3
$\begingroup$

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: $A_{\mu} \rightarrow A_{\mu}+\partial_{\mu}x$?

$\endgroup$
0

2 Answers 2

8
$\begingroup$

$A_{\mu}$ 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.

$\endgroup$
1
  • $\begingroup$ Now I see the trick. Thanks a lot! $\endgroup$ Commented May 24, 2020 at 13:23
5
$\begingroup$

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 $F^{\mu \nu}$.

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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