# Why is this expression infrared divergent

On P.217 of Quantum Field Theory by Peskin and Schroeder, it is stated that the Eq.(7.16) is infrared divergent and therefore a small photon mass $\mu$ is added to the photon propagator. The equation in question reads

$$-i\Sigma_2(p)=(ie)^2\int \frac{d^4k}{(2\pi)^4}\gamma^\mu\frac{i(k\!\!\!\big /+m_0)}{k^2-m_0^2+i\epsilon}\gamma_\mu\frac{-i}{(p-k)^2-\mu^2+i\epsilon}$$

It seems to me that the integral might be infrared divergent at $\vec{k}\rightarrow 0$, if one exchanges the momentum of the virtual eletron $k$ with that of the photon $(p-k)$, the above expression is then rewritten as

$$-i\Sigma_2(p)=(ie)^2\int \frac{d^4k}{(2\pi)^4}\gamma^\mu\frac{i(p\!\!\!\big /-k\!\!\!\big /+m_0)}{(p-k)^2-m_0^2+i\epsilon}\gamma_\mu\frac{-i}{k^2+i\epsilon}$$

First, I tried to carry out the integral in $k_0$. By choosing the contour which goes around the upper half of the complex plane, one picks the poles at $k^0=-|k|+i\epsilon$ and $k^0=k^0_{-}=p^0-\sqrt{p^{02}+|\vec{k}|^2-2\vec{p}\cdot\vec{k}}+i\epsilon$ (the latter corresponds to one of the two poles $k^0_{\pm}=p^0\pm\sqrt{p^{02}+|\vec{k}|^2-2\vec{p}\cdot\vec{k}}\mp i\epsilon$ at $k^2-2p\cdot k=0$). The resulting expression turns out to be the summation of two terms (considering that $p$ is on the mass shell)

$$-i\Sigma_2(p)\\=-ie^2\int \frac{d^3k}{(2\pi)^3}\left[\gamma^\mu\frac{i(p\!\!\!\big /-k\!\!\!\big /+m_0)}{2(p^0+|p|)|k|}\gamma_\mu\frac{-i}{-2|k|}|_{k^0=-|k|}+\gamma^\mu\frac{i(p\!\!\!\big /-k\!\!\!\big /+m_0)}{-2\sqrt{p^{02}+|\vec{k}|^2-2\vec{p}\cdot\vec{k}}}\gamma_\mu\frac{-i}{2p\cdot k}|_{k^0=k^0_-}\right]$$

So that the resulting integral of $\vec{k}$ goes like $\sim k^2dk(1/k^2+\text{no singularity})$, which does not diverge at $|k|\rightarrow 0$. What am I missing?

Also, if one evaluates the integral of $k^0$ by choosing the lower half of the complex plane (which seems also valid to me, since the module of $k^0$ becomes infinity as well and the integral at infinity approaches $2\pi k^{0}/k^{04} \rightarrow 0$), different poles will be picked but the resulting expression seems to be the same.

Edit: user110373's answer provided below explains that the divergence is infrared as it is due to the pole at $k^2=0$. I thought about it but did not understand well. It seems to me on P.199, when it discusses the infrared divergence of vertex function, the divergence disappears when $\mu$ stay finite (its origin connected to $k^2=0$ is not straightforward there). However, in the present case, the divergence is always there (for $k\rightarrow +\infty$) even when $\mu$ is finite.

• Just count powers. :) – user122066 Jul 8 '16 at 4:37
• Thanks for the hint, but I still didn't get it, would you please be more specific? – gamebm Jul 8 '16 at 9:11

If you look at the first part in your expression after integrating out $k^0$, you will find it divergent, i.e.
$$\lim_{L\to\infty}\int_{L^3} \frac{d^3k}{(2\pi)^3}\left[\gamma^\mu\frac{i(p\!\!\!\big /-k\!\!\!\big /+m_0)}{2(p^0+|p|)|k|}\gamma_\mu\frac{-i}{-2|k|}|_{k^0=-|k|}\right]=\infty$$
as one can direct count the power of $|k|$ in the integrand.
This is exactly your infrared divergence. Because the condition $k^2=0$ actually gives you a unbounded hyper surface instead of a point, the behaviour of your integral around $k^2=0$ should really be probed by the above term you wrote, which is obviously divergent.
• Many thanks for the explanation. I still have some doubts. My question is about how can I tell the origin of the divergence is NOT ultraviolet? In particular, if one inserts the (regulating) photo mass term $\mu$ into the propagator, then for the first term one has $2E_k=2\sqrt{\mu^2+|k|^2}$ instead of $2|k|$, however it seems that the divergence for large $|k|$ (you pointed out) persists(?) – gamebm Jul 9 '16 at 5:39
• The key point is that after introducing the fictitious photon mass we no longer have the divergence from $k^2=0$. The divergence caused by $k^2=0$(IR divergence) behaves differently than the divergence caused by the usual on shell condition(UV divergence):for $k^2=0$, the Lehmann-Kallen form no longer has isolated pole and that makes the residue ill-defined. In fact, a short calculation can show that if one does not use the fictitious mass for photon, the usual procedure for 'on shell scheme' does not yield a finite result. – user110373 Jul 9 '16 at 6:38
• Thanks again for the comments. I understood the "Lehmann-Kallen form" part. But for this specific example, following the arguments of the textbook until Eq.(7.29), it seems to me when one assumes $\mu=0$ in Eq.(7.19) or (7.28), one still gets (7.29) which is UV divergent (only depends on $\Lambda$ but not on $\mu$). Does this imply that the infrared divergence in this particular case, if any, does not really affect the discussion? – gamebm Jul 9 '16 at 7:03