3

The context: I'm calculating the cross section for a scalar particle to decay into a fermion-antifermion pair in Yukawa theory, at tree level.

In doing this, when calculating the amplitude from Feynman diagrams I get a term of the kind: $$\bar{u}_s(p)v_{s'}(p')$$ Where $s$ and $s'$ are the spins of the outgoing particles.

At first it seemed to me reasonable to sum over the spins and take the square modulus (I'll call this procedure A) but from what I saw on some solutions of similar exercises the correct procedure is to take the square modulus and then sum (procedure B).

Namely the choice is between

$|\sum_{s, s'} \bar{u}_s(p)v_{s'}(p')|^2 $ $\hspace{4cm}$ (Procedure A)

and

$\sum_{s, s'} |\bar{u}_s(p)v_{s'}(p')|^2 $ $\hspace{4cm}$ (Procedure B)

Physically this corresponds to summing probabilities instead of summing amplitudes. So my questions are:

1) What is the reason behind this? Is it that since we are summing over spins as a consequence of our ignorance on the outcome we have to consider a statistically mixed state instead of a pure state as the final one?

If this is the right answer I would also be glad if someone could expand on the matter to give me more insight and possibly make the choice of the procedure B more compelling.

2) Would I obtain the same result by summing the amplitudes and then squaring? I mean, would eventual "interference" terms cancel out by themselves for independent mathematical reasons or must the procedure be forced upon this calculation to obtain the correct result?

Qmechanic
  • 201,751
DR10
  • 487

1 Answers1

1
  1. What do we square before summing?

because, a particle with different spin configurations defines different (quantum mechanics) states. So you add the probability to be in one state (let's say spin up) and the probability to be in another (spin down). As soon as different states are involved you do have to sum the probability and not the amplitude. You sum amplitude only when you consider the very same state occurring via different modalities.

  1. Would I obtain the same result by summing the amplitudes and then squaring?

of course not, and it that case it would be wrong.

You were wondering why we do so with spin, but you could as well have the same question for momentum! As you know, in order to compute the cross-section, we square the amplitude and then sum (integrate) over the momentum configuration. We do so because different momenta define different quantum mechanics states even if you don't have the detector resolution to distinguish those states. I'm sure that you would find weird to integrate the amplitude over the momentum and then square the final result. That would be the same mistake as what you proposed for spin in the procedure A.

Paganini
  • 1,823