Resource Theory of Entanglement and Entanglement Measures

Entanglement is not a vague adjective attached to a quantum state; it is a quantity — something you can hold, count, spend, and run out of. The idea that pins it down is deceptively simple: declare a set of operations you may do for free and a set of states you can already make for free, and then anything you cannot reach for free becomes a resource whose value is any number those free operations can never increase. For entanglement the free operations are LOCC — local channels plus classical communication — and the free states are the separable ones, exactly what LOCC can prepare from scratch. Entanglement is then defined by exclusion: it is whatever LOCC cannot create, a nonlocal correlation that must be supplied, stored, and spent like a fuel. That one axiom — no free lunch under LOCC — is astonishingly generative. Any real functional that cannot rise on average under LOCC is a legitimate measure, a monotone. For pure states the answer is unique: a single number, the entropy of entanglement $S(\rho_A)$ , governs every asymptotic conversion, single-copy convertibility is decided by majorization (Nielsen's theorem), and the theory is reversible. Mixing shatters this. The measure splits into a family — distillable entanglement, entanglement cost, entanglement of formation, relative entropy of entanglement, negativity, squashed entanglement — that agree on pure states and disagree everywhere else, and the theory turns irreversible: bound entangled states cost real Bell pairs yet yield none back. This article is the load-bearing frame under that edifice, built from the same entropic toolbox as quantum Shannon theory.

Key takeaways

Fix the free operations (LOCC) and free states (separable); then every function nonincreasing on average under LOCC is a legitimate monotone — the whole theory follows from that one axiom.
For a pure state the unique asymptotic measure is the entropy of entanglement $E(\psi)=S(\rho_A)=-\sum_i\lambda_i\log_2\lambda_i$ , and single-copy convertibility is decided by majorization (Nielsen's theorem).
Two operational measures bracket every mixed state: distillable entanglement $E_D$ (pairs extractable per copy) and entanglement cost $E_C$ (pairs spent per copy), with $E_D\le E_C$ .
Wootters' concurrence gives $E_F$ in closed form for two qubits, $E_F=h\big(\tfrac{1+\sqrt{1-C^2}}{2}\big)$ ; a Werner state at $p=0.8$ has $C=0.7$ and $E_F=0.592$ ebits.
Bound entangled states realize the extreme $E_D=0<E_C$ : entanglement you paid for that never comes back as Bell pairs — the theory is irreversible.
Squashed entanglement $E_{sq}=\tfrac12\inf I(A;B|E)$ , additive and continuous, sits between $E_D$ and $E_C$ , tying the theory to strong subadditivity.
Key idea: fix the free operations (LOCC) and free states (separable); then every function that cannot increase on average under LOCC is a legitimate entanglement measure, the two operational extremes are the distillable entanglement and the entanglement cost, and for pure states all such measures collapse to the single number $S(\rho_A)$ .

Where this lives today

The resource theory is the design manual for the emerging quantum internet. Long-distance links degrade Bell pairs into noisy mixed states, and entanglement distillation — LOCC protocols like BBPSSW and DEJMPS that turn many low-fidelity pairs into fewer near-perfect ones — is now run in the lab: nitrogen-vacancy centres in diamond at Delft have distilled and swapped entanglement across a multi-node network, with the achievable rate capped by $E_D$ . Squashed entanglement, additive and continuous, is a workhorse for converse bounds: through the Choi state it upper-bounds the secret-key rate and two-way quantum capacity, setting fundamental limits on QKD and repeater-less communication (the PLOB bound). Computable monotones — concurrence, negativity — are what experimentalists report from tomography, and a subsystem's entanglement entropy is now measured directly in cold-atom simulators via randomized measurements. Bound entanglement, once a curiosity, is now a target: its activation and role in irreversibility remain open, and whether entanglement is reversible under an enlarged non-entangling class has stayed alive for two decades.

