# QED BRST Symmetry

This is a homework problem that I am confused about because I thought I knew how to solve the problem, but I'm not getting the result I should. I'll simply write the problem verbatim:

"Consider QED with gauge fixing $\partial _\mu A^\mu=0$ and without dropping the Fadeev-Popov ghost fields. Thus the gauge fixed Lagrangian is $$\mathcal{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}+\frac{1}{2}(\partial ^\mu A_\mu )^2+\overline{\psi}(i\gamma ^\mu D_\mu -m)\psi+\overline{c}(-\partial ^\mu \partial _\mu)c$$ Verify that the Lagrangian is invaraint under the BRST transformation $$\delta A_\mu=\epsilon \partial _\mu c,\delta \psi =0,\delta c=0,\delta \overline{c}=\epsilon \partial _\mu A^\mu"$$ Two questions. First of all, if we are working in the gauge where $\partial _\mu A^\mu =0$, then why has he left this term in the Lagrangian? Does he mean something different by "gauge fixing" than I am thinking? Second of all, I am not getting this Lagrangian to be invariant under the transformation listed. I find that the transformation of the gauge field gives an extra term of the form $$-\epsilon e\overline{\psi}(\gamma ^\mu \partial _\mu c)\psi$$ that doesn't cancel. This term arises from the $A_\mu$ contained in the covariant derivative. Did I screw up the computation somewhere? What's going on?

-

Let me try to briefly address OP's two questions(v3):

1. Recall that quantum mechanically in the path integral, the Lorenz gauge condition $\partial _\mu A^\mu\approx 0$ is only implemented in an appropriate quantum-averaged sense. Traditionally, there is a free gauge parameter $\xi$ in front of the gauge-fixing term $$\frac{1}{2\xi}(\partial ^\mu A_\mu )^2$$ in the Lagrangian density ${\cal L}$. Hence OP is implicitly assuming that $\xi=1$, the so-called Feynman - 't Hooft gauge. To enforce the Lorenz gauge condition strongly (in a Wick-rotated Euclidean path integral), one should go to the Landau gauge $\xi\to 0^{+}$.

2. The fermion $\psi$ is not invariant under BRST (or gauge) transformations as OP writes (v3), but transforms as $$\delta\psi~=~ie\epsilon c\psi.$$

-
With respect to (2), does this not imply that the Lagrangian is not invariant under the BRST transformation? Why would he ask me to show this if this is not in fact the case? –  Jonathan Gleason May 3 '12 at 13:20