Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 2

A bipartite density operator is a separable quantum state whenfor probabilities $p_j$ and local states. If no such convex decomposition exists, it is an entangled state.

Solved by gpt-5.6-sol high.

The positive partial transpose criterion says separability implies ${\rho }_{AB}^{T_B}\ge 0$. A nonpositive partial transpose therefore proves entanglement. Positive partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, but not in general higher dimensions.

Solved by gpt-5.6-sol high.

After permuting the computational basis, $\rho$ is the direct sum ofThe stated diagonal and trace conditions already give Hermiticity and trace one. Each $2\times 2$ block is positive semidefinite exactly when its determinant is nonnegative. Thus $\rho$ is a density matrix precisely when

Solved by gpt-5.6-sol high.

Partial transposition interchanges the positions occupied by $x$ and $y$. Positivity of ${\rho }^{T_B}$ therefore requiresBecause the system is $2\otimes 2$, the positive partial transpose criterion is necessary and sufficient. Combining these inequalities with validity of the original state gives

Solved by gpt-5.6-sol high.

For a bipartite input,Each second factor is positive and has trace equal to the probability of measurement outcome $k$. Dividing nonzero factors by their traces therefore writes the output as a convex combination of product states. Hence every measure-and-prepare channel is an entanglement-breaking channel.

Solved by gpt-5.6-sol high.

If $T$ is entanglement breaking, applying it to one half of $|\varphi ⟩⟨\varphi |$ immediately shows that its Choi matrix is separable.

Conversely supposeis separable. The Choi reconstruction formula for the normalized convention isThe trace-preserving condition ${\mathrm{Tr}}_1{C}_T=I/d$ implies ${\sum }_k{M}_k=I$, so $\{M_k\}$ is a POVM. Part (i) now proves that $T$ is entanglement breaking. Thus

Solved by gpt-5.6-sol high.