Entanglement Entropy of Pure States

Entanglement is the resource that separates quantum information from classical, and a resource is only manageable once it can be counted. For a pure state of a bipartite system $\mathcal H_A\otimes\mathcal H_B$ the counting is essentially unique and remarkably clean: throw away one half, look at the mixedness left behind, and measure that mixedness with the von Neumann entropy. The number that comes out is the entanglement entropy, the central scalar of pure-state entanglement theory and the fixed point against which every harder measure is calibrated. The whole construction rests on the Schmidt decomposition. Any bipartite pure state can be written $|\psi\rangle=\sum_i\sqrt{p_i}\,|i\rangle_A|i\rangle_B$ with $p_i\ge0$ and $\sum_i p_i=1$ , so the reduced state $\rho_A=\mathrm{Tr}_B|\psi\rangle\langle\psi|$ is diagonal with eigenvalues $\{p_i\}$ — and so, with the same eigenvalues, is $\rho_B$ . The entanglement entropy $S(\rho_A)=-\sum_i p_i\log_2 p_i$ is therefore just the Shannon entropy of the Schmidt coefficients: zero for a product state (one coefficient equal to one), maximal $\log_2 d$ when the weight spreads evenly over $d$ terms. What lifts this above an arbitrary functional is its operational meaning. Under local operations and classical communication (LOCC) two parties can never manufacture entanglement, only reshape it, and in the many-copy limit the reversible rate at which copies of $|\psi\rangle$ convert to and from Bell pairs is exactly $S(\rho_A)$ ebits per copy. That is the Bennett–Bernstein–Popescu–Schumacher theorem, and it makes $S(\rho_A)$ the canonical amount of entanglement, precisely as Shannon entropy is the canonical amount of information. On pure states every reasonable measure — entanglement of formation, distillable entanglement, the entropy of either subsystem — collapses onto this one number; the richness of the mixed-state theory that follows exists only because that collapse fails.

Key takeaways

For a bipartite pure state $|\psi\rangle_{AB}$ the entanglement entropy is the von Neumann entropy of either reduced density matrix, $S(\rho_A)=-\mathrm{Tr}(\rho_A\log_2\rho_A)$ .
The Schmidt decomposition gives $\rho_A$ and $\rho_B$ identical nonzero eigenvalues, so $S(\rho_A)=S(\rho_B)$ : entanglement is a property of the cut, shared symmetrically, never localized on the larger side.
It vanishes if and only if the state is a product (Schmidt rank 1) and reaches its maximum $\log_2 d$ for a maximally entangled state of two $d$ -level systems.
It is the unique pure-state entanglement measure: it equals the asymptotic LOCC rate of Bell-pair distillation and dilution, so it literally counts ebits.
The Rényi entropies $S_\alpha$ interpolate around it and are easier to measure ( $S_2$ via swap tests), but only $S=S_1$ carries the ebit interpretation.
Key idea: trace out one half and take the von Neumann entropy of what remains — $S(\rho_A)=-\mathrm{Tr}(\rho_A\log_2\rho_A)$ is the Shannon entropy of the Schmidt coefficients, equal on both sides, zero only for product states, and operationally the ebit content of the pure state.

Where this lives today

Entanglement entropy is now a working instrument across three very different frontiers. In many-body physics it is the diagnostic of choice: cutting a lattice into a region $A$ and its complement, gapped local Hamiltonians obey an area law $S\sim|\partial A|$ , while critical one-dimensional systems show the logarithmic violation $S\sim\tfrac{c}{3}\log L$ that reads off the central charge $c$ of the underlying conformal field theory. This scaling is not academic — it is exactly what bounds the bond dimension a matrix-product-state or DMRG simulation needs, so the entanglement entropy sets the practical cost of classically simulating a quantum material. In quantum experiment, the Rényi-2 entropy of equation (6) is measured directly: by preparing two copies of a state and performing a controlled-SWAP (or, more scalably, randomized single-qubit measurements), groups have extracted $S_2$ of few-site subsystems in ultracold-atom lattices (Islam and coworkers, 2015) and in trapped-ion and superconducting registers (Brydges and coworkers, 2019), turning entanglement entropy from a theorist's quantity into a laboratory observable. And in quantum gravity the same bipartite-cut entropy drives the Page curve of an evaporating black hole — radiation entropy rising then falling to preserve unitarity — with the Ryu–Takayanagi formula equating the entanglement entropy of a boundary region to the area of a minimal surface in the bulk, making $S(\rho_A)$ a bridge between information theory and spacetime geometry. The open frontier is the mixed-state analogue: no single entropy plays this role once $\rho_{AB}$ is mixed, which is what motivates the entanglement-of-formation and negativity measures this article leads into.

