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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.