{x}(t-t_{M})}{m}\Big{)}\Big{]}=$
$=N\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}+iG\Big{(}x-\frac{p_{x}t}{m},p_{x}\Big{)}]\times$
$\times\Big{[}C_{1}\Big{(}x-\frac{p_{x}(t-t_{M})}{m}\Big{)}+C_{2}\Big{(}x-\frac{p_{x}(t-t_{M})}{m}\Big{)}\Big{]}$
(124)

Similar to (122), under geometric considerations one can show that the time to
the back wall is $t_{R}=my_{R}/p_{0y}$, and so we can substitute this and
(122) into the above expression to get the expression for the wavefunction at
the back wall (wall where we should see the interference pattern emerge in
QM).

We want to know the probability distribution for finding a particle at
coordinate $x$ at this back wall ([14]). We have no need of the momentum
information, as the interference pattern is a shape that appears in space. So
what we can do is get the probability distribution in terms of $x$ is the
following:

$P(x)=\int_{-\infty}^{\infty}dp_{x}|\psi(x,p_{x},t_{R}-t_{M})|^{2}=\int_{-\infty}^{\infty}dp_{x}\bra{\psi(t_{R}-t_{M})}\ket{x,p_{x}}\bra{x,p_{x}}\ket{\psi(t_{R}-t_{M})}$

Plugging in (124) for $\bra{x,p_{x}}\ket{\psi(t_{R}-t_{M})}$ we get:

$P(x)=N\int_{-\infty}^{\infty}dp_{x}\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t_{R}}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}]^{2}\exp[-iG\Big{(}x-\frac{p_{x}t_{R}}{m},p_{x}\Big{)}]\times$
$\times\exp[iG\Big{(}x-\frac{p_{x}t_{R}}{m},p_{x}\Big{)}]\Big{[}C_{1}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}+C_{2}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}\Big{]}^{2}=$
$=N\int_{-\infty}^{\infty}dp_{x}\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t_{R}}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}]^{2}\times$
$\times\Big{[}C_{1}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}+C_{2}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}\Big{]}^{2}$

We will use a very special property of the $\theta$-Heavyside step function at
this point. We know that the following important property is true ([14]):

$(C_{1}+C_{2})^{2}=C_{1}+C_{2}$

Therefore:

$P(x)=N\int_{-\infty}^{\infty}dp_{x}\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t_{R}}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}]^{2}\times$
$\times\Big{[}C_{1}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}+C_{2}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}\Big{]}$

This last equation is very important, because it shows that no interference
pattern will emerge from the classical KvN wavefunction. If you look at the
above equation closely, you can see that the probability distribution of $x$
is a superposition of two probability distributions:

$P(x)=N\int_{-\infty}^{\infty}dp_{x}\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t_{R}}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}]^{2}C_{1}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}+$
$+N\int_{-\infty}^{\infty}dp_{x}\exp[-\frac{1}{2\sigma_{x}^{2}}\Big{(}{x-\frac{p_{x}t_{R}}{m}}\Big{)}^{2}-\frac{p_{x}^{2}}{2{\sigma_{p}}^{2}}]^{2}C_{2}\Big{(}x-\frac{p_{x}(t_{R}-t_{M})}{m}\Big{)}$

The first integral in the above expression is the probability at $x$ if only
the first slit $\Delta_{1}$ was opened and the second slit $\Delta_{2}$ was
closed, and the second integral in the above expression is the probability at
$x$ if only the second slit $\Delta_{2}$was opened and the first slit
$\Delta_{1}$ was closed. This can be seen from how $C_{1}$ filters out any
portion of the wavefunction that does not go through $\Delta_{1}$ and $C_{2}$
filters out any portion of the wavefunction that does not go through
$\Delta_{2}$. This means that the total probability distribution is the sum of
the probability distribution of each slit, which is exactly what you would
expect for particle instead of wave behavior (no interference effects).

Notice also that the phase of the wavefunction has completely disappeared from
the above equation. You will see this is important, because for interference
effects we will want the phase to play an active role. Waves must be out of
phase in order to produce interference effects, not have the phase cancel each
other out like in the above equations.

If you graph the above function, you will see that indeed no interference
fringes were produced (see Figure 2a). Koopman wavefunctions therefore
successfully predict the behavior of classical particles. Despite their
complex nature in a Hilbert space, they do no produce any interference
effects. All the interference effects completely disappear. It shows how
wavefunctions can describe classical physics.

##  6 Miscellaneous KvN Topics of Interest

###  6.1 Operational Dynamic Modeling

The history of physics is full of failed models, models that did not stand the
test of time. Physics is often carried out in a trial and error fashion, a
process that can be repetitive, messy, and slow. This raises the question if
discovery and model building in physics can be optimized, for the sake of
speeding up the progress of discovery.

One such optimization model is called Operation Dynamical Modeling (ODM) and
utilizes KvNM as an essential ingredient ([4]). The two main pillars of modern
physics is QM and CM. 222222Note, General Relativity and Special Relativity
are here considered to be part of CM. Since we have seen that both QM and CM
must obey the Ehrenfest Theorems, any particular model must conform to either
QM or CM to the best of our knowledge, and if it falls outside either pillar,
the model is quite likely false. If repeated measurements influence the
outcome of the experiment, one knows that there must be a noncommutative
algebra involved (i.e., there must be a nonzero commutator involved, like in
QM.) If repeated measurements do not generally alter the outcome of the
experiment in the fashion of QM, you can assume a commutator of 0 between all
variables involved.

Since the Ehrenfest Theorems deal with dynamical weighted averages
(expectation values), this is a statistical test of different models. If a
model does not obey the Ehrenfest Theorems, it is quite likely to be false,
and progress can be saved by either fixing the model so that it is inline with
the fundamental Ehrenfest Theorems or scrapping the model.

Not only does ODM allow quantum and classical physics, it also explains the
nexus of the two. Semi-classical systems are incorporated into ODM, as they
must also obey the Ehrenfest Theorems. By construction, one could create a
commutator that encompasses both the classical and quantum worlds, such as
[$\hat{q}$, $\hat{p}$] = $i\hbar\kappa$ where $\kappa$ is a constant that lies
between 0 (classical world) and 1 (purely quantum world). By taking the limit
of $\kappa$ goes to 0, one should be able to recover either CM.

This melding, in fact, results in the derivation of the semi-classical Wigner
quasiprobability distribution. This function is known to accurately describe
the interface between the quantum world and the classical world. As the
Schrödinger equation is generally written in terms of configuration space, the
Wigner quasiprobability distribution is important because it recasts QM into a
phase space theory. The Wigner function has a few odd properties, including
the fact that it can take on negative probabilities, but thanks to ODM some
light has been shed to the reasons why this might be the case (as we will
discuss later.)

We will begin with Stone’s Theorem again like we did in section 3, i.e.
$i\frac{\partial}{\partial t}\ket{\psi}=\hat{H}\ket{\psi}$, and try to create
an operator $\hat{H}_{QC}$ which will encompass both the classical and quantum
worlds using our newly constructed commutator [$\hat{q}$, $\hat{p}$] =
$i\hbar\kappa$, $0\leq\kappa\leq 1$.

We must have a unified quantum-classical algebra which gives us the Koopman-
von Neumann Algebra (43) when $\kappa=0$ and a quantum algebra for $\kappa=1$.
We desire:

$\lim\limits_{\kappa\to 0}\hat{q}_{QC}=\hat{q}_{c}$ $\lim\limits_{\kappa\to
0}\hat{p}_{QC}=\hat{p}_{c}$ $\lim\limits_{\kappa\to
0}\hat{H}_{QC}=\hbar\hat{K}$

In order to reflect the KvN Algebra, [5] proposed the following unified
Classical-Quantum Algebra:

$[\hat{q}_{QC},\hat{\theta}_{QC}]=[\hat{p}_{QC},\hat{\lambda}_{QC}]=i$ (125)

All other commutators will equal 0. For the unified dynamics, we first begin
with the Ehrenfest Theorems (equations (36) and (37) for reference) as before:

$\frac{d}{dt}\langle
q\rangle=\frac{d}{dt}\bra{\psi(t)}\hat{q}_{QC}\ket{\psi(t)}=\frac{1}{m}\langle
p\rangle=\frac{1}{m}\bra{\psi(t)}\hat{p}_{QC}\ket{\psi(t)}$
$\frac{d}{dt}\langle
p\rangle=\frac{d}{dt}\bra{\psi(t)}\hat{p}_{QC}\ket{\psi(t)}=\langle-V^{\prime}(q)\rangle=\bra{\psi(t)}-V^{\prime}(\hat{q}_{QC})\ket{\psi(t)}$

Following the same series of steps as in section 3, we will obtain the
differential equations:

$\kappa m\frac{\partial H}{\partial p_{QC}}+\frac{m}{\hbar}\frac{\partial
H}{\partial\theta_{QC}}=p_{QC}$ $\kappa\frac{\partial H}{\partial
q_{QC}}-\frac{1}{\hbar}\frac{\partial
H}{\partial\lambda_{QC}}=V^{\prime}(q_{QC})$

The general solution $H$ to the two above differential equation is simply:

$H(\hat{q}_{QC},\hat{p}_{QC},\hat{\theta}_{QC},\hat{\lambda}_{QC})=\frac{1}{\kappa}\Big{[}\frac{\hat{p}_{QC}^{2}}{2m}+V(\hat{q}_{QC})\Big{]}+F(\hat{p}_{QC}-\hbar\kappa\hat{\theta}_{QC},\hat{q}_{QC}+\hbar\kappa\hat{\lambda}_{QC})$

where the function $F$ is a function undetectable to experiment because one
can show that for any observable $A$, you have $[\hat{F},\hat{A}]=0$ ([4]).

The quantum operators can be constructed from classical ones, so [4] propose
that

$\hat{q}_{QC}=\hat{q}_{c}-\hbar\kappa\frac{\hat{\lambda}_{c}}{2}$
$\hat{p}_{QC}=\hat{p}_{c}+\hbar\kappa\frac{\hat{\theta}_{c}}{2}$
$\hat{\theta}_{QC}=\hat{\theta}_{c}$ $\hat{\lambda}_{QC}=\hat{\lambda}_{c}$

It is clear that the above expressions have a nice, smooth classical limit as
$\kappa\rightarrow 0$. All the variables operators become classical operators
in that limit.

