In Nielsen and Chuang's Quantum Computation and Quantum Information textbook, there is an example (on page 92) of a POVM containing elements:

$E_1=\frac{\sqrt{2}}{1+\sqrt{2}}|1\rangle \langle1|$,

$E_2=\frac{\sqrt{2}}{2(1+\sqrt{2})}(|0\rangle - |1\rangle)(\langle0| - \langle1|)$

$E_3=I - E_1 - E_2$

This POVM makes it possible for Bob to perform a measurement that distinguishes the states $|\psi_1\rangle = |0\rangle$ and $|\psi_2\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$ some of the time, but never makes an error of mis-identification.

Is there a general way to create a POVM that will allow Bob to do the same but for 3 states $|\phi_1\rangle$, $|\phi_2\rangle$ and $|\phi_3\rangle$?


1 Answer 1


Assuming that the discussion is restricted to single-qubit states then the answer to your question is no. The unambiguous discrimination of non-orthogonal states relies on using POVM elements that are proportional to elements of the orthogonal complements of the states in question:

$$ E_1 \propto |\phi_1^{\perp}\rangle\langle\phi_1^{\perp}|, \qquad \langle\phi_1^{\perp}|\phi_1\rangle = 0, $$ and so on. This choice of POVM element ensures that the input state $|\phi_1\rangle$ will never yield the outcome '1', and observing this measurement outcome therefore reveals the information that the input state wasn't $|\phi_1\rangle$. If you know that the input state belongs to the set $\{|\phi_1\rangle,|\phi_2\rangle\}$ this in turn implies that the state was $|\phi_2\rangle$.

Extending this to more than two non-orthogonal single-qubit states would require a POVM element that is proportional to an element of the union of the orthogonal complements of two different states. In other words: $$ E \propto |\phi^{\perp}\rangle\langle\phi^{\perp}|, \qquad \langle\phi^{\perp}|\phi_1\rangle = 0, \quad \langle\phi^{\perp}|\phi_2\rangle = 0, $$ but for qubits this implies that $|\phi_1\rangle = |\phi_2\rangle$ since for every single-qubit state there is exactly one orthogonal single-qubit state.

In general, when considering states of more qubits, a necessary and sufficient condition for being able to discriminate $n$ non-orthogonal states is that they are linearly independent. See arXiv:quant-ph/9807022 for a reference.


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