Wave Collapse and Degenerate States - A Quick Question

[BAnders1]

Let's say you have some quantum particle whose eigenvalues for some observable Q are either degenerate or non-degenerate. If you measure the observable and find it to be in a non-degenerate state, then you know that the wave function has collapsed onto this state. Now if you measure the observable and find it to be in a non-degenerate state, does this mean that the particle's wave function has collapsed to a superposition of all states sharing this eigenvalue? My answer before writing this was "I have no idea," but my answer afterwords was "Of course it does." I'd still like to see if my latter answer was wrong, because that would mean something interesting is going on.

 

[Demystifier]

It depends on details of the measurement procedure, but typically yes, it collapses to a superposition.

 

[Entropia]

Your latter answer is correct. When a quantum particle is in a non-degenerate state, it means that its wave function has collapsed onto a single eigenstate of the observable in question. However, if the particle is in a degenerate state, it means that its wave function has collapsed onto a superposition of all states sharing that eigenvalue. This is because in a degenerate state, there are multiple possible eigenstates that the particle could be in, and the wave function represents the probability amplitudes for each of these states. So, when measuring the observable, the particle's wave function collapses onto a superposition of all these possible states.