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?

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

  • 2
    $\begingroup$ What is the measurement you would like to implement? Do you want a projective measurement with outcomes |n1>, |n2>, and something orthogonal to it? Do you just want to be able to distinguish |n1> from |n2>? Should the measurement only work on the state you write, or in a general setting? Do you care about the post-measurement state? You need to be more specific. $\endgroup$ Sep 30, 2023 at 13:00
  • $\begingroup$ My apologies, the phrasing was indeed confusing. I hope it is clearer now. $\endgroup$
    – Abelaer
    Sep 30, 2023 at 15:33
  • $\begingroup$ Is the "concrete example" what you really want? Because otherwise, it is still quite unclear. Maybe you just dump the general part and stick to the "concrete example"? $\endgroup$ Sep 30, 2023 at 15:53

1 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.

  • $\begingroup$ Ok, yes that is a very good and simple counterexample. Thanks! $\endgroup$
    – Abelaer
    Sep 30, 2023 at 16:46
  • $\begingroup$ @Abelaer My feeling is the nature of the example rather points to a weakness in your question: E.g., if I just change the phases of the two states, the situation might change drastically, so it is highly sensitive to the decomposition you choose. Also, it is not a property of the state alone, or of n1 and n2 alone ... $\endgroup$ Sep 30, 2023 at 16:48
  • 1
    $\begingroup$ I am interested in doing this for some specific orthonormal states, and was hoping that there was a guarantee that such joint measurements always existed. I now see that they clearly don’t, so that I will have to come up with a specific way to exploit the structure of n1 and n2. There’s still hope, as the superposition I work with is already entangled, but it’s going to take some work then. Thanks for your help! $\endgroup$
    – Abelaer
    Oct 1, 2023 at 7:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.