Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  QED BRST Symmetry

+ 4 like - 0 dislike
925 views

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?

This post imported from StackExchange Physics at 2014-04-13 14:07 (UCT), posted by SE-user Jonathan Gleason
asked May 3, 2012 in Theoretical Physics by Jonathan Gleason (265 points) [ no revision ]
retagged Apr 13, 2014

1 Answer

+ 5 like - 0 dislike

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

This post imported from StackExchange Physics at 2014-04-13 14:07 (UCT), posted by SE-user Qmechanic
answered May 3, 2012 by Qmechanic (3,120 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-04-13 14:07 (UCT), posted by SE-user Jonathan Gleason

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysic$\varnothing$Overflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...