Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 51 / 2

Introduce the orthonormal local vectors $|+⟩=(|1⟩+|2⟩)/\sqrt{2}$ and $|-⟩=(|1⟩-|2⟩)/\sqrt{2}$. Reexpressing the second state's factors in this basis gives its Schmidt decompositionThe squared Schmidt coefficients are therefore $1/3,2/3$, whereas those of the first state are $1/2,1/2$.

To see directly why these distinguish the states, a local unitary operation $U_A\otimes U_B$ changes the reduced density matrix toThis is local-unitary invariance of reduced-state spectra: the unitary on the traced subsystem cancels, and conjugation on $A$ leaves its eigenvalues unchanged. Here the reductions areTheir spectra differ. Hence no pair of local unitary operations carries the first state to the second. More generally, normalized bipartite pure states are locally unitarily equivalent exactly when their squared Schmidt coefficients agree, including multiplicities; sufficiency follows by mapping one pair of Schmidt bases to the other.

If the preparation label $i$ is sampled with probability $p_i$ but is not supplied to the experimenter, the quantum state ensemble is represented byFor normalized states, this density operator is Hermitian, positive semidefinite and has trace one. The states in an ensemble need not be orthogonal or distinct. For any measurement in quantum mechanics effect $E$, averaging the Born rule over the preparation label givesThus the density operator summarizes all measurements performed on the prepared system alone. For an ensemble of mixed states ${\rho }_i$, the same reasoning gives $\rho ={\sum }_i{p}_i{\rho }_i$.

Yes: distinct density operators have a projective measurement with different outcome probabilities. Let $\rho ,\sigma$ be the two density operators and let $D=\rho -\sigma$. It is a nonzero Hermitian operator with trace zero. Its nonzero eigenvalues cannot all have the same sign, since their sum is zero. Let $P_{+}$ be the orthogonal projection onto its positive-eigenvalue subspace. Then the two-outcome projective measurement $\{P_{+},I-P_{+}\}$ satisfiesThe outcome therefore carries statistical information about which quantum ensemble was used.

For example, with equal prior probabilities, guess $\rho$ after $+$ and $\sigma$ after the other outcome. The success probability isSince the positive and negative eigenvalue sums of $D$ have equal absolute value, $\mathrm{Tr}(P_{+}D)=\Vert D{\Vert }_1/2$. ConsequentlyThis is the equal-prior measurement in the Holevo–Helstrom theorem, expressed using the trace distance. The conclusion is some information, not necessarily perfect one-shot discrimination: nonorthogonal density operators generally cannot be distinguished with certainty.

No measurement on the supplied system alone distinguishes ensembles with the same density operator. A completely general measurement in quantum mechanics has POVM effects $E_r\ge 0$ with ${\sum }_r{E}_r=I$. The outcome probability for a quantum ensemble isIf both ensembles have the same $\rho$, every outcome probability is identical. This proves the claim for all positive operator-valued measures, and includes projective measurements as a special case. An quantum ancilla in a fixed independent state does not help: the input density operators remain identical after tensoring with that ancilla, and the same argument applies to any joint measurement.

The premise concerns access only to the prepared state. An accessible preparation label or a correlated purification of a density operator is additional information outside that premise and can distinguish global preparations. Without such side information, different ensemble decompositions do not define distinguishable states.