The physical picture

The paradox that entanglement entropy resolves is this: a bipartite pure state is perfectly known — its global density matrix $|\psi\rangle\langle\psi|$ has one eigenvalue equal to one and all the rest zero, so its global entropy is exactly zero — yet the part you can locally touch looks random. Trace away the far half and the reduced state $\rho_A$ is generically mixed. The amount of that local randomness, measured in bits, is exactly the entanglement, because the information missing from $A$ has not been destroyed; it lives in the correlations between $A$ and $B$ . Entanglement entropy is best read as "how surprised half the system is about its own state."

Figure

Tracing out one half of a Bell pair leaves a maximally mixed qubit — the global state is pure with S_global = 0, yet ρ_A has eigenvalues ½ and ½ giving S(ρ_A) = S(ρ_B) = 1 ebit, so the local randomness is exactly the entanglement across the cut

The Schmidt picture makes it quantitative. A bipartite pure state is a weighted superposition of perfectly aligned pairs $|i\rangle_A|i\rangle_B$ ; the weights $p_i$ are a probability distribution, and the entropy is theirs. Put all the weight on a single pair and the halves are independent — a product state, zero entropy. Spread it evenly over $d$ pairs and learning $A$ 's label pins down $B$ 's perfectly — maximal correlation, maximal entropy $\log_2 d$ . The surprising symmetry $S(\rho_A)=S(\rho_B)$ is a corollary: a lone qubit entangled with an enormous environment carries the same entropy as that whole environment carries about the qubit, because both reductions share the identical Schmidt spectrum. For a two-term state everything collapses onto a single knob $p$ , and the entropy traces out the binary-entropy curve below — the working shape of the whole subject.

Figure

The binary entropy S = H₂(p) = −p log₂ p − (1−p) log₂(1−p) for a two-term state: zero at the product limits p = 0 and p = 1, rising to one full ebit S = 1 at the Bell point p = ½, and passing through S = 0.469 ebits at p = 0.9

The core equations

The definition is a two-step recipe: reduce, then take the entropy. Everything else is a consequence of these two moves. The entropy is measured in bits — with the base-2 logarithm one Bell pair carries exactly one unit, the ebit.

S(\rho_A)=-\mathrm{Tr}\big(\rho_A\log_2\rho_A\big)=-\sum_i \lambda_i\log_2\lambda_i \qquad (1)

where: $\rho_A=\mathrm{Tr}_B|\psi\rangle\langle\psi|$ — reduced density matrix of subsystem $A$ ; $\lambda_i$ — eigenvalues of $\rho_A$ , which form a probability distribution ( $\lambda_i\ge0$ , $\sum_i\lambda_i=1$ ); $\log_2$ — base-2 logarithm, so $S$ is in bits/ebits. Reading it: the entropy is a functional of the eigenvalues of $\rho_A$ alone, blind to the basis, so any unitary on $A$ leaves it untouched. The reason $\rho_A$ has such a clean spectrum is the Schmidt decomposition, which every bipartite pure state admits. It writes the state as a single sum over aligned orthonormal pairs, with as many terms as the Schmidt rank $r$ .

|\psi\rangle_{AB}=\sum_{i=1}^{r}\sqrt{p_i}\,|i\rangle_A\otimes|i\rangle_B \qquad (2)

where: $r$ — Schmidt rank, $1\le r\le\min(d_A,d_B)$ ; $p_i$ — squared Schmidt coefficients, a probability distribution; $|i\rangle_A,|i\rangle_B$ — orthonormal Schmidt bases of $A$ and $B$ . Reading it: the single index $i$ correlating the two halves is the entire content of the entanglement; $r=1$ is a product state. Because both reductions inherit the same weights $p_i$ , the entropy of the cut is symmetric — the single most useful structural fact about pure-state entanglement.

