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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:

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where each coefficient is independent of $p$:

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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?

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1 Answer 1

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  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.

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