## Gauge freedom and four-potentials in the STA form of Maxwell’s equation.

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

## Motivation.

In a recent video on the tensor structure of Maxwell’s equation, I made a little side trip down the road of potential solutions and gauge transformations. I thought that was worth writing up in text form.

The initial point of that side trip was just to point out that the Faraday tensor can be expressed in terms of four potential coordinates
\label{eqn:gaugeFreedomAndPotentialsMaxwell:20}
F_{\mu\nu} = \partial_\mu A_\nu – \partial_\nu A_\mu,

but before I got there I tried to motivate this. In this post, I’ll outline the same ideas.

## STA representation of Maxwell’s equation.

We’d gone through the work to show that Maxwell’s equation has the STA form
\label{eqn:gaugeFreedomAndPotentialsMaxwell:40}

This is a deceptively compact representation, as it requires all of the following definitions
\label{eqn:gaugeFreedomAndPotentialsMaxwell:60}
\grad = \gamma^\mu \partial_\mu = \gamma_\mu \partial^\mu,

\label{eqn:gaugeFreedomAndPotentialsMaxwell:80}
\partial_\mu = \PD{x^\mu}{},

\label{eqn:gaugeFreedomAndPotentialsMaxwell:100}
\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu,

\label{eqn:gaugeFreedomAndPotentialsMaxwell:160}
\gamma_\mu \cdot \gamma_\nu = g_{\mu\nu},

\label{eqn:gaugeFreedomAndPotentialsMaxwell:120}
\begin{aligned}
F
&= \BE + I c \BB \\
&= -E^k \gamma^k \gamma^0 – \inv{2} c B^r \gamma^s \gamma^t \epsilon^{r s t} \\
&= \inv{2} \gamma^{\mu} \wedge \gamma^{\nu} F_{\mu\nu},
\end{aligned}

and
\label{eqn:gaugeFreedomAndPotentialsMaxwell:140}
\begin{aligned}
J &= \gamma_\mu J^\mu \\
J^\mu &= \frac{\rho}{\epsilon} \gamma_0 + \eta (\BJ \cdot \Be_k).
\end{aligned}

## Four-potentials in the STA representation.

In order to find the tensor form of Maxwell’s equation (starting from the STA representation), we first split the equation into two, since
\label{eqn:gaugeFreedomAndPotentialsMaxwell:180}

The dot product is a four-vector, the wedge term is a trivector, and the current is a four-vector, so we have one grade-1 equation and one grade-3 equation
\label{eqn:gaugeFreedomAndPotentialsMaxwell:200}
\begin{aligned}
\grad \cdot F &= J \\
\end{aligned}

The potential comes into the mix, since the curl equation above means that $$F$$ necessarily can be written as the curl of some four-vector
\label{eqn:gaugeFreedomAndPotentialsMaxwell:220}

One justification of this is that $$a \wedge (a \wedge b) = 0$$, for any vectors $$a, b$$. Expanding such a double-curl out in coordinates is also worthwhile
\label{eqn:gaugeFreedomAndPotentialsMaxwell:240}
\begin{aligned}
&=
\lr{ \gamma_\mu \partial^\mu }
\wedge
\lr{ \gamma_\nu \partial^\nu }
\wedge
A \\
&=
\gamma^\mu \wedge \gamma^\nu \wedge \lr{ \partial_\mu \partial_\nu A }.
\end{aligned}

Provided we have equality of mixed partials, this is a product of an antisymmetric factor and a symmetric factor, so the full sum is zero.

