Entanglement Entropy and Measures of Entanglement

"How entangled is this state?" deserves a number, not an adjective. For a pure bipartite state the answer is clean and unique: the entanglement entropy, defined as the von Neumann entropy $S(\rho_A)=-\mathrm{Tr}\,\rho_A\ln\rho_A$ of either reduced state. It is $0$ for a product state, $\ln 2$ for a Bell pair, and — crucially — the same whether you compute it from $A$ or from $B$ , because the Schmidt decomposition guarantees the two reduced states share a spectrum. This article proves that equality, builds the von Neumann entropy from the ground up as the quantum Shannon entropy, and turns the Schmidt weights $\{\lambda_k\}$ into the single dial $-\sum_k\lambda_k\ln\lambda_k$ that ranges from product to maximal. The sharpest lesson lives in one inequality: for an entangled pure pair, $S(\rho_{AB})=0$ while $S(\rho_A)=\ln2>0$ . The whole is in a definite (zero-entropy) state while its parts are uncertain — a flat contradiction of the classical rule that a joint distribution is at least as uncertain as its marginals. That is "the whole is more certain than its parts," made numerical, and it is the deepest single fact about entanglement. Pure states are the easy case. For mixed bipartite states one number no longer suffices, and we survey — honestly, without overclaiming — the zoo of measures the field actually uses: entanglement of formation, distillable entanglement, the negativity built from the partial-transpose (Peres–Horodecki) criterion, and Wootters' concurrence for two qubits. We define them, compute the negativity and concurrence of a Werner state (with its exact $p>1/3$ entanglement threshold), and defer the deep measure-theory to graduate texts and to Quantum Discord and Non-Classical Correlations for the correlations that live beyond entanglement.

Key takeaways

For a pure bipartite state the entanglement is $S(\rho_A)=-\sum_k\lambda_k\ln\lambda_k$ (the Schmidt weights), and $S(\rho_A)=S(\rho_B)$ . It runs from $0$ (product) to $\ln2$ (Bell). For mixed states one number fails: negativity, concurrence, and entanglement of formation take over.

The core insight

Entanglement entropy = local mixedness. A pure global state whose part is mixed must be entangled, and how mixed the part is (its von Neumann entropy) is exactly how entangled the whole is. $S(\rho_{AB})=0 < S(\rho_A)$ is the signature no classical joint distribution can show.

Where this lives today

Entanglement entropy has become a central observable far beyond quantum information: in condensed matter its area law (entropy scaling with the boundary, not the volume, of a region) diagnoses ground-state phases and underlies the tensor-network methods pointed to in The Schmidt Decomposition; in quantum gravity the Ryu–Takayanagi formula equates it with a minimal surface area, and it sharpens the black-hole information problem via Bekenstein–Hawking entropy. The mixed-state measures — negativity especially, being efficiently computable — are the practical entanglement witnesses in experiments today. Correlations that survive beyond entanglement (present even in separable mixed states) are the province of Quantum Discord and Non-Classical Correlations.

Origins · 1 minvon neumann entropy and how this was discovered

von neumann entropy

1927–1948 · Göttingen, Princeton

John von Neumann wrote down $S=-\mathrm{Tr}\,\rho\ln\rho$ in 1927, in the same body of work that gave quantum mechanics its density-operator formalism — two decades before Claude Shannon's 1948 $H=-\sum p\log p$ founded information theory. The convergence is not a coincidence: Shannon reportedly asked von Neumann what to call his new quantity, and (in the well-worn anecdote) von Neumann advised "call it entropy — it already has that name in statistical mechanics, and besides, nobody really knows what entropy is, so you'll always have the advantage in a debate." Whatever the exact words, the mathematics is identical, and von Neumann's version — carrying the non-commutativity of quantum states in the operator $\rho$ — is the one that quantifies entanglement. Its role reached a modern crescendo when Bekenstein and Hawking found that a black hole's entropy scales with its horizon area, tying von Neumann's quantity to gravity itself.

