# How to construct a POVM to discriminate between 3 non-orthogonal states?

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 $$|\phi_3\rangle$$?

$$E_1 \propto |\phi_1^{\perp}\rangle\langle\phi_1^{\perp}|, \qquad \langle\phi_1^{\perp}|\phi_1\rangle = 0,$$ 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.