I am looking at some exercises in an online course in QFT and there is a question about the $U(1)$ gauge invariance of this operator: $$i\bar{\psi}\sigma^{\mu\nu}\gamma_5(\partial_{\mu}A_{\nu})\psi$$ Initially I thought this operator is not invariant since the $A_{\nu}$ is not gauge invariant itself under $U(1)$ instead its field strength tensor is. Though the correct answer in the solutions says that this operator is gauge invariant under $U(1)$. How can I see this? Is it because we can fix the gauge such that the terms like $\partial_{\mu}\partial_{\nu}x$ vanish, where: $A_{\mu} \rightarrow A_{\mu}+\partial_{\mu}x$?


2 Answers 2


$A_{\mu}$ is not gauge invariant, and $\partial_{\mu} A_{\nu}$ also isn't.

But its antisymmetric part is:

$$ \frac{1}{2} \left(\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}\right) = \frac{1}{2} F_{\mu \nu}. $$

Since in your expression you multiply by an antisymmetric $\sigma^{\mu \nu}$, you're allowed to anti-symmetrize the tensor $\partial_{\mu} A_{\nu}$, which makes the contraction gauge invariant.

  • $\begingroup$ Now I see the trick. Thanks a lot! $\endgroup$ Commented May 24, 2020 at 13:23

The set of matrices $\sigma^{\mu \nu } = \gamma^\mu \gamma^\nu - \gamma^\nu \gamma^\mu$ is defined in such a way that $\sigma^{\mu \nu } = - \sigma^{\nu \mu } $. Therefore, inside the parenthesis you really have the field strength tensor $F^{\mu \nu}$.

  • 3
    $\begingroup$ +1, but you were late by ~ 1 min :) $\endgroup$ Commented May 23, 2020 at 23:22
  • $\begingroup$ It happens, I was writing the answer on my mobile and it's not so easy to insert fomulas. $\endgroup$
    – Quillo
    Commented May 23, 2020 at 23:40

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