History

1927

— von Neumann defines

S=-\mathrm{Tr}\,\rho\ln\rho

for quantum states

1948

— Shannon's classical entropy

H=-\sum p\log p

1989

— Werner constructs separable-yet-correlated mixed states

1996

— Peres and the Horodeckis: partial-transpose entanglement criterion

1998

— Wootters gives the closed-form concurrence for two qubits

The physical picture

Take a pure entangled pair and look at just one half. By the partial-trace machinery, that half is described by a mixed density matrix $\rho_A$ — it has genuine classical-looking uncertainty, even though the pair as a whole is in a perfectly definite pure state. Where did the uncertainty come from? Entirely from the correlations with the other half that you threw away by looking locally. The amount of that induced uncertainty — the von Neumann entropy of $\rho_A$ — is the entanglement. A product state induces no local uncertainty ( $S=0$ ); a Bell state induces the maximum ( $S=\ln2$ , i.e. one full bit). This gives entanglement entropy a beautiful operational meaning and a symmetry: since the Schmidt decomposition makes $\rho_A$ and $\rho_B$ share eigenvalues, $A$ and $B$ are equally uncertain about each other. A qubit maximally entangled with an entire galaxy is exactly $\ln2$ entangled with it, no more — the small partner caps the entropy. Then comes the shock. Classically, a joint distribution is always at least as uncertain as either marginal: knowing less about the parts than the whole is impossible. Quantum mechanically, an entangled pure state has zero joint entropy and positive marginal entropy. The whole is perfectly known; the parts are not. This is not a paradox to be explained away — it is the defining quantitative feature of entanglement, and it is why von Neumann entropy, borrowed from thermodynamics, becomes the master tool of quantum information.

Figure

The entanglement lab — E(θ) for cosθ|00⟩+sinθ|11⟩, peaking at the Bell state, zero at products

Figure

The whole is more certain than its parts — zero joint entropy, positive marginal entropy

Derivation AEntanglement entropy and why S(ρ_A)=S(ρ_B)

We define the pure-state entanglement measure and prove the equality of the two halves' entropies. The proof is a one-liner once the Schmidt decomposition is in hand — which is exactly why we built that machinery first. The tools are the von Neumann entropy and the shared Schmidt spectrum from The Schmidt Decomposition.

Step 1. Define entanglement entropy. For a pure bipartite $|\Psi\rangle$ , set $E(|\Psi\rangle)\equiv S(\rho_A)=-\mathrm{Tr}\,\rho_A\ln\rho_A$ , where $\rho_A$ is the reduced state from The Reduced Density Matrix and Partial Trace. Why: the local mixedness of one half is the natural, basis-independent measure of how correlated it is with the other — a product state gives $\rho_A$ pure ( $E=0$ ), an entangled state gives $\rho_A$ mixed ( $E>0$ ).

Step 2. Insert the Schmidt spectrum. By The Schmidt Decomposition, $\rho_A=\sum_k\lambda_k|k\rangle\langle k|_A$ , so $S(\rho_A)=-\sum_k\lambda_k\ln\lambda_k$ . Why: the von Neumann entropy of a diagonal density matrix is the Shannon entropy of its eigenvalues; the Schmidt weights $\{\lambda_k\}$ are precisely those eigenvalues.

Step 3. The equality falls out. The same decomposition gives $\rho_B=\sum_k\lambda_k|k\rangle\langle k|_B$ with the identical weights, so $S(\rho_B)=-\sum_k\lambda_k\ln\lambda_k=S(\rho_A)$ . Why: $A$ and $B$ share one Schmidt spectrum, so they share one entropy — the two halves of a pure state are equally entangled, no matter how different their sizes.

Step 4. Bound the range. $E=0$ iff one $\lambda_k=1$ (product); $E$ is maximized at $\ln d$ when all $\lambda_k=1/d$ (maximally entangled), where $d=\min(d_A,d_B)$ . Why: the Shannon entropy of a probability vector is maximal at the uniform distribution and zero at a point mass — the flat Schmidt spectrum is the Bell state for qubits ( $E=\ln2$ ). *Verified in `verify_aq-qm-entanglement-entropy-measures.py`: $S(\rho_A)=S(\rho_B)$ at three angles; $E(0)=0$ , $E(\pi/4)=\ln2$ , $E(\pi/8)=0.4165$ nats $=0.6009$ bits; $E(\theta)$ maximal at $\theta=\pi/4$ .*

Go deeper · Derivation B · 1 minThe von Neumann entropy and the whole-more-certain-than-its-parts inequality

Derivation BThe von Neumann entropy and the whole-more-certain-than-its-parts inequality

Here we build the von Neumann entropy properly and expose the entanglement paradox as a violation of a classical inequality. This is the article's structural crown: a number that is impossible for any classical joint distribution. The tools are the density-operator formalism and subadditivity.

Step 1. Define $S(\rho)=-\mathrm{Tr}\,\rho\ln\rho$ . For eigenvalues $\{p_i\}$ of $\rho$ this is $-\sum_i p_i\ln p_i$ — Shannon's entropy applied to the spectrum. Why: $\rho$ replaces the classical probability distribution, and its eigenvalues are the "populations"; the entropy measures how spread out those populations are.

Step 2. Endpoints. $S=0$ iff $\rho$ is pure (one eigenvalue $1$ ); $S=\ln d$ iff $\rho=I/d$ is maximally mixed. Why: a pure state carries no ignorance; the flat spectrum carries the most — exactly as for Shannon entropy.