S(\rho_A)=S(\rho_B)=-\sum_{i=1}^{r}p_i\log_2 p_i \qquad (3)

where: $p_i$ — the shared nonzero eigenvalues of both reductions. Reading it: entanglement belongs to the partition, not to one subsystem; the extra dimensions of a larger side contribute only zero eigenvalues, which do not touch the entropy. The entropy is pinned between two extremes by nonnegativity of Shannon entropy and by the rank ceiling.

0\le S(\rho_A)\le \log_2\min(d_A,d_B) \qquad (4)

where: $d_A,d_B$ — Hilbert-space dimensions of $A$ and $B$ . Reading it: the floor $S=0$ is the product state; the ceiling $\log_2\min(d_A,d_B)$ is the maximally entangled state, saturated only by a flat Schmidt spectrum. The number is not merely a convenient functional; it is the pure-state measure, because it equals the reversible LOCC exchange rate with the standard currency of Bell pairs.

E(|\psi\rangle)=S(\rho_A)=\text{(distillation rate)}=\text{(dilution rate)} \qquad (5)

where: $E$ — asymptotic LOCC-interconversion rate with Bell pairs, in ebits per copy. Reading it: distillation (extracting Bell pairs) and dilution (spending them) both run at the same rate $S$ , so the conversion is asymptotically lossless — the signature of a unique measure. A one-parameter family, the Rényi entropies, surrounds $S$ and reduces to it in a limit; the members with integer $\alpha$ are what experiments actually reach.

S_\alpha(\rho_A)=\frac{1}{1-\alpha}\log_2\mathrm{Tr}(\rho_A^{\alpha}),\qquad \lim_{\alpha\to1}S_\alpha=S \qquad (6)

where: $\alpha>0$ — Rényi order; $\mathrm{Tr}(\rho_A^\alpha)=\sum_i\lambda_i^\alpha$ — the $\alpha$ -th moment of the Schmidt spectrum. Reading it: $S_\alpha$ decreases in $\alpha$ , so $S_2\le S_1=S$ ; the $\alpha=2$ member is fixed by the purity $\mathrm{Tr}(\rho_A^2)$ and is measurable without full tomography. For the ubiquitous two-term (two-qubit) state, all of the above collapses to the binary entropy function graphed above.

S(\rho_A)=H_2(p)\equiv -p\log_2 p-(1-p)\log_2(1-p) \qquad (7)

where: $H_2$ — binary entropy function; $p$ — larger squared Schmidt coefficient, $p=\cos^2\theta$ for $|\psi\rangle=\cos\theta\,|00\rangle+\sin\theta\,|11\rangle$ . Reading it: one knob $\theta$ sweeps the state from product ( $\theta=0$ ) through Bell ( $\theta=\pi/4$ , $S=1$ ) back to product ( $\theta=\pi/2$ ).

Derivation AReduced state diagonalizes in the Schmidt basis

The goal is to prove the workhorse identity of the whole subject: that tracing out $B$ from the Schmidt form produces a reduced matrix $\rho_A$ which is already diagonal, whose eigenvalues are precisely the squared Schmidt coefficients $p_i$ , so that the entanglement entropy is nothing more than the Shannon entropy of those coefficients. We start from the Schmidt decomposition, form the global projector, carry out the partial trace term by term, and let orthonormality do the work of collapsing the double sum to a single one.

Step 1. Form the global projector from the Schmidt form, $|\psi\rangle\langle\psi|=\sum_{i,j}\sqrt{p_i p_j}\,|i\rangle_A\langle j|_A\otimes|i\rangle_B\langle j|_B$ , a double sum over Schmidt indices $i$ and $j$ . Why: the partial trace acts only on the $B$ factor, so we need the state written with the $A$ and $B$ operators cleanly separated before tracing.

