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