Step 3. State subadditivity. For any bipartite $\rho_{AB}$ , $S(\rho_{AB})\le S(\rho_A)+S(\rho_B)$ , with equality iff $\rho_{AB}=\rho_A\otimes\rho_B$ . Why: correlations can only reduce the joint entropy below the sum of the parts; this is the quantum echo of $H(X,Y)\le H(X)+H(Y)$ , proved with the Klein inequality (we cite it — the full proof is standard graduate material).

Step 4. The paradox. Take an entangled pure pair: $S(\rho_{AB})=0$ (pure), but $S(\rho_A)=\ln2>0$ . So $S(\rho_{AB})<S(\rho_A)$ . Why: classically $H(X,Y)\ge H(X)$ always — the joint is at least as uncertain as a marginal. Quantum mechanics violates this: the whole is perfectly known while a part is not. Subadditivity still holds ( $0\le\ln2+\ln2$ ), but the classical lower bound is shattered. This is the sharpest numerical fingerprint of entanglement. *Verified: for the Bell state $S(\rho_{AB})=0$ , $S(\rho_A)=\ln2$ , so $S(\rho_{AB})<S(\rho_A)$ while subadditivity $S_{AB}\le S_A+S_B$ still holds.*

Limiting cases

State	$S(\rho_A)$	Entangled?	Measure notes
Product $\|00\rangle$	$0$	no	all measures vanish
$\cos\theta\|00\rangle+\sin\theta\|11\rangle$	$-\cos^2\theta\ln\cos^2\theta-\sin^2\theta\ln\sin^2\theta$	yes ( $\theta\notin\{0,\tfrac\pi2\}$ )	concurrence $=\sin2\theta$
Bell $\|\Phi^+\rangle$	$\ln2$	maximally	concurrence $=1$ , negativity $=\tfrac12$
Werner $p\|\Phi^+\rangle\langle\Phi^+\|+(1-p)\tfrac I4$	(mixed — use negativity)	iff $p>\tfrac13$	$C=\max(0,\tfrac{3p-1}{2})$

For pure states one number does everything. The moment the state is mixed, "entangled or not" itself becomes non-trivial, and the negativity or concurrence — not the entropy of a reduced state — is the right tool.

Common trap“Entanglement entropy measures how entangled any state is.”

Only for pure states. For a mixed bipartite $\rho_{AB}$ , the reduced-state entropy $S(\rho_A)$ conflates entanglement with plain classical mixedness — a completely separable, classically-correlated $\rho_{AB}$ can have large $S(\rho_A)$ . The maximally mixed two-qubit state $I/4$ has $S(\rho_A)=\ln2$ yet zero entanglement. So $S(\rho_A)$ is not an entanglement measure off the pure-state manifold. There you need genuine mixed-state measures — negativity (from the partial transpose), concurrence, entanglement of formation — which are constructed to vanish on all separable states. Using $S(\rho_A)$ as "the entanglement" of a mixed state is the single most common error in the subject.

Worked examples

1 · The entanglement dial E(θ) (confidence builder)

Take $|\Psi(\theta)\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle$ . Enumerate the reduced state: $\rho_A=\mathrm{diag}(\cos^2\theta,\sin^2\theta)$ , already diagonal, so the Schmidt weights are $\lambda=\{\cos^2\theta,\sin^2\theta\}$ . Apply $E=-\sum\lambda\ln\lambda$ at three angles. At $\theta=0$ : $\lambda=\{1,0\}$ , $E=0$ — a product state, no entanglement. At $\theta=\pi/8$ : $\cos^2(\pi/8)=0.8536$ , $\sin^2(\pi/8)=0.1464$ , giving $E=-0.8536\ln0.8536-0.1464\ln0.1464=0.4165$ nats $=0.6009$ bits. At $\theta=\pi/4$ : $\lambda=\{0.5,0.5\}$ , $E=\ln2=0.6931$ nats $=1$ bit — the Bell state, maximal. Boundary number: the dial runs from $0$ at the product endpoints to $\ln2$ at the symmetric point $\theta=\pi/4$ , and it is maximized exactly there (the derivative of $E(\theta)$ vanishes at $\pi/4$ ). One angle sweeps the entire range of pure-state two-qubit entanglement.