Step 2. Take the partial trace over $B$ by sandwiching with the orthonormal Schmidt basis, $\rho_A=\sum_k\langle k|_B\big(|\psi\rangle\langle\psi|\big)|k\rangle_B$ . The $B$ -matrix element is $\langle k|i\rangle_B\langle j|k\rangle_B=\delta_{ki}\delta_{jk}$ , which forces $i=j=k$ . Why: orthonormality of the Schmidt basis annihilates every off-diagonal ( $i\neq j$ ) term, and this is exactly why the Schmidt basis is special — it is the basis in which the reduction is automatically diagonal.

Step 3. Collect the survivors: $\rho_A=\sum_i p_i\,|i\rangle_A\langle i|_A$ . The reduced state is diagonal with the squared Schmidt coefficients as its eigenvalues. Why: no diagonalization step is needed — the Schmidt decomposition has already diagonalized both reductions simultaneously, which is what makes $S(\rho_A)=S(\rho_B)$ automatic.

Step 4. Evaluate the von Neumann entropy of a diagonal operator, which is just the Shannon entropy of its diagonal: $S(\rho_A)=-\mathrm{Tr}(\rho_A\log_2\rho_A)=-\sum_i p_i\log_2 p_i$ . Why: this closes the loop — the abstract operator entropy becomes an ordinary sum over a classical probability distribution, ready to compute by hand. *For example, for $|\Phi^+\rangle$ the coefficients are $p_1=p_2=\tfrac12$ , giving $S=-2\cdot\tfrac12\log_2\tfrac12=1$ ebit — exactly the Bell-pair unit.*

Go deeper · Derivation B · 1 minEntanglement concentration distills Bell pairs at rate S

Derivation BEntanglement concentration distills Bell pairs at rate S

The goal is to establish the operational meaning of the entropy — that $n$ identical copies of a partially entangled $|\psi\rangle$ can be converted by local operations and classical communication into roughly $nS(\rho_A)$ near-perfect Bell pairs, and back again at the same rate, so that $S$ is not an arbitrary functional but the literal ebit content of the state. The engine is the asymptotic equipartition property applied to the Schmidt distribution: for many copies, almost all the probability weight concentrates on a nearly uniform "typical" set whose size is set by the entropy.

Step 1. Tensor $n$ copies in the Schmidt basis, $|\psi\rangle^{\otimes n}=\sum_{\vec i}\sqrt{p_{i_1}\cdots p_{i_n}}\,|\vec i\rangle_A|\vec i\rangle_B$ , indexed by strings $\vec i=(i_1,\dots,i_n)$ whose weight is a product of independent Schmidt distributions. Why: independence across copies is what lets the law of large numbers act on the log-weights, turning a lopsided single-copy distribution into a nearly flat many-copy one.

Step 2. Invoke typicality: with probability approaching one, the string $\vec i$ is $\varepsilon$ -typical, meaning its weight satisfies $p_{\vec i}\approx 2^{-nS(\rho_A)}$ , and there are about $2^{nS(\rho_A)}$ such strings. Why: the entropy is precisely the exponential rate at which the typical set grows, so it directly counts how many nearly-equal-weight branches the many-copy state effectively occupies.

Step 3. Have each party locally project onto its typical set, $\Pi^{\rm typ}_A\otimes\Pi^{\rm typ}_B\,|\psi\rangle^{\otimes n}\approx 2^{-nS/2}\sum_{\vec i\ \rm typ}|\vec i\rangle_A|\vec i\rangle_B$ , which succeeds with probability near one and leaves a state that is nearly maximally entangled across a dimension $2^{nS}$ . Why: a projection on each side is a local operation, so it is free under LOCC, and it flattens the residual weights onto the uniform superposition characteristic of maximal entanglement.