To make the limit smooth as you go from the quantum Hamiltonian $\hat{H}$ to
$\hbar\hat{K}$ as $\kappa$ goes from 1 to 0, [4] dervied:

$\hat{H}_{QC}=\frac{1}{\kappa}[\frac{\hat{p}_{QC}^{2}}{2m}+V(\hat{q}_{QC})]-\frac{1}{2m\kappa}(\hat{p}_{QC}-\hbar\kappa\hat{\theta}_{QC})^{2}-\frac{1}{\kappa}V(\hat{q}_{QC}+\hbar\kappa\hat{\lambda}_{QC})$
(126)

Equivalently, you can write by plugging in the quantum-classical operators:

$\hat{H}_{QC}=\frac{\hbar}{m}\hat{p}_{c}\hat{\theta}_{c}+\frac{1}{\kappa}V(\hat{q}_{c}-\frac{\hbar\kappa}{2}\hat{\lambda}_{c})-\frac{1}{\kappa}V(\hat{q}_{c}+\frac{\hbar\kappa}{2}\hat{\lambda}_{c})$

The above expression can be shown to lead to the phase space formulation of
QM, the Wigner quasiprobability distribution. This derivation will be left for
the next section. The Wigner distribution is significant because it is known
to accurately model the ‘nexus’ between quantum and classical worlds ([4];
[5]). It also has a nice classical limit ([4]; [5]). Independently deriving
this distribution indicates that the Ehrenfest Theorems apply at all times,
including in the regime of quantum/classical boundary. ODM, therefore, is a
very successful melding of the two worlds.

ODM suggests that any physical theory will follow KvNM, QM, or the above
melding of the two. KvNM is fundamentally important because any novel
classical theory proposed must be consistent with it. We will see certain
applications of ODM and its successes in the pages that follow. Without KvNM,
there would be no unifying paradigm based in the Ehrenfest Theorems. ODM is a
methodology that will pave for more efficient discovery in Physics.

###  6.2 Wigner Quasiprobability Distribution

The Wigner Quasiprobability Distribution connects our understanding of the
quantum wavefunction to phase space. It is an important area of study
developed by Eugen Wigner in 1932 and independently formulated by another
eminent mathematical physicist, José Enrique Moyal. The Wigner Distribution
has a variety of applications, from quantum optics to quantum computing ([5]).
It has one significantly odd feature though. This distribution allows for
negative probabilities, which is axiomatically impossible by mainstream
probability theories ([5]). ODM sheds some light on this odd feature. The
Wigner Distribution can be derived from the unified dynamics of ODM, and we
will proceed to do so below.

It is very simple to derive the Wigner quasiprobability distribution from
ODM’s unified dynamics. Again, we hark back to Stone’s Theorem, in order to
try to figure out the generator of motion:

$i\hbar\frac{d}{dt}\ket{\psi(t)_{\kappa}}=\hat{H}_{QC}\ket{\psi(t)_{\kappa}}$

We will sandwich it from the left with $\ket{q_{c},\lambda_{c}}$ to get the
$\lambda_{c},q_{c}$ representation. Referring back to the rules for operators
acting on vector kets, we can see that:

$\bra{q,\lambda}\hat{q}_{c}\ket{\psi(t)}=[\bra{q,\lambda}\hat{q}_{c}]\ket{\psi(t)}=q\innerproduct{q,\lambda}{\psi(t)}$
(127)
$\bra{q,\lambda}\hat{\lambda}_{c}\ket{\psi(t)}=[\bra{q,\lambda}\hat{\lambda}_{c}]\ket{\psi(t)}=\lambda\innerproduct{q,\lambda}{\psi(t)}$
(128)

where $q$ and $\lambda$ are the eigenvalues. From Theorem IV of Appendix A, we
can say that $B=p$, $A=\lambda$, $\kappa=-1$ (since
$[\hat{\lambda},\hat{p}]=-i$) 232323This is because
$[\hat{\lambda},\hat{p}]=-[\hat{p},\hat{\lambda}]$ and as indicated earlier,
$[\hat{p},\hat{\lambda}]=i$. and keep in mind that the Classical Commutator is
$[\hat{q}_{c},\hat{p}_{c}]=0$. Therefore:

$\bra{q,\lambda}\hat{p}_{c}\ket{\psi(t)}=-i(-1)\frac{\partial}{\partial\lambda}\innerproduct{q,\lambda}{\psi(t)}=i\frac{\partial}{\partial\lambda}\innerproduct{q,\lambda}{\psi(t)}$
(129)

There is no partial derivative with respect to $q$ as the Classical Commutator
between $\hat{q}$ and $\hat{p}$ is 0. Theorem IV can also be applied in the
situation where $B=\theta$ in order to derive:

$\bra{q,\lambda}\hat{\theta}_{c}\ket{\psi(t)}=-i\frac{\partial}{\partial
q}\innerproduct{q,\lambda}{\psi(t)}$ (130)

Now we create the actual ‘sandwich’ from Stone’s Theorem:

$\bra{q,\lambda}i\hbar\frac{d}{dt}\ket{\psi(t)_{\kappa}}=\bra{q,\lambda}[\frac{\hbar}{m}\hat{p}_{c}\hat{\theta}_{c}+\frac{1}{\kappa}V(\hat{q}_{c}-\frac{\hbar\kappa}{2}\hat{\lambda}_{c})-\frac{1}{\kappa}V(\hat{q}_{c}+\frac{\hbar\kappa}{2}\hat{\lambda}_{c})]\ket{\psi(t)_{\kappa}}$

$i\hbar\frac{d}{dt}\bra{q,\lambda}\ket{\psi(t)_{\kappa}}=\bra{q,\lambda}\frac{\hbar}{m}\hat{p}_{c}\hat{\theta}_{c}\ket{\psi(t)_{\kappa}}+\bra{q,\lambda}\frac{1}{\kappa}V(\hat{q}_{c}-\frac{\hbar\kappa}{2}\hat{\lambda}_{c})\ket{\psi(t)_{\kappa}}+$
$-\bra{q,\lambda}\frac{1}{\kappa}V(\hat{q}_{c}+\frac{\hbar\kappa}{2}\hat{\lambda}_{c})\ket{\psi(t)_{\kappa}}$

Utilizing the previous results for Theorem I (see footnote 13), we can see:

$i\hbar\frac{d}{dt}\bra{q,\lambda}\ket{\psi(t)_{\kappa}}=\frac{\hbar}{m}\bra{q,\lambda}\hat{p}_{c}\hat{\theta}_{c}\ket{\psi(t)_{\kappa}}+\frac{1}{\kappa}V(q-\frac{\hbar\kappa}{2}\lambda)\bra{q,\lambda}\ket{\psi(t)_{\kappa}}-\frac{1}{\kappa}V(q+\frac{\hbar\kappa}{2}\lambda)\bra{q,\lambda}\ket{\psi(t)_{\kappa}}$

Finally utilizing the results from Theorem IV above:

$i\hbar\frac{d}{dt}\bra{q,\lambda}\ket{\psi(t)_{\kappa}}=-i^{2}\frac{\hbar}{m}\frac{\partial^{2}}{\partial\lambda\partial
q}\bra{q,\lambda}\ket{\psi(t)_{\kappa}}+\frac{1}{\kappa}V(q-\frac{\hbar\kappa}{2}\lambda)\bra{q,\lambda}\ket{\psi(t)_{\kappa}}+$
$-\frac{1}{\kappa}V(q+\frac{\hbar\kappa}{2}\lambda)\bra{q,\lambda}\ket{\psi(t)_{\kappa}}$

Let us move all the terms to the left hand side of the equation, so that the
left hand side equals 0 pm the right hand side. Introducing two new variables,
$u=q-\hbar\kappa\lambda/2$ and $v=q+\hbar\kappa\lambda/2$, we will rewrite the
whole equation so it has a prettier form. To eliminate the double partial
derivative, we can take the following steps:

$u-v=q-\frac{\hbar\kappa\lambda}{2}-q-\frac{\hbar\kappa\lambda}{2}=-\hbar\kappa\lambda$
$u+v=q-\frac{\hbar\kappa\lambda}{2}+q+\frac{\hbar\kappa\lambda}{2}=2q$

Therefore the partial derivative can be rewritten:

$\frac{\partial^{2}}{\partial\lambda\partial
q}=\frac{\partial^{2}}{\partial(\frac{u-v}{-\hbar\kappa})\partial(\frac{u+v}{2})}=-2\hbar\kappa\frac{\partial^{2}}{\partial(u-v)\partial(u+v)}=-2\hbar\kappa\Big{[}\frac{\partial^{2}}{\partial
u^{2}}-\frac{\partial^{2}}{\partial v^{2}}\Big{]}$

Recasting the entire equation in terms of $u$ and $v$:

$\Big{[}i\hbar\kappa\frac{\partial}{\partial
t}-\frac{(\hbar\kappa)^{2}}{2m}\Big{(}\frac{\partial^{2}}{\partial
u^{2}}-\frac{\partial^{2}}{\partial
v^{2}}\Big{)}-V(u)+V(v)\Big{]}\innerproduct{q,\lambda}{\psi(t)_{\kappa}}=0$
(131)

Here, $\innerproduct{q,\lambda}{\psi(t)_{\kappa}}$ is equal to the density
matrix $\rho_{\kappa}(u,v,t)$ up to a constant of proportionality $C$, i.e.,
$\rho_{\kappa}(u,v,t)=C\innerproduct{q,\lambda}{\psi(t)_{\kappa}}$.

We can transition to $q,p$ representation to derive the Wigner distribution
from the above using Theorem II with $A=(q,\lambda)$:

$\innerproduct{q,p}{\psi(t)_{\kappa}}=\bra{q,p}I\ket{\psi(t)_{\kappa}}=\int
d\lambda\bra{q,p}\ket{q,\lambda}\bra{q,\lambda}\ket{\psi(t)_{\kappa}}$

Note that we do not integrate over $q$ due to the commutator relationships
$[\hat{q},\hat{q}]=[\hat{q}_{c},\hat{p}_{c}]=0$, whereas
$[\hat{p}_{c},\hat{\lambda}_{c}]=i$. Since
$[\hat{q},\hat{q}]=[\hat{q}_{c},\hat{p}_{c}]=0$ it makes no sense to integrate
over $q$, as might be initially implied by looking at the expressions akin to
(31). Since the other commutator is nonzero, we will retain integration over
$\lambda$, however. We know that
$\rho_{\kappa}(u,v,t)=C\innerproduct{q,\lambda}{\psi(t)_{\kappa}}$, so we will
plug that in.

The question now is to figure out what $\bra{q,p}\ket{q,\lambda}$ is. We will
utilize Theorem IV and the commutator relationships in order to deduce this.
Since $[\hat{q},\hat{q}]=[\hat{q}_{c},\hat{p}_{c}]=0$, solving this expression
is equivalent to solving $\bra{p}\ket{\lambda}$, as the other terms commute
(i.e., the commutator of their operators is equal to 0) and are therefore
irrelevant to this problem. By Theorem IV, we set $A=p$, $B=\lambda$, and
$\kappa=1$. We know that this expression is therefore

$\innerproduct{q,p}{q,\lambda}=\innerproduct{p}{\lambda}=\frac{1}{\sqrt{2\pi}}e^{ip\lambda}$

Putting it all together:

$\innerproduct{q,p}{\psi(t)_{\kappa}}=\int
d\lambda\frac{C}{\sqrt{2\pi}}\rho_{\kappa}(q-\frac{\hbar\kappa\lambda}{2},q+\frac{\hbar\kappa\lambda}{2},t)e^{ip\lambda}$
(132)

The above expression shows us that the wavefunction
$\innerproduct{q,p}{\psi(t)_{\kappa}}$ is equal to the Wigner quasi-
probability distribution up to a constant of proportionality $C$. The Wigner
distribution is very accurate when it comes to describing the boundary of
Quantum and Classical ([4]; [5]), so this is no trivial derivation.