2 · Negativity and concurrence of a Werner state (exam level)

Take the Werner state $\rho(p)=p\,|\Phi^+\rangle\langle\Phi^+|+(1-p)\tfrac{I}{4}$ — a Bell pair degraded by white noise. Enumerate the partial-transpose (Peres–Horodecki) test: transpose the $B$ -indices of $\rho(p)$ and look for negative eigenvalues; their presence certifies entanglement. Apply: the partial transpose has three eigenvalues $\tfrac{1+p}{4}$ and one eigenvalue $\tfrac{1-3p}{4}$ , which turns negative exactly when $p>\tfrac13$ . The negativity $N=\sum|\text{neg. eigenvalues}|$ is therefore $N(p)=\max\!\big(0,\tfrac{3p-1}{4}\big)$ , and the concurrence works out to $C(p)=\max\!\big(0,\tfrac{3p-1}{2}\big)$ . Boundary number: at $p=\tfrac13$ both vanish — the Werner state crosses from entangled to separable exactly at $p=1/3$ ; at $p=1$ (pure Bell) $C=1$ (maximal), and at $p=\tfrac12$ , $C=\tfrac14$ , still entangled. We verified numerically that $p=0.5$ gives positive negativity, $p=0.2$ gives none, and the threshold sits precisely at $p=1/3$ . This is the textbook example of an entanglement measure detecting entanglement that the reduced-state entropy would completely miss.

Going further

For pure states the entanglement entropy is the complete measure; Nielsen's majorization theorem (noted in The Schmidt Decomposition) makes it the top of a whole ordering of "more entangled." For mixed states the field uses a family of inequivalent measures — entanglement of formation and distillable entanglement (operational, hard to compute), the negativity and concurrence used above (computable), all built to vanish on separable states; the deep measure theory is graduate material and the monogamy inequalities (Coffman–Kundu–Wootters) constrain how entanglement is shared. Correlations that persist even without entanglement are quantified by Quantum Discord and Non-Classical Correlations. The reduced states here are from The Reduced Density Matrix and Partial Trace, the spectra from The Schmidt Decomposition, and the maximal endpoint is realized by the Bell States and Maximally Entangled Pairs.

Nats or bits — which units for entanglement entropy?

Both are used. With the natural log, $S$ is in nats and a Bell state gives $\ln2\approx0.693$ ; with $\log_2$ , $S$ is in bits (or "ebits" for entanglement) and a Bell state gives exactly $1$ . They differ only by the factor $\ln2$ . Quantum information usually prefers bits (one Bell pair $=1$ ebit); statistical mechanics prefers nats. We quote both: $E(\pi/8)=0.4165$ nats $=0.6009$ bits.

Why is the reduced state mixed if the whole state is pure?

Because the purity is stored in the correlations between the halves, which the partial trace discards. Tracing out $B$ is exactly "forget everything about $B$ , including how it is correlated with $A$ " — and that forgetting is what turns the locally-viewed $A$ into a mixed state. The more entangled the pair, the more information lives in the correlations, and the more mixed each half becomes. This is proved as a theorem in The Reduced Density Matrix and Partial Trace.

What is the negativity, intuitively?

It measures how much the partial transpose of $\rho$ "fails to be a valid density matrix." Transposing one subsystem is a legitimate operation on separable states (it stays positive), but on entangled states it can produce negative eigenvalues — an impossible density matrix. The sum of the magnitudes of those negative eigenvalues is the negativity. Peres and the Horodeckis showed that for two qubits (and qubit-qutrit) a nonzero negativity is exactly the entanglement criterion; for larger systems it detects most, but not all, entanglement.

Is concurrence the same as entanglement entropy?

For a pure two-qubit state they carry the same information (each determines the other), but they are different functions: concurrence $C=\sin2\theta$ for $\cos\theta|00\rangle+\sin\theta|11\rangle$ , while the entropy is $-\cos^2\theta\ln\cos^2\theta-\sin^2\theta\ln\sin^2\theta$ . Concurrence's advantage is that Wootters extended it, in closed form, to mixed two-qubit states, where the entanglement entropy fails. So concurrence is the practical two-qubit measure; entanglement entropy is the conceptual pure-state one.

Does subadditivity ever fail for quantum states?

No — subadditivity $S(\rho_{AB})\le S(\rho_A)+S(\rho_B)$ always holds. What fails (relative to classical intuition) is the lower bound: classically $S(\rho_{AB})\ge\max(S_A,S_B)$ , and entanglement violates that. The subtler quantum inequality that does survive is the Araki–Lieb triangle inequality $|S_A-S_B|\le S_{AB}$ , and the celebrated strong subadditivity $S_{ABC}+S_B\le S_{AB}+S_{BC}$ — deep results underpinning quantum information theory, cited here and proved in graduate texts.

How does entanglement entropy connect to black holes?

Through area laws. In many-body ground states the entanglement entropy of a region scales with the area of its boundary rather than its volume — the same scaling as the Bekenstein–Hawking entropy of a black hole horizon. The Ryu–Takayanagi formula makes this precise in holographic theories, equating a region's entanglement entropy with the area of a minimal surface in a higher-dimensional spacetime. It is one of the most active bridges between quantum information and gravity, and it all begins with $-\mathrm{Tr}\,\rho\ln\rho$ applied to a reduced state.