# How to deal with cross-terms in squared slashed covariant derivative in QED?

I am trying to compute the square of the slashed covariant derivative (the sum of the partial derivative and vector potential contracted with the gamma matrices) in terms of the square of the standard covariant derivative plus a term depending on the electromagnetic tensor and the commutator of the gamma matrices:

$$\not D^2=D_μD^μ+ \frac e 2 F_{μν}σ^{μν}$$ where $$\not D = (\partial_μ -ieA_μ)γ^μ$$ $$σ_{μν}=\frac i2[γ^μ,γ^ν]$$ $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

However, I'm getting stuck with either $$\not A \not\partial$$ or $$A_μ \partial_\nu σ^{μν}$$ terms from the cross product which I've got not idea what to do with, since they're clearly absent in the second term.

• This is an easy calculation. I think you need to show more details of what you are doing if we are to see where you are going wrong. – mike stone Jan 1 at 20:24
• @mikestone well I developed the square then used the fact $γ^μγ^ν$ are the sum of their commutator and anti commutator. From the anti commutator I get the cross products of the D² part, from the commutator I get the sigma matrix, but this sigma matrix is multiplying both d*A terms which go into F and A*d terms which I'm stuck with. – Xelote Jan 1 at 20:59
• The commutator gives you $\sigma_{\mu \nu}F^{\mu\nu}$ and nothing else that I can see: $\sigma_{\mu\nu}\nabla_\mu\nabla_\nu=\sigma_{\mu\nu}[\nabla_\mu, \nabla_\nu]/2= \sigma_{\mu\nu}F_{\mu\nu}/2$. – mike stone Jan 1 at 21:45
• @mikestone I don't follow your last step. You have $∇_μ=d_μ-ieA_μ$, thus $[∇_μ,∇_ν]=-ie(d_μA_ν-d_νA_μ) - ie(A_μd_ν-A_νd_μ)$ Now ignoring the -ie coefficient the first parenthesis is F, but why is the second term 0? – Xelote Jan 1 at 22:28
• I'll write it out as an answer – mike stone Jan 1 at 22:47

The commutator $$[\nabla_\mu,\nabla_\nu]\phi$$ is
$$(\partial_\mu+A_\mu)(\partial_\nu+A_\nu)\phi- (\partial_\nu+A_\nu)(\partial_\mu+A_\mu)\phi\\ = \partial_{\mu\nu}\phi + A_\mu \partial_\nu\phi+(\partial_\mu A_\nu) \phi + A_\nu \partial_\mu \phi+ A_\mu A_\nu \phi-(\mu\leftrightarrow \nu)\\ = ((\partial_\mu A_\nu)- (\partial_\nu A_\mu))\phi$$ Notice that we have both $$A_\mu (\partial_\nu\phi)$$ and $$A_\nu(\partial_\mu \phi)$$ when we expand the product $$\nabla_\mu\nabla_\nu \phi$$ and they cancel when we subtract the second product $$\nabla_\nu\nabla_\mu\phi$$.
• Oh I feel dumb, I was doing my computation without an explicit field and thus applying $\partial_\mu (A_\nu \phi)$ only to A and not phi, so I was missing those terms. So I suppose the real lesson is to always remember those computations only make sense with a field there. – Xelote Jan 1 at 23:06