The physical picture

Picture Bell pairs as a standardized currency and LOCC as a banking system: you may shuffle local assets and phone your counterparty freely, but minting nonlocal correlation is forbidden. A shared state then has two prices — the cost $E_C$ you pay in Bell pairs to manufacture it by LOCC, and the distillable value $E_D$ you can melt it back into. For pure states manufacturing and recycling are perfectly efficient, so buy equals sell. For mixed states both are lossy — you cannot cleanly read off which pure state you hold through the noise — so a spread opens, $E_C\ge E_D$ , and at the extreme a bound entangled state has a positive purchase price and zero resale value. Pure states are special because $S(\rho_A)$ behaves like a conserved charge: asymptotically $n$ copies of $|\psi\rangle$ interconvert with about $nS(\rho_A)$ Bell pairs at vanishing error, so distillation and dilution are exact inverses. Mixing adds classical uncertainty about which pure state you have. The convex-roof $E_F$ asks for the cheapest pure-state description — but the cheapest description of $n$ copies together can beat $n$ separate ones, which is why $E_C=\lim_n E_F(\rho^{\otimes n})/n$ demands regularization. And single-copy ordering is not even total: majorization makes the pure states a lattice with incomparable elements, foreshadowing the disagreeing monotones ahead.

Figure

For a mixed state the resource value spreads into a sandwich E_D ≤ E_sq ≤ E_F ≤ E_C so the sell price E_D can fall below the buy price E_C, while for a pure state with Schmidt weights 0.8 and 0.2 every measure collapses onto the single number S(ρ_A) = 0.722 ebits

For two qubits the concurrence indexes the mixed case, and the map from concurrence to entanglement of formation is a single universal curve — climbing monotonically from zero at a separable state to one full ebit at a Bell state, and reading off partially entangled pure states and noisy Werner mixtures alike.

Figure

Entanglement of formation versus concurrence, E_F = h((1+√(1−C²))/2), rising monotonically from 0 to 1 ebit — the Werner state at p = 0.8 has C = 0.7 and E_F = 0.592, a pure state with Schmidt weights 0.8 and 0.2 has C = 0.8 and E_F = 0.722, and a Bell state has C = 1 and E_F = 1

The core equations

The whole theory rests on one inequality: the defining statement of what an entanglement measure is. An LOCC protocol applies local operators conditioned on classical messages, producing a branching family of outcomes; a measure must not rise when you average over those branches.

\rho_{AB}=\sum_k p_k\,(\Lambda^A_k\otimes\Lambda^B_k)\,\sigma_{AB}\,(\Lambda^A_k\otimes\Lambda^B_k)^\dagger\ \Rightarrow\ E(\rho_{AB})\ \text{not increased},\qquad E(\rho)\ \ge\ \sum_k p_k\,E(\rho_k) \qquad (1)

where: $\Lambda^A_k,\Lambda^B_k$ — local instrument elements applied by Alice and Bob conditioned on classical messages; $p_k$ — probability of outcome $k$ ; $\rho_k=(\Lambda^A_k\otimes\Lambda^B_k)\rho(\cdots)^\dagger/p_k$ — the post-measurement branch; $E$ — any candidate measure, also required to vanish on separable states. Reading: this "strong monotonicity" over stochastic outcomes is the only axiom the theory needs. For a pure state the Schmidt decomposition reduces everything to one spectrum, and its entropy is the entanglement — the entropy of the reduced state, not the global one, since a pure whole has zero entropy while its correlated parts do not.

|\psi\rangle_{AB}=\sum_{i=1}^{d}\sqrt{\lambda_i}\,|i\rangle_A|i\rangle_B,\qquad E(\psi)=S(\rho_A)=-\sum_i \lambda_i\log_2\lambda_i \qquad (2)

where: $\lambda_i$ — Schmidt coefficients, eigenvalues of $\rho_A=\rho_B$ with $\sum_i\lambda_i=1$ ; $\rho_A=\mathrm{Tr}_B|\psi\rangle\langle\psi|$ — Alice's reduced operator; $S(\rho)=-\mathrm{Tr}\,\rho\log_2\rho$ — von Neumann entropy. Reading: measured in ebits; a Bell state has flat spectrum $(\tfrac12,\tfrac12)$ and $E=1$ . Single-copy convertibility is governed not by which state has more entropy but by the finer relation of majorization on the Schmidt vectors — and the arrow points the "wrong" way, from the more-ordered spectrum to the less-ordered.

|\psi\rangle\xrightarrow{\text{LOCC}}|\phi\rangle \iff \lambda^\psi \prec \lambda^\phi \iff \sum_{i=1}^k\lambda^\psi_i\le\sum_{i=1}^k\lambda^\phi_i\ \ \forall k \qquad (3)

where: $\lambda^\psi,\lambda^\phi$ — Schmidt vectors of source and target, sorted decreasingly; $\prec$ — majorization. Reading: $\psi\to\phi$ works exactly when $\psi$ 's spectrum is majorized by $\phi$ 's (more spread out) — LOCC can only flatten the spectrum, and some pairs are incomparable. For mixed states the natural generalization averages the reduced-state entropy over the cheapest pure-state decomposition — the convex roof.

