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The generalized optical theorem is given by:

\begin{equation}\label{eq:optical_theorem} M(i\to f) - M^*(f\to i) = i \sum_X \int d\Pi_X (2\pi)^4 \delta^4(p_i-p_X)M(i\to X)M^*(f\to X).\tag{Box 24.1} \end{equation}

In https://arxiv.org/abs/2306.05976 eq. (3.21): $$\begin{equation} -2i\left(M(i\to f) - M^*(f\to i)\right) = \text{Disc}_{p^2} M \end{equation}\tag{3.21}$$ where the discontinuity is across the $s$-channel.

This relation is crucial because it relates the optical theorem with the discontinuity of an amplitude.

How can I prove it?

The Schwartz chapter 24 takes a different approach and considers the discontinuity across the energy axis $p^0$. How are the 2 discontinuities related?

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  • $\begingroup$ A very general way is to write down an expression for Feynman diagrams, then analyze poles of denominator and determine when the discontinuity appears. I assume that this arxiv.org/abs/1610.06090 reference is enough. The cornerstone is the Schwarz reflection principle + Cauchy pole theorem $\endgroup$
    – Artem Alexandrov
    Commented Jan 27 at 12:53
  • $\begingroup$ Hi Andrea. So the main issue is $p^0$ vs. $p^2$? $\endgroup$
    – Qmechanic ♦
    Commented Jan 29 at 8:35
  • $\begingroup$ I am failing to understand how I can relate the discontinuity across a given cut with the optical theorem. Does the optical theorem ensure that I have branch cuts in one-loop amplitudes and I have only poles at tree level? Regarding, the P^2 issue, I thought (P^0)^2 = P^2 in the center of mass frame. $\endgroup$
    – Andrea
    Commented Jan 30 at 10:30

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