$\rho$ meson decay rate In page 10 of this talk by Boris Kayser (http://www.pha.jhu.edu/groups/particle-theory/seminars/talks/S10/talk.kayser.pdf) it is said that although the final state of the decay $\rho^0 \to \pi^+ \pi^-$ is a coherent superposition of momentum states, the decay rate is the incoherent sum over them. 


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*Can someone explain me why this is the case? 

*and why it is analogous to the neutrino produced in the reaction $P \to D + \nu$ discussed in page 20? 
 A: Here's what I think the author is trying to get across: 
As stated on page 9 of the linked document, the $rule$ in quantum mechanics is that
"the rates to produce different final states that differ from one another in any way ... contribute to the total event rate incoherently."
So, for the example of $\rho$ decay, the total rate is an incoherent sum, i.e., the sum of the $squared$ amplitudes, over each possible decay angle.
Now, because the final state in $\rho$ decay is two pions in a p-wave, you might be tempted to think that you should instead calculate the amplitude
for the $\rho$ to decay into that p-wave. That would be a coherent sum of the decay amplitudes for each possible decay angle.
But you would be wrong, because of the above $rule$.
Similarly, in $P \rightarrow D + \nu$, the total rate is the sum of the $squared$ amplitudes for each possible neutrino type.
You might be tempted to think that, because the final state is a coherent sum of neutrino mass eigenstates, you should instead calculate the amplitude for $P$ to decay into that coherent sum.
But again you would be wrong, because of the $rule$. 
And there's no arguing with the rule. It is just how quantum mechanics works.
In both cases, the fact that the final state is a coherent sum is just a red herring.