Things get interesting if one imposes a $$\grad \cdot A = \partial_\mu A^\mu = 0$$ constraint on the potential. If we do so, then
\label{eqn:gaugeFreedomAndPotentialsMaxwell:260}

Observe that $$\grad^2$$ is the wave equation operator (often written as a square-box symbol.) That is
\label{eqn:gaugeFreedomAndPotentialsMaxwell:280}
\begin{aligned}
&= \partial^\mu \partial_\mu \\
&= \partial_0 \partial_0
– \partial_1 \partial_1
– \partial_2 \partial_2
– \partial_3 \partial_3 \\
\end{aligned}

This is also an operator for which the Green’s function is well known ([1]), which means that we can immediately write the solutions
\label{eqn:gaugeFreedomAndPotentialsMaxwell:300}
A(x) = \int G(x,x’) J(x’) d^4 x’.

However, we have no a-priori guarantee that such a solution has zero divergence. We can fix that by making a gauge transformation of the form
\label{eqn:gaugeFreedomAndPotentialsMaxwell:320}
A \rightarrow A – \grad \chi.

Observe that such a transformation does not change the electromagnetic field
\label{eqn:gaugeFreedomAndPotentialsMaxwell:340}

since
\label{eqn:gaugeFreedomAndPotentialsMaxwell:360}

(also by equality of mixed partials.) Suppose that $$\tilde{A}$$ is a solution of $$\grad^2 \tilde{A} = J$$, and $$\tilde{A} = A + \grad \chi$$, where $$A$$ is a zero divergence field to be determined, then
\label{eqn:gaugeFreedomAndPotentialsMaxwell:380}
=

or
\label{eqn:gaugeFreedomAndPotentialsMaxwell:400}

So if $$\tilde{A}$$ does not have zero divergence, we can find a $$\chi$$
\label{eqn:gaugeFreedomAndPotentialsMaxwell:420}
\chi(x) = \int G(x,x’) \grad’ \cdot \tilde{A}(x’) d^4 x’,

so that $$A = \tilde{A} – \grad \chi$$ does have zero divergence.

# References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

## Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

I’ve posted a second update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 9, my ungraded solutions for the second problem set, and some additional worked practise problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

## PHY1520H Graduate Quantum Mechanics. Lecture 6: Electromagnetic gauge transformation and Aharonov-Bohm effect. Taught by Prof. Arun Paramekanti

October 6, 2015 phy1520 , , ,

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 2 content.

### Particle with $$\BE, \BB$$ fields

We express our fields with vector and scalar potentials

\label{eqn:qmLecture6:20}
\BE, \BB \rightarrow \BA, \phi

and apply a gauge transformed Hamiltonian

\label{eqn:qmLecture6:40}
H = \inv{2m} \lr{ \Bp – q \BA }^2 + q \phi.

Recall that in classical mechanics we have

\label{eqn:qmLecture6:60}
\Bp – q \BA = m \Bv

where $$\Bp$$ is not gauge invariant, but the classical momentum $$m \Bv$$ is.

If given a point in phase space we must also specify the gauge that we are working with.

For the quantum case, temporarily considering a Hamiltonian without any scalar potential, but introducing a gauge transformation

\label{eqn:qmLecture6:80}
\BA \rightarrow \BA + \spacegrad \chi,

which takes the Hamiltonian from

\label{eqn:qmLecture6:100}
H = \inv{2m} \lr{ \Bp – q \BA }^2,

to
\label{eqn:qmLecture6:120}
H = \inv{2m} \lr{ \Bp – q \BA -q \spacegrad \chi }^2.

We care that the position and momentum operators obey

\label{eqn:qmLecture6:140}
\antisymmetric{\hat{r}_i}{\hat{p}_j} = i \Hbar \delta_{i j}.

We can apply a transformation that keeps $$\Br$$ the same, but changes the momentum

\label{eqn:qmLecture6:160}
\begin{aligned}
\hat{\Br}’ &= \hat{\Br} \\
\hat{\Bp}’ &= \hat{\Bp} – q \spacegrad \chi(\Br)
\end{aligned}

This maps the Hamiltonian to

\label{eqn:qmLecture6:101}
H = \inv{2m} \lr{ \Bp’ – q \BA -q \spacegrad \chi }^2,

We want to check if the commutator relationships have the desired structure, that is

\label{eqn:qmLecture6:180}
\begin{aligned}
\antisymmetric{r_i’}{r_j’} &= 0 \\
\antisymmetric{p_i’}{p_j’} &= 0
\end{aligned}

This is confirmed in \ref{problem:qmLecture6:1}.