Step 4. Identify the residue with Bell pairs: a maximally entangled state of dimension $2^m$ is local-unitary equivalent to $|\Phi^+\rangle^{\otimes m}$ with $m=nS(\rho_A)$ , so the rate is $m/n\to S(\rho_A)$ . Dilution reverses the process at the same cost plus negligible classical communication. Why: matching rates in both directions is the definition of asymptotic reversibility, which is what certifies $S$ as THE unique pure-state entanglement measure rather than one of many.

Go deeper · Derivation C · 1 minThe Rényi family and its von Neumann limit

Derivation CThe Rényi family and its von Neumann limit

The goal is to show that the von Neumann entropy is not an isolated definition but the $\alpha\to1$ member of a smooth one-parameter family of Rényi entropies $S_\alpha$ , and to derive that limit explicitly — a fact of real practical weight, because the integer-order members $S_2, S_3,\dots$ are polynomial in $\rho_A$ and therefore accessible to experiment through moments of the state, while the operationally meaningful $S=S_1$ sits at a point where the defining formula is naively $0/0$ . We expand around $\alpha=1$ and apply L'Hôpital's rule.

Step 1. At $\alpha=1$ the trace moment is $\mathrm{Tr}(\rho_A)=\sum_i\lambda_i=1$ , so $\log_2\mathrm{Tr}(\rho_A^\alpha)\to0$ while the prefactor $1/(1-\alpha)\to\infty$ : the expression $S_\alpha=\frac{1}{1-\alpha}\log_2\mathrm{Tr}(\rho_A^\alpha)$ is indeterminate of type $0/0$ once written as a ratio. Why: this is exactly the removable singularity that lets a single limit stitch the whole family together at the physically distinguished point $\alpha=1$ .

Step 2. Differentiate numerator and denominator in $\alpha$ . Using $\partial_\alpha\sum_i\lambda_i^\alpha=\sum_i\lambda_i^\alpha\ln\lambda_i$ , L'Hôpital gives $\lim_{\alpha\to1}S_\alpha=\frac{1}{\ln2}\cdot\frac{\sum_i\lambda_i\ln\lambda_i}{\sum_i\lambda_i}=-\sum_i\lambda_i\log_2\lambda_i=S$ . Why: the limit collapses precisely onto equation (1), proving the von Neumann entropy is the analytic continuation of the Rényi family and inherits its monotonicity, so $S_2\le S$ always. *For instance, for the spectrum $\{0.9,0.1\}$ , $S_2=-\log_2(0.9^2+0.1^2)=-\log_2 0.82=0.286$ , which is indeed below $S=H_2(0.9)=0.469$ , confirming that $S_\alpha$ decreases in $\alpha$ .*

Limiting cases

How the entropy behaves at the extremes and the useful special cases of the Schmidt spectrum:

Case	Result	Meaning
Product state	$\vert\psi\rangle=\vert a\rangle_A\otimes\vert b\rangle_B\Rightarrow S=0$	Single Schmidt coefficient $p_1=1$ ; $\rho_A$ is pure. Zero entanglement — the definition of separability for pure states.
Bell state	$\vert\Phi^+\rangle=\tfrac{1}{\sqrt2}(\vert00\rangle+\vert11\rangle)\Rightarrow S=1$	Flat two-term spectrum $\{\tfrac12,\tfrac12\}$ ; $\rho_A=\mathbb 1/2$ . The unit of entanglement — one ebit, the calibration point of every bipartite measure.
Maximally entangled qudits	$\vert\psi\rangle=\tfrac{1}{\sqrt d}\sum_{i=1}^{d}\vert i\rangle_A\vert i\rangle_B\Rightarrow S=\log_2 d$	Flat $d$ -term spectrum; $\rho_A=\mathbb 1/d$ . Saturates the upper bound; $d=2$ recovers the Bell state, $d=3$ gives $\log_2 3=1.585$ ebits.
Two-qubit family	$\vert\psi\rangle=\cos\theta\vert00\rangle+\sin\theta\vert11\rangle\Rightarrow S=H_2(\cos^2\theta)$	Interpolates smoothly: $S=0$ at $\theta=0,\pi/2$ and $S=1$ at $\theta=\pi/4$ , tracing the binary-entropy curve of the figure.
Many copies	$E(\vert\psi\rangle^{\otimes n})=nS+O(\sqrt n)$	Additive to leading order; the $O(\sqrt n)$ finite-size correction is why exact reversibility is an asymptotic statement.

