Entanglement Measures

Entanglement measures quantify different capabilities of a shared quantum state: preparing it, extracting Bell pairs, or bounding possible conversions. The choice depends on the subsystem split, allowed operations, and whether one copy or many copies are available. This page compares the principal finite-dimensional bipartite quantities, then explains what changes for multipartite states and modes.

Helpful background. Separable mixed states defines the states with no bipartite entanglement; Schmidt decomposition gives pure-state entanglement spectra; quantum entropy supplies the entropy identities used below.

Entanglement monotones and local operations

In the usual bipartite setting, Alice and Bob can act locally and exchange classical messages, but cannot send quantum systems or create shared entanglement for free. These LOCC operations define what a measure must respect. We use base- $2$ logarithms; a Bell pair is the unit resource, called one ebit. One-way and two-way classical communication can give different conversion rates, so the permitted direction must be specified.

Let $E(\rho_{AB})$ be a candidate bipartite quantity. Conventions differ, but the following properties organize most of the literature.

Vanishing on free states

Every separable state should have zero entanglement:

\begin{aligned} \rho_{AB} &= \sum_x p_x\, \rho_A^{(x)}\otimes\rho_B^{(x)} \\ &\Longrightarrow E(\rho_{AB})=0. \end{aligned}

If the converse also holds, the measure is faithful. Faithfulness is valuable but not universal among operational rates. Distillable entanglement, for example, vanishes on bound-entangled states even though those states are not separable.

Monotonicity under LOCC

For a deterministic LOCC channel $\Lambda$ ,

E\!\left(\Lambda(\rho_{AB})\right) \leq E(\rho_{AB}).

A stronger branch-resolved requirement applies to an LOCC instrument with outcome probabilities $p_i$ and normalized conditional states $\rho_i$ :

E(\rho_{AB}) \geq \sum_i p_i E(\rho_i).

One postselected branch may have more entanglement than the input. The average cannot. The probability matters: a rare highly entangled output is not a deterministic supply of that state.

Local-unitary invariance follows because a local unitary and its inverse are both LOCC:

E\!\left( (U_A\otimes U_B)\rho_{AB} (U_A^\dagger\otimes U_B^\dagger) \right) =E(\rho_{AB}).

Desirable properties are not identical

Several additional properties are useful, but they answer different mathematical or operational concerns.

Normalization: often $E(\Phi_d)=\log_2 d$ for a maximally entangled state of Schmidt rank $d$ .
Convexity: the quantity on a mixture is no larger than the average value on its components,

E\!\left(\sum_x p_x\rho_x\right) \leq \sum_x p_x E(\rho_x).

Additivity: independent resources satisfy

E(\rho\otimes\sigma) = E(\rho)+E(\sigma).

Asymptotic continuity: small state errors should not cause an extensive jump after the quantity is scaled by system size.
Computability: a definition should admit an exact formula, controlled numerical optimization, or certified bounds in the intended setting.

Convexity is a mathematical condition, distinct from monotonicity under physically discarding a locally stored classical register: the entanglement assigned to the flagged input need not equal the average component value. These properties do not come as one package. Logarithmic negativity is additive and monotone under a broad class of operations, but it is not convex. Entanglement of formation is convex, but its optimization is generally hard. An operational rate can be conceptually exact while computationally inaccessible.

Measures at a Glance

Quantity	Natural domain	Main question	Main limitation
Entropy of entanglement	Pure bipartite states	Reversible asymptotic conversion rate	Reduced-state entropy is not a mixed-state measure
Concurrence	Two qubits	Closed mixed-state proxy and route to $E_F$	The Wootters formula is special to two qubits
Negativity $\mathcal N$	Bipartite mixed states	Computable NPT-sensitive quantifier	Vanishes on PPT entangled states
Logarithmic negativity $E_{\mathrm N}$	Bipartite and many continuous-variable settings	Additive computable bound on distillation	PPT blind and not convex
Entanglement of formation $E_F$	Bipartite mixed states	Least average pure-state entanglement in a decomposition	Convex-roof optimization is usually hard
Entanglement cost $E_C$	Asymptotic bipartite preparation	Bell pairs consumed per target copy	Requires regularization
Distillable entanglement $E_D$	Asymptotic bipartite extraction	Bell pairs recoverable per noisy copy	Depends on communication class and is often unknown
Squashed entanglement $E_{\mathrm{sq}}$	Bipartite states	Additive information-theoretic monotone and bounds	Optimization over all extensions is hard

The Entanglement Diagnostic Table answers a neighboring question: how to establish that entanglement is present. Detection criteria, witnesses, and Bell tests are not automatically quantitative measures.

