Infrared divergence in electron self-energy

Question

On Peskin and Schroeder's QFT book, page 319, the book discussed various situations of QED divergence.

On the first paragraph of p.319, the book considered Taylor series of electron self-energy diagram:

where each coefficient is independent of $p$:

The book further said these coefficients are infrared divergent.

I am quite confused for above statements. We know that electron self-energy diagram value is: $$\int d^4 x\langle\Omega|T \psi(x) \bar{\psi}(0)| \Omega\rangle e^{i p \cdot x} =\frac{i}{ \not p-m_0-\Sigma(\not p)} \tag{7.23}$$ where $\Sigma(\not p)$ is value of 1-Particle-Irreducible self-energy diagrams.

1. I am troubled for why we can do this analytic expansion, and why the coefficient is independent of $p$. I find similar analysis in here, here and here. So this have been almost solved.

2. I am also troubled why these coefficients are infrared divergent? Which means when $\not p \rightarrow 0$, their have divergent? And why?

Answer 1

1. The infrared (IR) divergencies comes from internal massless propagators, i.e. photon propagators. An infrared regulator can be introduced e.g. by giving the photon a mass.

2. Now let's turn to ultraviolet (UV) divergencies. It's difficult to give an ironclad/watertight argument for analyticity of correlator functions. In this answer we will just try to amend the explanation given in Ref. 1.

3. The electron self-energy $\Sigma(p,m)$ in $d=4$ has superficial degree of divergence (SDOD) $D=1$. Each time we differentiate wrt. the fermion 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.

4. Lorentz covariance dictates that the tensor structure of $\Sigma$ comes from $\not p$.

5. We can therefore Taylor expand around $p=0=m$: $$ \Sigma(p,m)~=~ A_0\Lambda + \{a_0m + a_1 \not p\} \ln\Lambda + \text{finite terms}, \tag{10.6} $$ where $\Lambda$ is a UV momentum cut-off. The Taylor coeffcients are independent of $p$ and $m$ by definition, cf. OP's question.

6. If $m=0$, then the QED Lagrangian has chiral symmetry. Then the 1PI effective action (which contains $\Sigma$ at quadratic order) should also possess chiral symmetry, cf. Ref. 2. The $A_0$ term breaks chiral symmetry, and must hence be absent. Since it does not depend on $m$, it must also be absent for $m\neq 0$. Hence $\Sigma$ is actually only logarithmically divergent.

7. For a similar analysis of the divergent parts of the photon self-energy, see e.g. this Phys.SE post.

References:

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

2. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Section 16.4.