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

146 submissions , 123 unreviewed
3,961 questions , 1,408 unanswered
4,890 answers , 20,766 comments
1,470 users with positive rep
506 active unimported users
More ...

Calculating the beta function through wavefunction renormalization

+ 4 like - 0 dislike
173 views

The ways I'm used to calculating $\beta$ functions is by either the Callan-Symanzik equation or by taking the total derivative with respect to the renormalization scale of a coupling and simplifying.

In these notes by Manohar (pg 20), he calculates the QED $\beta$ function through a different approach which doesn't use the three point function at all. He considers the wave function renormalization of the photon through and gets a renormalized two point function, 

\begin{equation} 
- i \frac{ e ^2 }{ 2 \pi ^2 } ( p _\mu p _\nu - p ^2 g _{ \mu \nu } ) \left[ \int _0 ^1 d x x ( 1 - x ) \log \frac{ m ^2 - p ^2 x ( 1 - x ) }{ m ^2 + M ^2 x ( 1 - x ) } \right] 
\end{equation} 
He then says that the $ \beta $ function is given by acting on the coefficient of $ i ( p _\mu p _\nu - p ^2 g _{ \mu \nu } ) $. He of course gets the right anwer. But why does this work?

asked May 2, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
edited May 2, 2014 by JeffDror

1 Answer

+ 3 like - 0 dislike

Preceding answer suppressed because wrong. New version coming soon.

EDIT. New answer.

Due to quantum corrections, the propagator of the photon is some multiple $(1+F(M))$ of the freestandard propagator. At the one loop order, we have $F(M) = e(M)^{2}G(M)$ for some function $G$. The free standard propagator of the photon  in $1/p^{2}$ gives rise to a static potential between two electrons of charges $e$ separated by a large distance $r$ in $e^{2}/r$, this is the standard Coulomb law. Given the quantum corrections to the photon propagator, this static potential at long distances is modified in $e(M)^{2}(1+F(M))/r$. This shows that the function $e(M)^{2}(1+F(M))$ is a physical quantity which can be experimentally measured. In particular, it can not depend of the renormalization scale $M$ i.e. $\frac{d}{dM}(e(M)^{2}(1+F(M)))=0$.

At the one loop order, this implies

$\beta(e(M)) = M \frac{d}{dM} e(M) = - \frac{e(M)}{2} M \frac{d}{dM}F(M)$.

answered May 2, 2014 by 40227 (4,690 points) [ revision history ]
edited May 10, 2014 by 40227

You're answer seemed correct to me. What was wrong with it?

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
...