# Find a POVM to distinguish the two states

Suppose $$|\psi \rangle = |1\rangle$$ and $$| \phi \rangle = | + \rangle = \frac{1}{\sqrt{2}} (|0 \rangle + |1 \rangle )$$. Write a POVM that allows for imperfect distinguishability between two states.

POVM stands for Positive Operator-Values Measures, it's a set of operators $$\{E_m\}$$ where the probability of measuring result $$m$$ is $$Pr(m)=⟨ψ|E_m|ψ⟩$$ and $$∑_mE_m=I$$. These operators allow for more general forms of measurement, such as distinguishing non-orthogonal states. One can do this if $$⟨ψ|E_i|ψ⟩=0$$ but $$⟨ϕ|Ei|ϕ⟩>0$$ but $$<1$$ for some $$E_i$$ and vice-versa for $$E_j$$. This allows one to imperfectly identify if the system is in state $$| \psi \rangle$$ or $$| \phi \rangle$$ since there some measurements are impossible (probability of 0).

• What is POVM and how is "imperfect distinguishability" defined?
– Korf
Dec 14, 2020 at 23:40

FYI this is probably better suited on the quantum computing stackexchange. But anyway, you can try taking $$E_1 = c_1 |0\rangle \langle 0 |$$, $$E_2 = c_2 |-\rangle \langle - |$$ and $$E_3 = I - E_1 - E_2$$. For some suitably chosen constants $$c_1, c_2 > 0$$ this should give you a POVM which satisfies $$\langle \psi | E_1 |\psi\rangle = 0 \qquad \langle \phi | E_1 |\phi \rangle >0$$ and $$\langle \psi | E_2 |\psi\rangle > 0 \qquad \langle \phi | E_2 |\phi \rangle =0.$$