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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).
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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.
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  • $\begingroup$ 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? $\endgroup$ Commented Apr 13 at 10:52
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    $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Commented Apr 13 at 12:23

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