Common trapReduced-state entropy always measures entanglement

It measures entanglement only when the global state is pure. For a mixed bipartite state $\rho_{AB}$ , the reduced entropy $S(\rho_A)$ tallies total correlation — classical plus quantum — not entanglement. The cleanest counterexample is the separable classical mixture $\rho_{AB}=\tfrac12(|00\rangle\langle00|+|11\rangle\langle11|)$ : it contains exactly zero entanglement (it is a mixture of product states, preparable by local coin flips and a phone call), yet $S(\rho_A)=1$ . Applying equation (1) to a mixed global state and reporting the answer as "entanglement" is the single most common error in the subject. For mixed states you need entanglement of formation, distillable entanglement, or the PPT criterion — measures that all coincide with $S(\rho_A)$ on pure states and diverge from it, and from one another, once the global state is mixed.

Worked examples

1 · A partially entangled qubit pair, in full (confidence builder)

A source emits the state $|\psi\rangle=\sqrt{0.9}\,|00\rangle+\sqrt{0.1}\,|11\rangle$ . Find $\rho_A$ , its von Neumann entropy, and the Rényi-2 entropy, and say what the entropy means operationally.

Step 1. The state is already in Schmidt form with squared coefficients $p_1=0.9$ , $p_2=0.1$ , so tracing out $B$ (Derivation 1) gives the diagonal reduced state $\rho_A=0.9\,|0\rangle\langle0|+0.1\,|1\rangle\langle1|$ , i.e. $\rho_A=\mathrm{diag}(0.9,0.1)$ . No matrix algebra is needed because the Schmidt basis is the eigenbasis.

Step 2. Apply the binary entropy, equation (7): $S=-0.9\log_2 0.9-0.1\log_2 0.1$ . Numerically $-0.9\times(-0.152)=0.137$ and $-0.1\times(-3.322)=0.332$ , so $S=H_2(0.9)=\mathbf{0.469\ \text{ebits}}$ . It is well below one because the state leans heavily toward the product configuration $|00\rangle$ .

Step 3. The Rényi-2 entropy uses the purity $\mathrm{Tr}(\rho_A^2)=0.9^2+0.1^2=0.82$ , so $S_2=-\log_2 0.82=\mathbf{0.286\ \text{ebits}}$ . As guaranteed by monotonicity, $S_2<S$ ; the gap is the price of using an easy-to-measure moment instead of the true ebit content.

Step 4. Operationally, entanglement concentration (Derivation 2) says $1000$ such copies can be converted by LOCC into about $1000\times0.469\approx\mathbf{469}$ near-perfect Bell pairs, and those Bell pairs can be spent on teleportation or superdense coding, or diluted back into copies of $|\psi\rangle$ at the same rate. The state is worth 0.469 ebits per copy.

2 · A maximally entangled qutrit pair

Two parties share $|\psi\rangle=\tfrac{1}{\sqrt3}\big(|00\rangle+|11\rangle+|22\rangle\big)$ of two qutrits ( $d=3$ ). Find the entanglement entropy and compare it to the maximum a qutrit pair can hold and to a Bell pair.

Step 1. By symmetry the reduced state is $\rho_A=\tfrac13(|0\rangle\langle0|+|1\rangle\langle1|+|2\rangle\langle2|)=\mathbb 1_3/3$ , the maximally mixed qutrit with flat spectrum $\{\tfrac13,\tfrac13,\tfrac13\}$ .

Step 2. The entropy of a uniform distribution over three outcomes is $S=-3\times\tfrac13\log_2\tfrac13=\log_2 3=\mathbf{1.585\ \text{ebits}}$ , which saturates the bound $\log_2\min(d_A,d_B)=\log_2 3$ from equation (4).

