I'm confused with the relation between the fully resummed propagator in a given QFT and the corresponding S-matrix element.

According to the LSZ reduction formula ($\phi^4$ theory for definiteness here), the S-matrix element for coming the simple "scattering" process:

$$ p \rightarrow q$$

can be computed as:

$$ \frac{i\sqrt{Z}}{p^{2} - m^{2}}\frac{i\sqrt{Z}}{q^{2} - m^{2}}\langle p|S|q\rangle = \int d^4y\text{ } e^{ip\cdot y}\int d^4x\text{ } e^{-iq\cdot x} \langle\Omega|T \phi(y) \phi(x)|\Omega\rangle$$

But the Fourier transformed of the 2-point correlator is just

$$\int d^4y\text{ } e^{ip\cdot y} \langle\Omega|T \phi(x) \phi(0)|\Omega\rangle = \frac{iZ}{p^2-m^2} + \text{terms regular at $p^{2} = m^{2}$}$$

Following this, for the right pole structur on both sides, the S-matrix element on the left side must be something like

$$c \cdot \delta^{(4)}(p-q) \text{ }(q^2-m^2)$$

which would vanish on-shell. But in my mind, the amplitude should just contribute to the 1 in the S-Matrix and thus just give a Kronecker delta (or just the delta function as in the above term).

In the linked image, I visualized my problem in a diagram. There, the hatched bubbles denote the fully resummed propagator and "Amputated" denotes the S - matrix element up to an factor of $Z$, in the sense of Peskin,Schröder. I'm asking for the value of this S-matrix element or the corresponding amputated diagram.

Amputated diagram

  • $\begingroup$ yeah, but these are just finite terms.. the important thing here is the first pole which gives the right pole structure to the S-matrix $\endgroup$
    – Moe
    Mar 26, 2016 at 18:28
  • $\begingroup$ Then can you explain me why the propagator has this structure ? Because in every computation I've seen it, this is exactly the expression for the 2 point correlator. $\endgroup$
    – Moe
    Mar 26, 2016 at 18:36
  • $\begingroup$ I corrected my post and made it more obvious where my problem is. I'm referring here especially on Prof. Weigands QFT script, maybe you are familiar with it if you are from Heidelberg @ACuriousMind thphys.uni-heidelberg.de/~weigand/QFT2-14/SkriptQFT2.pdf p.69 and following. $\endgroup$
    – Moe
    Apr 7, 2016 at 21:38
  • $\begingroup$ so what are you telling me is that the fourier transformation with respect to both variables will yield me the right pole structure and the Correlator would have to analytical poles at $p^2 = m^2$ and $q^2 = m^2$? Because my problem here is simply that I dont get how I can compute this S-matrix element. Take for example the 4 point correlator. Here you get 4! fully resummed propagators and thus the S-matrix element tree level can be extracted with LSZ to be just $i \lambda * \text{momentum conservation}$. But for the 2 point fct I dont see how 2! propagators come in for the right pole structur $\endgroup$
    – Moe
    Apr 7, 2016 at 22:54
  • $\begingroup$ Oh god, I'm sorry, I see what the issue is now. I'm writing an answer. $\endgroup$
    – ACuriousMind
    Apr 7, 2016 at 23:22

1 Answer 1


There is no problem, you just have to go back to the definition of the terms in the LSZ formula to see that $\langle p \vert S \vert q \rangle\vert_\text{connected}$ is indeed just zero for the 1-particle process $p\to q$, and it is only this connected piece that appears in the LSZ formula you are trying to use to compute this amplitude. (In the notes you linked in a comment [pdf], it is chapter 2.4. where the split into connected and disconnected pieces happens.)

This is because the definition of "disconnected term" in the S-matrix is precisely "term in which nothing happens", i.e. a term that is represented by a Feynman diagram in which at least one particle just propagates without "participating in scattering". These are called "disconnected" because for all diagrams with more than 2 particles, they indeed correspond to disconnected graphs. For "2" particles, it's just a propagator, but this is a "disconnected piece" in the sense of the LSZ formula.

So there is no issue. Your reasoning is correct, the connected S-matrix is zero on-shell for this process - which is physically sensible, because the connected S-matrix entries represent interaction, something happening. A particle just propagating is not a scattering event.

  • $\begingroup$ perfect, thank you. I thought about it that way but i wasn't sure if it really counts as disconnected, because in the proof of LSZ in \hyperlink{thphys.uni-heidelberg.de/~weigand/QFT2-14/SkriptQFT2.pdf}{2.4} , for the 1 particle process you can still go through all the steps which eventually lead to the LSZ formula. But of course you can also just omit it right at the beginning and say "ok there is no scattering, this is disconnected in the LSZ sense". Thanks for your help. $\endgroup$
    – Moe
    Apr 8, 2016 at 1:48

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