Can you project on an orthogonal basis for a multipartite quantum system using only local operations and classical communication?

Question

Say Alice possesses one qubit, and Bob two, and that the joint state is $|\psi_{A, B_1, B_2}\rangle = \alpha|n_1\rangle + \beta |n_2\rangle$, where $|n_1\rangle$ and $|n_2\rangle$ are orthonormal basis states for the combined Hilbert space. If you have access to all qubits, there obviously is a measurement which projects on the basis that includes $|n_1\rangle$ and $|n_2\rangle$.

However, what if you only allow Alice and Bob to do local measurements on their qubits? Then, I assume that Bob would need to send at least one classical bit to Alice for a joint measurement to be possible. Do they also need local quantum registers? Can we say anything about when such a joint measurement is possible?

Perhaps a concrete example would be: Let's say Alice and Bob share the state $|\psi_{A, B_1, B_2}\rangle = \alpha|n_1\rangle + \beta |n_2\rangle$. Is it possible, using only LOCC, to end up in the joint state $|n_1\rangle$ with probability $|\alpha|^2$, and in $|n_2\rangle$ with probability $|\beta|^2$, so that afterwards both parties know what the shared state is?

I assume that the precise implementation of such a measurement will depend on the states $|n_1\rangle$ and $|n_2\rangle$, but is there a general rule for when such a joint measurement is possible?

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Crossposted to qc.se: https://quantumcomputing.stackexchange.com/questions/34323/can-you-project-on-an-orthogonal-basis-for-a-multipartite-system-using-only-loca

Answer 1

No, this (the concrete example) is impossible.

Take, for instance, $$ |\psi\rangle = \tfrac{1}{\sqrt{2}} |n_1\rangle + \tfrac{1}{\sqrt{2}}|n_2\rangle \ , $$ with $|n_{1,2}\rangle = \tfrac1{\sqrt{2}}(|00\rangle \pm |11\rangle)$.

Then, $|\psi\rangle$ is unentangled, so there is no way to end up with an entangled state $|n_i\rangle$ using LOCC.