Negativity and PPT Criterion

The positive-partial-transpose criterion is one of the most useful first tests for mixed-state entanglement. It starts from a simple operation on one subsystem: transpose the matrix indices belonging to that subsystem while leaving the other subsystem alone.

If the resulting operator is not positive semidefinite, the original state is entangled. This gives a practical entanglement diagnostic called the PPT test. The associated scalar measure, negativity, adds up the negative eigenvalue weight of the partial transpose.

The test is powerful but not universal: in $2\times2$ and $2\times3$ dimensions, PPT is equivalent to separability; in higher dimensions, some entangled states are PPT.

Helpful background. Separable mixed states supplies the convex-mixture definition; eigenvalues determine matrix positivity. This page treats finite-dimensional tensor products.

Partial transpose on one subsystem

Choose product bases $\{\lvert i\rangle_A\}$ and $\{\lvert \mu\rangle_B\}$. A bipartite density operator can be expanded as

$$\rho_{AB} = \sum_{i,j,\mu,\nu} \rho_{i\mu,j\nu}\, \lvert i\rangle\langle j\rvert_A \otimes \lvert \mu\rangle\langle \nu\rvert_B.$$

The partial transpose on subsystem $B$ is

$$\rho_{AB}^{T_B} = \sum_{i,j,\mu,\nu} \rho_{i\mu,j\nu}\, \lvert i\rangle\langle j\rvert_A \otimes \lvert \nu\rangle\langle \mu\rvert_B.$$

Only the $B$ matrix indices are transposed. Equivalently,

$$\left( \lvert i\rangle\langle j\rvert_A \otimes \lvert \mu\rangle\langle \nu\rvert_B \right)^{T_B} = \lvert i\rangle\langle j\rvert_A \otimes \lvert \nu\rangle\langle \mu\rvert_B.$$

The operation depends on a chosen product basis for writing the transpose, but the positivity or nonpositivity of $\rho^{T_B}$ is invariant under local changes of basis.

For a two-qubit matrix in the order $|00\rangle,|01\rangle,|10\rangle,|11\rangle$, view $\rho$ as a $2\times2$ array of $B$-blocks. Transpose each block in place, without interchanging the blocks. This implements $(\rho^{T_B})_{i\mu,j\nu}=\rho_{i\nu,j\mu}$.

For an active local unitary, the transformed partial transpose is

$$[(U_A\otimes U_B)\rho(U_A^\dagger\otimes U_B^\dagger)]^{T_B} =(U_A\otimes U_B^*)\rho^{T_B}(U_A^\dagger\otimes U_B^{\mathsf T}).$$

This is a unitary conjugation, explaining the invariance of its eigenvalues and negativity.

Why Separable States Are PPT

Suppose $\rho_{AB}$ is separable:

$$\rho_{AB} = \sum_k p_k\, \rho_A^{(k)}\otimes\rho_B^{(k)}, \qquad p_k\ge0, \qquad \sum_k p_k=1.$$

Then

$$\rho_{AB}^{T_B} = \sum_k p_k\, \rho_A^{(k)}\otimes \left(\rho_B^{(k)}\right)^T.$$

The transpose of a positive matrix is positive, so each term is positive semidefinite. A convex sum of positive semidefinite operators is positive semidefinite. Therefore

$$\rho_{AB}\ \text{separable} \quad\Longrightarrow\quad \rho_{AB}^{T_B}\ge0.$$

This implication is the heart of the test. If $\rho_{AB}^{T_B}$ has a negative eigenvalue, then $\rho_{AB}$ cannot be separable.

The PPT Criterion

A state is called PPT if its partial transpose is positive semidefinite:

$$\rho_{AB}^{T_B}\ge0.$$

The criterion has three standard levels:

Subsystem dimensions	Meaning of a positive partial transpose
$2\times2$	Necessary and sufficient for separability
$2\times3$ (or $3\times2$)	Necessary and sufficient for separability
$2\times4$, $3\times3$, and dimensions containing either case	Necessary, but not sufficient in general

Thus, for two qubits, a negative eigenvalue of $\rho^{T_B}$ proves entanglement, while nonnegative eigenvalues prove separability. For example, $2\times4$ and $3\times3$ systems admit PPT entangled states. Such states are examples of PPT entanglement, often discussed together with bound entanglement. Entanglement Distillation explains why PPT is an LOCC obstruction to extracting Bell pairs and separates that theorem from the still-open general NPT distillability problem.

