Why does the vacuum polarization vanish for massless fermion-loop and on-shell photon?

Question

In https://arxiv.org/abs/1708.01256 on p.8, it says:

> The contribution corresponding to the rightmost graph yields zero in the $\overline{\mathrm{MS}}$ scheme since we consider massless fermions, since at this order in collinear factorisation the
photon is implicitly on shell and so the vacuum polarisation is evaluated at zero virtuality.

The process in question is depicted in the Feynman diagram above. Can somebody explain to me why the vacuum polarisation (also called self-energy of the photon in other literature) should vanish in this case?

The polarisation function in the $\overline{\mathrm{MS}}$ scheme reads
$$\Pi^{\mu\nu}(p,\mu^2)=\frac{e^2}{2\pi^2}(p^\mu p^\nu-g^{\mu\nu}p^2)\int^1_0dxx(1-x)\ln\frac{\mu^2}{m^2-p^2x(1-x)}.$$

If the photon is on-shell and the fermions massless, then $0=p^2=m^2$. But then the logarithm is not well-defined anymore since
$$\ln\frac{\mu^2}{p^2x(x-1)}=\ln\frac{\mu^2}{x(x-1)}-\ln p^2\stackrel{p^2\rightarrow0}{\rightarrow}-\infty.$$

Can somebody show me **by calculation** how $\Pi^{\mu\nu}$ vanishes for an on-shell photon?

Comments

If the photon is real (i.e., on shell), then it may not obtain radiative correction.

Teschnically you must have two diagrams with the oppisite signes  (opposite directions of integration) that cancel each other.

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@VladimirKalitvianski How do the opposite signs arise?

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Wait a minute if $p^2 \to 0$ then $\ln p^2 \to -\infty$, so in your last equation it should be: $-\ln p^2 \to +\infty$ .

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@MathematicalPhysicist You are right. But the question remains. But I think I know the answer. The LSZ theorem says that only amputated diagrams contribute to S-matrix elements.

Answer 1

The answer is quite simple. The polarisation does not vanish. However, the LSZ theorem says that diagrams with external propagators do not contribute to the S-matrix. The diagram in question is such a diagram.