Entanglement Diagnostic Table

Choose an entanglement test by first fixing the subsystems and determining whether the joint state is pure, mixed, or known to be Gaussian. A test can be mathematically correct yet inconclusive for the wrong state class. The tables separate exact criteria from sufficient witnesses and from quantities that measure total correlation.

Helpful background. Separable mixed states defines what a test must exclude, while Schmidt decomposition gives the pure-state shortcut. Entropies below use base-two logarithms.

Bipartite entanglement tests

“Exact” means necessary and sufficient under the stated assumptions. “Sufficient” means detection proves entanglement, but a failed detection need not prove separability. The finite-dimensional rows assume a valid normalized density operator and a specified tensor-product split.

State and test	Conclusion	Scope and explanation
Pure bipartite vector: Schmidt rank greater than one	Exact entanglement test	Equivalent to a mixed reduction or positive reduced entropy; Schmidt rank.
Mixed two-qubit state: concurrence $C>0$	Exact entanglement test	The closed mixed-state formula is specific to two qubits; concurrence.
$2\times2$ or $2\times3$ state: $\rho^{T_B}\ge0$	Exact separability test	A negative eigenvalue instead proves entanglement; PPT.
Larger bipartite state: $\mathcal N>0$	Sufficient for entanglement	Zero negativity is inconclusive in general; PPT-entangled states exist already in $2\times4$ and $3\times3$.
Valid witness: $\operatorname{Tr}(W\rho)<0$	Sufficient for entanglement	Nonnegative on every separable alternative; statistical and calibration bounds must preserve the negative sign; witnesses.
One-mode versus $N$-mode Gaussian state: covariance PPT	Exact separability test	Finite covariances and Gaussianity required for sufficiency; not generally complete for two-mode versus two-mode cuts; Gaussian states.
Any two-mode state with finite second moments: EPR variance sum below two	Sufficient for entanglement	Uses $[q,p]=i$ and vacuum variance $1/2$; the unoptimized symmetric test is not necessary in general; EPR bound.

Correlation and multipartite questions

Quantity or setting	What it tells you	What it does not prove
Local entropy of a mixed joint state	Local mixedness	Entanglement: even the product $I_A/d_A\otimes I_B/d_B$ has mixed marginals.
Mutual information $I(A:B)$	Total correlation; zero exactly for product states	Nonzero mutual information is not necessarily entanglement; mutual information.
Pure-state entropy across a spatial cut	Bipartite entanglement and its size dependence	A complete phase classification or tensor-network error bound; many-body examples.
Multipartite pure state entangled across every cut	Genuine multipartite entanglement	The same implication fails for mixed states; separability classes.
Identical-particle exchange symmetry	Bosonic or fermionic state-space restriction	Useful entanglement between unnamed particles; specify modes, regions, or accessible algebras; particle cautions.

Pure Bipartite Shortcut

For a normalized pure state on $\mathcal H_A\otimes\mathcal H_B$, compute the Schmidt decomposition:

$$\lvert\psi\rangle = \sum_{r=1}^{R} s_r \lvert r_A\rangle \lvert r_B\rangle, \qquad s_r>0, \qquad \sum_{r=1}^{R}s_r^2=1.$$

The following statements are equivalent:

• $\lvert\psi\rangle$ is a product state across $A\vert B$.
• The Schmidt rank is $R=1$.
• The reduced state $\rho_A$ is pure.
• The reduced state $\rho_B$ is pure.
• The entanglement entropy is zero:

$$S(\rho_A) = - \operatorname{Tr}(\rho_A\log_2\rho_A) = 0.$$

Equivalently, the following statements are equivalent:

• $\lvert\psi\rangle$ is entangled across $A\vert B$.
• The Schmidt rank is $R>1$.
• The reduced state $\rho_A$ is mixed.
• The entanglement entropy is positive:

$$S(\rho_A) = - \sum_{r=1}^{R} s_r^2\log_2 s_r^2 > 0.$$

This is the cleanest entanglement-detection setting in ordinary finite-dimensional quantum mechanics. It is also the setting where it is easiest to overgeneralize. Once the global state is mixed, reduced-state mixedness and entropy no longer distinguish quantum entanglement from classical correlation.