What ODM reveals is that the Wigner “distribution” is therefore a type of
wavefunction for a quantum particle at a point in classical phase space. The
strange property of having negative probabilities can by explained in the ODM
framework as being due to the fact that the Wigner “distribution” is not a
distribution at all! It is a wavefunction. Wavefunctions, unlike probability
distributions, can take on negative values ([5]). The KvN wavefunction can
have negative or imaginary weights to it, whereas traditional Louiville theory
gives only positive, real weights to the same trajectories ([5]).

It is worth to highlight one more interesting result of the above analysis.
Although not directly physically observable, the operators $\hat{\theta}$ and
$\hat{\lambda}$ can be indirectly observed through the Wigner wavefunction.
They are a critical ingredient as the so-called ‘Bopp Operators’.

###  6.3 Time Dependent Harmonic Oscillator

The Harominc Oscillator is one of the most thoroughly studied models in all of
physics. Its application arises in many situations. We are especially
interested in the Time Dependent Harmonic Oscillator, where factors like the
mass or frequency might change with time. Because of the common mathematical
basis for QM and KvNM, it should come to no surprise that common sets of tools
can be used to investigate both. The Harmonic Oscillator is no exception. The
operatorial tools to investigate quantum oscillators can be used to explore
classical (KvN) oscillators and vise versa.

For a classical or quantum Time Dependent Harmonic Oscillator we need a
quadratic potential of the following form:

$V(\hat{q},t)=k(t)\frac{\hat{q}^{2}}{2}$

The above potential is modelled off of the classical spring potential. An
ideal block-mass and spring system can undergo harmonic oscillation. The
spring potential is given by $V=(1/2)kq^{2}$ with spring constant $k$ and
displacement $q$. Here, we add an extra layer of complexity by making the
spring constant time dependent. This is our model potential for all Time
Dependent Harmonic Oscillation, whether quantum or classical.

In the quantum case, we would plug in the above expression for the potential
term into the Schrödinger equation to solve for the Quantum Oscillator. In the
classical case, we plug in the derivative of the above expression into the
Koopman Generator (45):

$\frac{\partial}{\partial\hat{q}}V(\hat{q},t)=\frac{\partial}{\partial\hat{q}}k(t)\frac{\hat{q}^{2}}{2}=k(t)\hat{q}=V^{\prime}(\hat{q})$
$i\frac{\partial}{\partial
t}\ket{\psi(t)}=\Big{[}\frac{\hat{p}\hat{\theta}}{m}-V^{\prime}(\hat{q})\hat{\lambda}\Big{]}\ket{\psi(t)}=\Big{[}\frac{\hat{p}\hat{\theta}}{m}-k(t)\hat{\lambda}\hat{q}\Big{]}\ket{\psi(t)}.$

Similar to what we did before (50) in section 3, we can attack both sides with
the bra vector $\bra{q,p}$ to get:

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

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

$i\frac{\partial}{\partial
t}\psi(q,p,t)=\Big{[}\frac{\hat{p}\hat{\theta}}{m}-k(t)\hat{\lambda}\hat{q}\Big{]}\psi(q,p,t)$
(133)

where we colloquially say that $\hat{q}=q$, $\hat{p}=p$,
$\hat{\lambda}=-i\frac{\partial}{\partial p}$, and
$\hat{\theta}=-i\frac{\partial}{\partial q}$, just as we derived in section 3.
Notice as for the Path Integral (section 4.2), we choose to set the constant
$C$ to zero in the Koopman Generator for simplicity’s sake.

One way to solve for the equations of motion of a Harmonic Oscillator is to
use Ermakov-Lewis invariants ([11]). If you have a Hamiltonian of the form

$\hat{H}=\frac{1}{2}[\hat{p}^{2}+\Omega^{2}(t)\hat{q}^{2}],$

where $\Omega(t)$ is an arbitrary function with time dependence, you can use
the invariant to write a transformation where the time dependent term
$\Omega(t)$ completely disappears. This makes solving the problem much
simpler. Like it sounds, the Ermakov-Lewis invariant $\hat{I}$ is an operator
that does not change (stays constant in time). The invariant $\hat{I}$ obeys
the following rule for the quantum system ([11]; [12]):

$\frac{d\hat{I}}{dt}=\frac{\partial\hat{I}}{\partial
t}+\frac{1}{i\hbar}[\hat{I},\hat{H}]=0.$ (134)

$\hat{H}$ is the generator of motion, usually the quantum Hamiltonian, but in
the classical case it will be the Koopman Generator. Based on the Koopman
Algebra (43), [18] deduced the following expression for $\hat{I}$:

$\hat{I}=\alpha_{0}\hat{q}^{2}+\alpha_{1}\hat{\theta}^{2}+\alpha_{2}\hat{p}^{2}+\alpha_{3}\hat{\lambda}^{2}+\alpha_{4}\hat{q}\hat{p}+\alpha_{5}\hat{\theta}\hat{\lambda}$

where all the $\alpha_{j}$ are time dependent. This was a scientific guess
basically, as there is no simple way to derive this fact. Plugging the above
expression into (134), we get:

$\frac{\partial\alpha_{0}}{\partial
t}\hat{q}^{2}+\frac{\partial\alpha_{1}}{\partial
t}\hat{\theta}^{2}+\frac{\partial\alpha_{2}}{\partial
t}\hat{p}^{2}+\frac{\partial\alpha_{3}}{\partial
t}\hat{\lambda}^{2}+\frac{\partial\alpha_{4}}{\partial
t}\hat{q}\hat{p}+\frac{\partial\alpha_{5}}{\partial
t}\hat{\theta}\hat{\lambda}+\frac{1}{i\hbar}[\hat{I},\hat{K}]=0$ (135)

Evaluating $[\hat{I},\hat{K}]$, we first can do (13):

$[\hat{I},\hat{K}]=\Big{[}\hat{I},\frac{\hat{p}\hat{\theta}}{m}-k(t)\hat{\lambda}\hat{q}\Big{]}=\Big{[}\hat{I},\frac{\hat{p}\hat{\theta}}{m}\Big{]}-\Big{[}\hat{I},k(t)\hat{\lambda}\hat{q}\Big{]}$
(136)

Taking the first term and applying (11) and (13):

$\Big{[}\hat{I},\frac{\hat{p}\hat{\theta}}{m}\Big{]}=-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\hat{I}\Big{]}=-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{0}\hat{q}^{2}+\alpha_{1}\hat{\theta}^{2}+\alpha_{2}\hat{p}^{2}+\alpha_{3}\hat{\lambda}^{2}+\alpha_{4}\hat{q}\hat{p}+\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=$
$=-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{0}\hat{q}^{2}\Big{]}-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{1}\hat{\theta}^{2}\Big{]}-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{2}\hat{p}^{2}\Big{]}-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{3}\hat{\lambda}^{2}\Big{]}-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{4}\hat{q}\hat{p}\Big{]}-\Big{[}\frac{\hat{p}\hat{\theta}}{m},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=$
$=-\frac{\hat{p}}{m}\Big{[}\hat{\theta},\alpha_{0}\hat{q}^{2}\Big{]}-\Big{[}\frac{\hat{p}}{m},\alpha_{0}\hat{q}^{2}\Big{]}\hat{\theta}-\frac{\hat{p}}{m}\Big{[}\hat{\theta},\alpha_{1}\hat{\theta}^{2}\Big{]}-\Big{[}\frac{\hat{p}}{m},\alpha_{1}\hat{\theta}^{2}\Big{]}\hat{\theta}-\frac{\hat{p}}{m}\Big{[}\hat{\theta},\alpha_{2}\hat{p}^{2}\Big{]}-\Big{[}\frac{\hat{p}}{m},\alpha_{2}\hat{p}^{2}\Big{]}\hat{\theta}+$
$-\frac{\hat{p}}{m}\Big{[}\hat{\theta},\alpha_{3}\hat{\lambda}^{2}\Big{]}-\Big{[}\frac{\hat{p}}{m},\alpha_{3}\hat{\lambda}^{2}\Big{]}\hat{\theta}-\frac{\hat{p}}{m}\Big{[}\hat{\theta},\alpha_{4}\hat{q}\hat{p}\Big{]}-\Big{[}\frac{\hat{p}}{m},\alpha_{4}\hat{q}\hat{p}\Big{]}\hat{\theta}-\frac{\hat{p}}{m}\Big{[}\hat{\theta},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}-\Big{[}\frac{\hat{p}}{m},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}\hat{\theta}$
(137)

Utilizing our nifty identities (10) and (12) and Koopman Algebra again the
following can be said about the above commutators:

$\Big{[}\hat{\theta},\alpha_{0}\hat{q}^{2}\Big{]}=\alpha_{0}\hat{q}\Big{[}\hat{\theta},\hat{q}\Big{]}+\Big{[}\hat{\theta},\alpha_{0}\hat{q}\Big{]}\hat{q}=-2i\alpha_{0}\hat{q}$

$\Big{[}\hat{\theta},\alpha_{1}\hat{\theta}^{2}\Big{]}=\alpha_{1}\hat{\theta}\Big{[}\hat{\theta},\hat{\theta}\Big{]}+\Big{[}\hat{\theta},\hat{\theta}\Big{]}\alpha_{1}\hat{\theta}=0$

$\Big{[}\hat{\theta},\alpha_{2}\hat{p}^{2}\Big{]}=\alpha_{2}\hat{p}\Big{[}\hat{\theta},\hat{p}\Big{]}+\Big{[}\hat{\theta},\hat{p}\Big{]}\alpha_{2}\hat{p}=0$

$\Big{[}\hat{\theta},\alpha_{3}\hat{\lambda}^{2}\Big{]}=\alpha_{3}\hat{\lambda}\Big{[}\hat{\theta},\hat{\lambda}\Big{]}+\Big{[}\hat{\theta},\hat{\lambda}\Big{]}\alpha_{3}\hat{\lambda}=0$

$\Big{[}\hat{\theta},\alpha_{4}\hat{q}\hat{p}\Big{]}=\alpha_{4}\hat{q}\Big{[}\hat{\theta},\hat{p}\Big{]}+\Big{[}\hat{\theta},\hat{q}\Big{]}\alpha_{4}\hat{p}=-i\alpha_{4}\hat{p}$