Step 3. This exceeds one ebit: a single maximally entangled qutrit pair carries more entanglement than one Bell pair (by a factor $\log_2 3\approx1.585$ ), which is exactly why higher-dimensional entanglement is attractive for boosting channel capacities — one qudit pair does the work of $\log_2 d$ qubit pairs. S = log₂3 = 1.585 ebits, the qutrit maximum.

Physics of the portrait

Entanglement entropy is where three strands of twentieth-century physics braid together: Schmidt's linear algebra, von Neumann's statistical entropy of the density operator, and Schrödinger's realization that the whole can be sharp while its parts stay indefinite. The single name most bound to the formula is von Neumann's.

Origins · 1 minJohn von Neumann and how this was discovered

John von Neumann

1903–1957 · mathematical foundations of quantum mechanics and quantum statistics

John von Neumann built the mathematical scaffolding on which entanglement entropy stands. In his 1927 papers on quantum statistics and in the 1932 treatise Mathematische Grundlagen der Quantenmechanik, he introduced the density operator $\rho$ to describe statistical ensembles of quantum states and defined its entropy $S(\rho)=-\mathrm{Tr}(\rho\ln\rho)$ — the quantum analogue of the Gibbs and, later, Shannon entropies. He proved its key properties: it is nonnegative, vanishes exactly for pure states, is invariant under unitary evolution, and is concave. That the entropy of a reduced state of a globally pure system measures entanglement was not his emphasis — the resource view of entanglement is a product of the 1990s quantum-information era — but every piece of the machinery is his. The story around him is characteristically broad: the same person laid the axioms of quantum measurement, co-founded game theory, designed the stored-program computer architecture that bears his name, and worked on the Manhattan Project. Erwin Schrödinger supplied the conceptual other half in 1935, coining the word Verschränkung (entanglement) and stressing that a maximally entangled pair can be completely specified as a whole while neither part has a definite state of its own — precisely the situation in which $S(\rho_A)>0$ despite $S(\rho_{AB})=0$ . The operational capstone came decades later, when Bennett, Bernstein, Popescu and Schumacher showed in 1996 that von Neumann's reduced-state entropy is the reversible rate of Bell-pair interconversion, elevating a statistical quantity into the definitive currency of pure-state entanglement.

History

1907

— Erhard Schmidt proves the decomposition of any bipartite vector into a sum over aligned orthonormal pairs

1927

— von Neumann defines the entropy S = −Tr(ρ log ρ) of a density operator

1935

— Schrödinger coins "entanglement" and notes that the whole can be known while the parts cannot

1996

— Bennett, Bernstein, Popescu and Schumacher prove S is the reversible LOCC rate of Bell-pair interconversion

2003

— Vidal, Latorre, Rico and Kitaev sharpen entanglement-entropy scaling and area laws in many-body ground states

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information (10th anniv. ed.), Ch. 11–12 (entropy and entanglement) · Preskill, Lecture Notes for Physics 219, Ch. 4 & 10 · Wilde, Quantum Information Theory (2nd ed.), Ch. 11 & 19 (entropies and entanglement manipulation) · Watrous, The Theory of Quantum Information, Ch. 5–6 · Bennett, Bernstein, Popescu & Schumacher, Phys. Rev. A 53, 2046 (1996). Key anchor values to keep in mind: the partial trace yields $\rho_A=\sum_i p_i|i\rangle\langle i|$ ; the calibrations $S(\Phi^+)=1$ and $S=\log_2 d$ for the maximally entangled qudit pair (with $\log_2 3=1.585$ ); the binary-entropy values $H_2(0.5)=1$ and $H_2(0.9)=0.469$ ; the Rényi-2 value $S_2=-\log_2 0.82=0.286$ and the ordering $S_2<S$ ; and the $\alpha\to1$ L'Hôpital limit recovering the von Neumann entropy. The operational uniqueness (distillation rate $=$ dilution rate $=S$ ) is the Bennett–Bernstein–Popescu–Schumacher theorem.

