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I am reading through Peskin and Schroeder and I am having some confusion in their derivation of the effective action being the generating functional of 1PI diagrams. In particular, they make a comment on page 381 that says

“… the connected two-point function, that is, the exact propagator.”

On the previous page, however, they express the full two-point function as the sum of both connected and disconnected diagrams. This is where my confusion comes in.

Throughout the book (up to this point), they have used the term “two-point function” and “propagator” interchangeably, and in a very meaningful way. More specifically, in chapter 7 they spend a lot of time discussing the analytic structure of the two-point function and how it has a pole and residue which give the physical mass and field strength renormalization, respectively. They then perform the geometric sum of 1PI diagrams to obtain a separate expression for the two-point correlation function (which they also call the propagator), and they relate these two in order to obtain expressions relating field strength renormalization to the mass shift. The point here being that they seem to define the physical mass by the two-point correlation function, and then use 1PI diagrams to compute it perturbatively through this relation. In chapter 10, when they begin discussion of renormalized perturbation theory, the definition of the mass as being the location of a pole for the two-point correlation function is again used as a renormalization condition.

My confusion is this: is the propagator the full two-point correlation function or just the sum of connected diagrams? In phi 4 Wick’s theorem basically ensures that these are identical because the one-point functions vanished, but in more general theories they need not. Page 380 and 381 seem to distinguish between the two, but then in page 383 they say that the particle masses are zeros of the (Fourier transform of) the inverse of the connected two-point function, not the whole thing (Eqs. 11.89, 11.90, 11.92, and 11.97). I can see how the geometric sum of 1PI diagrams gives the connected correlation function, but not the whole thing, and it seems like the mass is defined by the whole thing. I also don’t see where this definition of the propagator being the connected two-point function comes from.

I apologize if this comes off as a rant more than a question, but I am a bit frustrated as I have been going crazy trying to connect the dots, not entirely sure if I am just missing something or if the language of the textbook is sloppy. This feels like an important distinction, not so much regarding what a “propagator” is, but regarding the relationship between 1PI diagrams and correlation functions (especially when it comes to renormalization conditions). Any and all help on clearing this up would be greatly appreciated

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    $\begingroup$ There is a connected and a disconnected two point function. Let $\langle \cdots \rangle_c$ be a connected correlation function, then $\langle T \phi(x) \phi(y)\rangle_c = \langle T \phi(x) \phi(y) \rangle - \langle \phi(x) \rangle \langle \phi(y) \rangle$. If the vev is zero $\langle \phi \rangle = 0$ they are the same (as you observed). The physical mass and normalization come from the pole/residue of the fourier transform of the connected two point function like P&S say although it's usually equivalent since the field operator is normally defined to have zero vev. $\endgroup$ Commented Jul 8 at 0:49
  • $\begingroup$ The fact that we care about connected diagrams falls out of the LSZ formula. The analytic structure of the fourier transform of the two point function can be seen with the Kahlen-Lehman spectral representation. Those are all exact statements. All the stuff about Feynman diagrams and resumming the geometric series is relating a perturbative loop calculation to the exact statements. $\endgroup$ Commented Jul 8 at 0:51
  • $\begingroup$ There is also an exact relation between the generator of connected correlated functions, and the effective action, which generates 1PI vertices. They are related by a Legendre transform. $\endgroup$ Commented Jul 8 at 0:54
  • $\begingroup$ I left these as comments since it is not clear to me what your question is (like you said what you wrote is a bit more of a rant than a question) but I thought there was a chance separating exact from approximate statements might help you. $\endgroup$ Commented Jul 8 at 1:35

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Apparently P&S are using the word exact/full propagator for connected 2-point function, but sometimes do not mention the qualifier connected explicitly.

For more information, see also e.g. this & this related Phys.SE posts and links therein.

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