Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 5

The generalized measurement postulate specifies operators $\{M_i\}$ satisfyingOn state $\rho$, outcome $i$ has probabilityand, when $p_i>0$, conditional stateThe effects $E_i=M_i^{†}{M}_i$ form a positive operator-valued measure.

Introduce a quantum ancilla with basis $|i⟩$ and defineThe completeness relation gives $V^{†}V=I$, so $V$ is an isometry and extends to a unitary operator on a sufficiently large system-plus-ancilla space. Prepare the ancilla in a fixed state, apply that unitary, and perform the projective measurement $\{I\otimes |i⟩⟨i|\}$. Outcome $i$ has probability $\Vert M_i|\psi ⟩{\Vert }^2$ and leaves the system in the normalized state $M_i|\psi ⟩$. By linearity the same holds for mixed states, implementing the generalized measurement.

LetThe three effectsform a POVM, because the largest eigenvalue of the sum of the two rank-one projectors is $1+\cos \theta$. Outcome 1 never occurs on $|\theta ⟩$, while outcome 2 never occurs on $|0⟩$. Thus conclusive outcomes are never wrong; $E_{?}$ records failure. This is unambiguous quantum state discrimination.

Arrange the amplitudes of $|{\varphi }_1⟩$ as the matrixThe second state has $C_2=C_{1}X$, where $X$ swaps the $A_2$ basis states. If Bob receives $A_1$, his two reduced density matrices areand are identical. No measurement on $A_1$ contains any information about the shared state.

If Bob instead receives $A_2$,They differ because $a^2+c^2<b^2+d^2$. Bob can therefore distinguish them with better-than-random success. For equal priors,so the optimal success probability is $\frac{1}{2}(1+D)$. It is generally below one, so a single copy does not permit certain identification.

Discarding the outcome of the complete projective measurement givesFor $m$ projectors, let $\omega =e^{2\pi i/m}$ and $U={\sum }_j{\omega }^j{P}_j$. Then the pinching identity isThe Concavity of Von Neumann entropy and its invariance under unitary operators implyEquality holds exactly when every conjugate in the average is the same, equivalentlyThus equality holds when the input already has no coherence between distinct measurement subspaces.