A step in the derivation of the magnetic moment 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?
 A: 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 :)
A: 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.
