D.71 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

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

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

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

so you need the com­mu­ta­tor

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

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

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

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

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

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

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

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,

$$m \frac{{\rm d}\langle v_i\rangle}{{\rm d}t} = \left\lang... ... q \left\langle \frac{\partial A_i}{\partial t} \right\rangle$$

re­quires the com­mu­ta­tor

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

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,

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

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:

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

Work out the sim­pler com­mu­ta­tor ap­pear­ing here first:

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

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

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

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:

$$\frac{q{\rm i}\hbar}{m} \left\{ ({\widehat p}_{\overline{... ...{\partial r_{\overline{\overline{\imath}}}} \right) \right\}$$

and that is

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

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 $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nabla$ $\times$ $\skew3\vec A$.