Another thing of interest is how are the wave functions altered by this change of variables? The wave functions must change in response to this transformation if the energies of the Hamiltonian are to remain the same.

Considering a plane wave specified by

\label{eqn:qmLecture6:200}
e^{i \Bk \cdot \Br},

where we alter the momentum by

\label{eqn:qmLecture6:220}
\Bk \rightarrow \Bk – e \spacegrad \chi.

This takes the plane wave to

\label{eqn:qmLecture6:240}
e^{i \lr{ \Bk – q \spacegrad \chi } \cdot \Br}.

We want to try to find a wave function for the new Hamiltonian

\label{eqn:qmLecture6:260}
H’ = \inv{2m} \lr{ \Bp’ – q \BA -q \spacegrad \chi }^2,

of the form

\label{eqn:qmLecture6:280}
\psi'(\Br)
\stackrel{?}{=}
e^{i \theta(\Br)} \psi(\Br),

where the new wave function differs from a wave function for the original Hamiltonian by only a position dependent phase factor.

Let’s look at the action of the Hamiltonian on the new wave function

\label{eqn:qmLecture6:300}
H’ \psi'(\Br) .

Looking at just the first action

\label{eqn:qmLecture6:320}
\begin{aligned}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta(\Br)} \psi(\Br)
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }
\psi(\Br)
+
\lr{
}
e^{i\theta}
\psi(\Br) \\
&=
e^{i\theta}
}
\psi(\Br).
\end{aligned}

If we choose

\label{eqn:qmLecture6:340}
\theta = \frac{q \chi}{\Hbar},

then we are left with

\label{eqn:qmLecture6:360}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta(\Br)} \psi(\Br)
=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA }
\psi(\Br).

Let $$\BM = -i \Hbar \spacegrad – q \BA$$, and act again with $$\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }$$

\label{eqn:qmLecture6:700}
\begin{aligned}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta} \BM \psi
&=
e^{i\theta}
\lr{ -i \Hbar i \spacegrad \theta – q \BA – q \spacegrad \chi } e^{i \theta} \BM \psi
+
e^{i\theta}
\lr{ -i \Hbar \spacegrad } \BM \psi \\
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad -q \BA + \spacegrad \lr{ \Hbar \theta – q \chi} } \BM \psi \\
&=
e^{i\theta} \BM^2 \psi.
\end{aligned}

Restoring factors of $$m$$, we’ve shown that for a choice of $$\Hbar \theta – q \chi$$, we have

\label{eqn:qmLecture6:400}
\inv{2m} \lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }^2 e^{i \theta} \psi = e^{i\theta}
\inv{2m} \lr{ -i \Hbar \spacegrad – q \BA }^2 \psi.

When $$\psi$$ is an energy eigenfunction, this means

\label{eqn:qmLecture6:420}
H’ e^{i\theta} \psi = e^{i \theta} H \psi = e^{i\theta} E\psi = E (e^{i\theta} \psi).

We’ve found a transformation of the wave function that has the same energy eigenvalues as the corresponding wave functions for the original untransformed Hamiltonian.

In summary
\label{eqn:qmLecture6:440}
\boxed{
\begin{aligned}
H’ &= \inv{2m} \lr{ \Bp – q \BA – q \spacegrad \chi}^2 \\
\psi'(\Br) &= e^{i \theta(\Br)} \psi(\Br), \qquad \text{where}\, \theta(\Br) = q \chi(\Br)/\Hbar
\end{aligned}
}

### Aharonov-Bohm effect

Consider a periodic motion in a fixed ring as sketched in fig. 1.

fig. 1. particle confined to a ring

Here the displacement around the perimeter is $$s = R \phi$$ and the Hamiltonian

\label{eqn:qmLecture6:460}
H = – \frac{\Hbar^2}{2 m} \PDSq{s}{} = – \frac{\Hbar^2}{2 m R^2} \PDSq{\phi}{}.

