# Degenerate states with quantum operators.

Lets say we have 2 commuting operators, $\Omega$ and $\Lambda$. The first operator has 3 separate eigenvalues, $\omega_1$, $\omega_2$ and $\omega_3$. The second operator has eigenvalues $\lambda_1$ and $\lambda_2=\lambda_3=\lambda$.

These commute and so they share eigenkets. The state of the system is: $$|\psi\rangle=\alpha|\omega_1\lambda_1\rangle+\beta|\omega_2\lambda\rangle+\gamma|\omega_3\lambda\rangle$$

I know what happens if we measure $\Lambda$ then $\Omega$. However what happens if we measure $\Omega$ and then $\lambda$?

I will assume $\Omega$ results in one of the last 2 eigenvalues, lets say $\omega_2$ and so the state collapses to:

$$|\psi\rangle=|\omega_2\lambda\rangle$$ and obviously we measure $\omega_2$. If we now measure $\Lambda$ we will obviously get $\lambda$.

My question is this: Does the state stay the same or does it become a superposition of both eigenkets which have the eigenvalue $\lambda$? If you measure the degenerate operator second does the system stay well defined in the other operator?

If you measure the non-degenerate operator first, then you get either $\lvert \omega_2\lambda\rangle$ or $\lvert \omega_3\lambda\rangle$ and that is for all purposes the new state of the system. Each of these states is already an eigenstate of $\Lambda$, so measuring $\Lambda$ does nothing to them. You can think of a measurement as applying the operator that projects onto the eigenspace belonging to the eigenvalue that you measure, and this projector does nothing to states already in said eigenspace.