0
$\begingroup$

I'm having problems understanding the relevance of time-reversal symmetry for scattering amplitudes when we have resonant states. Although my question is essentially conceptual, it was raised by an exercise in Weinberg's Lectures on Quantum Mechanics, so I hope I can cite it here:

The $\Lambda^0$ is particle of spin 1/2 and mass 1116 GeV/$c^2$. It decays only through the weak nuclear forces, into an isotopic spin-1/2 state of a nucleon and a pion. Find the phases of the amplitudes for decay into states with $l = 0$ and $l = 1$, in terms of the phase shifts for $s$-wave and $p$-wave pion-nucleon scattering with total angular momentum $j = 1/2$ and total isospin $t = 1/2$ at total energy 1116 GeV. This process does not conserve parity, but you can assume time-reversal invariance

So, here is my guess. I will have to consider pion-nucleon scattering in the conditions mentioned, and probably the $\Lambda^0$ will then appear as a resonant state. But it will decay, as mentioned, to the nucleon-pion state again. So my initial and final states are essentially the same in terms of particle content, and therefore time-reversal symmetry when applied to this process will relate states with the same particle content but different momenta, and the corresponding $S$-matrix elements. So I do not know what to make out of this, and in particular how to conclude anything about the decay of the $\Lambda^0$.

However, there is something I interpret as a clue, and that is the first sentence I highlighted: only through the weak interactions. Could it be that, if I considered the weak interactions to be absent, the $\Lambda^0$ would be a final state? Could I apply then something as the distorted Born approximation to find the amplitude for the decay, in terms of the mentioned phase shifts, if I apply time-reversal symmetry? This all seems very confusing because I do not understand how the $\Lambda^0$ is formed if I turn off the weak interactions, but this is until now my only idea, although I cannot make it more precise... So any help would be very much appreciated!

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.