This question is taken from this Physics Stack Exchange question by "A friendly helper" with it's answer given by "A friendly helper".


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



Answer 1Edit

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.


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