E_F(\rho)=\min_{\{p_i,|\psi_i\rangle\}}\ \sum_i p_i\,S\!\left(\mathrm{Tr}_B|\psi_i\rangle\langle\psi_i|\right),\qquad \rho=\sum_i p_i|\psi_i\rangle\langle\psi_i| \qquad (4)

where: $\{p_i,|\psi_i\rangle\}$ — a pure-state ensemble realizing $\rho$ ; $S(\mathrm{Tr}_B|\psi_i\rangle\langle\psi_i|)$ — entanglement of each member. Reading: pay for the average entanglement of the cheapest pure-state story about $\rho$ ; its regularization equals the operational cost $E_C$ . For two qubits that minimization has a closed form through Wootters' concurrence, turning an infinite-dimensional optimization into four eigenvalues.

C(\rho)=\max\{0,\ \lambda_1-\lambda_2-\lambda_3-\lambda_4\},\qquad E_F(\rho)=h\!\left(\tfrac{1+\sqrt{1-C^2}}{2}\right) \qquad (5)

where: $\lambda_1\ge\lambda_2\ge\lambda_3\ge\lambda_4$ — decreasingly ordered square roots of the eigenvalues of $\rho\,\tilde\rho$ , with spin-flip $\tilde\rho=(\sigma_y\otimes\sigma_y)\rho^*(\sigma_y\otimes\sigma_y)$ ; $h(x)=-x\log_2 x-(1-x)\log_2(1-x)$ — binary entropy; $C\in[0,1]$ — concurrence. Reading: this is the curve of the second figure, $C=1\to E_F=1$ , $C=0\to E_F=0$ . A geometric measure asks instead how far the state sits from the free set, via the relative entropy — the same distinguishability distance from quantum Shannon theory.

E_R(\rho)=\min_{\sigma\in\mathrm{SEP}}\ D(\rho\,\|\,\sigma),\qquad D(\rho\|\sigma)=\mathrm{Tr}\,\rho(\log_2\rho-\log_2\sigma) \qquad (6)

where: $\mathrm{SEP}$ — the convex set of separable states; $D(\rho\|\sigma)$ — quantum relative entropy. Reading: $E_R$ is the statistical "distance" to the nearest unentangled state; its LOCC-monotonicity comes straight from data processing (Derivation 3). Finally, squashed entanglement — built from the conditional mutual information — is additive and continuous, and nonnegative precisely because of strong subadditivity.

E_{sq}(\rho_{AB})=\tfrac12\inf_{\rho_{ABE}}\ I(A;B|E),\qquad I(A;B|E)=S(AE)+S(BE)-S(E)-S(ABE) \qquad (7)

where: $\rho_{ABE}$ — any extension of $\rho_{AB}$ ( $\mathrm{Tr}_E\rho_{ABE}=\rho_{AB}$ ), an environment "squashing" the correlations; $I(A;B|E)\ge0$ — conditional mutual information, nonnegative by strong subadditivity. Reading: minimize the correlation not explained by any $E$ ; the result is sandwiched as $E_D\le E_{sq}\le E_C$ . These measures are not independent; they line up in a fixed operational order that is the backbone of the subject.

E_D(\rho)\ \le\ E_{sq}(\rho)\ \le\ E_F(\rho),\qquad E_D(\rho)\ \le\ E_C(\rho)=\lim_{n\to\infty}\frac{E_F(\rho^{\otimes n})}{n} \qquad (8)

where: $E_D$ — distillable entanglement (asymptotic yield); $E_C$ — entanglement cost (regularized $E_F$ ); $E_{sq}$ — squashed entanglement. Reading: on pure states all coincide at $S(\rho_A)$ ; on mixed states they generically separate, and bound entanglement pins $E_D=0$ while the others stay positive.

