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!


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.