Suppose a particle that is under a quantum oscillator potential and is, initially, in the state $\Psi(x,0)=\frac{1}{\sqrt3}\phi_1(x)+\sqrt{\frac23}\phi_2(x)$, where $\phi_1(x)$ and $\phi_2(x)$ are eigenstates of the operator $\hat{H}$. Three consecutive measurements are performed: first, the energy $E$ is measured; then, the observable $A$ is measured, whose associated operator is $\hat{A}$ and which accomplishes that $[\hat{H},\hat{A}]=0$; finally, $E$ is measured again.

(a) Is the first measured value of $E$ the same as the one measured after?

(b) What would happen if $[\hat{H},\hat{A}]\neq0$?

The first value of $E$ depends on probability: the probability of getting $E_1$ is $\frac13$ and the probability of getting $E_2$ is $\frac23$. Now, as far as I know, after measuring an observable in a quantum system, the state immediately (that is, if the system is not allowed to evolute over time) after the measurement is an eigenstate. That would mean that


So if I measured it again, now the measured system's state would be $\phi_i$ so


because it is an eigenstate. So I think that the same result of $E$ would be obtained, no matter how many times I performed the same measurement. Is this reasoning right?

Now regarding the original question, knowing that $[\hat{H},\hat{A}]=0$ means that those observables are compatible, i.e. they can be measured simultaneously. What I don't know is if knowing that allows me to say the same as before: due to the compatibility of both operators, measuring $A$ between the two measurements of $E$ doesn't interfere, so I will keep on getting the same value of $E$ over and over again. I don't know if the fact that the commutator is $0$ can lead to that reasoning, as this would mean that every pair of commutable operators have the same eigenstates. Is this true or not?

And about question (b), in that case both measurements of $E$ would depend on the probability for sure, because the system was not left in an eigenstate after the measurement of $A$, so there is no way to know if the same value of $E$ is going to come out or not. I'm pretty sure about this one, but your confirmation would be great.



yes, you are correct, once you measure the energy, the state collapses to one of the two, with the associated probability that you have correctly identified.

Now, energy eigenkets are sometimes called ''stationary states", because they do not evolve in time, being eigenstates of the Hamiltonian (time evolution generator).

Secondly yes, in the case of non degeneracy (i.e. all eigenvectors have different associated eigenvalues) commuting operators share the same set of eigenvectors, and measuring A would have no effect at all, leaving the original state as it was.

And lastly, yes, if the commutator is not 0, then you you will get a superposition of states, and there is no way of determining the energy, unless more information is given.


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