Entropy of Entanglement

For a pure bipartite state $\lvert\psi\rangle_{AB}$ , define

\rho_A = \operatorname{Tr}_B \lvert\psi\rangle\langle\psi\rvert.

The entropy of entanglement is

E(\psi) = S(\rho_A) = S(\rho_B),

with base- $2$ entropy

S(\rho) = -\operatorname{Tr} \left(\rho\log_2\rho\right).

If the Schmidt probabilities are $\lambda_k$ , then

E(\psi) = -\sum_k\lambda_k\log_2\lambda_k.

It is zero for a product state and equals $\log_2 d$ for a maximally entangled state with $d$ equal Schmidt probabilities. For pure bipartite states in the many-copy, vanishing-error limit, this entropy gives both the rate at which Bell pairs can be distilled and the rate needed to create the state. This reversibility is why one number is so successful in that restricted setting.

Two cautions matter.

First, the full Schmidt spectrum, not only its entropy, governs deterministic single-copy LOCC conversion through majorization. Two pure states with equal entropy need not be exactly interconvertible in one shot.

Second, if $\rho_{AB}$ is mixed, $S(\rho_A)$ includes local thermal uncertainty, noise, and classical correlation. It is not a general mixed-state entanglement measure. A maximally mixed product state already has nonzero subsystem entropy and zero entanglement.

The entanglement-entropy derivation explains how the Schmidt probabilities determine these asymptotic rates.

Concurrence

C(\psi)=2\lvert ad-bc\rvert.

For a mixed two-qubit state, $C(\rho)$ is the minimum average pure-state concurrence over all ensemble decompositions of $\rho$ . The mixed-state concurrence calculation gives Wootters’ closed spin-flip formula and its stable numerical evaluation.

Here $0\leq C\leq1$ : zero is equivalent to separability, and one characterizes maximally entangled pure states, including local-unitary rotations of the Bell states. Its operational use is the exceptional two-qubit relation

E_F(\rho)=h_2\!\left(\frac{1+\sqrt{1-C(\rho)^2}}{2}\right),

where $h_2(x)=-x\log_2x-(1-x)\log_2(1-x)$ . This gives entanglement of formation in bits, not the distillation rate of an arbitrary mixed state. The relation to entanglement of formation supplies the structural connection. Generalized concurrences exist, but they are not one universal continuation of this formula; always state the system dimension and definition.

Negativity and Logarithmic Negativity

Let $\rho^{T_B}$ denote the partial transpose on subsystem B, and let $\lVert X\rVert_1=\operatorname{Tr}\sqrt{X^\dagger X}$ be the trace norm. Negativity and logarithmic negativity are

\begin{aligned} \mathcal N(\rho)&=\frac{\lVert\rho^{T_B}\rVert_1-1}{2},\\ E_{\mathrm N}(\rho)&=\log_2\lVert\rho^{T_B}\rVert_1 =\log_2\!\left(2\mathcal N(\rho)+1\right). \end{aligned}

Negativity sums the magnitudes of the negative eigenvalues of $\rho^{T_B}$ , so a nonzero value certifies entanglement. For a maximally entangled pair of local dimension $d$ , $\mathcal N(\Phi_d)=(d-1)/2$ while $E_{\mathrm N}(\Phi_d)=\log_2d$ . The negativity calculation develops the spectral interpretation and examples.

The logarithmic form is additive because the trace norm is multiplicative on tensor products. For a finite-dimensional state supplied in independent identical copies, it upper-bounds the asymptotic Bell-pair rate achievable by two-way LOCC with vanishing error. It need not equal that achievable rate.

Every PPT state has $\mathcal N=E_{\mathrm N}=0$ . In $2\times2$ and $2\times3$ systems, PPT is equivalent to separability; higher dimensions admit PPT entangled states, so zero negativity does not generally prove separability. This is a structural limitation, not a numerical defect. Partial transpose is a diagnostic transformation, not a physical channel; see the PPT criterion for the dimensional theorem.

Formation, Cost, and Distillation

Entanglement of formation

For a mixed state, consider every pure-state ensemble

\rho_{AB} = \sum_i p_i \lvert\psi_i\rangle\langle\psi_i\rvert.

Entanglement of formation is the convex roof

E_F(\rho_{AB}) = \inf_{\{p_i,\psi_i\}} \sum_i p_i\, S\!\left(\rho_A^{(i)}\right),

where

\rho_A^{(i)} = \operatorname{Tr}_B \lvert\psi_i\rangle\langle\psi_i\rvert.

