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.

Partial Transpose

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.

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:

$$\begin{array}{c|c} \text{dimension} & \text{meaning of PPT}\\ \hline 2\times2 & \text{PPT if and only if separable}\\ 2\times3 & \text{PPT if and only if separable}\\ \text{larger dimensions} & \text{PPT is necessary but not sufficient} \end{array}$$

Thus, for two qubits, a negative eigenvalue of $\rho^{T_B}$ proves entanglement, while nonnegative eigenvalues prove separability. For $3\times3$ and larger systems, PPT states can still be entangled. 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 with nonpositive partial transpose. 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.

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 remaining eigenvalues 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.

Limitations

The PPT test and negativity are extremely useful, but their scope should be stated carefully:

• A negative eigenvalue of $\rho^{T_B}$ always proves entanglement.
• For $2\times2$ and $2\times3$ systems, PPT is also sufficient for separability.
• In higher dimensions, PPT does not imply separability.
• Negativity is blind to PPT entangled states.
• The value of negativity is not a measure of total correlation.
• The partial transpose depends on a chosen subsystem split, just like entanglement itself.

The safest practical workflow is: specify the bipartite split, compute $\rho^{T_B}$, inspect its spectrum, and remember which dimensional theorem is being used.

Common Mistakes

• Treating PPT as sufficient for separability in all dimensions.
• Forgetting that partial transpose is taken with respect to one subsystem, not the whole matrix.
• Confusing partial transpose with partial trace.
• Calling a zero negativity state separable in dimensions where PPT entanglement can occur.
• Forgetting to specify the product basis and subsystem split before writing matrix indices.
• Treating the partial transpose as a physical time evolution or allowed quantum channel.

Cross-Links

• Separable Mixed States
• Entangled States
• Entanglement Measures
• Bell States
• Classical Correlation versus Entanglement
• Local Unitary Equivalence
• Entanglement Depends on a Decomposition
• Reduced Density Operators
• Partial Trace
• Schmidt Decomposition
• Entanglement Entropy
• Concurrence for Two Qubits
• Mutual Information
• Entanglement Witnesses
• LOCC Preview
• Density Operators
• Eigenvalues and Eigenvectors
• Certification of Entanglement explains how PPT and NPT conclusions are supported from finite tomographic data and how their trust assumptions compare with steering and Bell tests.

References

• A. Peres, “Separability Criterion for Density Matrices,” Physical Review Letters 77, 1413-1415, 1996.
• M. Horodecki, P. Horodecki, and R. Horodecki, “Separability of Mixed States: Necessary and Sufficient Conditions,” Physics Letters A 223, 1-8, 1996.
• G. Vidal and R. F. Werner, “Computable Measure of Entanglement,” Physical Review A 65, 032314, 2002.
• M. B. Plenio, “Logarithmic Negativity: A Full Entanglement Monotone That Is Not Convex,” Physical Review Letters 95, 090503, 2005.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865-942, 2009.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

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. Werner-state threshold. For $p=1/4$, $p=1/2$, and $p=1$, compute $\mathcal N(\rho_W)$.

Solution

Use

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

For $p=1/4$, the expression is negative, so $\mathcal N=0$. For $p=1/2$,

$$\mathcal N = \frac{3/2-1}{4} = \frac18.$$

For $p=1$,

$$\mathcal N = \frac{2}{4} = \frac12.$$

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.