3 votes

Are loops counted twice in Feynman diagrams?

Well, the self-loop propagator $D(z-z)$ in OP's Feynman diagram occupies 2 of the 4 legs of the 4-vertex. It is also responsible for a symmetry factor $S=2$ that the diagram should be divided with.
Qmechanic's user avatar
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3 votes

Are loops counted twice in Feynman diagrams?

Propagators correspond to line segments; since there are three line segments, there are three propagators. Two 'offshoots' of the vertex are covered by one propagator (the loop), but you'll notice ...
John Dumancic's user avatar
3 votes
Accepted

Difference between renormalizable and super-renormalizable theories

It should stressed be that Peskin & Schroeder are here using the old Dyson definitions of renormalizability. For a more general derivation of eq. (10.13), see e.g. this Phys.SE post, which also ...
Qmechanic's user avatar
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2 votes

Some integrals in QED Renormalisation

Your suspicion is justified. You are indeed missing a crucial approximation not mentioned in your post, turning a simple few-line calculation into an exercise in self-torture. The photon mass $m_\...
Hyperon's user avatar
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2 votes
Accepted

How do vacuum bubbles "dress" terms in the $S$-matrix numerator?

Your second equation is not the full contribution resulting from the Wick decomposition of the numerator up to order $\lambda$ but only the (physically relevant) tree contribution of order $\lambda$. ...
Hyperon's user avatar
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1 vote

How do vacuum bubbles "dress" terms in the $S$-matrix numerator?

The factorization between the numerator and denominator is perhaps best appreciated by remembering that an $n$-point function in the Heisenberg picture is $$ \langle \phi^{k_1}\ldots \phi^{k_n}\...
Qmechanic's user avatar
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1 vote

How do vacuum bubbles "dress" terms in the $S$-matrix numerator?

This is a subtle point, illustrated far better than I can in some books linked below. First, I'll try to sketch the cancellation of the vacuum diagrams - Consider the corrections up to second order: (...
catalogue_number's user avatar
1 vote

Some integrals in QED Renormalisation

Your complaint is not entirely justified, the first integrand is just a rational function, for which solution methods can easily be found: integration-rational-functions Although tedious, it can be ...
Jos Bergervoet's user avatar
1 vote
Accepted

How can I count diagrams (in this scalar QED example) at a particular order without drawing all the Feynman diagrams?

I believe that the approach you outline in your question is more-or-less all that Schwartz is looking for. In particular, with 8 vertices of the type $A_\mu \phi^* \partial_\mu \phi$ and $4$ external ...
QCD_IS_GOOD's user avatar
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1 vote
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Divergences in tree-level diagrams?

More generally in perturbation theory, a connected $n$-point tree-diagram $$(2\pi)^d\delta^d(\sum_{j=1}^np_j)~{\cal M}(p_1,\ldots,p_n)$$ in momentum space $$(p_1,\ldots,p_n)~\in~(\mathbb{R}^d)^n$$ ...
Qmechanic's user avatar
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1 vote
Accepted

Why do correlation functions involving composite fields require special analysis?

As already pointed out by @Prahar, the problem of the definition of a composite operator arises already in the free theory. Considering the Lagrangian of a free real scalar field, $$\mathcal{L}[\...
Hyperon's user avatar
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1 vote
Accepted

Understanding $W^{(n)}$, $\Gamma^{(n)}$, and $\Sigma$ in Feynman diagrams

The connected $n$-point function $$\langle \phi^{k_1}\ldots \phi^{k_n}\rangle^c_{J=0}~=~\left(\frac{\hbar}{i}\right)^{n-1} W_{c,n}^{k_1\ldots k_n}$$ is the sum of connected Feynman diagrams with $n$ ...
Qmechanic's user avatar
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1 vote

Difference between renormalizable and super-renormalizable theories

Qmechanic's answer is very good and tackles your main question well. I want to however add an important detail to your first question. Namely, we generally do not just consider $[\lambda]$ when ...
mika's user avatar
  • 46
1 vote
Accepted

Feynman rule: proton-proton-photon versus electron-electron-photon QED vertex

A fermionic field operator $\psi(x)$ describes particles $f$ (with charge $q_f$) as well as antiparticles $\bar{f}$ (with charge $-q_f$). The Feynman rule derived from $$\mathcal{L}_{\rm int}=-q_f\...
Hyperon's user avatar
  • 6,073
1 vote
Accepted

Why is $\frac{1}{2}\delta_Z(\partial_\mu \phi_r)^2 - \frac{1}{2}\delta_m \phi_r^2$ treated together in Feynman diagrams?

Why are these two terms treated as one You don't have to treat them as a single contribution. It is perfectly acceptable to consider two different vertices with factors of $ip^2\delta_Z$ and $-i\...
SolubleFish's user avatar
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