Peres–Horodecki criterion

Remarks

For part 1, write $\rho = \sum_i p_i \, \rho_i^A \otimes \rho_i^B$. Then $\rho^{T_B} = \sum_i p_i \, \rho_i^A \otimes (\rho_i^B)^T$, and the transpose of a positive semidefinite matrix is positive semidefinite. Part 2 uses a result of Størmer and Woronowicz on positive maps from $2 \times 2$ matrices to $2 \times 2$ or $3 \times 3$ matrices. Each such map is a completely positive map plus a completely positive map composed with the transpose.

Example: for the Bell state $\psi = (|00\rangle + |11\rangle)/\sqrt{2}$, the partial transpose of $|\psi\rangle\langle\psi|$ is $\tfrac12 F$, where $F$ is the swap operator. Its eigenvalues are $\tfrac12, \tfrac12, \tfrac12, -\tfrac12$, so the state is entangled.

The bound $mn \le 6$ is necessary. In dimensions $3 \times 3$ and $2 \times 4$ there are entangled states with positive semidefinite partial transpose. These states have bound entanglement: no entanglement can be distilled from them.

The spectrum of $\rho^{T_B}$ does not depend on the choice of basis of $H_B$, and $\rho^{T_A}$ has the same spectrum, since $\rho^{T_A} = (\rho^{T_B})^T$.

Going deeper

For a student

The criterion is the most used separability test, because it reduces to an eigenvalue computation. It is an instance of the positive-map criterion: $\rho$ is separable if and only if $(\id \otimes \Lambda)(\rho) \ge 0$ for every positive map $\Lambda$. The transpose is positive but not completely positive, which is what makes it detect entanglement. The negativity, built from the negative eigenvalues of $\rho^{T_B}$, is a computable entanglement measure based on this criterion.