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


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

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  • $\begingroup$ "It a pity that I can't accept my own answer" You can. You may have to wait a while first though, I don't know. $\endgroup$ – Michael Brown Jun 19 '13 at 2:01
  • $\begingroup$ You can/ . You don't even need to wait a while, because you have posted it using different accounts, . You can request the moderators to merge your accounts, ttoo, by the way. physics.stackexchange.com/contact $\endgroup$ – Abhimanyu Pallavi Sudhir Dec 16 '13 at 3:13
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    $\begingroup$ that first euqation where you write the self energy in terms of $A$ and $B$ is because Lorentz invariance and not gauge invariance. $A=B$ on the other hand is indeed because of the ward identity $\endgroup$ – Yossarian Mar 8 '16 at 13:26

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