A question about two-point 1-particle-irreducible diagram

Question

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?

Answer 1

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 $n>2$".

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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)).

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