W States

Ask how three qubits can be entangled and you expect a single answer with a knob on it — a little, a lot, maximal. For two qubits that is right: one number, the Schmidt rank or entanglement entropy, orders every pure state from product to Bell. The surprise of three qubits is that the knob splits in two. In 2000 Dür, Vidal and Cirac proved that every genuinely tripartite entangled pure state can be turned, with nonzero probability, into one of exactly two canonical states by invertible local operations — the GHZ state or the W state — and that no local operation whatsoever carries one class into the other. Tripartite entanglement is not a scale but a pair of species. The W state is the gentler species. Concretely $|W_N\rangle$ is the totally symmetric state of $N$ qubits carrying a single shared excitation — one flipped spin, one photon — that refuses to sit on any one site and spreads equally over all of them. Its signature is robustness: trace out any one qubit and the survivors stay entangled, with a fixed nonzero concurrence for $N=3$ . GHZ has the opposite temperament — a maximal, all-or-nothing global correlation that a single loss annihilates. Where GHZ concentrates its entanglement into one fragile collective resource, W distributes it thinly and durably across every pair: W has zero three-tangle yet the maximum possible pairwise concurrence, GHZ the reverse. This article builds that dichotomy from the definition up — the reduced states, the concurrence $2/N$ , the monogamy saturation, and the three-tangle argument that proves the classes can never meet.

Key takeaways

The $N$ -qubit W state is the equal-weight symmetric superposition of every basis state with exactly one excitation: $|W_3\rangle=\tfrac{1}{\sqrt3}(|001\rangle+|010\rangle+|100\rangle)$ .
W and GHZ are the two inequivalent classes of genuine three-qubit entanglement under stochastic LOCC — no local operation connects them, so "more entangled" is not a total order for $N\ge3$ .
W entanglement is loss-tolerant: trace out one qubit of $|W_3\rangle$ and the pair is still entangled, concurrence $2/3$ ; the same operation on GHZ leaves a fully separable pair.
Every pair in $|W_N\rangle$ has concurrence $C_{ij}=2/N$ , and $\sum_j C_{1j}^2=4(N-1)/N^2$ exactly saturates the Coffman–Kundu–Wootters monogamy bound.
The three-tangle vanishes, $\tau(W)=0$ , while $\tau(\mathrm{GHZ})=1$ : W carries no genuine three-party entanglement, only distributed pairwise entanglement — the physics of single magnons, single-photon Dicke states, and weak collective excitation.
Key idea: A single excitation symmetrically delocalized over $N$ qubits produces entanglement that is pairwise and loss-tolerant — the SLOCC-inequivalent partner of the collective, fragile GHZ entanglement.

Where this lives today

W states are now standard laboratory objects. In trapped ions, Häffner and coworkers created W states of up to eight ions in 2005 with full tomographic readout; today symmetric Dicke and W states are engineered routinely in ions, in photonic setups that delocalize a single photon over many modes, and on superconducting processors preparing Dicke states of ten or more qubits. The robustness Dür–Vidal–Cirac identified is why: because W entanglement survives the loss of a party, it is a natural resource for loss-tolerant tasks — heralded entanglement distribution across a quantum network, quantum memories built from collective spin-wave excitations (the DLCZ protocol), and anonymous leader election, where measuring the single excitation of a shared $|W_N\rangle$ picks a uniformly random winner. Certifying a state as genuinely W-class uses tailored fidelity witnesses like $\mathcal{W}=\tfrac23\mathbb{1}-|W_3\rangle\langle W_3|$ , whose threshold no GHZ-class or biseparable state can beat. And the classification stays a frontier: beyond three qubits the number of SLOCC classes becomes infinite, so the clean W-versus-GHZ dichotomy is the well-charted base camp of a larger landscape.

The physical picture

Picture one quantum of excitation — a single flipped spin, one photon — dropped onto $N$ identical sites. It cannot decide which site it occupies, so quantum mechanically it occupies all of them at once, coherently and with equal weight. That delocalized single excitation is the W state. Because the excitation is genuinely shared rather than collectively agreed, removing one site only removes the small chance it was there — the rest still share it, and their entanglement survives. Contrast GHZ, where all $N$ qubits agree on one classical bit superposed with its complement, $|0\cdots0\rangle+|1\cdots1\rangle$ : that agreement is all-or-nothing, and a single loss collapses everyone. The first figure sets the two species side by side with the numbers that separate them.

