# Why the divergent part of the 1-loop correction for the photon propagator *cannot* depend on the fermion mass $m$?

Last week my QFT II professor claimed that the divergent part of the diagram of the one-loop correction to the photon propagator by means of a fermion cannot depend on the fermion's mass.

I haven't been able to find out the reason of this. I have checked that indeed the divergent part of the diagram doesn't depend on $$m^2$$ (where $$m$$ is the mass of the fermion), but I don't underestand why this must be like this, or how could we know this prior to the calculation.

My guesses are:

• The divergence occurs when $$k\rightarrow\infty$$ (UV-divergent), and so the mass of the fermion can be neglected.
• The fermion having a mass in the divergente term would imply adding a mass to the photon, which would break gauge symmetry (I'm not sure if this makes any sense).

Let us sketch an argument:

1. The photon vacuum polarization $$\Pi^{\mu\nu}(p,m)$$ in $$d=4$$ has superficial degree of divergence (SDOD) $$D=2$$. Each time we differentiate wrt. the photon momentum $$p^{\mu}$$ or the fermion mass $$m$$, we effectively gain 1 more fermion/boson propagator, and hence lower the SDOD by (at least) 1 unit, cf. Ref. 1. (We implicitly assume that there are no divergent subdiagrams.)

2. Lorentz covariance dictates that the tensor structure $$\Pi^{\mu\nu}$$ comes from either $$p^{\mu}p^{\nu}$$ or $$g^{\mu\nu}$$.

We can therefore Taylor expand around $$(p,m)=(0,0)$$:

\begin{align} \Pi^{\mu\nu}(p,m)~=~& Ag^{\mu\nu}\Lambda^2+ g^{\mu\nu} \{B_1m+B_0\} \Lambda +C p^{\mu}p^{\nu}\ln\Lambda\cr & +g^{\mu\nu}\{D p^2 +E_2 m^2 +E_1m +E_0\} \ln\Lambda \cr &+ \text{finite terms},\end{align} where $$\Lambda$$ is a UV momentum cut-off. Hm. It seems we need one more input.

1. From the Ward identity, we know that the vacuum polarization should be transversal, $$\Pi^{\mu\nu}(p,m)~=~(g^{\mu\nu}p^2-p^{\mu}p^{\nu})\Pi(p,m).$$

Therefore

$$\Pi(p,m)~=~D\ln\Lambda + \text{finite terms}.$$

This answers OP's question: The divergent terms do not depend on the fermion mass $$m$$.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 10.1, p. 319.
• Hey @Qmechanic , I do not see why you do not add a $D(m^2) p^\mu p^\nu \Lambda^2$ term. Could you please elaborate on that? Commented Apr 13 at 10:52
• Hey @Gabriel Ybarra Marcaida. Thanks for the feedback. A term quadratic in external momenta corresponds to a diagram with SDOD $D=0$, cf. Ref. 1. Commented Apr 13 at 12:23