# A step in the derivation of the magnetic momentum of the electron in Zee's QFT book

In chapter III.6 of his Quantum Field Theory in a Nutshell, A. Zee sets out to derive the magnetic moment of an electron in quantum electrodynamics. He starts by replacing in the Dirac equation the derivative $\partial_\mu$ by the covariant derivative $D_\mu = \partial_\mu - i e A_\mu$, where $A_\mu$ is a (classical) external electromagnetic field. We have $$(i \gamma^\mu D_\mu - m) \psi ~=~ 0.$$ From that he derives $$\left(D_\mu D^\mu - \frac{e}{2} \sigma^{\mu \nu} F_{\mu\nu} + m^2\right) \psi~=~ 0,$$ where, as usual, $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$ and $\sigma^{\mu \nu}$ are the commutators of the Dirac $\gamma$ matrices: $$\sigma^{\mu \nu}=\frac{i}{2}[\gamma^\mu, \gamma^\nu].$$

My problem deals with a apparently simple step that Zee uses in the derivation. He claims that $$(i/2) \sigma^{\mu \nu}[D_\mu, D_\nu] = (e/2) \sigma^{\mu \nu} F_{\mu\nu}.$$ However, I get $$[D_\mu, D_\nu] = -ie\partial_\mu A_\nu + ie\partial_\nu A_\mu -ieA_\mu\partial_\nu + ie A_\nu \partial_\mu = -ieF_{\mu\nu} -ie A_\mu\partial_\nu + ie A_\nu \partial_\mu,$$ but I do not see right now why the last two terms vanish when multiplied by $\sigma^{\mu \nu}$. I even tried to use the explicit expressions for $\sigma^{\mu \nu}$ and got a nonzero value. I have the feeling that I am missing something really simple here. Does somebody see what I did wrong?

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When you do the commutator you have to remember that it acts on something. That means you'll have (don't care for the convention of i and e):

$[D_\mu,D_\nu]\phi=(-[\partial_\mu,A_\nu]-[A_\mu,\partial_\nu])\phi$

Now you have to take into account the chain rule for differentiation! The first commutator evaluates to:

$-(\partial_\mu A_\nu)\phi-A_\nu(\partial_\mu \phi)+A_\nu(\partial_\mu \phi)=-(\partial_\mu A_\nu)\phi$

Doing that for the second commutator as well gives the wanted result.

Cheers, A friendly helper

Edit: Thanks for the Latex fix :)

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Right. I guess I was blinded by the appearance of $\partial_\mu A_\nu - \partial_\nu A_\mu$ so I forgot that the $\partial$s were still operating on everything to their right. Thanks. –  Alejandro Luque Nov 22 '12 at 22:36

You've forgotten that the expression $[D_{\mu},D_{\nu}]$ is an operator, so the derivatives act on everything to their right. It is easiest to work things out if you actually operate your expression on an arbitrary test function $f(x)$. So in your last equation, for example, the first term on the right-hand side of the first equality sign becomes $$-i e \partial_{\mu}A_{\nu} \to -i e \partial_{\mu}(A_{\nu}f(x)) = -i e (\partial_{\mu}A_{\nu})f(x) -i e A_{\nu}(\partial_{\mu} f(x)).$$ Carry out this procedure in full and the unwanted terms should cancel.

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This is basically identical to the answer by A friendly helper and both are correct. I am accepting his answer because he was slightly quicker but many thanks anyway. –  Alejandro Luque Nov 22 '12 at 22:38
yup, a friendly helper posted while i was still in the middle of writing this, so he/she deserves it :) –  Mark Mitchison Nov 23 '12 at 0:50