Figure

One excitation shared over three qubits is the W state W₃, an equal superposition of 001, 010 and 100 with amplitude 1/√3 each; GHZ instead makes all three agree, 000 with 111. Trace out qubit C and the W pair AB keeps concurrence C = 2/3, while the GHZ pair collapses to a separable mixture, C = 0

Robustness is bought at a price the second figure makes quantitative. Because entanglement obeys monogamy — a qubit can only share so much — spreading a single excitation over more sites must thin the entanglement each pair receives. In the W state that thinning is exact and universal: every one of the $\binom N2$ pairs carries the same concurrence $C_{ij}=2/N$ , falling as $1/N$ . Yet this is optimal sharing, not inefficiency: the squared concurrences add up to precisely the monogamy ceiling $4\det\rho_1$ , so W distributes the maximum total pairwise entanglement the laws permit, with nothing left for a genuine three-way part.

Figure

Pairwise concurrence C_ij = 2/N in the N-qubit W state falls as 1/N — 2/3 at N = 3, 1/2 at N = 4, → 0 as N grows — while the squared sum Σ C₁ⱼ² = 4(N−1)/N² always equals the monogamy ceiling 4 det ρ₁, saturated for every N, with three-tangle τ = 0 throughout

The core equations

The W state is defined by a symmetry, not a dynamical equation: it is the unique (up to phase) totally symmetric state of $N$ qubits with exactly one excitation. Writing that out gives the master definition.

|W_N\rangle=\frac{1}{\sqrt N}\sum_{j=1}^{N}|0\rangle^{\otimes(j-1)}|1\rangle_j\,|0\rangle^{\otimes(N-j)} \qquad (1)

where: $N$ — number of qubits; $|1\rangle_j$ — the single excitation on qubit $j$ ; $1/\sqrt N$ — normalization fixing $\langle W_N|W_N\rangle=1$ . Reading: sum over the $N$ ways to place one excitation with equal weight — manifestly invariant under permuting the qubits. For $N=3$ this is the state every discussion of tripartite entanglement returns to.

|W_3\rangle=\tfrac{1}{\sqrt3}\big(|001\rangle+|010\rangle+|100\rangle\big) \qquad (2)

where: $|abc\rangle$ — computational basis of qubits $A,B,C$ , read left to right. Reading: three equally weighted terms, each with a single $1$ ; the amplitude on any chosen qubit is $1/\sqrt3$ , so the probability is $1/3$ . The same object is the single-excitation member of the Dicke family, tying W states to atomic and spin physics.

|W_N\rangle=|D^{(1)}_N\rangle,\qquad |D^{(k)}_N\rangle=\binom{N}{k}^{-1/2}\!\!\sum_{\mathrm{perm}}\big|\,\underbrace{1\cdots1}_{k}\,\underbrace{0\cdots0}_{N-k}\big\rangle \qquad (3)

where: $|D^{(k)}_N\rangle$ — symmetric Dicke state with $k$ excitations over $N$ qubits; $k=1$ gives the W state. Reading: W is the "one magnon" rung of the Dicke ladder; GHZ, by contrast, is not a Dicke state. Everything distinctive about W entanglement is read off its reduced states: tracing out all but two qubits leaves a rank-two mixture of a Bell state and the vacuum.

\rho_{ij}=\frac{2}{N}\,|\Psi^+\rangle\langle\Psi^+|+\frac{N-2}{N}\,|00\rangle\langle00|,\qquad |\Psi^+\rangle=\tfrac{1}{\sqrt2}\big(|01\rangle+|10\rangle\big) \qquad (4)

where: $\rho_{ij}$ — reduced state of any pair $i\neq j$ ; $2/N$ — probability the excitation sits on this pair (as a Bell state); $(N-2)/N$ — probability it is elsewhere, both qubits in $|0\rangle$ . Reading: the Bell component keeps the pair entangled after loss and never vanishes for finite $N$ . Wootters' formula turns that reduced state into the concurrence of every pair — the same for all by symmetry, and the number the second figure plotted.

C_{ij}\big(|W_N\rangle\big)=\frac{2}{N}\qquad\text{for all }i\neq j \qquad (5)

where: $C_{ij}$ — concurrence between qubits $i$ and $j$ . Reading: pairwise entanglement is uniform and falls as $1/N$ — $2/3$ at $N=3$ , $1/2$ at $N=4$ — and drives the monogamy and three-tangle results below. Monogamy of entanglement — the Coffman–Kundu–Wootters inequality — bounds how much of qubit $1$ 's entanglement can be parcelled out to partners. The W state is the case of equality.

