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I am trying to show that $$(i\gamma^\mu\partial_\mu -e\gamma^\mu A_\mu)^2 = (i\partial_\mu -eA_\mu)^2 -\frac{e}{2} \sigma^{\mu\nu}F_{\mu\nu},$$ where $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$.

So far I have $$\begin{align} (i\gamma^\mu\partial_\mu -e\gamma^\mu A_\mu)^2 &= \gamma^\mu\gamma^\nu(i\partial_\mu-e A_\mu)(i\partial_\nu -e A_\nu) =(i\partial_\mu)^2+(eA_\mu)^2 - ie\gamma^\mu\gamma^\nu(\partial_\mu A_\nu + A_\mu\partial_\nu) \\ &=(i\partial_\mu-eA_\mu)^2+ie\partial_\mu A^\mu +ie A_\mu\partial^\mu - \frac{1}{2}ie(\gamma^\mu\gamma^\nu+2\eta^{\mu\nu}-\gamma^\nu\gamma^\mu)(\partial_\mu A_\nu + A_\mu\partial_\nu) \\ &= (i\partial_\mu-eA_\mu)^2- \frac{1}{2}ie(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu)(\partial_\mu A_\nu + A_\mu\partial_\nu) \\ &= (i\partial_\mu-eA_\mu)^2 - e\sigma^{\mu\nu}(\partial_\mu A_\nu + A_\mu\partial_\nu), \end{align}$$ which is close, but not quite there. I can show that $\sigma^{\mu\nu}\partial_\mu A_\nu = \frac{1}{2} \sigma^{\mu\nu}F_{\mu\nu}$, so I would have the correct result if the $A_\mu\partial_\nu$ term vanishes. But I don't have a good argument for why that would be the case. Why would that be?

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In your last expression, the derivative in $\partial_\mu A_\nu$ is acting both in $A$ and in whatever terms lie in the right of it. In other words, your last line acting on a function $K$ is: \begin{align}\sigma^{\mu\nu}(\partial_\mu A_\nu+A_\mu\partial_\nu)K=&\sigma^{\mu\nu}(\partial_\mu (A_\nu K)+A_\mu\partial_\nu K)\\=&\sigma^{\mu\nu}(\partial_\mu A_\nu)K+\sigma^{\mu\nu}(A_\nu\partial_\mu+A_\mu\partial_\nu)K\\ =&\sigma^{\mu\nu}(\partial_\mu A_\nu)K\\ =&\frac{1}{2}\sigma^{\mu\nu}F_{\mu\nu}K \end{align} hence, as an operator identity, you have \begin{equation} \sigma^{\mu\nu}(\partial_\mu A_\nu+A_\mu\partial_\nu)=\frac{1}{2}\sigma^{\mu\nu}F_{\mu\nu} \end{equation}

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