Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading a text book where they show the electron has spin 1/2 using Dirac's equation. At one point in the derivation they define $\pi=P-qA/c$ where $P$ is the momentum operator and A is the vector potential. They then claim that $\pi\times \pi=iq\hbar B / c$ where B is the magnetic field. Apparently $\nabla\times A=B$ as we are assuming the scalar potential is static.

My question is what happened to the $A\times P$ term in $\pi\times\pi$, why is that set to zero?

share|cite|improve this question
I'm guessing you meant spin-$\frac{1}{2}$? – David Z Jun 29 '12 at 3:52
Thanks, yes I did. I have corrected it. – Virgo Jun 29 '12 at 4:09
up vote 3 down vote accepted

$A\times P$ – more precisely, an expression proportional to $A\times P + P\times A$ – wasn't set to zero. It was properly evaluated and the result gave the $iq\hbar B/c$ term.

Note that if $\pi$ were a vector of $c$-numbers rather than operators, $\pi\times \pi$ would be equal to zero. That's how the cross product behaves. So any term in the cross product $\pi\times \pi$ that is nonzero must be proportional to the nonzero commutators between the components of $\pi$. Now, all three components of $P$ commute with each other; and all three components of $A$ (which depend on the vector $x$) commute with each other. So all the terms in $\pi\times \pi$ must arise from the commutators of components of $P$, essentially a derivative with respect to $x$, and components of the vector potential $A$. By rotational symmetry, it's clear that one must get a multiple of $\nabla\times A$ in this way. And by the way, $B = \nabla\times A$ holds exactly even if there is a time-dependent scalar potential!

Let me write the calculation here: $$ \pi\times\pi = \epsilon_{ijk} \pi_j\pi_k = \frac 12\epsilon_{ijk} [\pi_j,\pi_k]=\dots $$ Here, I could have replaced the product $\pi_j\pi_k$ by one-half of the commutator because it's multiplied by a $jk$-antisymmetric epsilon symbol, anyway. Continue: $$ \dots = \epsilon_{ijk} [P_j,-qA_k/c] = \dots $$ Here, I used the distributive law for the commutator, realized that $[P_j,P_k]=0$ and $[A_j,A_k]=0$, so only the mixed commutators contribute something that is nonzero and these mixed terms are there twice, $[P,A]$ and $[A,P]$ with the opposite sign (cancelled by the opposite sign of the epsilon symbol), so it's enough to write one of them and erase the factor of $1/2$ again.

Now, $[P_j,Y]\equiv -i\hbar \partial_j Y$ so we have $$\dots = -i\hbar \epsilon_{ijk} \partial_j(-qA_k/c) = \frac{iq\hbar}{c} \epsilon_{ijk}\partial_j A_k = \frac{iq\hbar}{c} B_i. $$ QED.

share|cite|improve this answer
Thanks for your clear explanation. – Virgo Jun 29 '12 at 7:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.