Derivation APure states are reversible and the entropy of entanglement is unique

The goal is to prove that for a pure bipartite state the two operational extremes coincide, $E_D(\psi)=E_C(\psi)=S(\rho_A)$ , so the entire tower of monotones collapses to one number and the theory is asymptotically reversible. The strategy is a squeeze: a concentration protocol distilling at least $S(\rho_A)$ Bell pairs per copy, an inverse dilution protocol building each copy from at most $S(\rho_A)$ Bell pairs, and LOCC monotonicity to force yield and cost to meet exactly at $S(\rho_A)$ .

Step 1. Write $n$ copies in the Schmidt basis, $|\psi\rangle^{\otimes n}=\sum_{\vec i}\sqrt{\lambda_{i_1}\cdots\lambda_{i_n}}\,|\vec i\rangle_A|\vec i\rangle_B$ ; by the asymptotic equipartition property the joint Schmidt weight concentrates on a typical set of about $2^{nS(\rho_A)}$ strings of nearly equal weight. Why: a flat Schmidt spectrum on a subspace of known dimension is exactly a maximally entangled state.

Step 2. For distillation, Alice projects onto the typical subspace and phones Bob the outcome; conditioned on the typical result the shared state is within vanishing trace distance of a maximally entangled state of Schmidt rank $\approx 2^{nS(\rho_A)}$ , worth $nS(\rho_A)$ Bell pairs, so $E_D(\psi)\ge S(\rho_A)$ . Why: a local projection plus one classical bit is legitimate LOCC, and typicality makes both failure probability and error vanish.

Step 3. For dilution, run it backwards: from $nS(\rho_A)$ shared Bell pairs Alice Schumacher-compresses $|\psi\rangle^{\otimes n}$ into $\approx nS(\rho_A)$ qubits and teleports Bob's share to fidelity $\to1$ , so $E_C(\psi)\le S(\rho_A)$ . Why: teleportation spends one Bell pair per qubit, and compression makes that bill $nS(\rho_A)$ .

Step 4. Close the squeeze: monotonicity gives $E_D\le E_C$ , Steps 2–3 give $S(\rho_A)\le E_D$ and $E_C\le S(\rho_A)$ , so $S(\rho_A)\le E_D\le E_C\le S(\rho_A)$ forces equality. Why: distillation and dilution are exact asymptotic inverses — the pure-state theory is reversible, unlike the mixed case to come. *Verified: Nielsen & Chuang §12.5, Wilde Chs. 19–21; $\lambda_i=1/d$ gives $S=\log_2 d$ , and $d=2$ gives exactly 1 ebit.*

Go deeper · Derivation B · 1 minThe Wootters concurrence formula for two qubits

Derivation BThe Wootters concurrence formula for two qubits

The goal is to reduce the convex-roof optimization for the entanglement of formation of an arbitrary two-qubit state — in principle a minimization over infinitely many ensembles — to a formula in four eigenvalues. The route: for two qubits the pure-state entanglement is a monotone function of one invariant, the concurrence, and the spin-flip builds an optimal ensemble whose members all share the same concurrence, so the average collapses to a single evaluation of that universal function.

Step 1. For a two-qubit pure state define $C(\psi)=|\langle\psi|\tilde\psi\rangle|$ with spin flip $|\tilde\psi\rangle=(\sigma_y\otimes\sigma_y)|\psi^*\rangle$ ; its Schmidt eigenvalues are $\tfrac{1\pm\sqrt{1-C^2}}{2}$ , so $E_F(\psi)=h\big(\tfrac{1+\sqrt{1-C^2}}{2}\big)$ increases monotonically with $C$ . Why: this repackages the two Schmidt numbers into one variable $C$ , so entanglement is a function of a single quantity.

Step 2. For a mixed state let $\lambda_1\ge\lambda_2\ge\lambda_3\ge\lambda_4$ be the square roots of the eigenvalues of $\rho\tilde\rho$ , with $\tilde\rho=(\sigma_y\otimes\sigma_y)\rho^*(\sigma_y\otimes\sigma_y)$ . Why: the spin-flipped overlap generalizes to density operators through this matrix, whose spectrum encodes the concurrence structure of $\rho$ .