$\Big{[}\hat{\theta},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=\alpha_{5}\hat{\theta}\Big{[}\hat{\theta},\hat{\lambda}\Big{]}+\Big{[}\hat{\theta},\hat{\theta}\Big{]}\alpha_{5}\hat{\lambda}=0$
$\Big{[}\frac{\hat{p}}{m},\alpha_{0}\hat{q}^{2}\Big{]}=\alpha_{0}\hat{q}\Big{[}\hat{p}/m,\hat{q}\Big{]}+\Big{[}\hat{p}/m,\hat{q}\Big{]}\alpha_{0}\hat{q}=0$

$\Big{[}\frac{\hat{p}}{m},\alpha_{1}\hat{\theta}^{2}\Big{]}=\alpha_{1}\hat{\theta}\Big{[}\frac{\hat{p}}{m},\hat{\theta}\Big{]}+\Big{[}\frac{\hat{p}}{m},\hat{\theta}\Big{]}\alpha_{1}\hat{\theta}=0$

$\Big{[}\frac{\hat{p}}{m},\alpha_{2}\hat{p}^{2}\Big{]}=\frac{\alpha_{2}}{m}\hat{p}\Big{[}\hat{p},\hat{p}\Big{]}+\Big{[}\hat{p},\hat{p}\Big{]}\frac{\alpha_{2}}{m}\hat{p}=0$

$\Big{[}\frac{\hat{p}}{m},\alpha_{3}\hat{\lambda}^{2}\Big{]}=\frac{\alpha_{3}}{m}\hat{\lambda}\Big{[}\hat{p},\hat{\lambda}\Big{]}+\Big{[}\hat{p},\hat{\lambda}\Big{]}\frac{\alpha_{3}}{m}\hat{\lambda}=\frac{2i\alpha_{3}}{m}\hat{\lambda}$

$\Big{[}\frac{\hat{p}}{m},\alpha_{4}\hat{q}\hat{p}\Big{]}=\frac{\alpha_{4}}{m}\hat{q}\Big{[}\hat{p},\hat{p}\Big{]}+\Big{[}\hat{p},\hat{q}\Big{]}\frac{\alpha_{4}}{m}\hat{p}=0$

$\Big{[}\frac{\hat{p}}{m},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=\frac{\alpha_{5}}{m}\hat{\theta}\Big{[}\hat{p},\hat{\lambda}\Big{]}+\Big{[}\hat{p},\hat{\theta}\Big{]}\frac{\alpha_{5}}{m}\hat{\lambda}=\frac{i\alpha_{5}}{m}\hat{\theta}$

Ergo, the above expression (137) reduces to:

$\Big{[}\hat{I},\frac{\hat{p}\hat{\theta}}{m}\Big{]}=2i\alpha_{0}\frac{\hat{p}\hat{q}}{m}+i\alpha_{4}\frac{\hat{p}^{2}}{m}-2i\alpha_{3}\frac{\hat{\lambda}\hat{\theta}}{m}-i\alpha_{5}\frac{\hat{\theta}^{2}}{m}$
(138)

Now, we do the same for $[\hat{I},k(t)\hat{\lambda}\hat{q}]$:

$\Big{[}\hat{I},k(t)\hat{\lambda}\hat{q}\Big{]}=-\Big{[}k(t)\hat{\lambda}\hat{q},\hat{I}\Big{]}=-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{0}\hat{q}^{2}+\alpha_{1}\hat{\theta}^{2}+\alpha_{2}\hat{p}^{2}+\alpha_{3}\hat{\lambda}^{2}+\alpha_{4}\hat{q}\hat{p}+\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=$
$=-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{0}\hat{q}^{2}\Big{]}-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{1}\hat{\theta}^{2}\Big{]}-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{2}\hat{p}^{2}\Big{]}+$
$-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{3}\hat{\lambda}^{2}\Big{]}-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{4}\hat{q}\hat{p}\Big{]}-\Big{[}k(t)\hat{\lambda}\hat{q},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=$

$=-k(t)\hat{\lambda}\Big{[}\hat{q},\alpha_{0}\hat{q}^{2}\Big{]}-\Big{[}k(t)\hat{\lambda},\alpha_{0}\hat{q}^{2}\Big{]}\hat{q}-k(t)\hat{\lambda}\Big{[}\hat{q},\alpha_{1}\hat{\theta}^{2}\Big{]}+$
$-\Big{[}k(t)\hat{\lambda},\alpha_{1}\hat{\theta}^{2}\Big{]}\hat{q}-k(t)\hat{\lambda}\Big{[}\hat{q},\alpha_{2}\hat{p}^{2}\Big{]}-\Big{[}k(t)\hat{\lambda},\alpha_{2}\hat{p}^{2}\Big{]}\hat{q}+$
$-k(t)\hat{\lambda}\Big{[}\hat{q},\alpha_{3}\hat{\lambda}^{2}\Big{]}-\Big{[}k(t)\hat{\lambda},\alpha_{3}\hat{\lambda}^{2}\Big{]}\hat{q}-k(t)\hat{\lambda}\Big{[}\hat{q},\alpha_{4}\hat{q}\hat{p}\Big{]}+$
$-\Big{[}k(t)\hat{\lambda},\alpha_{4}\hat{q}\hat{p}\Big{]}\hat{q}-k(t)\hat{\lambda}\Big{[}\hat{q},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}-\Big{[}k(t)\hat{\lambda},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}\hat{q}$

We can perform a similar series of steps to achieve:

$\Big{[}\hat{q},\alpha_{0}\hat{q}^{2}\Big{]}=0$

$\Big{[}k(t)\hat{\lambda},\alpha_{0}\hat{q}^{2}\Big{]}=0$

$\Big{[}\hat{q},\alpha_{1}\hat{\theta}^{2}\Big{]}=\alpha_{1}\hat{\theta}\Big{[}\hat{q},\hat{\theta}\Big{]}+\Big{[}\hat{q},\hat{\theta}\Big{]}\alpha_{1}\hat{\theta}=2i\alpha_{1}\hat{\theta}$

$\Big{[}k(t)\hat{\lambda},\alpha_{1}\hat{\theta}^{2}\Big{]}=0$

$\Big{[}\hat{q},\alpha_{2}\hat{p}^{2}\Big{]}=0$

$\Big{[}k(t)\hat{\lambda},\alpha_{2}\hat{p}^{2}\Big{]}=k(t)\alpha_{2}\hat{p}\Big{[}\hat{\lambda},\hat{p}\Big{]}+\Big{[}\hat{\lambda},\hat{p}\Big{]}k(t)\alpha_{2}\hat{p}=-2ik(t)\alpha_{2}\hat{p}$

$\Big{[}\hat{q},\alpha_{3}\hat{\lambda}^{2}\Big{]}=0$

$\Big{[}k(t)\hat{\lambda},\alpha_{3}\hat{\lambda}^{2}\Big{]}=0$

$\Big{[}\hat{q},\alpha_{4}\hat{q}\hat{p}\Big{]}=0$

$\Big{[}k(t)\hat{\lambda},\alpha_{4}\hat{q}\hat{p}\Big{]}=k(t)\alpha_{4}\hat{q}\Big{[}\hat{\lambda},\hat{p}\Big{]}+\Big{[}\hat{\lambda},\hat{q}\Big{]}k(t)\alpha_{4}\hat{p}=-ik(t)\alpha_{4}\hat{q}$

$\Big{[}\hat{q},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=\alpha_{5}\hat{\theta}\Big{[}\hat{q},\hat{\lambda}\Big{]}+\Big{[}\hat{q},\hat{\theta}\Big{]}\alpha_{5}\hat{\lambda}=i\alpha_{5}\hat{\lambda}$

$\Big{[}k(t)\hat{\lambda},\alpha_{5}\hat{\theta}\hat{\lambda}\Big{]}=0$

Therefore, putting it all together:

$\Big{[}\hat{I},k(t)\hat{\lambda}\hat{q}\Big{]}=-2i\alpha_{1}k(t)\hat{\lambda}\hat{\theta}+2ik(t)\alpha_{2}\hat{p}\hat{q}+ik(t)\alpha_{4}\hat{q}^{2}-ik(t)\alpha_{5}\hat{\lambda}^{2}$
(139)

Utilizing equations (136), (138), (139) with (135), we can compute the
invariant to be:

$\frac{\partial\alpha_{0}}{\partial
t}\hat{q}^{2}+\frac{\partial\alpha_{1}}{\partial
t}\hat{\theta}^{2}+\frac{\partial\alpha_{2}}{\partial
t}\hat{p}^{2}+\frac{\partial\alpha_{3}}{\partial
t}\hat{\lambda}^{2}+\frac{\partial\alpha_{4}}{\partial
t}\hat{q}\hat{p}+\frac{\partial\alpha_{5}}{\partial
t}\hat{\theta}\hat{\lambda}+\frac{1}{i\hbar}\Big{(}2i\alpha_{0}\frac{\hat{p}\hat{q}}{m}+i\alpha_{4}\frac{\hat{p}^{2}}{m}-2i\alpha_{3}\frac{\hat{\lambda}\hat{\theta}}{m}-i\alpha_{5}\frac{\hat{\theta}^{2}}{m}+$
$+2i\alpha_{1}k(t)\hat{\lambda}\hat{\theta}-2ik(t)\alpha_{2}\hat{p}\hat{q}-ik(t)\alpha_{4}\hat{q}^{2}+ik(t)\alpha_{5}\hat{\lambda}^{2}\Big{)}=0$
(140)

Combining like terms:

$\Big{(}\frac{\partial\alpha_{0}}{\partial
t}-\frac{k(t)\alpha_{4}}{\hbar}\Big{)}\hat{q}^{2}+\Big{(}\frac{\partial\alpha_{1}}{\partial
t}-\frac{\alpha_{5}}{\hbar
m}\Big{)}\hat{\theta}^{2}+\Big{(}\frac{\partial\alpha_{2}}{\partial
t}+\frac{\alpha_{4}}{\hbar
m}\Big{)}\hat{p}^{2}+\Big{(}\frac{\partial\alpha_{3}}{\partial
t}+\frac{k(t)\alpha_{5}}{\hbar}\Big{)}\hat{\lambda}^{2}+$
$+\Big{(}\frac{\partial\alpha_{4}}{\partial t}+\frac{2\alpha_{0}}{\hbar
m}-\frac{2k(t)\alpha_{2}}{\hbar}\Big{)}\hat{q}\hat{p}+\Big{(}\frac{\partial\alpha_{5}}{\partial
t}-\frac{2\alpha_{3}}{\hbar
m}+\frac{2k(t)\alpha_{1}}{\hbar}\Big{)}\hat{\theta}\hat{\lambda}=0$ (141)

It must equal zero, but each of the operators we know is nonzero, so the only
way for this equation to be true is if all the coefficients above equal zero.
Setting all the coefficients equal to zero and solving the system of equations
will lead you to the dynamics of the (here classical) Harmonic Oscillator.