Now assume that there is a magnetic field squeezed into the point at the origin, by virtue of a flux at the origin

\label{eqn:qmLecture6:480}
\BB = \Phi_0 \delta(\Br) \zcap.

We know that

\label{eqn:qmLecture6:500}
\oint \BA \cdot d\Bl = \Phi_0,

so that

\label{eqn:qmLecture6:520}
\BA = \frac{\Phi_0}{2 \pi r} \phicap.

The Hamiltonian for the new configuration is

\label{eqn:qmLecture6:540}
\begin{aligned}
H
&= – \lr{ -i \Hbar \spacegrad – q \frac{\Phi_0}{2 \pi r } \phicap }^2 \\
&= – \inv{2 m} \lr{ -i \Hbar \inv{R} \PD{\phi}{} – q \frac{\Phi_0}{2 \pi R } }^2.
\end{aligned}

Here the replacement $$r \rightarrow R$$ makes use of the fact that this problem as been posed with the particle forced to move around the ring at the fixed radius $$R$$.

For this transformed Hamiltonian, what are the wave functions?

\label{eqn:qmLecture6:560}
\psi(\phi)’
\stackrel{?}{=}
e^{i n \phi}.

\label{eqn:qmLecture6:580}
\begin{aligned}
H \psi
&= \inv{2 m}
\lr{ -i \Hbar \inv{R} (i n) – q \frac{\Phi_0}{2 \pi R } }^2 e^{i n \phi} \\
&=
\underbrace{\inv{2 m}
\lr{ \frac{\Hbar n}{R} – q \frac{\Phi_0}{2 \pi R } }^2}_{E_n} e^{i n \phi}.
\end{aligned}

This is very unclassical, since the energy changes in a way that depends on the flux, because particles are seeing magnetic fields that are not present at the point of the particle.

This is sketched in fig. 2.

fig. 2. Energy variation with flux.

we see that there are multiple points that the energies hit the minimum levels

## Question:

Show that after a transformation of position and momentum of the following form

\label{eqn:qmLecture6:600}
\begin{aligned}
\hat{\Br}’ &= \hat{\Br} \\
\hat{\Bp}’ &= \hat{\Bp} – q \spacegrad \chi(\Br)
\end{aligned}

all the commutators have the expected values.

The position commutators don’t need consideration. Of interest is the momentum-position commutators

\label{eqn:qmLecture6:620}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{x}_k’}
&=
\antisymmetric{\hat{p}_k – q \partial_k \chi}{\hat{x}_k} \\
&=
\antisymmetric{\hat{p}_k}{\hat{x}_k} – q \antisymmetric{\partial_k \chi}{\hat{x}_k} \\
&=
\antisymmetric{\hat{p}_k}{\hat{x}_k},
\end{aligned}

and the momentum commutators

\label{eqn:qmLecture6:640}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{p}_j’}
&=
\antisymmetric{\hat{p}_k – q \partial_k \chi}{\hat{p}_j – q \partial_j \chi} \\
&=
\antisymmetric{\hat{p}_k}{\hat{p}_j}
– q \lr{ \antisymmetric{\partial_k \chi}{\hat{p}_j} + \antisymmetric{\hat{p}_k}{\partial_j \chi} }.
\end{aligned}

That last sum of commutators is

\label{eqn:qmLecture6:660}
\begin{aligned}
\antisymmetric{\partial_k \chi}{\hat{p}_j} + \antisymmetric{\hat{p}_k}{\partial_j \chi}
&=
– i \Hbar \lr{ \PD{k}{(\partial_j \chi)} – \PD{j}{(\partial_k \chi)} } \\
&= 0.
\end{aligned}