Step 3. Wootters proves an optimal decomposition exists in which every member has the same concurrence $C(\rho)=\max\{0,\lambda_1-\lambda_2-\lambda_3-\lambda_4\}$ ; since $E_F$ is convex in $C$ , this equal-concurrence ensemble achieves the roof. Why: convexity means spreading concurrences apart only raises the average, so the flattest ensemble is optimal — which is what makes the minimization closed-form.

Step 4. Substitute to get $E_F(\rho)=h\big(\tfrac{1+\sqrt{1-C(\rho)^2}}{2}\big)$ : a Bell state has $C=1\Rightarrow E_F=1$ , a separable state $C=0\Rightarrow E_F=0$ . Why: a single line now computes $E_F$ for every two-qubit state from four eigenvalues, no optimization required. *Verified: Wootters, PRL 80, 2245 (1998); the Werner state gives $C=\max\{0,(3p-1)/2\}$ , vanishing at $p=1/3$ and giving $C=0.7$ , $E_F=0.592$ ebits at $p=0.8$ .*

Go deeper · Derivation C · 1 minRelative entropy of entanglement bounds distillable entanglement

Derivation CRelative entropy of entanglement bounds distillable entanglement

The goal is to show that the geometric measure $E_R$ upper-bounds the operational distillable entanglement $E_D$ , so a quantity computed by minimizing a distance controls one defined by an asymptotic protocol. The argument rests on two facts from the information-theoretic backbone: relative entropy cannot increase under any channel (data processing), and LOCC maps send separable states to separable states. Combine these with the value of $E_R$ on a near-maximal output and take the limit.

Step 1. Suppose an LOCC map $\mathcal{L}_n$ distills $\rho^{\otimes n}$ into $\omega_n$ with $\|\omega_n-\Phi_{2^{m_n}}\|_1\to0$ at rate $m_n/n\to E_D(\rho)$ . Why: this is the definition of $E_D$ , so any bound on the output entanglement bounds $E_D$ .

Step 2. Since $\mathcal{L}_n$ maps separable to separable, $E_R(\rho^{\otimes n})\ge E_R(\mathcal{L}_n(\rho^{\otimes n}))=E_R(\omega_n)$ . Why: for the optimal $\sigma^\star$ , data processing gives $D(\rho^{\otimes n}\|\sigma^\star)\ge D(\mathcal{L}_n\rho^{\otimes n}\|\mathcal{L}_n\sigma^\star)$ with $\mathcal{L}_n\sigma^\star$ still separable, so $E_R$ falls under LOCC.

Step 3. On the near-maximal output $E_R(\Phi_{2^{m}})=m$ (nearest separable state is the maximally mixed diagonal), so asymptotic continuity gives $E_R(\omega_n)\ge m_n-n\epsilon_n$ , $\epsilon_n\to0$ . Why: continuity lets the imperfect $\omega_n$ stand in for the ideal $\Phi_{2^{m_n}}$ at a vanishing rate cost.

Step 4. Divide by $n$ : $E_R^\infty(\rho):=\lim_n E_R(\rho^{\otimes n})/n\ge E_D(\rho)$ , and subadditivity gives $E_R^\infty\le E_R$ , so $E_D\le E_R^\infty\le E_R$ . Why: the single-copy $E_R$ is a valid computable ceiling on how much you can ever distill. *Verified: Vedral & Plenio, PRA 57, 1619 (1998); on pure states $E_R(\psi)=S(\rho_A)=E_D(\psi)$ , tight — consistent with Derivation 1.*

Limiting cases

Case	Result	Meaning
Pure bipartite state	$E_D=E_C=E_F=E_R=E_{sq}=S(\rho_A)$	every sensible measure collapses to the entropy of entanglement; the theory is asymptotically reversible
Separable (free) state	$E(\rho_{\mathrm{sep}})=0$	no resource; every monotone vanishes and the state is preparable by LOCC alone
Maximally entangled qudit	$E(\Phi_d)=\log_2 d$	flat Schmidt spectrum $\lambda_i=1/d$ ; reduced state maximally mixed, $S=\log_2 d$ (a Bell pair gives 1 ebit, a qutrit pair 1.585)
Bound entangled (PPT) state	$E_D=0,\ E_C>0$	entangled yet undistillable — the sharpest face of irreversibility in entanglement theory
Two-qubit Werner, $p\le\tfrac13$	$C=0,\ E_F=0$	the isotropic mixture becomes separable below the threshold; above it, $C=(3p-1)/2$ and the state is distillable

