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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,852 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

What is the significance of the simplification of the equal-time commutation relations when we take the Feynman propagator for gauge parameter $\lambda = 1$?

+ 2 like - 0 dislike
381 views

Classical electromagnetism (with no sources) follows from the actions$$S = \int d^4x\left(-{1\over4}F_{\mu\nu}F^{\mu\nu}\right),\text{ where }F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$The Lagrangian for $A_\mu$, including a gauge fixing term, is$$\mathcal{L} = -{1\over4}F^2 - {\lambda\over2}(\partial_\mu A^\mu)^2.$$Computation of the equal time commutators $[\dot{A}_\mu(\textbf{x}), A_\nu(\textbf{y})]$ and $[\dot{A}_\mu(\textbf{x}), \dot{A}_\nu(\textbf{y})]$ for general $\lambda$ gets$$[\dot{A}_\mu (\textbf{x}), {A}_\nu (\textbf{y})] = i\left(g^{\mu\nu} - {{\lambda - 1}\over{\lambda}}g^{\mu0}g^{\nu0}\right)\delta^3(\textbf{x} - \textbf{y}),$$$$[\dot{A}_\mu (\textbf{x}), \dot{A}_\nu (\textbf{y})] = i{{\lambda - 1}\over{\lambda}}(g^{\mu 0}g^{\nu k} + g^{\nu 0}g^{\mu k})\partial_k \delta^3(\textbf{x} - \textbf{y}).$$Taking $\lambda = 1$, we have$$[\dot{A}_\mu (\textbf{x}), {A}_\nu (\textbf{y})] = ig^{\mu v}\delta^3(\textbf{x} - \textbf{y}), \text{ }[\dot{A}_\mu (\textbf{x}), \dot{A}_\nu (\textbf{y})] = 0.$$My question is, what is the significance of the simplifying that happens here with the equal-time commutators when we take $\lambda = 1$?

asked Oct 16, 2015 in Theoretical Physics by Brian Ng (10 points) [ no revision ]

What do you mean? The whole point gauge symmetry is that your choice of gauge doesn't affect the physics at all, so you are free to exploit gauge freedom to simplify calculations.

1 Answer

+ 1 like - 0 dislike

Comparing this to electromagnetism with source (in slightly different notation) with the Lagrangian

\[\mathcal L = -\frac{1}{4}F^2_{\mu\nu} - \frac{1}{2\xi}(\partial_{\mu}A_{\mu})^2 -J_{\mu}A_{\mu}\]
and the corresponding time-ordered Feynman propagator for a photon

\[i \Pi^{\mu\nu}((p) = \frac{-i}{p^2 + i\varepsilon}\left [ g^{\mu\nu} - (1-\xi)\frac{p^{\mu}p^{\nu}}{p^2}\right]\]

it can be seen that the parameter $\lambda$ and $\xi$ are related as $\xi = \frac{1}{\lambda}$.

So the cas $\lambda = 1$ corresponds to the Feynman-'t Hooft gauge with $\xi = 1$. The only advantage of this is that it makes the propagator consisting of only one term, but there is as said in the comments no physical meaning of this.

answered Dec 5, 2015 by anonymous [ no revision ]

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$\hbar$ysicsOverflo$\varnothing$
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).
To avoid this verification in future, please log in or register.




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

Your rights
...