D.72 Electromagnetic commutators

The purpose of this note is to identify the two commutators of chapter 13.1; the one that produces the velocity (or rather, the rate of change in expectation position), and the one that produces the force (or rather the rate of change in expectation linear momentum). All basic properties of commutators used in the derivations below are described in chapter 4.5.4.

The Hamiltonian is

\begin{displaymath}
H = \frac{1}{2m}\left({\skew 4\widehat{\skew{-.5}\vec p}}-...
...= \frac{1}{2m}\sum_{j=1}^3({\widehat p}_j-qA_j)^2 + q \varphi
\end{displaymath}

when the dot product is written out in index notation.

The rate of change in the expectation value of a position vector component $r_i$ is according to chapter 7.2 given by

\begin{displaymath}
\frac{{\rm d}\langle r_i \rangle}{{\rm d}t} =
\left\langle \frac{{\rm i}}{\hbar} [H,r_i] \right\rangle
\end{displaymath}

so you need the commutator

\begin{displaymath}[H,r_i]= \left[\frac{1}{2m}\sum_{j=1}^3({\widehat p}_j-qA_j)^2+q\varphi,r_i\right]
\end{displaymath}

Now the term $q\varphi$ can be dropped, since functions of position commute with each other. On the remainder, use the fact that each of the two factors ${\widehat p}_j-qA_j$ comes out at its own side of the commutator, to give

\begin{displaymath}[H,r_i]= \frac{1}{2m}\sum_{j=1}^3
\bigg\{
({\widehat p}_j...
...i] + [{\widehat p}_j-qA_j,r_i]({\widehat p}_j-qA_j)
\bigg\}
\end{displaymath}

and then again, since the vector potential is just a function of position too, the $qA_j$ can be dropped from the commutators. What is left is zero unless $j$ is the same as $i$, since different components of position and momentum commute, and when $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$, it is minus the canonical commutator, (minus since the order of $r_i$ and ${\widehat p}_i$ is inverted), and the canonical commutator has value ${\rm i}\hbar$, so

\begin{displaymath}[H,r_i]= -\frac{1}{m}{\rm i}\hbar({\widehat p}_i-qA_i)
\end{displaymath}

Plugging this in the time derivative of the expectation value of position, you get

\begin{displaymath}
\frac{{\rm d}\langle r_i \rangle}{{\rm d}t} = \frac{1}{m}
\left\langle {\widehat p}_i-qA_i \right\rangle
\end{displaymath}

so the normal momentum $mv_i$ is indeed given by the operator ${\widehat p}_i-qA_i$.

On to the other commutator! The $i$-​th component of Newton’s second law in expectation form,

\begin{displaymath}
m \frac{{\rm d}\langle v_i\rangle}{{\rm d}t} =
\left\la...
...
\left\langle \frac{\partial A_i}{\partial t} \right\rangle
\end{displaymath}

requires the commutator

\begin{displaymath}[H,{\widehat p}_i-qA_i]=
\left[\frac{1}{2m}\sum_{j=1}^3({\widehat p}_j-qA_j)^2+q\varphi,p_i-qA_i\right]
\end{displaymath}

The easiest is the term $q\varphi$, since both $\varphi$ and $A_i$ are functions of position and commute. And the commutator with ${\widehat p}_i$ is the generalized fundamental operator of chapter 4.5.4,

\begin{displaymath}[q\varphi,p_i]={\rm i}\hbar q\frac{\partial\varphi}{\partial r_i}
\end{displaymath}

and plugging that into Newton’s equation, you can verify that the electric field term of the Lorentz law has already been obtained.

In what is left of the desired commutator, again take each factor ${\widehat p}_j-qA_j$ to its own side of the commutator:

\begin{displaymath}
\frac{1}{2m}
\sum_{j=1}^3
\bigg\{
({\widehat p}_j-qA...
...[{\widehat p}_j-qA_j,p_i-qA_i]({\widehat p}_j-qA_j)
\bigg\}
\end{displaymath}

Work out the simpler commutator appearing here first:

\begin{displaymath}[{\widehat p}_j-qA_j,p_i-qA_i]= - q [p_j,A_i] -q[A_j,p_i]
=...
...tial r_j}
- {\rm i}\hbar q\frac{\partial A_j}{\partial r_i}
\end{displaymath}

the first equality because momentum operators and functions commute, and the second equality is again the generalized fundamental commutator.

Note that by assumption the derivatives of $\skew3\vec A$ are constants, so the side of ${\widehat p}_j-qA_j$ that this result appears is not relevant and what is left of the Hamiltonian becomes

\begin{displaymath}
\frac{q{\rm i}\hbar}{m}
\sum_{j=1}^3
\left\{
\frac{\...
...partial A_j}{\partial r_i}
\right\}
({\widehat p}_j-qA_j)
\end{displaymath}

Now let ${\overline{\imath}}$ be the index following $i$ in the sequence $123123\ldots$ and ${\overline{\overline{\imath}}}$ the one preceding it (or the second following). Then the sum above will have a term where $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$, but that term is seen to be zero, a term where $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\overline{\imath}}$, and a term where $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\overline{\overline{\imath}}}$. The total is then:

\begin{displaymath}
\frac{q{\rm i}\hbar}{m}
\left\{
({\widehat p}_{\overli...
...artial r_{\overline{\overline{\imath}}}}
\right)
\right\}
\end{displaymath}

and that is

\begin{displaymath}
- \frac{q{\rm i}\hbar}{m}
\left\{
({\widehat p}_{\over...
...nabla\times\skew3\vec A\right)_{\overline{\imath}}
\right\}
\end{displaymath}

and the expression in brackets is the $i$-​th component of $({\skew 4\widehat{\skew{-.5}\vec p}}-q\skew3\vec A)$ $\times$ $\left(\nabla\times\skew3\vec A\right)$ and produces the $q\vec{v}$ $\times$ $\skew2\vec{\cal B}$ term in Newton’s equation provided that $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nabla$ $\times$ $\skew3\vec A$.