# Self energy, 1PI, and tadpoles

I'm having a hard time reconciling the following discrepancy:

Recall that in passing to the effective action via a Legendre transformation, we interpret the effective action $\Gamma[\phi_c]$ to be the generating functional of 1-particle irreducible Green's functions $\Gamma^{[n]}$. In particular, the 2-point function is the reciprocal of the connected Green's function,

$$\tilde \Gamma^{[2]}(p)=i\big(\tilde G^{[2]}(p)\big)^{-1}=p^2-m^2-\Sigma(p)$$

which is the dressed propagator.

But, the problem is this: in the spontaneously broken $\phi^4$ theory, the scalar meson (quantum fluctuations around the vacuum expectation value) receives self energy corrections from three diagrams:

$-i\Sigma(p^2)=$ + +

Note that the last diagram (the tadpole) is not 1PI, but must be included (see e.g. Peskin & Schroeder p. 361). In the MS-bar renormalization scheme, the tadpole doesn't vanish.

If the tadpole graph is included in $\Sigma$, and hence in $\tilde{G}$ and $\tilde\Gamma$, then $\tilde\Gamma$ cannot be 1PI. If the tadpole is not included, then $\tilde G$ is not the inverse of the dressed propagator (that's strange, too). What's going on?

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You may try to study e.g. people.fas.harvard.edu/~xiyin/section4.pdf from page 2. –  Luboš Motl Sep 10 '12 at 5:28
That reference is very nice -- it nicely connects Wilsonian effective action to the 1PI effective action. I'm still missing details concerting the tadpoles. –  QuantumDot Sep 10 '12 at 22:30
I have not composed an answer, but if you're in a hurry, the tadpoles only contributes because it is a degenerate diagram with zero momentum going through the line. You need to deal with the linear field counterterm, which cancels the tadpoles, and this is annoying formally, so I need to pick a scheme and be consistent to make a real answer, but the 1PI proof goes through fine when all the lines carry nontrivial momentum, so no tadpoles, so this is the only exception, and in a good scheme, there are no exceptions because the tadpoles are cancelled by linear-in-field counterterm fixing the VEV. –  Ron Maimon Sep 12 '12 at 7:18
@RonMaimon This is intriguing! Would you write up an answer with added details? Thanks!! –  QuantumDot Sep 12 '12 at 13:18
@QuantumDot: I'll do it, but I probably will be beaten to it--- it's well known stuff. You can work it out yourself by going over the proof that the effective action is 1PI diagrams (it requires composing diagrams with momentum flow through lines, the zero momentum on a line is an exceptional case because of tadpoles) and working out the tadpole cancellation with the counterterm linear in the field, but to write an answer means picking a consistent subtraction scheme, and this requires a block of free time that I don't have at the moment (soon). –  Ron Maimon Sep 12 '12 at 18:21

(I came across this question while looking through the "unanswered" category. Not sure why a 9-month-old question appeared in the list, but there still seems to be a room for my contribution.)

The quantum effective action $\Gamma[\phi]$ generates 1PI diagrams, and a tree-level analysis on it amounts to a full analysis on the original action $S[\phi]$.

The sum of all 2-point 1PI diagrams is what contributes (in addition to $p^{2}-m^{2}$, which is also present in the original action) to the quadratic part of $\Gamma[\phi]$. If there were no term linear in $\phi$ in $\Gamma[\phi]$, this would be the self energy itself; however, if $\Gamma[\phi]$ contained a linear term, one has to do an additional tree-level calculation to obtain the self energy from $\Gamma[\phi]$.

As for Feynman diagrams in the original post, the first two contribute to $\Gamma[\phi]$, and the third one appears when calculating the self energy from $\Gamma[\phi]$.

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Ok, the answer is fairly independent of renormalization. The paradox is an effect of perturbation theory.

The definition of the effective action $\Gamma(\phi)=\phi\, j(\phi)-W(j(\phi))$ is such that tadpoles vanish when $\phi$ is evaluated at the all-orders minimum $\bar\phi$ (which is the all-orders VEV in the above broken $\phi^4$ example):

$$\frac{\delta \Gamma(\phi)}{\delta \phi}\bigg|_{\phi=\bar\phi}=j(\phi)\big|_{\phi=\bar\phi}=0$$

But in perturbation theory (for calculating scattering amplitudes and other Green's functions), only the tree-level VEV is being used. Evaluated at the tree-level minimum of the potential, the all-orders source doesn't vanish, leaving non-zero tadpoles. Therefore, $\Gamma$ is the generator of 1PI Green's function only up to tadpoles.

(what a waste of my hard-earned 100 pts) :(

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This is not a satisfactory answer--- it is just repeating the comments. It's true that the tadpoles vanish at the minimum, but you can compute the Legendre transform at any point you want in principle. There is a bit of a problem regarding the Legendre transform for $\phi$ constant inbetween the two minima, because these $\phi$'s don't correspond to a J. But you can fix this in finite volume. The main confusion is about the convention in Peskin and Schroder of putting the 1PI diagrams in the propagator, which is not where they belong. I'll answer (but I have a biology job now, less time). –  Ron Maimon Oct 1 '12 at 7:02
@RonMaimon No matter, thanks for thinking about it, and answering it in the comments. I wasn't astute enough to gather from the comments that the effect was a purely perturbative one, unrelated to renormalization. Most other books (incl. Itzykson+Zuber, and Srednicki) put 1PI in the propagator, too. –  QuantumDot Oct 1 '12 at 17:56
I misspoke--- I meant putting tadpoles in the propagator, not putting 1PI in the propagator (the statement I wrote doesn't make sense--- the propagator is the inverse of the 1PI 1-1 diagrams) --- normally you don't put tadpoles in the propagator, they are a vertex. I didn't answer in the comments, this is why I didn't put an answer, because you need to move the 1PI tadpoles into and out of the propagator in a consistent way, you need to figure out the field shift caused by the tadpole diagram as you vary the unphysical renormalized "m^2" at the point $\phi=0$, and I didn't get to do that yet. –  Ron Maimon Oct 1 '12 at 18:49