Sign of residues of S-matrix elements for bound states

Question

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?

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!