What state does a system collapse to after measuring a degenerate eigenvalue? $\newcommand{\ket}[1]{|#1\rangle}$
Let $\hat A$ be some observable, and $\ket n$ and $\ket m$ two degenerate eigenstates with eigenvalue $a$, such that
$$\hat A \ket n=a\ket n,$$
$$\hat A \ket m=a\ket m,$$
$$\hat A(\sin\theta\ket n+\cos\theta\ket m)=a(\sin\theta\ket n+\cos\theta\ket m).$$
If $\hat A$ is measured with outcome $a$, what state will the system collapse to? Since there is a whole subspace that is degenerate, I suppose we cannot know precisely the final state (we do not have a CSCO). I suppose a density matrix is not suitable either. Is the fact that the final state will have the form $\sin\theta\ket n+\cos\theta\ket m$ the only thing we can say?
Edit: if the system is in a state that contains $\ket n$ and $\ket m$ as $\ket n +\ket m$ can we say that the system will collapse to $(\ket n +\ket m)/\sqrt2$?
 A: In terms of projective measurements, we just project the state.
So consider a measurement operator $\hat{A}$ with eigenvalues $\{a\}$. Let $\hat{P}_a$ be the projector onto the space spanned by the eigenvectors with eigenvalue $a$. In other words, if $|n_a\rangle$ are eigenvectors of $\hat{A}$ with eigenvalue $a$, then
$$ \hat{P}_a = \sum_{n} | n_a\rangle\langle n_a| $$
Now given some state that is in a superposition of degenerate states:
$$ |\psi \rangle = \sum_{n} c_{n} | n_a \rangle $$
After a projective measurement of $\hat{A}$, the system will be in the suitably normalized projected state $\hat{P}_a|\psi\rangle$. So it is in
$$ \hat{P}_a |\psi\rangle = \sum_n |n_a\rangle\langle n_a |\sum_{n'} c_{n'}|n'_a\rangle = \sum_{nn'}c_{n'}|n_a\rangle \underbrace{\langle n_a| n'_a\rangle}_{\delta_{nn'}} $$
$$ \hat{P}_a |\psi\rangle = \sum_n c_n | n_a\rangle $$
So the state does not change. This is expected since the state $|\psi\rangle$ is still simply an eigenvector of $\hat{A}$ with eigenvalue $a$.
