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.

  • 1
    $\begingroup$ Can you not look at the expression Weinberg calculates and just check (admittedly in a potentially tedious calculation) whether it is gauge invariant? $\endgroup$
    – ACuriousMind
    Mar 6, 2019 at 20:33
  • $\begingroup$ 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! $\endgroup$
    – fewfew4
    Mar 6, 2019 at 21:07

1 Answer 1


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.

The calculation Weinberg does is only valid in this gauge, and making a gauge transformation will give you a different supercurrent. This is alright as long as the conservation equation holds as an expectation value between two physical states.


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