Negativity

The negativity is the most computationally friendly entanglement measure: no minimization, no convex roof — just the eigenvalues of a partially transposed density matrix. Its price is that it cannot see bound entanglement, but for everything else it is a fast, faithful LOCC monotone.

The partial transpose

Write a bipartite density operator in a product basis,

\rho = \sum_{ij,\,kl} \rho_{ij,kl}\,|i\rangle\langle k|_A \otimes |j\rangle\langle l|_B.

The partial transpose with respect to $B$ transposes only the $B$ indices:

\rho^{T_B} = \sum_{ij,\,kl} \rho_{ij,kl}\,|i\rangle\langle k|_A \otimes |l\rangle\langle j|_B,

i.e. $(\rho^{T_B})_{ij,kl} = \rho_{il,kj}$ . The partial transpose is still Hermitian and still has unit trace, but it need not be a valid density operator: it can have negative eigenvalues. Those negative eigenvalues are the entanglement signature.

Definition of negativity

The negativity is half the amount by which the trace norm of $\rho^{T_B}$ exceeds $1$ :

\mathcal{N}(\rho) = \frac{\|\rho^{T_B}\|_1 - 1}{2} = \sum_{i:\,\mu_i < 0} |\mu_i|,

where $\mu_i$ are the eigenvalues of $\rho^{T_B}$ and $\|\cdot\|_1$ is the trace norm (sum of absolute eigenvalues). Because $\rho^{T_B}$ has unit trace, $\sum_i \mu_i = 1$ , so the trace norm equals $1 + 2\sum_{\mu_i<0}|\mu_i|$ , and the second equality follows. In words: negativity is the total magnitude of the negative eigenvalues of the partial transpose.

A frequently used companion is the logarithmic negativity

E_{\mathcal{N}}(\rho) = \log_2 \|\rho^{T_B}\|_1 = \log_2\big(2\mathcal{N}(\rho) + 1\big),

which is additive and upper-bounds the distillable entanglement.

Worked example: the Werner state

The two-qubit Werner state mixes a Bell state with white noise,

\rho = p\,|\Phi^+\rangle\langle\Phi^+| + (1-p)\,\frac{I}{4}, \qquad |\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle),\ \ 0 \le p \le 1.

Its partial transpose has a degenerate spectrum. Three eigenvalues equal $\tfrac{1+p}{4}$ and one equals $\tfrac{1-3p}{4}$ . The single eigenvalue $\tfrac{1-3p}{4}$ is negative precisely when $p > \tfrac13$ , giving

\mathcal{N}(\rho) = \max\!\left(0,\ \frac{3p - 1}{4}\right).

So the Werner state is entangled (by negativity, hence by PPT) exactly for $p > \tfrac13$ — the same threshold the PPT criterion will give in the next lesson. At $p = \tfrac23$ the negative eigenvalue is $\tfrac{1-2}{4} = -\tfrac14$ , so $\mathcal{N} = \tfrac14$ .

Try it

For the Werner state at $p = \tfrac23$ , the four eigenvalues of $\rho^{T_B}$ are $\tfrac{1+p}{4}$ (three times) and $\tfrac{1-3p}{4}$ (once). Build that list and return the negativity — the sum of the magnitudes of the negative eigenvalues.

Run your code to see the quantum state.