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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?

Jonathan Gleason
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1 Answers1

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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 $$\delta\psi~=~ie\epsilon c\psi.$$

Qmechanic
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