Quantum binary hypothesis testing

Quantum binary hypothesis testing chooses a two-outcome measurement to distinguish two candidate density operators, trading the probabilities of the two kinds of error.

For $k$ equiprobable quantum code states decoded by a POVM $\{E_m\}$, testing whether the classical message label agrees with the decoded output gives a joint binary POVM with $F_0={\sum }_m|m⟩⟨m|\otimes E_m$. Its acceptance probability is the decoding success $1-ϵ$ on the joint classical-quantum state and $1/k$ on the product of its marginals. Applying quantum relative entropy monotonicity to this test gives the one-shot classical-quantum coding converse.

Quantum channel discrimination feeds a common input into one of several candidate quantum channels and measures the output. An entangled quantum ancilla permits inputs ${\rho }_{HR}$ and compares $(T_j\otimes {\mathrm{id}}_R)({\rho }_{HR})$; optimal binary discrimination is governed by the diamond norm of the weighted channel difference.

The four unitary operators $I,Z,X,iY$ can be identified with one use when an input qubit is entangled with a reference ancilla qubit. Applying the unknown operation to one half of $|{\varphi }_{+}⟩$ produces the four orthogonal Bell states, which a Bell-basis measurement distinguishes exactly. For operators $V{U}_j{V}^{†}$ in any known orthonormal basis, prepare $(V\otimes I)|{\varphi }_{+}⟩$ and apply $V^{†}$ after the unknown call. This reduces to the same quantum channel discrimination protocol. Direct expansion proves the four output identities; equivalently, their overlaps are $\mathrm{Tr}(U_j^{†}{U}_k)/2={\delta }_{jk}$.

For the three qutrit unitary operators $U_j=I-2|j⟩⟨j|$, prepare $|s⟩=(|0⟩+|1⟩+|2⟩)/\sqrt{3}$, query once, apply the known inversion about the mean $D=2|s⟩⟨s|-I$, and use a computational-basis measurement to guess $j$. Since $D^2=I$, the processing is a unitary transformation. Direct multiplication gives $D{U}_j|s⟩=(5|j⟩-{\sum }_{k\ne j}|k⟩)/(3\sqrt{3})$, establishing success probability $25/27$ for each label. This calculation establishes the achieved probability, without requiring an optimality claim.

For the transposition map $\Theta :M_d\to M_d$, $\Theta (X)=X^T$, the induced trace norm is one, while the diamond norm isThe first equality follows because transposition preserves every singular value and hence the trace norm.

For the lower bound, apply $\Theta \otimes {\mathrm{id}}_d$ to the normalized maximally entangled state projector $\rho =|{\Phi }_d⟩⟨{\Phi }_d|$. Its partial transpose is $F/d$, with $F$ the swap operator. All $d^2$ singular values of $F$ are one, so $\Vert F/d{\Vert }_1=d$ while $\Vert \rho {\Vert }_1=1$.

For the upper bound, partial transpose preserves the Frobenius norm $\Vert \cdot {\Vert }_2$. On matrices of size $d^2$,A $d$-dimensional ancilla suffices in the definition of the diamond norm, proving equality. ^[ref]

For $d>1$, transposition is not completely positive: $F/d$ has a negative eigenvalue on the antisymmetric subspace. Thus this is a mathematical example of norm amplification by an entangled ancilla; transposition itself is not a physically implementable quantum channel.

For hypotheses ${\rho }_0,{\rho }_1$ with priors $p,1-p$, the optimal success probability isAn optimal measurement projects onto the positive and negative spectral subspaces of the weighted difference.

This is the minimum-error optimum in binary quantum state discrimination.

For equiprobable qubit states with Bloch vectors $r,s$, their density-matrix difference is $(r-s)\cdot \sigma /2$, with eigenvalues $\pm |r-s|/2$. An effect $0\le E\le I$ declaring the first state yields success $1/2+\mathrm{Tr}[E({\rho }_r-{\rho }_s)]/2$. Selecting its positive-eigenvalue eigenspace maximizes this trace and proves the displayed result. Measure along $r-s$ and attach the corresponding state labels; if the vectors coincide, no measurement improves on guessing.