We’ve shown that

\label{eqn:qmLecture6:680}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{x}_k’} &= \antisymmetric{\hat{p}_k}{\hat{x}_k} \\
\antisymmetric{\hat{p}_k’}{\hat{p}_j’} &= \antisymmetric{\hat{p}_k}{\hat{p}_j}.
\end{aligned}

All the other commutators clearly have the desired transformation properties.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Question: Gauge transformed probability current ([1] pr. 2.37 (b))

For the gauge transformed Schrodinger equation

\label{eqn:gaugeTxCurrent:20}
\inv{2m} \BPi(\Bx) \cdot \BPi(\Bx) \psi(\Bx, t) + e \phi(\Bx) \psi(\Bx, t) = i \Hbar \PD{t}{}\psi(\Bx, t),

where

\label{eqn:gaugeTxCurrent:40}
\BPi(\Bx) = -i \Hbar \spacegrad – \frac{e}{c} \BA(\Bx),

find the probability current defined by

\label{eqn:gaugeTxCurrent:60}

Equation \ref{eqn:gaugeTxCurrent:20} and its conjugate are

\label{eqn:gaugeTxCurrent:22}
\begin{aligned}
\inv{2m} \BPi \cdot \BPi \psi + e \phi \psi &= i \Hbar \PD{t}{\psi} \\
\inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj &= -i \Hbar \PD{t}{\psi^\conj}
\end{aligned}

which can be used immediately in a chain rule expansion of the probability time derivative

\label{eqn:gaugeTxCurrent:80}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
i \Hbar \psi^\conj \PD{t}{\psi} +
i \Hbar \psi \PD{t}{\psi^\conj} \\
&=
\psi^\conj \lr{ \inv{2m} \BPi \cdot \BPi \psi + e \phi \psi } –
\psi \lr{ \inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj
} \\
&=
\inv{2m} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
}.
\end{aligned}

We have a difference of conjugates, so can get away with expanding just the first term

\label{eqn:gaugeTxCurrent:100}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi
&=
\psi^\conj
\psi \\
&=
\psi^\conj
\lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA } \cdot \lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA }
\psi \\
&=
\psi^\conj
\lr{
+ \frac{e^2}{c^2} \BA^2
}
\psi.
\end{aligned}

Note that in the directional derivative terms, the gradient operates on everything to its right, including $$\BA$$. Also note that the last term has no imaginary component, so it will not contribute to the difference of conjugates.

This gives

\label{eqn:gaugeTxCurrent:120}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi – \psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
&=
\psi^\conj
\lr{
-\Hbar^2 \spacegrad^2 \psi + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi + \spacegrad \cdot (\BA \psi) }
} \\
\psi
\lr{
-\Hbar^2 \spacegrad^2 \psi^\conj – \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi^\conj + \spacegrad \cdot (\BA \psi^\conj) }
} \\
&=
\frac{i \Hbar e}{c}
\lr{
\psi^\conj
+
\psi
}
\end{aligned}

The first term is recognized as a divergence

\label{eqn:gaugeTxCurrent:140}
\begin{aligned}
&=
+

\end{aligned}

The second term can also be factored into a divergence operation

\label{eqn:gaugeTxCurrent:160}
\begin{aligned}
\psi^\conj
+
\psi
&=
}
+\lr{
} \\
&= 2 \spacegrad \cdot \lr{ \BA \psi \psi^\conj } \\
\end{aligned}

Putting all the pieces back together we have

\label{eqn:gaugeTxCurrent:180}
\begin{aligned}
\PD{t}{\rho}
&=
\inv{2m i \Hbar} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
} \\
&=
\inv{2m i \Hbar} \lr{
-\Hbar^2
+ \frac{ i \Hbar e}{c} 2 \BA \psi \psi^\conj
} \\
&=
\lr{
\frac{i \Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{e}{m c} \BA \psi \psi^\conj
}.
\end{aligned}

