Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 2

A bipartite density operator is a separable quantum state when it has a convex combination decompositioninto product states. A state for which no such decomposition exists is an entangled state.

The positive partial transpose criterion states that every separable quantum state satisfiesConsequently, a negative eigenvalue of the partial transpose proves entanglement. Positivity of the partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, equivalently $3\otimes 2$, but it is not sufficient in general higher dimensions.

Each rank-one density operator isTaking their tensor product and averaging over the $4^d$ independent choices of the fourth roots of unity givesThis is the claimed expansion in matrix elements.

The average of $z_j{z}_k^{\ast }{z}_{j^{\prime }}^{\ast }{z}_{k^{\prime }}$ vanishes unless the exponent of every independent fourth root is balanced modulo four. The surviving index patterns are $j=k,j^{\prime }=k^{\prime }$ and $j=j^{\prime },k=k^{\prime }$. Their intersection $j=k=j^{\prime }=k^{\prime }$ has been counted twice. Hence, writingthe phase average isSolving for the projector onto the maximally entangled state gives

At the proposed boundary $p_0=1/(d+1)$, the identity from part (iv) yieldsBoth $\sigma$ and $D/d$ are convex combinations of product states, so ${\rho }_{p_0}$ is a separable quantum state. For $0\le p\le p_0$, the state ${\rho }_p$ is a convex combination of ${\rho }_{p_0}$ and the maximally mixed product state $I/d^2$, and is therefore separable.

For the converse, the partial transpose of the maximally entangled projector is $F/d$, where $F$ is the swap operator. ThereforeOn the antisymmetric subspace, $F$ has eigenvalue $-1$, so the corresponding eigenvalue of ${\rho }_p^{T_B}$ iswhich is negative exactly when $p>1/(d+1)$. The positive partial transpose criterion then proves that ${\rho }_p$ is entangled. Thus