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$\newcommand{\ket}[1]{|#1\rangle}$ 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 $(\ket n +\ket m)/\sqrt2$?

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  • $\begingroup$ If the system were in a state $| m \rangle$, what would it collapse to after the measurement? $\endgroup$
    – nwolijin
    Mar 2, 2021 at 9:31
  • $\begingroup$ @nwolijin I would say it would collapse to a general linear combination of $|n\rangle$ and $|m\rangle$. That is, we can only say it will be a state belonging to that subspace. That is what I am asking. $\endgroup$
    – Asier R.
    Mar 2, 2021 at 9:39
  • $\begingroup$ my initial guess would be that the system stays in the same state. Imagine that we have two observables, say energy and spin and that the basis in the Hilbert space is given by two states: $|\uparrow\rangle$ and $|\downarrow \rangle$ that are degenerate in energy. Now assume that initially the system is in $|\psi \rangle = \alpha_\uparrow |\uparrow\rangle + \alpha_\downarrow |\downarrow\rangle$, s.t. $|\alpha_\uparrow|^2+|\alpha_\downarrow|^2=1$. $\endgroup$
    – nwolijin
    Mar 2, 2021 at 10:10
  • $\begingroup$ Measuring the spin obviously result in the collapse of the wave function either into $|\uparrow\rangle$ or $|\downarrow \rangle$ with probabilities $|\alpha_\uparrow|^2$ and $|\alpha_\downarrow|^2$ correspondingly. Assuming measuring energy results in a collapse of the wave function to an arbitrary eigenstate of energy $|\psi' \rangle = \beta_\uparrow |\uparrow\rangle + \beta_\downarrow |\downarrow\rangle$, results in different probabilities for measuring spin and energy, depending on the order. $\endgroup$
    – nwolijin
    Mar 2, 2021 at 10:10
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    $\begingroup$ Knowing the observed eigenvalue $a$ s not enough to determine the collapsed state. You also need to know the original state $\phi$. The collapsed state is the projection of $\phi$ onto the subspace of eigenstates with eigenvalue $a$. $\endgroup$
    – WillO
    Mar 2, 2021 at 14:28

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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 $a$.

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