I'm self-studying QFT and am trying to see the relation between Cutkosky rules and the optical theorem, which are presented together as consequences of unitarity in almost every elementary presentation I can find. I think I roughly have a clue what's going on but was hoping someone could confirm/refute!

As I understand it: unitarity of the S-matrix implies the generalised optical theorem, which in the case of forward-scattering allows us to express the imaginary part of a scattering amplitude as a sum over all possible intermediate states. By considering perturbation theory, we see that the imaginary part of a loop corresponds to a sum of tree-level amplitudes.

How exactly this relates to Cutkosky's rule is where I get slightly lost. Zee says in justifying the Cutkosky rules:

"Thus finally, the generalized optical theorem tells that the imaginary part of the Feynman amplitude is given by a sum over intermediate states. The particles in the intermediate state are of course physical and on shell. We are to sum over all possible states allowed by quantum numbers and by the kinematics."

Is it therefore the case that the Cutkosky approach of summing over cuts, putting the propagators on-shell is essentially equivalent to the sum over intermediate states (obviously on-shell) which the optical theorem tells us we can do -- thus meaning the the optical theorem implies the existence of something like the cutting rules? If not, what does the optical theorem actually have to do with the cutting rules, beyond that it involves the imaginary part of a scattering amplitude which the rules help us to compute?

  • 1
    $\begingroup$ See also Peskin & Schroeder for the Cutkosky rules. They actually say (236) that using the cutting rules it is possible to prove the optical theorem to all orders of perturbation theory. $\endgroup$ Mar 6, 2021 at 10:17
  • $\begingroup$ Thanks for the reference -- actually, I had seen this and unfortunately it confused me too! Why do we need to prove the optical theorem to all orders in perturbation theory if simple considerations about unitarity prove that the optical theorem is true non-perturbatively? By saying it's true "to all orders in perturbation theory" does this mean it holds at each order, which is obviously a more powerful statement? Thanks so much for your reply! $\endgroup$
    – Facieod
    Mar 6, 2021 at 10:30
  • $\begingroup$ Yes, the generalised optical theorem holds at each order in perturbation theory $\endgroup$ Mar 6, 2021 at 10:32

1 Answer 1


The case is somewhat the opposite. The cutting rules are far more general and can be used to prove the optical theorem at each order in perturbation theory. Furthermore, the cutting rules mean that the generalised optical theorem is a generic statement for all diagrammatic amplitudes, not just S-matrix elements. Additionally, the statement that the imaginary part of loop amplitudes come from intermediate particles going on shell is a consequence of the analytic structure of the Feynman propagator, without having to know anything about the intermediate states present in the theory.

The optical theorem is then kind of a special case of the cutting rules. For example (without having to prove everything to all orders), you can apply the cutting rules to a one-loop diagram for $1\to2\to1$ decay loop, make a change of variables and end up with an integral over your Lorentz-invariant phase space (see e.g. Srednicki, Chapter 25) which yields the mass times decay width - precisely what the optical theorem tells you. You can obtain the cutting rules directly via contour integration, and the method of order reduction by discarding products of propagators is really a statement about their causal (pole) structure (as in Veltman's derivation).

  • $\begingroup$ Thanks so much for this really helpful answer, that has really helped to elucidate what is going on here. The only thing that's bothering me is -- in this case, why are the cutting rules often cited as a consequence of unitarity? I have seen them derived e.g. Veltman making no reference to it. $\endgroup$
    – Facieod
    Mar 6, 2021 at 18:58
  • 1
    $\begingroup$ @Facieod good question. I view it as "generalised unitarity" (also updated the answer) $\endgroup$ Mar 7, 2021 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.