The infimum is essential because a density operator has many ensemble decompositions. $E_F$ asks for the least average pure-state entanglement compatible with the mixed state, not the entanglement of one preferred preparation story.

Entanglement cost

The asymptotic Bell-pair cost is the regularized quantity

E_C(\rho) = \lim_{n\to\infty} \frac{1}{n} E_F\!\left(\rho^{\otimes n}\right).

Thus $E_F$ is a single-block convex-roof quantity, while $E_C$ is the many-copy preparation rate. They should not be silently identified.

Distillable entanglement

Suppose an LOCC protocol converts $n$ copies of $\rho$ into $m_n$ nearly perfect Bell pairs, with the error tending to zero as $n\to\infty$ . The distillable entanglement is the best achievable rate:

E_D(\rho) = \sup_{\{\Lambda_n\}} \liminf_{n\to\infty} \frac{m_n}{n}.

The notation must specify whether $\Lambda_n$ permits one-way or two-way classical communication. Those rates can differ.

For bipartite pure states,

E_D(\psi) = E_C(\psi) = S(\rho_A).

For mixed states, formation and extraction can be irreversible:

0 \leq E_D(\rho) \leq E_C(\rho).

Some entangled mixed states have $E_D=0$ and are called bound entangled. Consequently, distillable entanglement is operationally sharp but not faithful. Exact values of $E_D$ are unknown for many ordinary-looking state families, so computable upper and lower bounds are often the scientifically honest result.

Entanglement distillation develops explicit recurrence and hashing protocols. Their success probabilities and output yields provide achievable lower bounds on the optimized rate.

Squashed Entanglement

An extension of $\rho_{AB}$ is any state $\rho_{ABE}$ satisfying

\operatorname{Tr}_E\rho_{ABE} = \rho_{AB}.

Squashed entanglement minimizes conditional mutual information over every such extension:

E_{\mathrm{sq}}(A:B)_\rho = \frac{1}{2} \inf_{\rho_{ABE}} I(A:B\mid E)_\rho,

where

I(A:B\mid E) ={} S(AE)+S(BE) -S(E)-S(ABE).

The extension asks how much of the observed $A$ – $B$ correlation can be explained by a third system. The infimum removes correlation that can be mediated through $E$ and retains an irreducibly quantum contribution.

In finite dimensions, squashed entanglement is LOCC monotone, convex, additive, asymptotically continuous, monogamous, and faithful. It reduces to entanglement entropy on pure states. It also lies between important operational quantities:

E_D(\rho) \leq E_{\mathrm{sq}}(\rho) \leq E_C(\rho) \leq E_F(\rho).

The middle bound follows by applying $E_{\mathrm{sq}}\le E_F$ to $n$ copies, using additivity of $E_{\mathrm{sq}}$ , and taking the regularized limit defining $E_C$ .

Its principal weakness is computability. The optimization ranges over extensions with no immediately fixed auxiliary dimension, and exact values are known only in restricted settings. Squashed entanglement is often most useful as a theorem-building quantity or a source of certified bounds, not as a routine number extracted from tomography.

Worked Comparison: A Noisy Bell Pair

For the two-qubit singlet mixed with white noise,

\rho_W(p)=p\lvert\Psi^-\rangle\langle\Psi^-\rvert +(1-p)\frac{I_4}{4}, \qquad 0\leq p\leq1,

both marginals are $\rho_A=\rho_B=I_2/2$ . Thus $S(\rho_A)=1$ for every $p$ , including the separable maximally mixed state at $p=0$ . Subsystem entropy alone would incorrectly assign that state the same entanglement as the singlet at $p=1$ .

The concurrence and partial-transpose calculations give

\begin{aligned} C(\rho_W)&=\max\!\left\{0,\frac{3p-1}{2}\right\} =2\mathcal N(\rho_W),\\ E_{\mathrm N}(\rho_W)&=\log_2\!\left(1+2\mathcal N(\rho_W)\right). \end{aligned}

Both quantities become positive at the same threshold $p>1/3$ , but their scales and operational roles differ. Concurrence determines $E_F$ ; logarithmic negativity supplies an additive computable upper bound on distillation. Neither number is “the true amount” independently of the question.

For $p>1/3$ , this two-qubit state is NPT and distillable by two-way LOCC, but the exact asymptotic distillation rate is not given by either $C$ or $\mathcal N$ . This is typical: detection can be easy while operational rate calculation remains hard.

For experimental data, an entanglement witness can establish nonseparability without full tomography. A witness expectation is not automatically a calibrated amount of entanglement. A measure computed from a reconstructed state also needs statistical uncertainty: finite samples and imposing positivity can bias a plug-in estimate. These issues are developed in entanglement witnesses.

