I am currently studying S-matrices in 2D QFT and came across a statement I do not fully understand.

If one parametrizes 2-particle S-matrix elements with relative rapidity $\theta$, crossing-symmetry relation $$ S(\theta) = S(i\pi-\theta) $$ ensures that if there is a pole in point $\theta_0$ lying on imaginary axis, there is another one in a point $i\pi -\theta_0$. From quantum mechanics I expect that both of these S-matrix poles correspond to some bound states in the theory (as the energy in center-of-mass system is less than the sum of masses of colliding particles). However, it is stated that only one of these poles has the following interpretation, and it is determined by the sign of the residue of S-matrix in this point.

Can someone provide a detailed explanation why a sign of residue is relevant in determining if a pole corresponds to a bound state?

  • $\begingroup$ Where did you come across this statement? $\endgroup$
    – Martin C.
    Mar 5, 2020 at 13:08
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    $\begingroup$ @MartinC. for example, here in Zamolodchikov's lectures on Izing model (see statement near the end of page 3) webhome.weizmann.ac.il/home/fnfal/papers/Ising/lecture10.pdf A simple example for the use of this statement is sine-Gordon model, where the spectrum of bound states is known, but only half of S-matrix poles (with positive residues in energy) corresponds to these bound states (see this note for reference projecteuclid.org/download/pdf_1/euclid.cmp/1103900986 ) $\endgroup$ Mar 5, 2020 at 20:34
  • $\begingroup$ Hm that is a bit cryptic. Are you familiar with Levinson’s theorem? The way it is usually proven is by applying the residue theorem to a certain integral involving the Jost function. I’m afraid that vague guess is the best I can do. $\endgroup$
    – Martin C.
    Mar 6, 2020 at 7:18

1 Answer 1


The situation becomes a bit clearer if we add particle type to the scattering matrix. For particle types $\alpha,\beta,\gamma\cdots$, the scattering between particles of type $\alpha$ and $\beta$ would then be described by $S_{\alpha\beta}(\theta)$. For each bound state, there is a pole in $S_{\alpha\beta}(\theta)$, let's say a $\theta = i\theta_{\alpha\beta}$ corresponding to this bound state which has positive imaginary residue. So far so good.

Denoting anti-particles by $\bar\alpha,\bar\beta, \cdots$, the crossing relation becomes $$ S_{\bar \alpha\beta}(\theta) = S_{\alpha\beta}(i\pi - \theta). $$ If $\bar\alpha\beta$ form a bound state, then $S_{\bar\alpha\beta}$ has a pole at $\theta = i\theta_{\bar\alpha\beta}$ with positive residue. It follows that $S_{\alpha\beta}(\theta)$ has an additional pole located at $\theta = i\pi - i\theta_{\bar\alpha\beta}$. This additional pole has negative residue because the sign of $\theta$ is changed.

So, as Zamolodchikov writes in the lecture you linked:

In fact, it is possible to show that the poles with positive residues in the variable $−i\theta$ correspond to the s-channel bound states, while the poles with negative residues inthat variable correspond to the u-channel images of these bound states.

In other words, the poles with negative residue do correspond to (u-channel scattering containing) bound states, but we already have accounted for them through the s-channel poles. These additional poles do not add new bound states to the spectrum.

At least this is how I understand the statement of yours.

Let me know if anything is unclear or wrong!

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    $\begingroup$ Thank you very much for the answer! Your argument helped me to understand why not all poles correspond to true bound states. But it is still unclear for me, why (as you supposed in the first paragraph) true bound states are distinguished be positivity of residue. Does the residue sign have some explicit physical meaning? Say, in 1D quantum mechanics scattering-on-the-potential problem sign of residue depends on the parity of bound state wave function (as written, for example, in Landau-Lifshitz). Can one possibly think of the similar argument in this situation? $\endgroup$ Jan 7, 2021 at 16:39
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    $\begingroup$ Thanks for the quick response! That is a really good question and I fear I have not found a clear answer yet... In many sources it is suggested this is due to unitarity of the $S$-matrix but not fully explained. I'll let you know when I found a straight answer. I think it is related to the fact that the residue is essentially given by the propagator of the bound state which has certain positivity properties (see physics.stackexchange.com/a/294233/276737). $\endgroup$
    – Cream
    Jan 8, 2021 at 16:16

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