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  Ward Identity makes QED logarithmic divergent?

+ 6 like - 0 dislike

Quick question regarding superficial degrees of freedom and Ward identities.

For instance in Peskin and Schroeder it is stated that the photon-self energy is superficially quadratically UV divergent but due to the Ward identity it is only logarithmically divergent. I don't see this argument.

The self-energy is given by

$\Pi^{1-loop}=(g^{\mu\nu}p^2-p^\mu p^\nu)\Pi(p^2)$

How does the Ward identity, or in other words, gauge invariance kill of the divergences?

Best, A friendly helper

This post imported from StackExchange Physics at 2014-04-15 16:38 (UCT), posted by SE-user A friendly helper
asked Nov 1, 2012 in Theoretical Physics by A friendly helper (320 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

Ok, I answer it myself. The reason is as follows; Based on gauge invariance the self-energy at one loop has to look like $$\Pi=(g^{\mu\nu}p^2 A -p^\mu p^\nu B)$$ where A and B are the explicit divergences not yet determined. However, in an explicit loop computation the first term does only arise with a divergence in D=2 whereas the second with a divergence in D=4 and not worse. But in order for gauge invariance to be true $A=B$ has to hold, i.e. the divergence is actually only in four dimensions and not in two.

Edit: It a pity that I can't accept my own answer :D

This post imported from StackExchange Physics at 2014-04-15 16:38 (UCT), posted by SE-user A friendly helper
answered Apr 19, 2013 by A friendly helper (320 points) [ no revision ]

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