For instance, if we solve for $\alpha_{1}$ we will get the equation:

$\frac{\partial^{2}\alpha_{1}}{\partial
t^{2}}+2k(t)\alpha_{1}=\frac{1}{2\alpha_{1}}\Big{(}\frac{\partial\alpha_{1}}{\partial
t}\Big{)}^{2}+\frac{C}{\alpha_{1}}$

And if we define an arbitrary function called $\rho(t)$ so that
$\alpha_{1}=\frac{\rho^{2}(t)}{2}$, we will have

$\frac{d^{2}\rho}{dt^{2}}+k(t)\rho=C/\rho^{3}$

which is an equation we call the Ermakov equation ([11]; [12]). Using the
Ermakov equation, all other coefficients can be rewritten so that the
invariant has the following form:

$\hat{I}=\frac{1}{2}\Big{[}\frac{\hat{q}}{\rho}+(\frac{d\rho}{dt}\hat{q}-\rho\hat{p})^{2}+\frac{\hat{\lambda}^{2}}{\rho^{2}}+(\frac{d\rho}{dt}\hat{\lambda}+\rho\hat{\theta})^{2}\Big{]}.$

Notice that the above classical Ermakov-Lewis invariant cannot be directly
observed. This is because operators like $\hat{\theta}$ and $\hat{\lambda}$
are not directly tied to physically observable quantities ([18]).
Nevertheless, it serves as an important tool in solving Harmonic Oscillation
problems.

We can define following the unitary transformation

$\hat{T}=\exp[i\frac{ln(\rho)}{2}(\hat{q}\hat{p}+\hat{p}\hat{q})]\exp[-\frac{i}{2\rho}\frac{d\rho}{dt}\hat{q}^{2}],$

which for the quantum case transforms the time dependent problem described
above to a time independent problem ([11]; [12]) by the following:

$\frac{1}{2}[\hat{p}^{2}+\hat{q}^{2}]=\hat{T}I\hat{T}^{\dagger}.$

In the classical case, we can define the following two unitary transformations
([18]):

$\hat{T}_{1}=\exp[\frac{i}{\rho}\frac{d\rho}{dt}\hat{q}\hat{\lambda}],$
$\hat{T}_{2}=\exp[i\frac{ln(\rho)}{2}(\hat{q}\hat{\theta}+\hat{\theta}\hat{q})]\exp[-i\frac{ln(\rho)}{2}(\hat{p}\hat{\lambda}+\hat{\lambda}\hat{p})].$

We can define the new wavefunction
$\Psi(q,p,t)=\hat{T}_{2}\hat{T}_{1}\psi(q,p,t)$ and we can rewrite our
equation of interest (133):

$i\frac{\partial}{\partial
t}\Psi(q,p,t)=\frac{1}{\rho^{2}}\Big{[}\frac{\hat{p}\hat{\theta}}{m}-\hat{\lambda}\hat{q}\Big{]}\Psi(q,p,t).$
(142)

where the values of the operators are exactly the same as in (133). The
original wavefunction equation got transformed into a wavefunction equation
without the time dependent potential term in front of the position operator.
This can be much more easily solved to deduce the statistical motion of the
classical particles.

Note that this method was originally used in the operatorial form for the
Schrödinger equation ([11]). We are able to successfully implement this
procedure for KvNM as well. If you graph the phase space diagrams for a
selected $k(t)$, you can see that it is indeed the graph of an oscillator
([18]). KvNM is successful at bridging methods that involve quantum-like
operators with classical physics.

###  6.4 Aharonov–Bohm Effect

The Aharonov-Bohm Effect is a surprising phenomena that emerges from the
quantum wavefunction interacting with the magnetic potential. In classical
electromagnetic theory, one can take the electric $\vec{E}$ and magnetic
$\vec{B}$ fields and decompose them into various derivatives of electric and
magnetic potentials:

$\vec{E}=-\vec{\nabla}\phi,$ $\vec{B}=\vec{\nabla}\crossproduct\vec{A},$

where $\phi$ is the electric scalar potential and $\vec{A}$ is the magnetic
vector potential. Both of these potentials have the unique property that they
make $\vec{E}$ and $\vec{B}$ gauge invariant, in other words, that a ‘shift’
$\chi$ will not produce different magnetic or electric fields. In other words,
the new potentials $\vec{A}^{\prime}$ and $\phi^{\prime}$ defined by

$\vec{A}^{\prime}=\vec{A}+\vec{\nabla}\chi,$

and

$\phi^{\prime}=\phi-\frac{\partial\chi}{\partial t},$

will produce the same magnetic and electric fields as $\vec{A}$ and $\phi$.
These potentials have historically been seen as useful mathematical constructs
with little physical bearing on reality in themselves. The Aharonov-Bohm
effect soon put this assumption to the test.

The Aharonov-Bohm effect is a a quantum interference, much like that in
Young’s Double Slit Experiment, produced by the magnetic potential changing
the phases of the quantum wavefunctions. When different wavefunctions with
difference phases meet, they interfere with themselves. The surprising thing
is that the interference pattern does not depend on the magnetic field
$\vec{B}$ but directly on the magnetic potential $\vec{A}$. In other words, if
the magnetic field $\vec{B}=\vec{\nabla}\crossproduct\vec{A}$ is zero but the
potential is nonzero, an effect will still be observed. This suggests that the
magnetic potential $\vec{A}$ is more fundamental than the magnetic field
$\vec{B}$, which have always been seen as being on equal footing in classical
theory.

Aharonov and Bohm described the theory behind the phenomena in 1959, but the
idea was not experimentally verified for the magnetic potential until after
their deaths. The experiment used a solenoid, which is the model we will use
here for the magnetic potential. Recall that a solenoid is a cylindrical
device made of winding wire which produces a magnetic field inside the
solenoid and zero magnetic field outside the solenoid.

Figure 4: Aharanov-Bohm Experiment set up. Reproduced from [1].

The generic setup of the experiment for either type of potential is depicted
in Figure 4. A coherent beam of electrons would be split into two branches,
travel through an electric or magnetic potential where $\vec{E}=\vec{0}$ or
$\vec{B}=\vec{0}$, and the two branches would then be recombined. Each
branch’s wavefunction took a slightly different path, so when the two branches
are recombined an interference pattern should emerge because of a phase
difference acumulated along each path. This is a semiclassical way to
visualize the experiment (recall that in QM proper, the particle actually
takes all possible paths between source and sink in the allotted time.)

To understand mathematically how the phase shift happens, first recall the
polar form of the wavefunction (3) and what the Action represents (51). We can
write that:

$\psi=R\exp[\frac{i}{\hbar}\int_{0}^{t}T-Vdt].$ (143)

The above wavefunction can be written for a region with no potential energy
(free particle case) as:

$\psi_{0}=R\exp[\frac{i}{\hbar}\int_{0}^{t}Tdt].$

And so (143) can be rewritten in the generic case as:

$\psi=\psi_{0}\exp[-\frac{i}{\hbar}\int_{0}^{t}Vdt].$

For a wavefunction going through an electric potential, from CM we know we can
write that the potential energy is $V=e\phi$, where $e$ is the charge of the
particle’s wavefunction and $\phi$ is the scalar potential. And so the generic
wavefunction can be written as the following for the case with electric
potential:

$\psi=\psi_{0}\exp[-\frac{ie}{\hbar}\int_{0}^{t}\phi dt].$ (144)

The above formulas can be used to describe what should happen in the
experiment in the presence of the electric potential.

Refer to Figure 4. Starting off, the beam of coherent electrons has
wavefunction $\psi_{0}$ in the absence of an external potential. It is split
into two, and both branches travel through a region of potential $\phi$, so
each wavefunction gains the form of (144). The branching could be represented
as $\psi_{1}^{0}$ and $\psi_{2}^{0}$ upon reaching the area with electric
potential at time $t_{1}$. After some evolution they leave the potential at
time $t_{2}$, and they are then recombined. The first branch travelled along
potential $\phi_{1}$ and the second branch travelled along $\phi_{2}$, in this
semiclassical picture. You can write the following superposition for the final
wavefunction:

$\psi_{f}=\psi_{1}+\psi_{2}=\psi_{1}^{0}\exp[-\frac{i}{\hbar}C_{1}]+\psi_{2}^{0}\exp[-\frac{i}{\hbar}C_{2}],$

where we define the phases

$C_{1}=e\int_{t_{1}}^{t_{2}}\phi_{1}dt,$

and

$C_{2}=e\int_{t_{1}}^{t_{2}}\phi_{2}dt.$

Since $C_{1}$ and $C_{2}$ are different, we have gained a phase shift of
$(C_{1}-C_{2})/\hbar$ 242424This comes straight from the Euler formula. Recall
that:  $e^{i(\theta+C)}=cos(\theta+C)+isin(\theta+C)$ . Here, the $C$ is the
phase. If the phase of two waves are different, we should expect interference.
There is absolutely no classical force $\vec{F}=e\vec{E}$ exerted on the
electrons as there is no electric field, but a measurable effect should still
be produced on the electron’s behavior ([1]).

A similar analysis can be done with the magnetic vector potential. Following
[1], we can generalize (144) to include the magnetic potential. In relativity,
you can write the electromagnetic potential as the four vector
$A^{\mu}=(\phi/c~{}\vec{A})=(\phi/c~{}A_{x}~{}A_{y}~{}A_{z})$ where as before,
$\phi$ is the electric potential and $\vec{A}$ is the magnetic potential. The
position four vector can be represented is
$x^{\mu}=(ct~{}\vec{x})=(ct~{}x~{}y~{}z)$. Rewriting the position four vector
as the differential $dx^{\mu}$, we can generalize the potential integral to
be:

$\frac{e}{\hbar}\oint\,A^{\mu}dx_{\mu}=\frac{e}{\hbar}\oint\,\Big{(}\phi
dt-\frac{\vec{A}}{c}\cdot d\vec{x}\Big{)}.$ (145)

The path of integration has to be closed in spacetime ([1]). Let us assume
that the variable $t$ is a constant in order to understand how interference
could arise strictly from the magnetic potential. The above integral suggests
that we will expect the following phase shift to hold true:

$(C_{1}-C_{2})/\hbar=-\frac{e}{c\hbar}\oint\,\vec{A}\cdot d\vec{x}$

We can set up a similar situation as to before.