Negativity

Negativity turns the partial-transpose test into a number. Let $\lambda_i$ be the eigenvalues of $\rho^{T_B}$. The negativity is

$$\mathcal N(\rho) = \sum_{\lambda_i<0} \lvert\lambda_i\rvert.$$

Equivalently,

$$\mathcal N(\rho) = \frac{ \left\lVert\rho^{T_B}\right\rVert_1 -1 }{2},$$

where $\lVert X\rVert_1=\operatorname{Tr}\sqrt{X^\dagger X}$ is the trace norm.

The logarithmic negativity is

$$E_{\mathcal N}(\rho) = \log_2 \left\lVert\rho^{T_B}\right\rVert_1 = \log_2\left(2\mathcal N(\rho)+1\right).$$

Negativity is zero for all separable states. It is positive for states whose partial transpose has a negative eigenvalue (NPT). It is not a complete entanglement measure in dimensions where PPT entangled states exist.

Bell-State Example

Consider

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \left( \lvert00\rangle+\lvert11\rangle \right).$$

The density operator is

$$\rho_{\Phi^+} = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right).$$

Partial transposition on $B$ gives

$$\rho_{\Phi^+}^{T_B} = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert01\rangle\langle10\rvert + \lvert10\rangle\langle01\rvert + \lvert11\rangle\langle11\rvert \right).$$

In the product basis

$$\lvert00\rangle,\quad \lvert01\rangle,\quad \lvert10\rangle,\quad \lvert11\rangle,$$

this has eigenvalues

$$\frac12,\quad \frac12,\quad \frac12,\quad -\frac12.$$

Therefore the Bell state is entangled, with

$$\mathcal N(\Phi^+) = \frac12, \qquad E_{\mathcal N}(\Phi^+)=1.$$

Pure Two-Qubit Schmidt Form

For

$$\lvert\psi\rangle = \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle, \qquad 0\le\theta\le\frac{\pi}{4},$$

the eigenvalues of $\rho^{T_B}$ are

$$\cos^2\theta,\quad \sin^2\theta,\quad \cos\theta\sin\theta,\quad -\cos\theta\sin\theta.$$

Thus

$$\mathcal N(\psi) = \cos\theta\sin\theta.$$

For two-qubit pure states, this is one half of the concurrence:

$$\mathcal N(\psi) = \frac{C(\psi)}{2}.$$

This simple relation is special to two-qubit pure states.

Pure states with more Schmidt terms

For $|\psi\rangle=\sum_i s_i|ii\rangle$, partial transpose produces eigenvalues $s_i^2$ and pairs $\pm s_i s_j$ for each $i<j$. The two-dimensional block on $|ij\rangle,|ji\rangle$ is $\begin{pmatrix}0&s_i s_j\\s_i s_j&0\end{pmatrix}$. Hence

$$\mathcal N(\psi)=\sum_{i<j}s_i s_j=\frac{(\sum_i s_i)^2-1}{2}, \qquad E_{\mathcal N}(\psi)=2\log_2\sum_i s_i.$$

For a pure state, logarithmic negativity equals the order-$1/2$ Rényi entropy of either marginal, generally different from von Neumann entanglement entropy. A maximally entangled state of Schmidt rank $d$ has $\mathcal N=(d-1)/2$ and $E_{\mathcal N}=\log_2d$.

Werner-State Example

For the two-qubit Werner state

$$\rho_W(p) = p\lvert\Psi^-\rangle\langle\Psi^-\rvert + \frac{1-p}{4}I, \qquad 0\le p\le1,$$

the partial transpose has one eigenvalue

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

The other three eigenvalues are all $(1+p)/4$ and are nonnegative. Hence

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

The state is NPT, and therefore entangled, exactly when

$$p>\frac13.$$

For two qubits this agrees with the separability threshold. It also matches the concurrence threshold, although the numerical values of concurrence and negativity differ:

$$C(\rho_W) = \max\left\{ 0,\frac{3p-1}{2} \right\}.$$

Partial Transpose Is Not a Physical Channel

The transpose map sends positive matrices to positive matrices, but it is not completely positive. This is why applying it to only one subsystem can reveal entanglement.

