gauge transformation

Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

October 20, 2015 phy1520 No comments , , , , , , , , , , , ,

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 No comments , , ,

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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

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

and apply a gauge transformed Hamiltonian

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

Recall that in classical mechanics we have

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

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

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

which takes the Hamiltonian from

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

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

We care that the position and momentum operators obey

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

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

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

This maps the Hamiltonian to

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

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

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

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

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

where we alter the momentum by

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

This takes the plane wave to

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

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

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

of the form

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

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

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

Looking at just the first action

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

If we choose

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

then we are left with

\begin{equation}\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).
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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.
\end{equation}

When \( \psi \) is an energy eigenfunction, this means

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

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
\begin{equation}\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}
}
\end{equation}

Aharonov-Bohm effect

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

fig. 1. particle confined to a ring

fig. 1. particle confined to a ring

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

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

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

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

We know that

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

so that

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

The Hamiltonian for the new configuration is

\begin{equation}\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}
\end{equation}

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?

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

\begin{equation}\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}
\end{equation}

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.

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

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

all the commutators have the expected values.

Answer

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

\begin{equation}\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}
\end{equation}

and the momentum commutators

\begin{equation}\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}
\end{equation}

That last sum of commutators is

\begin{equation}\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}
\end{equation}

We’ve shown that

\begin{equation}\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}
\end{equation}

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.

Gauge transformed probability current

September 17, 2015 math and physics play No comments , , , ,

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

For the gauge transformed Schrodinger equation

\begin{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),
\end{equation}

where

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

find the probability current defined by

\begin{equation}\label{eqn:gaugeTxCurrent:60}
\PD{t}{\psi} + \spacegrad \cdot \Bj.
\end{equation}

Answer

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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{
-\Hbar^2 \spacegrad^2 + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad + \spacegrad \cdot \BA }
+ \frac{e^2}{c^2} \BA^2
}
\psi.
\end{aligned}
\end{equation}

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

\begin{equation}\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) }
} \\
&\quad –
\psi
\lr{
-\Hbar^2 \spacegrad^2 \psi^\conj – \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi^\conj + \spacegrad \cdot (\BA \psi^\conj) }
} \\
&=
-\Hbar^2 \lr{ \psi^\conj \spacegrad^2 \psi – \psi \spacegrad^2 \psi^\conj } \\
&\quad +
\frac{i \Hbar e}{c}
\lr{
\psi^\conj
\BA \cdot \spacegrad \psi + \psi^\conj \spacegrad \cdot (\BA \psi)
+
\psi
\BA \cdot \spacegrad \psi^\conj + \psi \spacegrad \cdot (\BA \psi^\conj)
}
\end{aligned}
\end{equation}

The first term is recognized as a divergence

\begin{equation}\label{eqn:gaugeTxCurrent:140}
\begin{aligned}
\spacegrad \cdot \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
&=
\psi^\conj \spacegrad \cdot \spacegrad \psi
+
\spacegrad \psi \cdot \spacegrad \psi^\conj

\psi \spacegrad \cdot \spacegrad \psi^\conj

\spacegrad \psi^\conj \cdot \spacegrad \psi \\
&= \psi^\conj \spacegrad^2 \psi – \psi \spacegrad^2 \psi^\conj.
\end{aligned}
\end{equation}

The second term can also be factored into a divergence operation

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

Putting all the pieces back together we have

\begin{equation}\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
} \\
&=
\spacegrad \cdot
\inv{2m i \Hbar} \lr{
-\Hbar^2
\lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{ i \Hbar e}{c} 2 \BA \psi \psi^\conj
} \\
&=
\spacegrad \cdot
\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}
\end{equation}

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

\begin{equation}\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,
\end{equation}

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

References

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

Gauge transformation of free particle Hamiltonian

September 15, 2015 math and physics play No comments , , , , , ,


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Question:

Given a gauge transformation of the free particle Hamiltonian to

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

where

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

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 \( \phi = \phi(\Bx), \BA = \BA(\Bx) \).

Answer

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

\begin{equation}\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}
\end{equation}

The time evolution of the Heisenberg picture position operator is therefore

\begin{equation}\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}
\end{equation}

For the \( \Bp^2 \) commutator we have

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

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

Computing the remaining commutator, we’ve got

\begin{equation}\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}
\end{equation}

so

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

Assembling these results gives

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

as asserted in the text.

Kinetic Momentum commutators

\begin{equation}\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}
\end{equation}

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

Quantum Lorentz force

For the force equation we have

\begin{equation}\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}
\end{equation}

For the \( \phi \) commutator consider one component

\begin{equation}\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}
\end{equation}

or
\begin{equation}\label{eqn:gaugeTx:280}
\inv{i \Hbar} \antisymmetric{\BPi}{e \phi}
=
– e \spacegrad \phi
=
e \BE.
\end{equation}

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

\begin{equation}\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}
\end{equation}

or

\begin{equation}\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}
\end{equation}

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

\begin{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}
}.
}
\end{equation}

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.