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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.