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 , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why does the vacuum polarization vanish for massless fermion-loop and on-shell photon?

+ 3 like - 0 dislike
1848 views

In https://arxiv.org/abs/1708.01256 on p.8, it says:

> The contribution corresponding to the rightmost graph yields zero in the $\overline{\mathrm{MS}}$ scheme since we consider massless fermions, since at this order in collinear factorisation the
photon is implicitly on shell and so the vacuum polarisation is evaluated at zero virtuality.

The process in question is depicted in the Feynman diagram above. Can somebody explain to me why the vacuum polarisation (also called self-energy of the photon in other literature) should vanish in this case?

The polarisation function in the $\overline{\mathrm{MS}}$ scheme reads
$$\Pi^{\mu\nu}(p,\mu^2)=\frac{e^2}{2\pi^2}(p^\mu p^\nu-g^{\mu\nu}p^2)\int^1_0dxx(1-x)\ln\frac{\mu^2}{m^2-p^2x(1-x)}.$$

If the photon is on-shell and the fermions massless, then $0=p^2=m^2$. But then the logarithm is not well-defined anymore since
$$\ln\frac{\mu^2}{p^2x(x-1)}=\ln\frac{\mu^2}{x(x-1)}-\ln p^2\stackrel{p^2\rightarrow0}{\rightarrow}-\infty.$$

Can somebody show me **by calculation** how $\Pi^{\mu\nu}$ vanishes for an on-shell photon?

asked Feb 24, 2020 in Q&A by twening (70 points) [ revision history ]

If the photon is real (i.e., on shell), then it may not obtain radiative correction.

Teschnically you must have two diagrams with the oppisite signes  (opposite directions of integration) that cancel each other.

@VladimirKalitvianski How do the opposite signs arise?

Wait a minute if $p^2 \to 0$ then $\ln p^2 \to -\infty$, so in your last equation it should be: $-\ln p^2 \to +\infty$ .

@MathematicalPhysicist You are right. But the question remains. But I think I know the answer. The LSZ theorem says that only amputated diagrams contribute to S-matrix elements.

1 Answer

+ 2 like - 0 dislike

The answer is quite simple. The polarisation does not vanish. However, the LSZ theorem says that diagrams with external propagators do not contribute to the S-matrix. The diagram in question is such a diagram.

answered Feb 25, 2020 by twening (70 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$\varnothing$ysicsOverflow
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
...