# $U(1)$ gauge invariance

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

$$A_{\mu}$$ 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.

• Now I see the trick. Thanks a lot! 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 $$F^{\mu \nu}$$.

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