# Ward Identity makes QED logarithmic divergent?

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.
• 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 Mar 8, 2016 at 13:26