Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 15C (Quantum Information and Computation)

Let ${\rho }_j=|{\alpha }_j⟩⟨{\alpha }_j|$. For a two-outcome measurement with effect $0\le M\le I$ interpreted as guess zero,The positive-eigenspace projector of ${\rho }_0-{\rho }_1$ maximizes the last trace, giving the Helstrom formulaOn the span of the two states, ${\rho }_0-{\rho }_1$ has eigenvaluesThereforeand the positive/negative eigenspace measurement attains equality.

The right side equals one exactly when $|⟨{\alpha }_0|{\alpha }_1⟩|=0$. Thus two pure states are perfectly distinguishable exactly when they are orthogonal.

For a chosen unit vector $|\psi ⟩$, the two possible outputs have inner productPart (a) says they can be perfectly distinguished exactly when this inner product is zero. Such an input exists exactly when

If $Uv=\lambda v$ with $v\ne 0$, unitarity givesHence $|\lambda |=1$. Every eigenvalue of a unitary matrix therefore lies on the unit circle.

The three eigenvalues areThus $N(U)$ is the filled triangle with verticesin the complex plane. Its upper edge is the horizontal chord joining the last two vertices, and the other two edges join those points to $1$. In particular, the origin lies outside this triangle.

The two eigenvalues are the points at arguments $0$ and $\gamma$. The counterclockwise arc from $1$ to $e^{i\gamma }$ has length $\gamma$, while the complementary clockwise arc has length $2\pi -\gamma$. The shorter of the two contains the spectrum. ThereforeThe two geometric cases are respectively the short counterclockwise arc and the short complementary clockwise arc; at $\gamma =\pi$ both semicircles have length $\pi$.

For points on the unit circle, their convex hull contains the origin exactly when they are not all contained in an open semicircle. Equivalently, the shortest closed arc containing them has length at least $\pi$. Applying this to $V=U_2^{†}{U}_1$ and usinggivesPart (b) now proves the claimed perfect-discrimination criterion.