Mixed-State Diagnostics

For mixed states, the definition of separability is a convex-decomposition condition:

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

In finite dimension, the state is entangled if no such decomposition exists. In infinite dimension the separable set includes trace-norm limits of finite such mixtures. This makes mixed-state entanglement a global property of the density operator, not a property of either marginal alone.

The common diagnostics have different logical status:

• Concurrence: for two qubits, $C(\rho)>0$ if and only if $\rho$ is entangled.
• PPT in low dimensions: for $2\times2$ and $2\times3$, $\rho^{T_B}\ge0$ if and only if $\rho$ is separable.
• Negativity: $\mathcal N(\rho)>0$ proves entanglement, but $\mathcal N(\rho)=0$ does not prove separability in dimensions where PPT entangled states exist.
• Witnesses: $\operatorname{Tr}(W\rho)<0$ proves entanglement for a valid witness $W$, but $\operatorname{Tr}(W\rho)\ge0$ for one witness usually proves only that this witness did not detect the state.

These tests are often combined. For example, in a two-qubit experiment, one might reconstruct an approximate density matrix, test positivity and trace normalization, compute concurrence, check PPT, and report a witness value with error bars. The diagnostic conclusion should match the assumptions: “entangled if the reconstructed two-qubit state and error model are valid” is stronger and more honest than a context-free claim that an apparatus “measured entanglement.”

Many-Body and Field-Theoretic Uses

In many-body physics, entanglement diagnostics are often used less as yes-or-no separability tests and more as structural probes. Given a pure state $\lvert\Psi\rangle$ and a spatial region or site subset $A$, the subsystem entropy

$$S_A = - \operatorname{Tr}(\rho_A\log_2\rho_A), \qquad \rho_A = \operatorname{Tr}_{\bar A} \lvert\Psi\rangle\langle\Psi\rvert$$

measures bipartite entanglement across the cut $A\vert\bar A$. Its scaling can reveal short-range entanglement, critical behavior, matrix-product-state efficiency, or topological contributions. The diagnostic is powerful precisely because it uses the subsystem cut as part of the question.

For mixed many-body states, thermal states, open systems, and continuum fields, entropy of a region is not automatically an entanglement measure. Mutual information, negativity, relative entropy, algebraic methods, and model-specific witnesses may be more appropriate. The Entanglement in QFT Preview discusses the extra complications from local algebras, UV behavior, and gauge constraints.

Gaussian and Continuous-Variable States

Continuous-variable systems add an additional layer: the Hilbert spaces are infinite-dimensional, and common laboratory states are often described by quadratures and covariance matrices rather than finite coefficient arrays.

For Gaussian states, first moments and the covariance matrix

$$V_{ij} = \frac12 \langle \Delta R_i\Delta R_j + \Delta R_j\Delta R_i \rangle$$

determine the state. Partial transposition reverses the momentum quadratures of the transposed subsystem, leaving its position quadratures unchanged. For two-mode Gaussian states, covariance-matrix PPT criteria give a practical entanglement test. For one mode against any finite number of modes, Gaussian PPT is equivalent to separability. A fixed symmetric Duan variance inequality is only sufficient; the necessary-and-sufficient Gaussian formulation allows the required optimization and local normal forms. The units and normalization of the quadratures matter.

This is why the table says “Gaussian state” rather than “continuous-variable state” without qualification. Non-Gaussian states can have entanglement that is invisible to purely covariance-based diagnostics.

Identical Particles

For identical particles, the most common false diagnostic is:

$$\text{antisymmetric wavefunction} \quad\Longrightarrow\quad \text{entanglement}.$$

That implication is not a safe physical statement. Antisymmetry is a kinematic constraint on fermionic states. Entanglement becomes a physical question only after one has specified accessible subsystems or algebras: modes, spin-orbitals, spatial regions, detector channels, occupation-number partitions, or other operationally meaningful degrees of freedom.

