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 , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why is baryon or lepton violation in standard model is a non-perturbative effect?

+ 1 like - 0 dislike
2263 views

The baryon number B or lepton number L violation in the standard model arise from triangle anomaly. Right? Triangle diagrams are perturbative diagrams. Then why the B or L violation in Standard model is said to be a non-perturbative effect? I'm confused.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user Roopam
asked Jul 31, 2014 in Theoretical Physics by Roopam (145 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

It is a non-perturbative effect because it is 1-loop exact.

The triangle diagram is actually the least insightful method to think about this, in my opinion. The core of the matter is the anomaly of the chiral symmetry, which you can also, for example, calculate by the Fujikawa method examining the change of the path integral measure under the chiral transformation. You can obtain quite directly that the anomaly is proportional to

$$\int \mathrm{Tr} (F \wedge F)$$

which is manifestly a global, topological term, (modulo some intricacies) it is the so-called second Chern class and takes only values of $8\pi^2k$ for integer $k$. It is, by the Atiyah-Singer index theorem (this can also be seen by Fujikawa), essentially the difference between positive and negative chiral zero modes of the Dirac operator. This is obviously a discontinuous function of $A$ (or $F$), which is already bad for something which, if it were perturbative, should be a smooth correction to something, and it is also the number describing which instanton vacuum sector we are in, see my answer here. Since perturbation theory takes place around a fixed vacuum, this is not a perturbative effect, since it is effectively describing a tunneling between two different vacuum sectors.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user ACuriousMind
answered Jul 31, 2014 by ACuriousMind (910 points) [ no revision ]
ACuriousMind Why do you say that "The triangle diagram is actually the least insightful method to think about this"? Doesn't the chiral anomaly arise due to triangle diagrams? Is it the same case for baryonic or leptonic current anomaly?

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user Roopam
@Roopam: "Due" is a difficult word here. If you insist on thinking about everything in terms of diagram, then yes, it arises due to them. But it is an anomaly of a symmetry, an effect produced by the non-invariance of the path integral measure under the symmetry, perfectly derivable without any reference to Feynman diagrams. I think it is not insightful because the triangle is "just another diagram" when the effect product is fundamentally different from "just another perturbative correction". It's just my opinion though, if you like the diagram best, it's your choice.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user ACuriousMind

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$\hbar$ysi$\varnothing$sOverflow
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
...