From \ref{eqn:gaugeTxCurrent:60}, the probability current must be

\label{eqn:gaugeTxCurrent:200}
\Bj
=
\frac{\Hbar}{2 i m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
– \frac{e}{m c} \BA \psi \psi^\conj,

or
\label{eqn:gaugeTxCurrent:220}
\boxed{
\Bj
=
\frac{\Hbar}{m} \textrm{Im} \lr{ \psi^\conj \spacegrad \psi }
– \frac{e}{m c} \BA \psi \psi^\conj.
}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Question:

Given a gauge transformation of the free particle Hamiltonian to

\label{eqn:gaugeTx:20}
H = \inv{2 m} \BPi \cdot \BPi + e \phi,

where

\label{eqn:gaugeTx:40}
\BPi = \Bp – \frac{e}{c} \BA,

calculate $$m d\Bx/dt$$, $$\antisymmetric{\Pi_i}{\Pi_j}$$, and $$m d^2\Bx/dt^2$$, where $$\Bx$$ is the Heisenberg picture position operator, and the fields are functions only of position $$\phi = \phi(\Bx), \BA = \BA(\Bx)$$.

The final results for these calculations are found in [1], but seem worth deriving to exercise our commutator muscles.

### Heisenberg picture velocity operator

The first order of business is the Heisenberg picture velocity operator, but first note

\label{eqn:gaugeTx:60}
\begin{aligned}
\BPi \cdot \BPi
&= \lr{ \Bp – \frac{e}{c} \BA} \cdot \lr{ \Bp – \frac{e}{c} \BA} \\
&= \Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp + \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2.
\end{aligned}

The time evolution of the Heisenberg picture position operator is therefore

\label{eqn:gaugeTx:80}
\begin{aligned}
\ddt{\Bx}
&= \inv{i\Hbar} \antisymmetric{\Bx}{H} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\BPi^2} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp
+ \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2 } \\
&= \inv{i\Hbar 2 m}
\lr{
\antisymmetric{\Bx}{\Bp^2}
– \frac{e}{c} \antisymmetric{\Bx}{ \BA \cdot \Bp + \Bp \cdot \BA }
}
.
\end{aligned}

For the $$\Bp^2$$ commutator we have

\label{eqn:gaugeTx:100}
\antisymmetric{x_r}{\Bp^2}
=
i \Hbar \PD{p_r}{\Bp^2}
=
2 i \Hbar p_r,

or
\label{eqn:gaugeTx:120}
\antisymmetric{\Bx}{\Bp^2}
=
2 i \Hbar \Bp.

Computing the remaining commutator, we’ve got

\label{eqn:gaugeTx:140}
\begin{aligned}
\antisymmetric{x_r}{\Bp \cdot \BA + \BA \cdot \Bp}
&= x_r p_s A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s p_s x_r \\
&= \lr{ \antisymmetric{x_r}{p_s} + p_s x_r } A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s \lr{ \antisymmetric{p_s}{x_r} + x_r p_s } \\
&= \antisymmetric{x_r}{p_s} A_s + {p_s A_s x_r – p_s A_s x_r} \\
&\quad+ {x_r A_s p_s – x_r A_s p_s} + A_s \antisymmetric{x_r}{p_s} \\
&= 2 i \Hbar \delta_{r s} A_s \\
&= 2 i \Hbar A_r,
\end{aligned}

so

\label{eqn:gaugeTx:160}
\antisymmetric{\Bx}{\Bp \cdot \BA + \BA \cdot \Bp} = 2 i \Hbar \BA.

Assembling these results gives

\label{eqn:gaugeTx:180}
\boxed{
\ddt{\Bx} = \inv{m} \lr{ \Bp – \frac{e}{c} \BA } = \inv{m} \BPi,
}

as asserted in the text.