If $\rho_{AB}$ is entangled, the formal operation

$$I_A\otimes T_B$$

may produce an operator that is not a density operator. That failure of positivity is not a physical state transformation; it is a diagnostic signature.

This also explains why partial transpose is different from a partial trace. The partial trace produces a valid reduced state. The partial transpose is a mathematical test on the joint density operator.

Interpreting small negative eigenvalues

Partial transpose preserves Hermiticity and trace. Check these first, alongside the positivity and normalization of the original density matrix. With exact data, any negative eigenvalue certifies entanglement. With numerical or tomographic data, an eigenvalue of order the calculation’s uncertainty does not justify the same conclusion; propagate errors or use a justified tolerance. Entanglement certification develops the statistical version.

Negativity vanishes on every PPT state, including PPT entangled states in higher dimensions. It is therefore an entanglement monotone that is not faithful in all dimensions: zero does not universally mean separable. Logarithmic negativity is additive under tensor products; ordinary negativity instead obeys $2\mathcal N(\rho\otimes\sigma)+1=[2\mathcal N(\rho)+1][2\mathcal N(\sigma)+1]$.

References

• M. Horodecki, P. Horodecki, and R. Horodecki, “Separability of mixed states: necessary and sufficient conditions”, Physics Letters A 223, 1–8 (1996), DOI: 10.1016/S0375-9601(96)00706-2.
• A. Peres, “Separability criterion for density matrices”, Physical Review Letters 77, 1413–1415 (1996).
• M. B. Plenio, “Logarithmic negativity: a full entanglement monotone that is not convex”, Physical Review Letters 95, 090503 (2005).
• G. Vidal and R. F. Werner, “Computable measure of entanglement”, Physical Review A 65, 032314 (2002).

Exercises

1. Separable states are PPT. Show directly that a product density operator $\rho_A\otimes\rho_B$ has positive partial transpose. Then extend the argument to separable mixtures.

Solution

For a product state,

$$(\rho_A\otimes\rho_B)^{T_B} = \rho_A\otimes\rho_B^T.$$

Since $\rho_B$ is positive semidefinite, $\rho_B^T$ is positive semidefinite. Therefore $\rho_A\otimes\rho_B^T$ is positive semidefinite. A separable state is a convex sum of such product density operators, and a convex sum of positive semidefinite operators is positive semidefinite.

2. Bell-state negativity. Starting from $\rho_{\Phi^+}^{T_B}$ in the Bell-state example, identify the eigenvector with negative eigenvalue.

Solution

The only nontrivial block acts on the span of $\lvert01\rangle$ and $\lvert10\rangle$:

$$\frac12 \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}.$$

The antisymmetric vector

$$\frac{1}{\sqrt2} \left( \lvert01\rangle-\lvert10\rangle \right)$$

has eigenvalue $-1/2$. Therefore the negativity is $1/2$.

3. Two Bell pairs. Regard $|\Phi^+\rangle_{A_1B_1}|\Phi^+\rangle_{A_2B_2}$ as a state across $A_1A_2:B_1B_2$. Compute its negativity and logarithmic negativity. Why is ordinary negativity not additive?

Solution

There are four Schmidt coefficients, each $1/2$. Thus $\mathcal N=(4-1)/2=3/2$ and $E_{\mathcal N}=\log_2 4=2$. Each individual Bell pair has negativity $1/2$, so their sum is $1$, not $3/2$. It is the trace norm $2\mathcal N+1$ that multiplies, and its logarithm that adds.

4. Transposing subsystem A instead of B. Show that $\rho^{T_A}$ and $\rho^{T_B}$ have the same eigenvalues.

Solution

The full transpose of $\rho^{T_B}$ is

$$\left(\rho^{T_B}\right)^T = \rho^{T_A}.$$

A matrix and its transpose have the same characteristic polynomial, so they have the same eigenvalues. Therefore it does not matter whether one tests positivity of $T_A$ or $T_B$; the spectra agree.