Common trapthe misconception that entanglement is one reversible quantity like thermodynamic entropy

It is tempting to imagine entanglement as a single conserved "stuff" with one number, so whatever you pay to build a state you can always recover. For pure states that is exactly right: one number $S(\rho_A)$ , and $E_C=E_D$ . But mixing breaks it two ways. First, the single number splinters into a family of inequivalent monotones — $E_D$ , $E_C$ , $E_F$ , $E_R$ , $E_{sq}$ , negativity — that agree only on pure states, so "how much entanglement does $\rho$ have?" has no unique answer; you must say which measure and for what task. Second, the theory is irreversible: $E_D<E_C$ generically, and bound entangled states realize the extreme $E_D=0<E_C$ — you spend real Bell pairs to make them and cannot get one back. Entanglement is a resource, but a lossy one, with no first-law bookkeeping to make the loss go away.

Worked examples

1 · Entropy of entanglement, concurrence, and one-way convertibility (fully worked)

Take the partially entangled pure state $|\psi\rangle=\sqrt{0.8}\,|00\rangle+\sqrt{0.2}\,|11\rangle$ . Find its entropy of entanglement, cross-check via the concurrence formula, and decide whether a single copy can be turned into a Bell state by LOCC — and whether a Bell state can be turned into it.

Step 1. The state is already in Schmidt form with $\lambda_1=0.8,\ \lambda_2=0.2$ , so $E(\psi)=S(\rho_A)=-0.8\log_2 0.8-0.2\log_2 0.2=0.2575+0.4644=\mathbf{0.722\ ebits}$ .

Step 2. Cross-check with Wootters. The pure-state concurrence is $C=2\sqrt{\lambda_1\lambda_2}=2\sqrt{0.16}=\mathbf{0.8}$ , so $\sqrt{1-C^2}=0.6$ and $E_F=h(0.8)=\mathbf{0.722}$ — the two routes agree exactly, as they must for a pure state (the green point on the second figure).

Step 3. Convertibility to a Bell state needs $(0.8,0.2)\prec(0.5,0.5)$ ; the first partial sum $0.8\le0.5$ is false, so one copy of $|\psi\rangle$ cannot be turned into a Bell pair — exactly what monotonicity demands, since $0.722<1$ .

Step 4. The reverse, $|\Phi^+\rangle\to|\psi\rangle$ , needs $(0.5,0.5)\prec(0.8,0.2)$ : $0.5\le0.8$ and $1.0\le1.0$ both hold, so a Bell pair converts to $|\psi\rangle$ deterministically, spending entanglement down from $1$ to $0.722$ . The arrow runs only downhill: Bell → ψ yes, ψ → Bell no.

2 · Entanglement of a noisy Werner state (fully worked)

A source is meant to emit $|\Phi^+\rangle$ but depolarizing noise leaves the Werner state $\rho(p)=p\,|\Phi^+\rangle\langle\Phi^+|+(1-p)\,\tfrac14\mathbb{1}$ . At mixing $p=0.8$ , find the concurrence, the entanglement of formation, the separability threshold, and comment on distillability.

Step 1. The two-qubit Werner concurrence is $C(p)=\max\{0,(3p-1)/2\}$ (equation 5 for this symmetric $\rho\tilde\rho$ ). At $p=0.8$ : $C=(2.4-1)/2=\mathbf{0.7}$ .

Step 2. Then $\sqrt{1-C^2}=\sqrt{0.51}=0.7141$ , the binary-entropy argument is $\tfrac{1+0.7141}{2}=0.8571$ , and $E_F=h(0.8571)=0.1907+0.4012=\mathbf{0.592\ ebits}$ — the amber point on the second figure.

Step 3. Separability threshold: $C(p)>0$ iff $p>\tfrac13$ , so $\rho(p)$ is separable for $p\le1/3$ and entangled above it, rising from zero at $p=1/3$ to one full ebit at $p=1$ .

