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I have a simple question about 1-particle-irreducible (1PI) diagram, I know I misunderstood something trivial but I just can not figure it out.

Following Introduction to quantum field theory by Peskin and Schroeder, the (connected) dressed propagator (see (7.43) on P.228) of a scalar field can be obtained by considering the contribution of two-point 1PI, $M^2$, namely, $$\frac{i}{p^2-m_0^2}+\frac{i}{p^2-m_0^2}(-iM^2)\frac{i}{p^2-m_0^2}+\cdots=\frac{i}{p^2-m_0^2-M^2}.$$

Now, by making use of the effective potential $\Gamma$, one may systematically introduce the $n$-point 1PI diagram. For instance, the three-point connected diagram can be built in terms of a three-point 1PC and three 2-point connected propagators. To be specific, see (11.95) on P.382, $$\frac{\delta^3E[J]}{\delta J_x\delta J_y\delta J_z}=i\int d^4u d^4v d^4w D_{xu}D_{yv}D_{zw}\frac{\delta^3\Gamma}{\delta\phi_u^{cl}\delta\phi_v^{cl}\delta\phi_w^{cl}}.$$

For a two-point diagram, it seems that a similar relation can be established. In fact, it seems that Eq.(11.93), $$\frac{\delta}{\delta J(z)}=i\int d^4w D(z,w)\frac{\delta}{\delta \phi_{cl}(w)},$$ can be interpreted that each additional derivative $\frac{\delta}{\delta J(z)}$ pulls out a connected propagator $D(z,w)$ from the diagram. The corresponding two-point 1PC is $$\frac{\delta^2\Gamma}{\delta\phi_{cl}(x)\phi_{cl}(y)}=D^{-1}(x,y),$$ its Fourier transform gives $$\tilde{D}^{-1}(p)=-i(p^2-m^2-M^2(p^2)).$$

My confusion is why the above $\tilde{D}^{-1}(p)$ is not simply the $-iM^2$ discussed previously, what does the extra $(p^2-m^2)$ stand for?

Qmechanic
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gamebm
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1 Answers1

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The effective 1PI action $\Gamma$ generates the sum of all proper diagrams, which are:

  1. Connected
  2. Truncated
  3. Irreducible on the internal lines. This means that if we cut any one of the internal lines of the diagram, it remains connected.

The truncated Green's function is obtained by multiplying the Green's function with the exact inverse propagators corresponding to the external lines: $$G^{(n)}_{\text {trunc}}(p_1,\dots ,p_n)=\Delta'(p_1)^{-1}\Delta'(p_2)^{-1}\cdots \Delta'(p_n)^{-1} G^{(n)}(p_1,\dots ,p_n)\qquad (\sum _i p_i =0).$$ Ref. [1] gives the above definition for $n>2$ (cfr. eq. (6-70)). In fact, for $n=2$, sticking to the above recipe gives: $$G^{(2)}_{\text {trunc}} (p)=\Delta'(p)^{-1} \Delta'(p)\Delta'(p)^{-1}=\Delta'(p)^{-1}=\Gamma_2(p),$$ so that $G^{(2)}_{\text {trunc}}$ would be just the proper function. I believe that the theorem is that "the $\Gamma$ functional generates proper function for $n>2$".

The connection of truncated functions with $S$-matrix elements is found by nothing that, near the pole of the propagator $p^2\approx m^2$: $$\Delta'(p^2)^{-1}\approx iZ^{-1}(p^2-m^2).$$ A glance to the LSZ formulas shows that the connection of truncated Green functions with connected $S$-matrix elements is: $$\langle p_1 ,p_2,\dots ,p_n\,\text {out}\vert q_1,q_2,\dots,q_m \text{in}\rangle = (2\pi)^4\delta^4(\sum _i p_i -\sum _j q_j)\times \\ \times Z^{\frac{n+m}{2}}G^{(n+m)}_{\text {trunc}}(-p_1,-p_2,\dots,-p_n,q_1,q_2,\dots ,q_m)$$ where all the momenta are on shell (cfr. Ref. [1], after eq. (6-70)).

[1] Itzykson&Zuber, "Quantum Field Theory", 6-2-2.

pppqqq
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  • Note that a term like $\frac{1}{p^2-m^2}M^2 \frac{1}{p^2-m^2}M^2 \frac{1}{p^2-m^2}$ is not 1PI, since you can cut the diagram beetween the two insertions $M^2$ and obtain two non-trivial diagrams. – pppqqq Jan 28 '17 at 12:19
  • I still have one doubt, considering the case of two point Green function. Since the prpoer two point diagram $p^2-m_0^2-M^2=(p^2-m^2)/Z$, substituting into the LSZ formula to replace the truncated S matrix, the last factor of the right hand side of Eq.(7.42), the $Z$ cancels the $(\sqrt{Z})^2$, so the resultant expression is almost the same as the dressed two point Green function, Eq.(7.9), except the factor $Z$ is missing. What have I missed? Many thanks. – gamebm Jan 28 '17 at 13:35
  • Hi, while trying to answer your second question, I substantially altered the body of the answer (v1), so if you don't feel satisfied with it, feel free to un-accept it. I think that my previous answer was essentially correct but a with some subtleties, see the "EDIT". – pppqqq Jan 28 '17 at 15:19
  • OKAY, sorry for the multiple edits. I checked on Itzykson&Zuber and I realized that they define "truncated", and therefore "proper" functions only for $n>2$ points. I suppose therefore that the correct proposition is that $\Gamma$ generates the proper $n$-point functions for $n>2$, while for $n=2$ gives the inverse exact propagator. – pppqqq Jan 28 '17 at 16:01
  • Your modified answer sorts thing up for me, many thanks. Now I understand the question I asked in the comments is a stupid one, I confused the Green function with S-matrix element, though they sometimes are both represented by Feynman diagram, they are different quantities. The original question, in your modified answer, is more in accordance with the textbook definition and all, as well as consistent with your previous answer. However, I felt that the original answer you posted, is more to the essence of my doubt (though naive). – gamebm Jan 29 '17 at 02:44
  • To me, there is indeed a difference between the intuitive graphical interpretations of $\delta^2\Gamma/\delta\phi\delta\phi$ and $\delta^3\Gamma/\delta\phi\delta\phi\delta\phi$. Only the former contains the extra contribution of a bare propagator, which gives rise to $p^2-m_0^2$ as you pointed out. Again, I really appreciate your help. – gamebm Jan 29 '17 at 02:48
  • Let us continue this discussion in chat. – pppqqq Jan 29 '17 at 09:52
  • @pppqqq may I ask how do you derive the expression for the truncated Green's function? –  Aug 26 '18 at 17:38
  • @pppqqq I've asked this question here so if you want to answer it feel free (nobody gave an answer yet) https://physics.stackexchange.com/questions/424952/truncated-n-point-functions –  Aug 26 '18 at 17:39
  • @TaylorTiger that's just the definition of truncated Green's function (graphically, dividing by the exact propagators amounts to amputate the external legs) – pppqqq Aug 27 '18 at 05:50