Derivatives of Fields in E&M

Question

In QED the field strength tensor $F_{\mu\nu}$ is given by the commutator of the covariant derivatives $$D_\mu=\partial_\mu-ieA_\mu$$ where $A_\mu$ is the gauge field. Explicitly we have

$$[D_\mu,D_\nu]\psi=ieF_{\mu\nu}\psi$$

Using this relation one can derive the standard result

$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$

However I only get this result if terms of the form $A_\mu\partial_\nu$ vanish. Is there some mathematical reason why this happens (some cancellation somewhere) or the action of the derivative on the field, in which case what is the physical reason behind the assumption of these terms vanishing?

Answer 1

There is no extra requirement for theses terms to vanish, everything works out if the calculation is done correctly. The most probable thing to forget is the product rule for the derivatives: \begin{align} [D_{\mu},D_{\nu}]\psi&=D_{\mu}D_{\nu}\psi-D_{\nu}D_{\mu}\psi\\&=D_{\mu}(\partial_{\nu}-ieA_{\nu})\psi-D_{\nu}(\partial_{\mu}-ieA_{\mu})\psi\\&=(\partial_{\mu}-ieA_{\mu})(\partial_{\nu}\psi-ieA_{\nu}\psi)-(\partial_{\nu}-ieA_{\nu})(\partial_{\mu}\psi-ieA_{\mu}\psi)\\&=\partial_{\mu}\partial_{\nu}\psi-ie\partial_{\mu}(A_{\nu}\psi)-ieA_{\mu}\partial_{\nu}\psi-e^{2}A_{\mu}A_{\nu}\psi-(\nu\leftrightarrow\mu)\\&=\partial_{\mu}\partial_{\nu}\psi-ie\partial_{\mu}A_{\nu}\psi-ieA_{\nu}\partial_{\mu}\psi-ieA_{\mu}\partial_{\nu}\psi-e^{2}A_{\mu}A_{\nu}\psi-(\nu\leftrightarrow\mu)\\&=-ieF_{\mu\nu}\psi \end{align}

Where the last equality used that all terms symmetric in $\mu \leftrightarrow\nu$ vanish. (Just the second term isn't.)

(btw.: How do I align equations?)

Answer 2

The Lagrangian is a Lorentz scalar. The terms you refer to show up like $\phi(\partial_{\mu} \phi)$, these are Lorentz vectors and cannot show up in the Lagrangian. All vector indices are contracted as in $\phi(\partial_{\mu} \phi)(\partial^{\mu} \phi)$.