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
5,054 questions , 2,207 unanswered
5,345 answers , 22,721 comments
1,470 users with positive rep
818 active unimported users
More ...

  When does the correlator of a string of fields and the current vanish "sufficiently fast" at infinity and Ward's identity?

+ 4 like - 0 dislike
1173 views

One consequence of the Ward identity (cf. Di Francesco et al) is that it means variation of correlators under infinitesimal transformation is zero. This can be seen by integrating the ward identity, and using the Gauss divergence theorem to integrate over the surface at infinity, and setting the surface term to zero.

$$ \delta_{\omega} \langle j^{\mu}_a \Phi(x_1) \ldots \Phi(x_n) \rangle =\int \frac{\partial}{\partial x^{\mu}}\langle j^{\mu}_a \Phi(x_1) \ldots \Phi(x_n) \rangle = \int_{\Sigma} ds_{\mu} \langle j^{\mu}_a \Phi(x_1) \ldots \Phi(x_n) \rangle \\ =0$$.

DiFrancesco says the integrand goes to zero at infinity because the divergence of the correlator vanishes at away from the points $x_1 \ldots x_n$. How does the vanishing of divergence mean vanishing of the correlator? In general when can we assume a correlator to vanish at infinity "sufficiently fast"?

In this particular case, from familiar examples of the free boson and fermion, I know that $j$ is proportional to the gradient of the fields, so does this mean we are considering field solutions which attain a constant value? How is it different from solitonic solutions?

I have seen surface term to be vanishing sufficiently fast in many places in QFT, cannot seem to remember them now ,but this seems to be quite an adhoc assumption, is it because you want the drama of the theory to happen only in a finite range, and you cannot consider particles at inifinity?

This post imported from StackExchange Physics at 2014-06-21 09:00 (UCT), posted by SE-user ramanujan_dirac
asked Jun 19, 2014 in Theoretical Physics by ramanujan_dirac (235 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:
p$\hbar$ysicsOver$\varnothing$low
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
...