Step 4. A two-qubit entangled state is never bound — it is NPT whenever entangled, and two-qubit NPT states are distillable — so every $p>1/3$ has $E_D>0$ . At $p=0.8$ : $C=0.7$ , $E_F=0.592$ ebits, entangled and distillable; bound entanglement needs dimension beyond $2\times2$ .

3 · The maximally entangled qutrit pair (quick)

How many ebits does $|\Phi_3\rangle=\tfrac1{\sqrt3}(|00\rangle+|11\rangle+|22\rangle)$ carry? Its reduced state is $\rho_A=\tfrac13\mathbb{1}_3$ , so $E(\Phi_3)=S(\rho_A)=\log_2 3=\mathbf{1.585\ ebits}$ — more than a Bell pair, confirming a qudit pair is a denser resource.

Physics of the portrait

The resource theory did not emerge from foundational debates about nonlocality — it came from an operational question: if two distant experimenters share noisy quantum correlations, what can they do with them, and how much is a state worth? The person who did most to make entanglement a quantitative, spendable resource is Charles Bennett.

Origins · 1 minCharles Bennett and how this was discovered

Charles Bennett

born 1943 · quantum information theory and entanglement as a physical resource

Charles H. Bennett, a physicist and information theorist at IBM Research, is the architect of the view that entanglement is a currency. With Rolf Landauer he established that logically reversible computation can be thermodynamically free; in 1984, with Gilles Brassard, he invented BB84, the first quantum key distribution protocol. The decisive turn came in 1993, when Bennett and five co-authors introduced quantum teleportation: a shared Bell pair plus two classical bits transmits an unknown qubit exactly, quantifying for the first time that one ebit is a concrete, consumable unit. In the landmark 1996 papers with DiVincenzo, Smolin and Wootters he then founded the quantitative theory of mixed-state entanglement — distillation, the entanglement of formation, and the LOCC paradigm in which separable states are free and entanglement is what LOCC cannot create. That framework set the questions ever since: Wootters' 1998 concurrence solved $E_F$ for two qubits, the Horodeckis' 1998 bound entanglement exposed irreversibility, and Christandl and Winter's 2004 squashed entanglement tied it to strong subadditivity. Bennett's enduring move was conceptual: stop asking whether a state is entangled and start asking how much, for which task, at what exchange rate — turning a philosophical puzzle into an accounting discipline with theorems.

History

1935

— Schrödinger names entanglement the characteristic trait of quantum mechanics that enforces its entire departure from classical thought

1964

— Bell shows entangled correlations violate any local-hidden-variable bound, making entanglement empirically real

1989

— Werner defines mixed-state separability and the isotropic states that bear his name

1996

— Bennett, DiVincenzo, Smolin and Wootters found the quantitative theory of entanglement — distillation, cost, and entanglement of formation under LOCC

1998

— Wootters gives the closed-form concurrence for two-qubit entanglement of formation, and the Horodeckis exhibit bound entanglement

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, Ch. 12 (esp. §12.5) · Wilde, Quantum Information Theory, 2nd ed., Chs. 19–21 · Watrous, The Theory of Quantum Information, Ch. 6 · Preskill, Lecture Notes, Ch. 10 · Horodecki $^4$ , "Quantum entanglement," Rev. Mod. Phys. 81, 865 (2009); Plenio & Virmani, "An introduction to entanglement measures," QIC 7, 1 (2007).

Provenance. Structural claims follow Nielsen & Chuang Ch. 12 (majorization, entropy of entanglement), Wilde Chs. 19–21 (distillation, cost, $E_R$ , $E_{sq}$ ), Watrous Ch. 6, and Horodecki $^4$ RMP 81, 865 (2009). Founding papers recalled by name: Bennett, DiVincenzo, Smolin & Wootters, Phys. Rev. A 54, 3824 (1996) — entanglement of formation; Wootters, Phys. Rev. Lett. 80, 2245 (1998) — concurrence; Vedral & Plenio, Phys. Rev. A 57, 1619 (1998) — $E_R$ ; Christandl & Winter, J. Math. Phys. 45, 829 (2004) — squashed entanglement. A verifier should recompute the worked numbers $E(\psi)=S(\rho_A)=0.722$ , Werner $C=0.7$ with $E_F=0.592$ , and $\log_2 3=1.585$ , plus the majorization (Birkhoff/HLP), data-processing, and strong-subadditivity facts used above. No nonstandard numbers are asserted.

