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

Question

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).

Answer 1

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 propagator, and hence lower the SDOD by 1 unit, cf. Ref. 1.

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 $p=0$:

$$\begin{align} \Pi^{\mu\nu}(p,m)~=~&A(m) g^{\mu\nu}\Lambda^2\cr &+ \{B(m) g^{\mu\nu}p^2 + C(m) p^{\mu}p^{\nu}\} \ln\Lambda \cr &+ \text{finite terms},\end{align}\tag{1} $$ where $\Lambda$ is a UV momentum cut-off. Then by differentiation: $$\begin{align} \frac{\partial\Pi^{\mu\nu}(p,m)}{\partial m}~=~&A^{\prime}(m) g^{\mu\nu}\Lambda^2\cr &+ \{B^{\prime}(m) g^{\mu\nu}p^2 + C^{\prime}(m) p^{\mu}p^{\nu}\} \ln\Lambda \cr &+ \text{finite terms}.\end{align}\tag{2} $$ The LHS of eq. (2) has SDOD $D=1$, so the singular terms$^1$ on the RHS of eq. (2) must vanish. This answers OP's question: The divergent terms do not depend on the fermion mass $m$.

Let us mention for completeness that from a Ward identity, we know that the vacuum polarization should be transversal, so in fact $$ A~=~0, \qquad B + C ~=~0.$$

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 10.1, p. 319.

--

$^1$ Note that already in eq. (1) we may from section 1 assume w.l.o.g. that $B(m)$ and $C(m)$ do not depend on $m$.