• 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.


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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,037 questions , 2,191 unanswered
5,345 answers , 22,705 comments
1,470 users with positive rep
816 active unimported users
More ...

  Why is the Yang-Mills Comparator unitary?

+ 4 like - 0 dislike

In chapter 15.2 of Peskin, the comparator is defined, as some object $U\left(y,\,x\right)$ which transforms as: $$ U\left(y,\,x\right) \mapsto V\left(y\right) U\left(y,\,x\right) \left[V\left(x\right)\dagger\right] $$

where $V\left(x\right)\in SU\left(2\right)^{\mathbb{R}^4}$.

This object is introduced mainly in order to be able to define the covariant derivative, but its main defining property is that the Fermion field transforms then as:

$$ U\left(y,\,x\right)\psi\left(x\right)\mapsto V\left(y\right)U\left(y,\,x\right)\psi\left(x\right)$$ so we basically have a way to make the Fermion field transform as if it were at $y$ instead of at $x$.

At some stage Peskin states it is reasonable to that this comparator be unitary. My question is why? What breaks down, and where, if we don't assume that?

This post imported from StackExchange Physics at 2014-05-04 11:19 (UCT), posted by SE-user Psycho_pr
asked May 1, 2014 in Theoretical Physics by PPR (135 points) [ no revision ]

1 Answer

+ 0 like - 0 dislike

If the comparator was not unitary, you could not expand it in terms of Hermitian generators of $SU(2)$, which is required in order to construct the non-Abelian covariant derivative.

This post imported from StackExchange Physics at 2014-05-04 11:19 (UCT), posted by SE-user Frederic Brünner
answered May 3, 2014 by Frederic Brünner (1,130 points) [ 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:
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