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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
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2 Answers 2

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I'm going to give an explanation at the one loop level (which is the order of the diagrams given in the question).

At one loop, the effective action is given by $$ \Gamma[\phi]=S[\phi]+\frac{1}{2l}{\rm Tr}\log S^{(2)}[\phi],$$ where $S[\phi]$ is the classical (microscopic) action, $l$ is an ad hoc parameter introduced to count the loop order ($l$ is set to $1$ in the end), $S^{(n)}$ is the $n$th functional derivative with respect to $\phi$ and the trace is over momenta (and frequency if needed) as well as other indices (for the O(N) model, for example).

The physical value of the field $\bar\phi$ is defined such that $$\Gamma^{(1)}[\bar\phi]=0.$$ At the meanfield level ($O(l^0)$), $\bar\phi_0$ is the minimum of the classical action $S$, i.e. $$ S^{(1)}[\bar \phi_0]=0.$$ At one-loop, $\bar\phi=\bar\phi_0+\frac{1}{l}\bar\phi_1$ is such that $$S^{(1)}[\bar \phi]+\frac{1}{2l}{\rm Tr}\, S^{(3)}[\bar\phi].G_{c}[\bar\phi] =0,\;\;\;\;\;\;(1)$$ where $G_c[\phi]$ is the classical propagator, defined by $S^{(2)}[\phi].G_c[\phi]=1$. The dot corresponds to the matrix product (internal indices, momenta, etc.). The second term in $(1)$ corresponds to the tadpole diagram at one loop. Still to one-loop accuracy, $(1)$ is equivalent to $$ S^{(1)}[\bar \phi_0]+\frac{1}{l}\left(\bar\phi_1.\bar S^{(2)}+\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}\right)=0,\;\;\;\;\;\;(2) $$ where $\bar S^{(2)}\equiv S^{(2)}[\bar\phi_0] $, etc. We thus find $$\bar \phi_1=-\frac{1}{2}\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c. \;\;\;\;\;\;(3)$$

Let's now compute the inverse propagator $\Gamma^{(2)}$. At a meanfield level, we have the meanfield propagator defined above $G_c[\bar\phi_0]=\bar G_c$ which is the inverse of $S^{(2)}[\bar\phi_0]=\bar S^{(2)}$. This is what is usually called the bare propagator $G_0$ in field theory, and is generalized here to broken symmetry phases.

What is the inverse propagator at one-loop ? It is given by $$\Gamma^{(2)}[\bar\phi]=S^{(2)}[\bar\phi]+\frac{1}{2l}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2l}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}, \;\;\;\;\;\;(4)$$ where we have already used the fact that the field can be set to $\bar\phi_0$ in the last two terms at one-loop accuracy. These two terms correspond to the first two diagrams in the OP's question. However, we are not done yet, and to be accurate at one-loop, we need to expand $S^{(2)}[\bar\phi]$ to order $1/l$, which gives $$\Gamma^{(2)}[\bar\phi]=\bar S^{(2)}+\frac{1}{l}\left(\bar S^{(3)}.\bar\phi_1+\frac{1}{2}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}\right). \;\;\;\;\;\;$$ Using equation $(3)$, we find $$\bar S^{(3)}.\bar\phi_1= -\frac{1}{2}\bar S^{(3)}.\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c,$$ which corresponds to the third diagram of the OP. This is how these non-1PI diagrams get generated in the ordered phase, and they correspond to the renormalization of the order parameter (due to the fluctuations) in the classical propagator.

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I had also eventually realized the resolution to my problem lies in a careful treatment of the loop expansion. This response most clearly gives the answer. –  QuantumDot May 7 at 13:08
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@QuantumDot: I had the same problem quite recently, and I figured that it might be useful to write a detailed answer somewhere. In fact, it seems to me that the standard definition of 1PI (or 1PR) is misleading (but ok in the standard cases): it should be "a 1PI self-energy diagram is a diagram that cannot be cut into two self-energy pieces by cutting any internal line". Your third diagram can be cut into two pieces, but these two are not self-energy diagrams, and thus it has to be included (that easier to see by thinking of what the Dyson equation does). –  Adam May 7 at 13:32
    
Indeed, the standard definition is a bit misleading. –  QuantumDot May 8 at 15:57
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(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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