The PPT Criterion

The PPT criterion — also called the Peres–Horodecki criterion — is the single most useful separability test. It is dead simple to apply (transpose a subsystem, check positivity) and, in the smallest cases, it is exact.

The criterion

Peres (necessity). If a bipartite state $\rho$ is separable, then its partial transpose is positive semidefinite: $\rho^{T_B} \geq 0$ .

Equivalently, the contrapositive is the working tool: if $\rho^{T_B}$ has any negative eigenvalue, then $\rho$ is entangled. A state with $\rho^{T_B}\geq0$ is called PPT; one with a negative partial-transpose eigenvalue is NPT.

The proof of necessity is a one-liner. A separable state is a convex mixture of products, $\rho = \sum_k p_k\,\rho_k^A \otimes \sigma_k^B$ . Partial transposition acts only on $B$ , so $\rho^{T_B} = \sum_k p_k\,\rho_k^A \otimes (\sigma_k^B)^{T}$ . Each transposed density operator $(\sigma_k^B)^{T}$ is again a valid (positive) density operator, so $\rho^{T_B}$ is a convex mixture of positive operators — hence positive. Entanglement is the only thing that can break this.

When is PPT also sufficient?

In general, NPT $\Rightarrow$ entangled is only a one-way implication: there exist entangled states that are nonetheless PPT (the bound-entangled states of the next lesson). The famous exception is small dimensions.

Horodecki (sufficiency in $2\times2$ and $2\times3$ ). For two qubits ( $2\times2$ ) and for a qubit–qutrit ( $2\times3$ ), $\rho$ is separable if and only if $\rho^{T_B}\geq0$ .

So for two qubits the criterion is complete: PPT $\Leftrightarrow$ separable, and NPT $\Leftrightarrow$ entangled. Beyond $2\times3$ this equivalence fails, and PPT becomes a necessary but not sufficient test.

Worked example: Werner state threshold

Recall the Werner state $\rho = p|\Phi^+\rangle\langle\Phi^+| + (1-p)I/4$ . Its partial transpose has eigenvalues $\tfrac{1+p}{4}$ (three-fold) and $\tfrac{1-3p}{4}$ (once). The smallest is

\lambda_{\min} = \frac{1 - 3p}{4},

which is negative exactly when $p > \tfrac13$ . Therefore:

$p \leq \tfrac13$ : PPT, hence (two qubits) separable.
$p > \tfrac13$ : NPT, hence entangled.

This matches the negativity result of the previous lesson, which was nonzero precisely for $p>\tfrac13$ — negativity is just the magnitude of this single negative eigenvalue. At $p = \tfrac12$ we get $\lambda_{\min} = \tfrac{1 - 1.5}{4} = -\tfrac18 < 0$ , so the state is entangled.

Try it

For the Werner state at $p = \tfrac12$ , compute the smallest partial-transpose eigenvalue $\lambda_{\min} = \tfrac{1-3p}{4}$ and decide whether the state is entangled. Return true if it is entangled (NPT), false if it is PPT/separable.

Run your code to see the quantum state.