Prerequisites. Schmidt Decomposition · Separable versus Entangled States · Bell States and EPR Pairs. Leads to. Entanglement of Formation · Concurrence and Wootters' Formula · PPT Criterion and the Peres–Horodecki Test.

Ask the tutor: Give me any bipartite pure state and I will pull out its Schmidt spectrum, compute its entanglement entropy in ebits, and tell you how many Bell pairs it is worth.

Why does the same entropy come out whether I trace over A or over B?

Because the Schmidt decomposition diagonalizes both reductions at once with the same list of nonzero eigenvalues $\{p_i\}$ . The larger subsystem simply carries extra dimensions with eigenvalue zero, and a zero eigenvalue contributes $0\log_2 0=0$ to the entropy. Since the von Neumann entropy is a function of eigenvalues alone, $S(\rho_A)$ and $S(\rho_B)$ must agree exactly. This is the mathematical content of the statement that entanglement is a property of the cut between the two halves, shared symmetrically, and never sitting more on the bigger side than the smaller one.

Is entanglement entropy the same thing as von Neumann entropy?

They are the same formula applied in a special situation. The von Neumann entropy $S(\rho)=-\mathrm{Tr}(\rho\log_2\rho)$ is defined for any density operator and measures its mixedness. It becomes the entanglement entropy only when $\rho$ is the reduced state of a bipartite system whose global state is pure — then, and only then, the mixedness of the part quantifies the entanglement across the cut. Feed the von Neumann formula a globally mixed state and it still returns a number, but that number now blends classical and quantum correlations and is no longer an entanglement measure.

What exactly is an ebit?

An ebit is one unit of bipartite entanglement, defined as the entanglement contained in one maximally entangled two-qubit state — a Bell pair such as $(|00\rangle+|11\rangle)/\sqrt2$ , for which $S(\rho_A)=1$ . It is the entanglement analogue of the bit. Saying a state $|\psi\rangle$ "has $0.469$ ebits" means that in the many-copy limit you can reversibly trade it for Bell pairs at a rate of $0.469$ pairs per copy by local operations and classical communication. The ebit is the currency; entanglement entropy is the exchange rate.

Can the entanglement entropy exceed one?

Yes, whenever the subsystems are larger than qubits. The ceiling is $\log_2\min(d_A,d_B)$ , so two qutrits ( $d=3$ ) can reach $\log_2 3\approx1.585$ ebits, two ququarts ( $d=4$ ) can reach $2$ ebits, and so on. The value one is special only because it belongs to the maximally entangled state of the smallest nontrivial case, a pair of qubits. Larger local dimensions simply have more Schmidt terms over which the weight can spread, and a flatter, wider distribution carries more entropy.

Why do experiments measure the Rényi-2 entropy instead of the von Neumann entropy?

Because $S_2=-\log_2\mathrm{Tr}(\rho_A^2)$ depends only on the purity $\mathrm{Tr}(\rho_A^2)$ , and the purity is the expectation value of the SWAP operator acting on two identical copies of the state. That means a single interference measurement on two copies — a controlled-SWAP test, or a set of randomized local measurements — yields $S_2$ without ever reconstructing the full density matrix, whereas the von Neumann entropy $S=S_1$ needs the entire eigenvalue spectrum and hence full tomography, which scales exponentially. Since the Rényi entropies bracket and reduce to $S$ , measuring $S_2$ gives a rigorous, experimentally cheap lower bound on the true entanglement.

Does a large entanglement entropy make a state more useful?

For pure states, more entanglement entropy does mean more of the LOCC-interconvertible resource — more Bell pairs per copy for teleportation, dense coding, or key distribution — so in that narrow accounting sense, higher is richer. But usefulness is task-dependent: some protocols want exactly one ebit, and a maximally entangled state is actually a poor input for certain measurement-based schemes that need structured, not maximal, entanglement. And the whole equivalence breaks for mixed states, where a high reduced entropy can reflect classical noise rather than any usable entanglement at all. Entanglement entropy answers "how many ebits," which is the right question only when ebits are what the task consumes.