Starting off, the beam of electrons has wavefunction $\psi_{0}$ in the absence
of an external magnetic potential. It is split into two, and both branches
travel through a region of potential $\vec{A}$. Each wavefunction gains the
form of (144) except we replace $\exp[-(ie/\hbar)\int\phi dt]$ with
$\exp[-(ie/\hbar c)\int\vec{A}\cdot d\vec{x}]$ as described above. In other
words:

$\psi=\psi_{0}\exp[-\frac{ie}{\hbar c}\int\vec{A}\cdot d\vec{x}].$ (146)

The branching could be represented as $\psi_{1}^{0}$ and $\psi_{2}^{0}$ upon
reaching the area with magnetic potential at time $t_{1}$. After some
evolution they leave the potential at time $t_{2}$, and they are then
recombined. You can write the following superposition:

$\psi_{f}=\psi_{1}+\psi_{2}=\psi_{1}^{0}\exp[-\frac{i}{\hbar}C_{1}]+\psi_{2}^{0}\exp[-\frac{i}{\hbar}C_{2}],$

where we define

$C_{1}=\frac{e}{c}\int_{\vec{x}(t_{1})}^{\vec{x}(t_{2})}\vec{A}\cdot
d\vec{x}^{\prime},$

and

$C_{2}=-\frac{e}{c}\int_{\vec{x}(t_{1})}^{\vec{x}(t_{2})}\vec{A}\cdot
d\vec{x}^{\prime},$

where we assume equal path length for each branch. If the path length is
indeed kept the same, then you will have phase difference
$(C_{1}-C_{2})/\hbar=(e/c\hbar)\int\vec{A}\cdot d\vec{x}$ in the end ([1]).
Note that this is true even if the magnetic force
$\vec{F}=q\vec{v}\times\vec{B}$ is zero because $\vec{B}=\vec{0}$. This is a
purely quantum effect.

The electric Aharonov-Bohm effect described above has been theorized, but no
experimental data exists for it as of yet. We will focus on the magnetic
potential therefore. No further space will be dedicated to the electric
potential, but all considerations from here on out will be framed in the
context of the magnetic potential.

One may ask the obvious question, in the descriptions above, if all possible
paths are taken in the Path Integral, would not some paths go through the
magnetic field, explaining the outcome of the experiment? Since the Feynman
Path Integral defines a set starting point and endpoint in a set time period,
it is actually possible to show that you can set up a barrier (described in
[1]) so that no wavefunction directly interacts with the inside of the
solenoid in the magnetic case. The changes we see, therefore, must be due to
the nonzero magnetic potential instead of the magnetic field itself.

KvNM, having the mathematical form of quantum theory, might be one of the best
tools to show that the Aharonov-Bohm effect does not exist in CM ([14]). This
is a purely quantum phenomenon. Interference is not something expected in the
domain of CM, but is expected under QM. This can be traced to the behavior of
the generator of motion, the quantum Hamiltonian, having the following form
under QM in the presence of the magnetic potential for a free particle:

$\hat{H}=\frac{1}{2m}\Big{(}\hat{p}-\frac{e}{c}\vec{A}\Big{)}^{2}$

In the presence of the magnetic potential, the free particle Hamiltonian has
the $\hat{p}$ operator replaced with $\hat{p}-(e/c)\vec{A}$. This is described
as a minimal coupling rule ([9]). If KvNM describes CM precisely, we should
expect that the generator of motion, the Koopman Generator (45), will not be
affected by the magnetic potential even if we treat it mathematically in the
exact same ways as we treat the quantum Hamiltonian. In the following pages,
we will put this to the test.

First, imagine two cylinders, one within the other (following [9]; [14]). Let
them be infinite in height. The smaller cylinder on the inside will represent
the solenoid which houses a magnetic field. The outside of the solenoid (area
inside the larger cylinder but outside of the smaller cylinder) will have no
magnetic field, but there will be a magnetic potential. Since we are dealing
with cylinders, it would be useful to rewrite the generators of motion in
terms of cylindrical coordinates. The conversion of cartesian to cylindrical
coordinates is given by:

$x=r\cos(\theta),$ $y=r\sin(\theta),$

and

$z=z,$ (147)

where $r$ is the distance from the $z$ axis and $\theta$ is the angle in the
$xy$ plane.

With this notation, the quantum Hamiltonian for a free particle can be
written:

$\hat{H}=-\frac{\hbar^{2}}{2m}\vec{\nabla}^{2}=-\frac{\hbar^{2}}{2m}\Big{(}\frac{\partial^{2}}{\partial
x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial
z^{2}}\Big{)}=-\frac{\hbar^{2}}{2m}\Big{(}\frac{\partial^{2}}{\partial
r^{2}}+\frac{1}{r}\frac{\partial}{\partial
r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}+\frac{\partial^{2}}{\partial
z^{2}}\Big{)}.$

This is now in cylindrical coordinates. The eigenvalue problem we want to
solve is:

$\hat{H}\psi(r,\theta,z)=E\psi(r,\theta,z).$

where $E$ are the energy eigenvalues. The above expression is often termed the
time independent Schrödinger equation, as it uses the quantum Hamiltonian to
extract energy values with no explicit time derivative.

Since the operators $\partial/\partial\theta$ and $\partial/\partial z$
commute with the above form of $\hat{H}$, we can choose solutions of the form
([9]):

$\psi(r,\theta,z)=\frac{1}{2\pi}\exp[\frac{ip_{z}^{0}z}{\hbar}]\exp[in\theta]R(r),$
(148)

where $p_{z}^{0}$ is a constant and $n$ is an integer. Plugging in this
wavefunction into the time independent Schrödinger equation will give you
solutions in the form of Bessel Functions:

$\frac{d^{2}R(r)}{dr^{2}}+\frac{1}{r}\frac{dR(r)}{dr}+\Big{(}\frac{2mE}{\hbar^{2}}-\frac{{p_{z}^{0}}^{2}}{\hbar^{2}}-\frac{n^{2}}{r^{2}}\Big{)}R(r)=0.$

Bessel Functions are a class of functions that govern cylindrical harmonics.

If you turn on the magnetic potential, we will now have to solve

$\hat{H}_{A}\psi(r,\theta,z)_{A}=E_{A}\psi(r,\theta,z)_{A}.$

In order to achieve this, we have the following minimal coupling rule in
cylindrical coordinates:

$\hat{p}_{\theta}=-i\hbar\frac{\partial}{\partial\theta}\Rightarrow\hat{p}_{\theta}-\frac{e}{c}\frac{\Phi}{2\pi}=-i\hbar\frac{\partial}{\partial\theta}-\frac{e}{c}\frac{\Phi}{2\pi}$
(149)

The quantum Hamiltonian therefore becomes:

$\hat{H}_{A}=-\frac{\hbar^{2}}{2m}\Big{(}\frac{\partial^{2}}{\partial
r^{2}}+\frac{1}{r}\frac{\partial}{\partial
r}+\frac{1}{r^{2}}\Big{(}\frac{\partial}{\partial\theta}-\frac{ie}{ch}{\Phi}\Big{)}^{2}+\frac{\partial^{2}}{\partial
z^{2}}\Big{)},$

where $\Phi$ is the flux (constant) through an area perpendicular to the axis
of the smaller cylinder. This area will be larger than the smaller cylinder’s
cross-section. Notice that $\hat{p}_{\theta}$ is not present in the
wavefunction, so the form of (148) will remain unaltered. The energy values in
the presence of the magnetic potential, as you will see, will be different.

As before, substituting (148) into the magnetic potential version of the time
independent Schrödinger equation will give you the following Bessel function

$\frac{d^{2}R(r)}{dr^{2}}+\frac{1}{r}\frac{dR(r)}{dr}+\Big{(}\frac{2mE}{\hbar^{2}}-\frac{{p_{z}^{0}}^{2}}{\hbar^{2}}-\frac{(n-\alpha)^{2}}{r^{2}}\Big{)}R(r)=0,$

where $\alpha=\frac{e\Phi_{B}}{ch}$. If you compare the two Bessel functions,
you can immediately see that the presence of the magnetic potential shifts the
function’s behavior. For the quantum case, the behavior of the system is
impacted by the presence of the vector potential.

Now we will do the same procedure for $\hat{L}$ as the generator of motion
(33). For the free particle in 3 dimensions, we know that

$\hat{L}=-i\frac{p_{x}}{m}\frac{\partial}{\partial
x}-i\frac{p_{y}}{m}\frac{\partial}{\partial
y}-i\frac{p_{z}}{m}\frac{\partial}{\partial z},$

based off of (33). Just as in the quantum case, we want this to be in
cylindrical coordinates. We will therefore need to convert the cartesian
momenta $p_{x}$, $p_{y}$, and $p_{z}$ into $p_{r}$, $p_{\theta}$, and $p_{z}$.
The latter are the canonical momenta, which can be calculated directly from
the Lagrangian $L$.252525Do not confuse the Lagrangian $L$ with the operator
$\hat{L}$. These are two different mathematical entities. As before, any
cartesian coordinates can be converted to cylindrical coordinates via (147),
which we will make use of.

To start off, we write down the Lagrangian of motion. Recall from section 4.1,
the Lagrangian is just the kinetic energy minus the potential energy, $L=T-V$.
Since the potential is zero, the Lagrangian is just equal to the total kinetic
energy, so that

$L=T=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}),$ (150)

where we use the standard dot notation, so that any variable $\gamma$ has a
time derivative $\dot{\gamma}=d\gamma/dt$. We can take time derivatives of
(147) to get:

$x=r\cos(\theta)\Rightarrow\dot{x}=\dot{r}\cos(\theta)-r\sin(\theta)\dot{\theta},$
$y=r\sin(\theta)\Rightarrow\dot{y}=\dot{r}\sin(\theta)+r\cos(\theta)\dot{\theta},$

with the $z$ coordinate unaffected. Plugging these terms into (150), we get
our Lagrangian in cylindrical coordinates:

$L=\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}mr^{2}\dot{\theta}^{2}+\frac{1}{2}m\dot{z}^{2}$

Recalling from Lagrangian Dynamics that the canonical momentum associated with
coordinate $\sigma$ is given by

$p_{\sigma}=\frac{\partial L}{\partial\dot{q_{\sigma}}}$

Therefore applying the above formula for coordinates $r$, $\theta$, and $z$,
we have:

$p_{r}=\frac{\partial L}{\partial\dot{r}}=m\dot{r}$ $p_{\theta}=\frac{\partial
L}{\partial\dot{\theta}}=mr^{2}\dot{\theta}$ $p_{z}=\frac{\partial
L}{\partial\dot{z}}=m\dot{z}$

If we apply the same rule to the cartesian Lagrangian (150) we get:

$p_{x}=\frac{\partial L}{\partial\dot{x}}=m\dot{x},$ $p_{y}=\frac{\partial
L}{\partial\dot{y}}=m\dot{y},$

with the $z$ coordinate momentum unaltered. Using all the above information,
we can rewrite the cartesian momenta in terms of the cylindrical momenta,
using the momenta expressions above:

$p_{x}=m\dot{x}=m[\dot{r}\cos(\theta)-r\sin(\theta)\dot{\theta}]=p_{r}\cos(\theta)-\frac{1}{r}p_{\theta}\sin(\theta),$
$p_{y}=m[\dot{r}\sin(\theta)+r\cos(\theta)\dot{\theta}]=p_{r}\sin(\theta)+\frac{1}{r}p_{\theta}\cos(\theta).$

We therefore can rewrite the generator of motion $\hat{L}$ using the above
information in terms of cylindrical coordinates:

$\hat{L}=-i\frac{p_{x}}{m}\frac{\partial}{\partial
x}-i\frac{p_{y}}{m}\frac{\partial}{\partial
y}-i\frac{p_{z}}{m}\frac{\partial}{\partial
z}=-i\frac{p_{r}}{m}\frac{\partial}{\partial
r}-i\frac{p_{\theta}}{mr^{2}}\frac{\partial}{\partial\theta}-i\frac{p_{z}}{m}\frac{\partial}{\partial
z}-i\frac{p_{\theta}^{2}}{mr^{3}}\frac{\partial}{\partial p_{r}}.$ (151)

For the case with no magnetic potential, we will be solving the analogue to
the time independent Schrödinger equation,

$\hat{L}\psi=\tilde{E}\psi,$

where $\psi$ is the KvN classical wavefunction and $\tilde{E}$ is the
eigenvalue of the above expression. Note that $\tilde{E}$ is certainly not the
same as the Energy like in the Schrödinger equation. We observe that $\hat{L}$
commutes with $-i\frac{\partial}{\partial\theta}$, $-i\frac{\partial}{\partial
z}$, $p_{\theta}$, and $p_{z}$. We can therefore deduce that the wavefunction
will have the form

$\psi=\frac{1}{2\pi}R(r,p_{r})\delta(p_{\theta}-p_{\theta}^{0})\delta(p_{z}-p_{z}^{0})\exp[in\theta]\exp[i\lambda_{z}^{0}z].$
(152)

If we plug this wavefunction into (151), we will get ([9]):

$\Big{(}-\frac{i}{m}p_{r}\frac{\partial}{\partial
r}+\frac{p_{\theta}^{0}n}{mr^{2}}-\frac{i{p_{\theta}^{0}}^{2}}{mr^{3}}\frac{\partial}{\partial
p_{r}}+\frac{\lambda_{z}^{0}p_{z}^{0}}{m}-\tilde{E}\Big{)}R(r,p_{r})=0.$ (153)

We have a similar expression for the magnetic potential case:

$\hat{L}_{A}\psi_{A}=\tilde{E}_{A}\psi_{A}$

Just as in the quantum case, we will utilize the minimal coupling rule (149).
The only difference is that the representation of the operator
$\hat{p}_{\theta}$ is no longer going to be a derivative with respect to
$\theta$, and instead we will have $\hat{p}_{\theta}=p_{\theta}$ for the KvNM
case ([9]). Making the substitution

$\hat{p}_{\theta}\Rightarrow\hat{p}_{\theta}-\frac{e}{c}\frac{\Phi}{2\pi},$
(154)

we will get the following two expressions:

$\hat{L}=-i\frac{p_{r}}{m}\frac{\partial}{\partial
r}-i\frac{1}{mr^{2}}\Big{(}{p}_{\theta}-\frac{e}{c}\frac{\Phi}{2\pi}\Big{)}\frac{\partial}{\partial\theta}-i\frac{p_{z}}{m}\frac{\partial}{\partial
z}-i\frac{1}{mr^{3}}\Big{(}{p}_{\theta}-\frac{e}{c}\frac{\Phi}{2\pi}\Big{)}^{2}\frac{\partial}{\partial
p_{r}},$ (155)

$\psi=\frac{1}{2\pi}R(r,p_{r})\delta(p_{\theta}-p_{\theta}^{0}-\frac{e\Phi}{2\pi
c})\delta(p_{z}-p_{z}^{0})\exp[in\theta]\exp[i\lambda_{z}^{0}z].$ (156)

Plugging (156) into (155) we will get:

$\Big{(}-\frac{i}{m}p_{r}\frac{\partial}{\partial
r}+\frac{p_{\theta}^{0}n}{mr^{2}}-\frac{i{p_{\theta}^{0}}^{2}}{mr^{3}}\frac{\partial}{\partial
p_{r}}+\frac{\lambda_{z}^{0}p_{z}^{0}}{m}-\tilde{E}_{A}\Big{)}R(r,p_{r})=0$
(157)

If you compare (157) (case with magnetic potential) to (153) (case without
magnetic potential) you will notice they happen to be the same equation. This
shows us that the same solutions will exist for classical mechanics regardless
of the presence or absence of the gauge potential. As demonstrated, the gauge
potential will however affect quantum systems ([9]; [14]).

On a philosophical level, the above analysis raises the question what do the
gauge potentials physically mean? Historically, it was simply seen as a
mathematical tool or convenience in the study of Electromagnetism. The
electromagnetic field and its potentials were fully equivalent descriptions of
reality, but the potentials were only seen as a mathematical construct with
useful properties. This might be a historic example of physicists not taking
the mathematics of their theory seriously enough. For example, Albert Einstein
famously invented the theories of special relativity by acknowledging that the
speed of light $c$ is the same in all noninertial reference frames. This is a
fact tacitly present in Maxwell’s equations, but physicists at the time did
not treat the mathematics seriously enough and thought the equations must only
be true relative to a hypothetical ‘fluid’ filling all of space and time, the
aether, an idea Einstein showed to be wrong. [1] say in their paper “a further
interpretation of the potentials is needed in the quantum mechanics”, so
perhaps this is an area for further scientific investigation and discovery in
the future.

##  7 Other Research

Wavefunctions in a complex Hilbert phase space can exactly model CM in the
form of statistical physics. Both QM and CM can be articulated in a common
mathematical language utilizing Hermitian operators to recover expectation
values of observables. In the pages above we covered a host of different
applications of the KvN wavefunction inside classical physics. We hardly
exhausted the possible applications of KvNM, however. There are many areas of
research we have not even touched upon.

Some recent examples of the widespread applicability of the KvNM formulism:

  1. 1.

Why the usage of finite-dimensional Hilbert Spaces in QM and KvNM is often
ill-advised

A recent trends in QM research has been to try utilize QM with finite
dimensions instead of the regular infinite Hilbert space. This is done to aid
in classical computer modeling of quantum systems and an interest in the
possibility of rooting QM in a finite instead of an infinite space. For
instance, approximations of the quantum Hamiltonian tend to be finite matrices
([2]). [2] demonstrate that this will unavoidably violate the Ehrenfest
Theorems. This is a serious shortcoming of finite dimensional treatments of
QM, but there are a small number of ways that the issues might be circumvented
([2]).

  2. 2.

KvNM has been used to study aspects of String Theory and the Gravitational
Principle of Equivalence

The KvN Path Integral is utilized in [7] to study the properties of Classical
Yang-Mills Theories (a subset of String Theories.) The Path Integral provides
geometrical tools like exterior derivatives, forms, etc. The Classical Path
Integral might shed light on the geometrical aspects of these string theories
([7]). [19] investigates the validity of Einstein’s principle of mass
equivalence for both classical KvNM and quantum systems. Fisher information is
discussed and how it relates to insights from KvNM and QM. It is demonstrated
that the weak principle of equivalence holds in QM just as it does in KvNM
([19]).

  3. 3.

ODM can be used to describe relativistic particles

The ODM paradigm is upheld in relativistic settings. The following two
Ehrenfest-like theorems were postulated ([6]) based on relativistic principles

$\frac{d}{dt}\langle\hat{x}^{k}\rangle=\langle c\gamma^{0}\gamma^{k}\rangle.$
$\frac{d}{dt}\langle\hat{p}_{k}\rangle=\langle
ce\partial_{k}\hat{A}_{\nu}\gamma^{0}\gamma^{\nu}\rangle.$

If the quantum commutator is chosen, you can follow a series of steps similar
to those in section 3 to derive the famous Dirac equation, which governs
relativistic quantum particles ([6]). If the Koopman Algebra is chosen and you
constrain the system from making antiparticles, the same series of steps as
deriving the KvNM will give you the classical Spohn equation ([6]). This is
quite a surprising result - ODM can study spin 1/2 relativistic particles.

  4. 4.

ODM has been used to study Time’s Arrow and stochastic behavior from
determinism

A lot of processes are reversible in physics, but our intuitions of the world
are that it is irreversible and things tend towards decay. How does
irreversibility arise out of the reversible? [15] utilizes the KvN Path
Integral to formulate a functional model for how irreversibility arises out of
the reversible laws we have.

  5. 5.

Reformulation of Electromagnetism

[17] provides a reformulation of electromagnetic theory that draws inspiration
from the KvN theory.

  6. 6.

KvN wavefunctions were utilized to understand the interaction of classical
with quantum systems

The problem of modeling quantum and classical system interaction is an old
one. Utilizing the KvN classical wavefunction, however, had led to new models
of how the interaction occurs. Not only does it describe the classical
system’s effect on the quantum system, but uniquely it models how the quantum
system will impact the classical system in return ([3]). It also overcomes
difficulty in some other models. For instance, the quantum density matrix will
always be positive-definite in this model ([3]).

Among other areas of inquiry. Doubtlessly, many more areas of research that
will benefit from KvNM and a paradigm like ODM.

It is important to have many “tools” in science when investigating novel
phenomena. Feynman famously said that the reason he had solved problems others
could not, is that his toolbox had a different set of tools than everyone
else. KvNM is yet an another novel tool for you, the reader, to have in your
toolkit. KvNM has and will continue to shed light on new and exciting
phenomena in science.

## References

  * [1] Yakir Aharonov and David Bohm  “Significance of Electromagnetic Potentials in the Quantum Theory”  In _Physical Review_ 115, 1959, pp. 485  DOI: https://doi:10.1103/PhysRev.115.485
  * [2] Denys I. Bondar, Renan Cabrera and Herschel A. Rabitz  “Conceptual inconsistencies in finite-dimensional quantum and classical mechanics.”  In _Physical Review A: Atomic, Molecular, and Optical Physics_ 88.1, 2013, pp. 012116  DOI: https://doi.org/10.1103/PhysRevA.88.012116
  * [3] Denys I. Bondar, Francois Gay-Balmaz and Cesare Tronci  “Koopman wavefunctions and classical-quantum correlation dynamics”  In _Proceedings of the Royal Society A_ 475.2229, 2019, pp. 20180879  DOI: https://doi.org/10.1098/rspa.2018.0879
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##  Appendix A Useful Mathematics for Hilbert Space Physics

Note: A lot of the following useful mathematics was found in the Appendix of
[4].