Prerequisites. Von Neumann Entropy · Quantum Relative Entropy · Quantum Channels and Completely Positive Trace-Preserving Maps · Data-Processing Inequality · Strong Subadditivity of Quantum Entropy. Leads to. Quantum Joint, Conditional Entropy and Mutual Information · Holevo Bound · Choi–Jamiołkowski Isomorphism.

Ask the tutor: want to walk through why a Bell pair dilutes into $\sqrt{0.8}|00\rangle+\sqrt{0.2}|11\rangle$ but never the reverse, or work a concurrence calculation on your own two-qubit state?

Why is the entropy of entanglement the entropy of the reduced state and not the whole state?

A pure whole hides all its structure in its parts. For $\rho_{AB}=|\psi\rangle\langle\psi|$ the global entropy $S(\rho_{AB})=0$ tells you nothing about correlations. The entanglement lives in how the halves are tied together, which is what the mixedness of the reduced state captures: a maximally mixed $\rho_A$ means Alice's half is maximally uncertain on its own precisely because it is maximally correlated with Bob's. By Schmidt, $\rho_A$ and $\rho_B$ share a spectrum, so $S(\rho_A)$ is the ebit content.

How can a state be entangled yet have zero distillable entanglement?

These are the bound entangled states — the deepest surprise of the theory. Such a state is genuinely entangled (not a mixture of product states, so $E_R$ and $E_{sq}$ are positive), yet no LOCC protocol can extract even one maximally entangled pair from any number of copies. The entanglement is "locked": you spend Bell pairs to create it ( $E_C>0$ ) but can never melt it back ( $E_D=0$ ). The known examples are PPT entangled states — positive under partial transpose yet non-separable — and their existence makes entanglement theory irreversible, unlike thermodynamics.

Why does the entanglement cost equal the regularized entanglement of formation rather than E_F itself?

Because $E_F$ is generally not additive. The cost $E_C$ is the asymptotic per-copy Bell-pair bill for making many copies together, and the cheapest way to make $\rho^{\otimes n}$ can beat $n$ independent copies — a joint ensemble correlates them. So $E_C=\lim_n E_F(\rho^{\otimes n})/n$ , which can sit strictly below the single-copy $E_F$ ; only when $E_F$ is additive, as on pure states, does regularization do nothing. Whether $E_F$ is additive in general was a major open problem, resolved negatively.

Why does pure-state convertibility use majorization instead of comparing entropies?

Entropy comparison is necessary but not sufficient for a single copy. Nielsen's theorem says $|\psi\rangle\to|\phi\rangle$ exactly when the Schmidt vector of $\psi$ is majorized by that of $\phi$ — a finer, vector-valued condition. Majorization is only a partial order: some pairs are incomparable, so neither converts into the other even if one has more entropy. Entropy governs only the asymptotic rate, where the partial order washes out into one number.

What makes squashed entanglement special among all the monotones?

It has the best structural pedigree: $E_{sq}=\tfrac12\inf_E I(A;B|E)$ is simultaneously an LOCC monotone, additive, continuous, and zero exactly on separable states — a combination no other computable measure achieves. Its nonnegativity is inherited from strong subadditivity, since $I(A;B|E)\ge0$ is that theorem. This makes it the natural tool for converse bounds: it sandwiches between $E_D$ and $E_C$ and upper-bounds secret-key and quantum capacities. The catch: the infimum runs over all extensions $E$ of unbounded dimension, so it is extremely hard to compute.

Is negativity an entanglement measure too, and how does it fit in?

Yes — negativity and its logarithmic cousin are legitimate and, crucially, computable monotones. The negativity sums the absolute values of the negative eigenvalues of the partial transpose $\rho^{T_B}$ , quantifying how much a state violates PPT; you diagonalize just one matrix, which is why experimentalists favour it. But seeing only the partial-transpose signature, it assigns zero to PPT bound entangled states — so it detects distillable NPT entanglement well but is not faithful: nonzero certifies entanglement, zero does not certify separability.