# Is the supercurrent gauge invariant?

If we consider $${\cal N}=1$$ renormalizeable chiral gauge theories, specifically discussed in section 27.4 of Weinberg's Quantum Theory of Fields, Supersymmetry book, should the supercurrent be gauge invariant?

Any observable of a gauge invariant theory must also be gauge invariant. However, the supercurrent is a Grassmann variable, and thus is not observable.

I ask this because in order to calculate the supercurrent, Weinberg argues to calculate it in a specific gauge to make the analysis easier. This is only possible if the supercurrent itself is gauge invariant.

• Can you not look at the expression Weinberg calculates and just check (admittedly in a potentially tedious calculation) whether it is gauge invariant? Mar 6, 2019 at 20:33
• Yes that is a good point. I checked and it does not appear to be gauge invariant, nor covariant. But that puts into question the whole procedure of working in a special gauge! Mar 6, 2019 at 21:07

It is true that the supercurrent in this case is not gauge invariant. Weinberg writes the supercurrent in the gauge where at the point $$x^{\mu}=X^{\mu}$$, we have that the gauge connection vanishes, so $$A_{\mu}^a(X)=0$$. In this gauge, one term in the supercurrent reads $$S^{\mu}\supset-\frac{1}{4}f^a_{\rho\sigma}[\gamma^{\rho},\gamma^{\sigma}]\gamma^{\mu}\lambda^a$$
Where $$f^a_{\rho\sigma}=\partial_{\rho}A_{\sigma}^a-\partial_{\sigma}A_{\rho}^a$$ is the non-abelian gauge curvature tensor (remember that we are working in the gauge that $$A_{\mu}^a(X)=0$$), and $$\lambda^a$$ is a Majorana fermion supersymmetric pair to the gauge field $$A^{a}_{\mu}$$. This term is enough to show that the supercurrent is not gauge invariant.