For example, a single determinant has definite occupations and no occupation-mode entanglement in its occupied-orbital basis. The same determinant can be entangled across a different orbital split, while some multi-term determinant expansions simplify to one determinant after a rotation. Orbital examples show why determinant counts and subsystem entropies answer different questions. The diagnostic therefore belongs to the chosen mode, region, or algebra, not to the particle labels that were introduced only to write antisymmetric wavefunctions.

A two-qubit decision example

Take $\rho_v=v\lvert\Psi^-\rangle\langle\Psi^-\rvert+(1-v)I_4/4$, with $0\le v\le1$. Both local states are $I/2$ for every $v$, so their entropy cannot decide whether the state is entangled. Partial transposition instead has eigenvalues $(1+v)/4$ three times and $(1-3v)/4$ once. Therefore the state is separable exactly for $v\le1/3$.

At $v=0.6$, the negative eigenvalue is $-0.2$, negativity is $0.2$, and concurrence is $0.4$. Three exact two-qubit criteria agree. A particular witness could still miss this state if it is aimed at a different entangled family; that failure would not undo the PPT result.

For experimental estimates, a tiny negative eigenvalue can be numerical or statistical error. Check physicality of the reconstructed state and uncertainty in the test, rather than counting any negative floating-point number as a detection. Computational examples show how to separate tolerances from physical assumptions.

Exercises

1. Pure-state shortcut. For

$$\lvert\Phi^+\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2},$$

find the Schmidt rank and the one-qubit entanglement entropy.

Solution

The state is already in Schmidt form with coefficients

$$s_1=s_2=\frac{1}{\sqrt2}.$$

The Schmidt rank is $R=2$, so the state is entangled across the two-qubit split. The reduced density operator is

$$\rho_A = \frac12 \lvert0\rangle\langle0\rvert + \frac12 \lvert1\rangle\langle1\rvert.$$

Therefore

$$S(\rho_A) = - 2 \left( \frac12\log_2\frac12 \right) = \log_2 2.$$

2. Mixed marginals are not enough. Consider

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert.$$

The one-qubit reductions are maximally mixed. Does that prove entanglement?

Solution

No. The state is explicitly a convex mixture of product states:

$$\rho_{\mathrm{cc}} = \frac12 \bigl( \lvert0\rangle\langle0\rvert \otimes \lvert0\rangle\langle0\rvert \bigr) + \frac12 \bigl( \lvert1\rangle\langle1\rvert \otimes \lvert1\rangle\langle1\rvert \bigr).$$

It is separable. Its mixed marginals show that each subsystem alone is uncertain, and its joint density operator shows classical correlation, but neither fact proves entanglement for a mixed global state.

3. PPT logic. In a two-qubit problem, what does a negative eigenvalue of $\rho^{T_B}$ prove? What does $\rho^{T_B}\ge0$ prove?

Solution

A negative eigenvalue of $\rho^{T_B}$ proves that $\rho$ is entangled. For two qubits, the converse is also complete: $\rho^{T_B}\ge0$ proves that $\rho$ is separable. This second statement relies on the $2\times2$ dimension. It is not true in all larger bipartite dimensions.

4. Witness logic. A valid entanglement witness $W$ gives $\operatorname{Tr}(W\rho)=-0.03$ for a state $\rho$. What can be concluded? What if the same witness gives $+0.02$?

Solution

For exact expectations and a valid witness, $\operatorname{Tr}(W\rho)<0$ proves that $\rho$ is entangled. For estimated expectations, an uncertainty bound must keep the entire allowed interval below zero: $-0.03\pm0.01$ does so, whereas $-0.03\pm0.04$ does not. If the same witness gives $+0.02$, this witness has not detected entanglement. That result alone does not prove that $\rho$ is separable.

References

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