Multipartite Cautions

For three or more parties, begin by naming a partition. The same state can have different entanglement across $A\vert BC$ , $B\vert AC$ , $C\vert AB$ , and reduced two-party states.

For the GHZ state,

\lvert\mathrm{GHZ}\rangle = \frac{ \lvert000\rangle+\lvert111\rangle }{\sqrt2},

the entropy across every one-versus-two split is one bit. Yet every two-qubit reduced state is separable:

\rho_{AB} = \frac{1}{2} \left( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right).

Pairwise concurrence therefore vanishes even though the global state has genuine tripartite entanglement.

For the W state,

\lvert W\rangle = \frac{ \lvert001\rangle +\lvert010\rangle +\lvert100\rangle }{\sqrt3},

two-party reductions retain pairwise entanglement. GHZ and W states are inequivalent under stochastic LOCC, and one scalar cannot encode that distinction, every bipartition, pairwise sharing, genuine multipartite structure, and conversion behavior simultaneously.

Multipartite work should report a family of quantities: entropies or negativities across named cuts, reduced-state measures, separability class, witnesses, and task-specific monotones. See Multipartite Systems, GHZ States, W States, and Monogamy of Entanglement.

Continuous Variables and Indistinguishable Particles

In continuous-variable systems, logarithmic negativity is popular because Gaussian partial transpose can be analyzed through covariance matrices. Entropies and trace norms may nevertheless require finite-energy assumptions, domain control, or converged truncations. Always specify the mode decomposition, quadrature convention, logarithm base, and cutoff treatment. For the covariance-matrix calculations, see Two-Mode Entanglement and Gaussian States Preview.

For identical particles, particle labels do not automatically define operational subsystems. Entanglement may instead refer to modes, orbitals, regions, or commuting observable algebras. Exchange symmetry alone is not a usable resource measure. See Identical-Particle Entanglement Cautions.

Exercises

Normalization on a Bell pair. For $\lvert\Phi^+\rangle=(\lvert00\rangle+\lvert11\rangle)/\sqrt2$ , calculate entanglement entropy, concurrence, negativity, logarithmic negativity, and entanglement of formation. What are $E_C$ and $E_D$ in the asymptotic pure-state setting?

Solution

Tracing out either qubit gives

\rho_A=\frac{I_2}{2},

S(\rho_A)=1.

For amplitudes $a=d=1/\sqrt2$ and $b=c=0$ ,

C=2\lvert ad-bc\rvert=1.

The partial transpose has eigenvalues

\left\{ \frac12,\frac12,\frac12,-\frac12 \right\},

hence

\mathcal N=\frac12, \qquad E_{\mathrm N} = \log_2(1+2\mathcal N) =1.

The Wootters relation gives $E_F=1$ . Pure-state entanglement manipulation is asymptotically reversible, so

E_C=E_D=1

ebit per copy. The raw negativity is $1/2$ , illustrating that normalization conventions differ even when the measures agree operationally on maximality.

A partially entangled pair. Let

\lvert\psi_\lambda\rangle = \sqrt{\lambda}\lvert00\rangle + \sqrt{1-\lambda}\lvert11\rangle.

Take $0\leq\lambda\leq1$ . Find $E$ , $C$ , $\mathcal N$ , and $E_{\mathrm N}$ . Check the product and maximally entangled limits.

Solution

The Schmidt probabilities are $\lambda$ and $1-\lambda$ , so

E(\psi_\lambda) = h_2(\lambda).

The concurrence is

C(\psi_\lambda) = 2\sqrt{\lambda(1-\lambda)}.

The partial-transpose eigenvalues are

\lambda,\quad 1-\lambda,\quad \sqrt{\lambda(1-\lambda)}, \quad -\sqrt{\lambda(1-\lambda)}.

Therefore

\mathcal N(\psi_\lambda) = \sqrt{\lambda(1-\lambda)},

and

E_{\mathrm N}(\psi_\lambda) = \log_2 \left( 1+2\sqrt{\lambda(1-\lambda)} \right).

At $\lambda=0$ or $1$ , all four quantities vanish. At $\lambda=1/2$ ,

E=C=E_{\mathrm N}=1, \qquad \mathcal N=\frac12.

The measures order this one-parameter pure-state family consistently, but their numerical scales are not the same. For any pure bipartite state, logarithmic negativity is the Rényi entropy of order $1/2$ of the reduced state, $E_{\mathrm N}=2\log_2\sum_k\sqrt{\lambda_k}$ , rather than its von Neumann entropy. At $\lambda=0.1$ , the distillation rate is $h_2(0.1)\simeq0.4690$ ebits per copy, while $E_{\mathrm N}=\log_2(1.6)\simeq0.6781$ is a strictly larger bound.