Theorem I: If you have a function of operator $\hat{A}$, i.e. $f(\hat{A})$,
then the following eigenvalue expressions are true:

$f(\hat{A})\ket{A}=f(A)\ket{A}$ $\bra{A}f(\hat{A})=\bra{A}f(A)$

This should purposefully resemble the prior eigenvalue problem you are
familiar with, $\hat{A}\ket{A}=A\ket{A}$. This theorem is true for
multivariable operators, for instance,

$A(\hat{q},\hat{p})\ket{q,p}=A(q,p)\ket{q,p}$

.

Theorem II (Resolution of Identity): The following theorem says that

$I=\int dA\ket{A}\bra{A}$

where $I$ is the identity operator, functionally equivalent to multiplying an
object by 1. Since it is just equal to 1, you can introduce the Resolution of
Identity in a variety of calculations. As always $A$ is an arbitrary
placeholder variable, and you can make it any variable of interest.

Theorem III (Cauchy-Schwartz Inequality): It simply states that the following
is always true for complex vectors in Hilbert Space:

$\innerproduct{C}{C}\innerproduct{D}{D}\geq|\innerproduct{C}{D}|^{2}$

where $C$ and $D$ are generic placeholder variables.

Theorem IV: If the commutator relationship between two generic operators
$\hat{A}$ and $\hat{B}$ is $[\hat{A},\hat{B}]=i\kappa$, then the following
statements hold true:

$\bra{A}\hat{B}\ket{\psi}=-i\kappa\frac{\partial}{\partial
A}\innerproduct{A}{\psi}$
$\bra{B}\hat{A}\ket{\psi}=i\kappa\frac{\partial}{\partial
B}\innerproduct{B}{\psi}$

Recall that $\innerproduct{B}{\psi}=\psi(B)$ and
$\innerproduct{A}{\psi}=\psi(A)$.

Theorem V: If the commutator relationship between two generic operators
$\hat{A}$ and $\hat{B}$ is $[\hat{A},\hat{B}]=i\kappa$, then the following
statements hold true:

$\innerproduct{A}{B}=\frac{1}{\sqrt{2\pi\kappa}}e^{iAB/\kappa}$ (158)

This is not difficult to demonstrate. First, take Theorem IV and realize that
$\ket{\psi}$ is a general vector that can be substituted for a more specific
vector. This is a common practice in QM. Let us say that $\ket{\psi}=\ket{B}$.
Then by Theorem IV we have:

$\bra{A}\hat{B}\ket{B}=B\innerproduct{A}{B}=-i\kappa\frac{\partial}{\partial
A}\innerproduct{A}{B}$

You can solve this common differential equation for the function
$\innerproduct{A}{B}$ to get:

$\innerproduct{A}{B}=Ce^{iAB/\kappa}$

where $C$ is a constant of integration. $C$ can be found in the following way:
Start with equation (18)

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

and use Theorem II with $A=B$ 262626In other words, since $A$ is just a
generic placeholder for Theorem II, we will use $I=\int dB\ket{B}\bra{B}$ to
get:

$\innerproduct{A}{A^{\prime}}=\delta(A-A^{\prime})=\bra{A}I\ket{A^{\prime}}=\int
dB\bra{A}\ket{B}\bra{B}\ket{A^{\prime}}$

Since we know what $\innerproduct{A}{B}$ is, we plug it in to get:

$\innerproduct{A}{A^{\prime}}=\delta(A-A^{\prime})=\int
dB[Ce^{iAB/\kappa}][Ce^{-iA^{\prime}B/\kappa}]=C^{2}\int
dBe^{i(A-A^{\prime})B/\kappa}$

From equation (22) of section 2, we can easily see that

$\int dBe^{i(A-A^{\prime})B/\kappa}=2\pi\kappa\delta(A-A^{\prime})$

Ergo,

$\delta(A-A^{\prime})=C^{2}2\pi\kappa\delta(A-A^{\prime})$ $C^{2}2\pi\kappa=1$
$C^{2}=\frac{1}{2\pi\kappa}$ $C=\frac{1}{\sqrt{2\pi\kappa}}$

We now know $C$ and have successfully demonstrated equation (158).

Theorem VI: As strange as it sounds, the mathematics of Hilbert Space defines
derivatives with respect to operators. If you have
$[\hat{A}_{k},\hat{B}]=Constant$, then a function of operators
$f(\hat{A}_{1},\hat{A}_{2},...,\hat{A}_{n})$ will obey the following rule:

$[f(\hat{A}_{1},\hat{A}_{2},...,\hat{A}_{n}),\hat{B}]=\sum_{k=1}^{n}[\hat{A}_{k},\hat{B}]\frac{\partial
f(\hat{A}_{1},\hat{A}_{2},...,\hat{A}_{n})}{\partial\hat{A}_{k}}$

Notice the partial derivative with respect to an operator. The next theorem
shows how to change this odd operator derivative to regular derivatives that
we know and love.

Theorem VII The following is the relationship between derivatives with respect
to operators and the regular derivatives we are most generally familiar with:

$\frac{\partial
f(\hat{A}_{1},\hat{A}_{2},...,\hat{A}_{n})}{\partial\hat{A}_{k}}=\frac{1}{(2\pi)^{n}}\int\prod_{l=1}^{n}d\epsilon_{l}~{}d\mu_{l}~{}exp[i\sum_{q=1}^{n}\mu_{q}(\hat{A}_{q}-\epsilon_{q})]\frac{\partial
f(\epsilon_{1},\epsilon_{2},...,\epsilon_{n})}{\partial\epsilon_{k}}$

where on the left you can see the derivative with respect to an operator and
on the right you can see the derivative with respect to a scalar variable.

Theorem VIII: An integral of a Gaussian function has the following form:

$\int_{-\infty}^{\infty}e^{-ax^{2}+bx+c}dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^{2}}{4a}+c}$

where $a$, $b$, and $c$ can be real or imaginary constants.

##  Appendix B The Principle of Causality and Stone’s Theorem

A unitary operator $\hat{U}(t_{2},t_{1})$ always has the following properties:

  1. 1.

Group product: $\hat{U}(t_{3},t_{2})\hat{U}(t_{2},t_{1})=\hat{U}(t_{3},t_{1})$

  2. 2.

Reversibility:
$\hat{U}(t_{2},t_{1})^{\dagger}=\hat{U}(t_{2},t_{1})^{-1}=\hat{U}(t_{1},t_{2})$

  3. 3.

Identity: $\hat{U}(t_{1},t_{1})=1$

The job of the unitary operator $\hat{U}(t_{2},t_{1})$ is to bring you from
$t_{1}$ to $t_{2}$. The group product encapsulates the concept of cause and
effect, as you move from cause to effect through time. Causality is central to
the philosophy of science, and we should not be surprised that it pops up
inside the mathematics of physics.

The three properties of the unitary operator above imply Stone’s Theorem:

If the three properties hold true, then the following relationship is also
true:

$i\frac{\partial}{\partial t}\hat{U}(t_{2},t_{1})=\hat{G}\hat{U}(t_{2},t_{1})$

where $\hat{G}$ is a Hermitian operator. If we assume that the wavefunction
ket evolves via the unitary operator

$\ket{\psi(t_{2})}=\hat{U}(t_{2},t_{1})\ket{\psi(t_{1})},$

then we recover the Schrödinger-like equation we spoke of in earlier sections:

$i\frac{\partial}{\partial
t}\hat{U}(t_{2},t_{1})\ket{\psi(t_{1})}=\hat{G}\hat{U}(t_{2},t_{1})\ket{\psi(t_{1})}\Rightarrow
i\frac{\partial}{\partial t}\ket{\psi(t_{2})}=\hat{G}\ket{\psi(t_{2})}$

This is the reason why Stone’s Theorem becomes an important ingredient in
section 3.

##  Appendix C Exponated Operators and the Propogator

Although it might seem odd at first, exponated operators are allowed within
Quantum Mechanics. The Maclaurin series for the $f(x)=e^{x}$ is given by:

$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+\frac{1}{1!}x+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+\frac{1}{4!}x^{4}+\cdot\cdot\cdot$

Let us say that you have a generic operator $\hat{A}$, like a square matrix.
As long as $\hat{A}^{n}$ is defined for any real number $n$, then you can say:

$e^{\hat{A}}=\sum_{n=0}^{\infty}\frac{\hat{A}^{n}}{n!}=1+\frac{1}{1!}\hat{A}+\frac{1}{2!}\hat{A}^{2}+\frac{1}{3!}\hat{A}^{3}+\frac{1}{4!}\hat{A}^{4}+\cdot\cdot\cdot$

The Maclaurin series defines exponated operators as an infinite sum.

If we have an expression such as $e^{\hat{A}t}$ where $\hat{A}$ is time-
independent, we can also derive a derivative rule for it:

$e^{\hat{A}t}=\sum_{n=0}^{\infty}\frac{(\hat{A}t)^{n}}{n!}=1+\frac{1}{1!}\hat{A}t+\frac{1}{2!}(\hat{A}t)^{2}+\frac{1}{3!}(\hat{A}t)^{3}+\frac{1}{4!}(\hat{A}t)^{4}+\cdot\cdot\cdot$
$\frac{d}{dt}e^{\hat{A}t}=\frac{d}{dt}1+\frac{d}{dt}\frac{1}{1!}\hat{A}t+\frac{d}{dt}\frac{1}{2!}(\hat{A}t)^{2}+\frac{d}{dt}\frac{1}{3!}(\hat{A}t)^{3}+\cdot\cdot\cdot=$
$=\frac{1}{1!}\hat{A}+\frac{1}{2!}\hat{A}^{2}(2t)+\frac{1}{3!}\hat{A}^{3}(3t^{2})+\frac{1}{4!}\hat{A}^{4}(4t^{3})+\cdot\cdot\cdot=\hat{A}\Big{[}1+\frac{1}{2!}\hat{A}(2t)+\frac{1}{3!}\hat{A}^{2}(3t^{2})+\frac{1}{4!}\hat{A}^{3}(4t^{3})+\cdot\cdot\cdot\Big{]}=$
$=\hat{A}\Big{[}1+\frac{1}{2\cdot 1}\hat{A}(2t)+\frac{1}{3\cdot 2\cdot
1}\hat{A}^{2}(3t^{2})+\cdot\cdot\cdot\Big{]}=\hat{A}\Big{[}1+\frac{1}{1!}\hat{A}t+\frac{1}{2!}(\hat{A}t)^{2}+\frac{1}{3!}(\hat{A}t)^{3}+\cdot\cdot\cdot\Big{]}=\hat{A}e^{\hat{A}t}$
$\therefore\frac{d}{dt}e^{\hat{A}t}=\hat{A}e^{\hat{A}t}$ (159)

The above relation is true even if a real or complex constant is placed in
front of the exponated term $\hat{A}t$. You can redo the exercise to convince
yourself of this. We will use this important fact in order to demonstrate the
important relationship (52) is fully equivalent to the Schrödinger equation:

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

A similar derivation can be done for (66), to demonstrate that it is fully
equivalent to Stone’s Theorem (Appendix B).