# LSZ formula and connected Green functions

My question is relatively simple. In the LSZ formalism, it is said that S-matrix elements correspond to on-shell limits of Green's functions. On the other hand, what people usually do is that they consider S-matrix elements to be on-shell limits of connected Green's functions, by which I mean that the diagrammatic expansion of the S-matrix only features diagrams with one connected components. Is there any way to bridge this apparent inconsistency?

I'm aware that one could make an argument about underlying events, and "double counting interactions". Personally, I don't think these arguments are really relevant, at least not at the S-matrix level.

Peskin and Schroeder say (section 7.2, page 227):

"We will consider explicitly the fully connected Feynman diagrams contributing to the correlator. By a similar analysis, it is easy to confirm that disconnected diagrams should be disregarded because they do not have the singularity structure, with the product of four poles, indicated on the right hand side of (7.42)."

However I can find plenty of examples of disconnected diagrams with non-zero on-shell limits. This statement looks so obviously wrong to me, that I fear I'm missing something.

• I don't really understand what your problem here is - P&S aren't saying the disconnected diagrams are zero, they're saying they don't have the right pole structure to be part of the connected S matrix, which is presumably what's on the r.h.s. of (7.42). The only inconsistency is that people aren't always careful to specify explicitly when they're talking only about the connected part of the S-matrix. – ACuriousMind May 23 at 10:21
• I assumed that their mention of the pole structure of disconnected diagrams was associated with the fact that in LSZ one takes the on-shell limit of these diagrams. So if you have a 4 point diagram in which the externals are connected by a simple propagator 2 by 2 (this is a disconnected diagram), this certainly is zero in the on-shell limit, because it does not have the pole structure on the r.h.s. of (7.42). Otherwise, if not for the on-shell limit, I don't see why the mention of the pole structure. – Tanatofobico May 23 at 10:41
• But even if I were to concede that they are just choosing diagrams because they have some particular pole structure, I would still argue that there are disconnected diagrams that have the pole structure on the r.h.s. of (7.42) – Tanatofobico May 23 at 10:42

The LSZ formula for a scalar field $$\phi$$ with $$n$$ out-states and $$r$$ in-states is $$\langle p_1,\dots,p_n\vert S \vert q_1,\dots q_r = \left(\mathrm{i}Z^{-1/2}\right)^{n+r}\prod_i (-p_i^2 + m^2)\prod_j (-q_j^2 + m^2)\left\langle \prod_k \phi(p_k)\prod_l \phi(q_l)\right\rangle + \text{disconnected pieces}$$ where the expectation value $$\langle \dots \rangle$$ is the time-ordered vacuum expectation value and the "disconnected pieces" part represents the disconnected part of the S-matrix where the amplitude factors into the product of two smaller amplitudes.

The first term is often called the "connected part of the S-matrix" and the usual expansion into Feynman diagrams is an expansion of $$\left\langle \prod_k \phi(p_k)\prod_l \phi(q_l)\right\rangle$$. Note that "(dis)connected" up to this point does not inherently refer to diagrams, we're just calling the pieces of the S-matrix that come from products of S-matrix elements with fewer in/out states "disconnected" and the other term "connected". The diagrammatic expansion is explicitly an expansion of the correlation function (or "Green's function") in the connected part.

In general, the correlation function can be expanded as a sum of terms with different pole structures in $$p^2_i$$ and $$q^2_j$$, a multi-variable version of a Laurent series. It is easy to see that only terms with at least the poles $$\prod_i \frac{1}{p_i^2 - m^2}\prod_j \frac{1}{q_j^2 - m^2}$$ contribute to the connected part of the S-matrix, because anything with fewer poles gets just zeroed by one of the $$(p_i^2 - m^2)$$ or $$(q_j^2 - m^2)$$ that are in front of it in the LSZ formula. It turns out that no terms with more poles than this can occur, so only the term with the $$\prod_i \frac{1}{p_i^2 - m^2}\prod_j \frac{1}{q_j^2 - m^2}$$ pole structure contributes to the connected part of the S-matrix.

If you now do the expansion of the correlation function in terms of Feynman diagrams, you find that the diagrams that have this pole structure are precisely the ones that do not have a part where one connected component is just a line between an in- and an out-state. For processes with $$n+r \leq 5$$, this is equivalent to all fully connected diagrams, i.e. diagrams with just one connected component. For processes with more states, these can be diagrams that are composed of two or more connected components, each of which is a non-trivial diagram in its own right. There is no universal agreement over whether these diagrams count as part of the "disconnected pieces" term in the LSZ formula - if they do, then the formula as written is wrong, if they don't, then the usual claim that "only the fully connected diagrams contribute to the connected S-matrix" is wrong for general $$n,r$$.

But in any case, it doesn't matter: It is still true that in order to compute all total S matrix elements, you only need to compute all fully connected diagrams, since the partially connected diagrams are just the product of two or more smaller fully connected diagrams. So by starting at $$n = 2, r = 2$$ and iteratively incrementing $$n$$ and $$r$$, computing the fully connected diagrams at each step suffices to know all S matrix elements - you just have to remember the contributions from the smaller pieces at $$n+r > 5$$.

• I would like to address a couple points: a) in actual cross-section calculations, everyone will agree that you only have to consider connected components. No one computes amplitudes including disconnected diagrams. So it is not relevant that disconnected diagrams can be decomposed into connected ones. What enters a cross-section that is used at LEP or LHC are connected diagrams only, that is, in practice the S-matrix is assumed to only have connected pieces in the on-shell limit. My understanding is that you would agree this is not consistent with the treatment you just presented. – Tanatofobico May 23 at 11:32
• b) You say that on-shell limits of Green's functions yield only connected diagrams, but I don't think that is the case. A Green function might contain a diagram which itself can be decomposed into separate connected components. Each of these components (unless we are talking about components that are single propagators), will also have that singular structure and be non-zero in the on-shell limit – Tanatofobico May 23 at 11:34
• Thank you for the answer by the way. Hopefully I'm not missing something obvious – Tanatofobico May 23 at 11:35
• As an example: the 8-point Green function in scalar theory contains a diagram which is the product of two vertices each of which connects 4 of the external particles. This thing has exactly eight poles, the exact number we need, even if it is disconnected – Tanatofobico May 23 at 11:37
• @Tanatofobico I see; you are right and I've edited the answer to address that concern. – ACuriousMind May 23 at 12:55