### Kinetic Momentum commutators

\label{eqn:gaugeTx:200}
\begin{aligned}
\antisymmetric{\Pi_r}{\Pi_s}
&=
\antisymmetric{p_r – e A_r/c}{p_s – e A_s/c} \\
&=
{\antisymmetric{p_r}{p_s}}
– \frac{e}{c} \lr{ \antisymmetric{p_r}{A_s} + \antisymmetric{A_r}{p_s}}
+ \frac{e^2}{c^2} {\antisymmetric{A_r}{A_s}} \\
&=
– \frac{e}{c} \lr{ (-i\Hbar) \PD{x_r}{A_s} + (i\Hbar) \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \lr{ -\PD{x_r}{A_s} + \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \epsilon_{t s r} B_t,
\end{aligned}

or
\label{eqn:gaugeTx:220}
\boxed{
\antisymmetric{\Pi_r}{\Pi_s}
=
\frac{i e \Hbar}{c} \epsilon_{r s t} B_t.
}

### Quantum Lorentz force

For the force equation we have

\label{eqn:gaugeTx:240}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&= \ddt{\BPi} \\
&= \inv{i \Hbar} \antisymmetric{\BPi}{H} \\
&= \inv{i \Hbar 2 m } \antisymmetric{\BPi}{\BPi^2}
+ \inv{i \Hbar } \antisymmetric{\BPi}{e \phi}.
\end{aligned}

For the $$\phi$$ commutator consider one component

\label{eqn:gaugeTx:260}
\begin{aligned}
\antisymmetric{\Pi_r}{e \phi}
&=
e \antisymmetric{p_r – \frac{e}{c} A_r}{\phi} \\
&=
e \antisymmetric{p_r}{\phi} \\
&=
e (-i\Hbar) \PD{x_r}{\phi},
\end{aligned}

or
\label{eqn:gaugeTx:280}
\inv{i \Hbar} \antisymmetric{\BPi}{e \phi}
=
=
e \BE.

For the $$\BPi^2$$ commutator I initially did this the hard way (it took four notebook pages, plus two for a false start.) Realizing that I didn’t use \ref{eqn:gaugeTx:220} for that expansion was the clue to doing this more expediently.

Considering a single component

\label{eqn:gaugeTx:300}
\begin{aligned}
\antisymmetric{\Pi_r}{\BPi^2}
&=
\antisymmetric{\Pi_r}{\Pi_s \Pi_s} \\
&=
\Pi_r \Pi_s \Pi_s – \Pi_s \Pi_s \Pi_r \\
&=
\lr{ \antisymmetric{\Pi_r}{\Pi_s} + {\Pi_s \Pi_r} }
\Pi_s
– \Pi_s
\lr{ \antisymmetric{\Pi_s}{\Pi_r} + {\Pi_r \Pi_s} } \\
&= i \Hbar \frac{e}{c} \epsilon_{r s t}
\lr{ B_t \Pi_s + \Pi_s B_t },
\end{aligned}

or

\label{eqn:gaugeTx:320}
\begin{aligned}
\inv{ i \Hbar 2 m} \antisymmetric{\BPi}{\BPi^2}
&= \frac{e}{2 m c } \epsilon_{r s t} \Be_r
\lr{ B_t \Pi_s + \Pi_s B_t } \\
&= \frac{e}{ 2 m c }
\lr{
\BPi \cross \BB
– \BB \cross \BPi
}.
\end{aligned}

Putting all the pieces together we’ve got the quantum equivalent of the Lorentz force equation

\label{eqn:gaugeTx:340}
\boxed{
m \frac{d^2 \Bx}{dt^2} = e \BE + \frac{e}{2 c} \lr{
\frac{d\Bx}{dt} \cross \BB
– \BB \cross \frac{d\Bx}{dt}
}.
}

While this looks equivalent to the classical result, all the vectors here are Heisenberg picture operators dependent on position.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.