The subsystem-entropy trap. For the noisy singlet $\rho_W(p)$ in the worked comparison, derive the potentially negative eigenvalue of $\rho_W^{T_B}$ , the negativity, and the concurrence threshold. Explain why $S(\rho_A)$ fails.

Solution

The partial transpose of the singlet projector has eigenvalues

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

Adding white noise shifts each eigenvalue by $(1-p)/4$ and multiplies the singlet contribution by $p$ . Three eigenvalues become $(1+p)/4$ , while the fourth is

\mu_{\min} = \frac{1-p}{4} -\frac{p}{2} = \frac{1-3p}{4}.

Thus

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

The two-qubit concurrence formula gives

C = \max\!\left\{ 0,\frac{3p-1}{2} \right\}.

Both become positive exactly when $p>1/3$ . However,

\rho_A=\frac{I_2}{2}

for every $p$ , so $S(\rho_A)=1$ even at $p=0$ , where $\rho_W=I_4/4$ is separable. The marginal entropy measures local mixedness, not mixed-state entanglement.

Additivity of logarithmic negativity. Use trace-norm multiplicativity to prove that $E_{\mathrm N}$ is additive on tensor products.

Solution

Partial transpose distributes over tensor products:

(\rho\otimes\sigma)^{T_{BB'}} = \rho^{T_B}\otimes\sigma^{T_{B'}}.

For finite-dimensional matrices (and more generally trace-class operators),

\lVert X\otimes Y\rVert_1 = \lVert X\rVert_1 \lVert Y\rVert_1.

Therefore

\begin{aligned} E_{\mathrm N}(\rho\otimes\sigma) &= \log_2 \left\lVert \rho^{T_B}\otimes\sigma^{T_{B'}} \right\rVert_1 \\ &= \log_2 \left( \lVert\rho^{T_B}\rVert_1 \lVert\sigma^{T_{B'}}\rVert_1 \right) \\ &= E_{\mathrm N}(\rho) + E_{\mathrm N}(\sigma). \end{aligned}

Negativity itself is not additive because the affine transformation between $\mathcal N$ and $E_{\mathrm N}$ converts products of trace norms into

1+2\mathcal N(\rho\otimes\sigma) = \left(1+2\mathcal N(\rho)\right) \times \left(1+2\mathcal N(\sigma)\right).

Squashed entanglement of a pure state. Show that $E_{\mathrm{sq}}(A:B)_\psi=S(\rho_A)$ when $\rho_{AB}=\lvert\psi\rangle\langle\psi\rvert$ is pure.

Solution

Any extension of a pure state on $AB$ must factor:

\rho_{ABE} = \lvert\psi\rangle\langle\psi\rvert_{AB} \otimes\sigma_E.

Consequently,

\begin{aligned} S(AE)&=S(A)+S(E),\\ S(BE)&=S(B)+S(E),\\ S(ABE)&=S(E). \end{aligned}

Substitution into conditional mutual information gives

\begin{aligned} I(A:B\mid E) &= S(A)+S(B) \\ &= 2S(A), \end{aligned}

because a pure bipartite state has $S(A)=S(B)$ . Every extension gives the same value, so the infimum does too:

E_{\mathrm{sq}}(A:B)_\psi = \frac12 I(A:B\mid E) = S(\rho_A).

Pairwise versus genuine tripartite entanglement. Trace out qubit $C$ from the GHZ state. Show that $\rho_{AB}$ is separable even though the entropy across $A\vert BC$ is one bit. What does this demonstrate about multipartite measures?

Solution

The GHZ projector contains two diagonal terms and two coherence terms:

\rho_{\mathrm{GHZ}} = \frac12\big( \lvert000\rangle\langle000\rvert + \lvert111\rangle\langle111\rvert + \lvert000\rangle\langle111\rvert + \lvert111\rangle\langle000\rvert \big).

The off-diagonal terms vanish under $\operatorname{Tr}_C$ because $\langle0\vert1\rangle=0$ . Hence

\rho_{AB} = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right),

which is an explicit convex mixture of product states. Its concurrence and negativity are zero.

By contrast,

\rho_A=\frac{I_2}{2}, \qquad S(\rho_A)=1,

so the pure global state has one ebit across $A\vert BC$ . Pairwise measures can therefore vanish while genuine global entanglement remains. Multipartite characterization must name partitions and usually requires several quantities or a separability class, not one scalar.

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