\sum_{j=2}^{N} C_{1j}^2\;\le\;4\det\rho_1=\mathcal{C}_{1(2\cdots N)}^2,\qquad \tau_{123}=\mathcal{C}_{1(23)}^2-C_{12}^2-C_{13}^2 \qquad (6)

where: $C_{1j}$ — pairwise concurrences; $\mathcal{C}_{1(2\cdots N)}$ — concurrence of qubit $1$ against the rest; $\rho_1$ — its reduced state; $\tau_{123}$ — the three-tangle, the residual genuine three-party entanglement. Reading: the left side is what the pairs hold, the right what qubit $1$ holds in total; for W they are equal, so $\tau=0$ . The two classes are separated by an operational relation, not a number — SLOCC equivalence: two states share a class exactly when an invertible local operator maps one to the other.

|\psi\rangle\sim_{\mathrm{SLOCC}}|\phi\rangle\iff |\psi\rangle=(A\otimes B\otimes C)\,|\phi\rangle,\quad A,B,C\in GL(2,\mathbb{C}) \qquad (7)

where: $A,B,C$ — invertible ( $\det\neq0$ ) single-qubit operators; $\sim_{\mathrm{SLOCC}}$ — convertible with nonzero probability by local operations and classical communication. Reading: invertibility is the whole point — it makes the three-tangle an invariant up to a positive factor, and hence forbids $W\leftrightarrow\mathrm{GHZ}$ .

Derivation ATracing out one qubit leaves the pair entangled

The goal is to prove the defining physical property of the three-qubit W state — its robustness under particle loss — by computing the two-qubit state that remains when the third qubit is discarded and showing it is entangled with concurrence exactly $2/3$ . We group the terms by the value of the lost qubit, take the partial trace to get a rank-two mixed state, apply Wootters' formula, and contrast each step with what the same operations do to GHZ, so the robustness gap is unmistakable.

Step 1. Group the three terms of $|W_3\rangle$ by the value of qubit $C$ : two of them have $C=0$ , one has $C=1$ , giving $|W_3\rangle=\tfrac{1}{\sqrt3}(|01\rangle+|10\rangle)_{AB}|0\rangle_C+\tfrac{1}{\sqrt3}|00\rangle_{AB}|1\rangle_C$ . Why: the partial trace over $C$ sums the AB blocks weighted by the orthonormal $C$ -states, so sorting the terms by $C$ makes the trace a one-line read-off.

Step 2. Trace out $C$ by summing $\langle 0|_C(\cdots)|0\rangle_C+\langle 1|_C(\cdots)|1\rangle_C$ . The $C=0$ block gives $\tfrac23|\Psi^+\rangle\langle\Psi^+|$ and the $C=1$ block gives $\tfrac13|00\rangle\langle00|$ , so $\rho_{AB}=\tfrac23|\Psi^+\rangle\langle\Psi^+|+\tfrac13|00\rangle\langle00|$ . Why: the two $C$ -branches are orthogonal, so they add incoherently — a classical mixture of "excitation here as a Bell pair" and "excitation on the lost qubit."

Step 3. Read the concurrence. For an X-state $p|\Psi^+\rangle\langle\Psi^+|+(1-p)|00\rangle\langle00|$ , Wootters' formula gives $C=p$ : the $|00\rangle$ term is annihilated by the spin-flip $\tilde\rho=(\sigma_y\otimes\sigma_y)\rho^*(\sigma_y\otimes\sigma_y)$ and only the Bell component survives, so with $p=\tfrac23$ , $C(\rho_{AB})=\tfrac23$ . Why: concurrence sees only genuine two-qubit coherence, and the sole coherence left is the Bell block.

Step 4. Contrast GHZ. Tracing one qubit from $\tfrac{1}{\sqrt2}(|000\rangle+|111\rangle)$ gives $\tfrac12(|00\rangle\langle00|+|11\rangle\langle11|)$ , diagonal with no coherence, so concurrence $0$ — separable. Why: GHZ's correlation lives entirely in the three-way coherence $|000\rangle\langle111|$ , which the partial trace destroys; W's lives in pairwise coherences that partly survive. *Verified: $\rho_{AB}=\tfrac23|\Psi^+\rangle\langle\Psi^+|+\tfrac13|00\rangle\langle00|$ , $C=2/3$ , GHZ pair separable — matches Dür–Vidal–Cirac (2000) and a direct eigenvalue computation.*

Go deeper · Derivation B · 1 minEvery pair in the N-qubit W state has concurrence two over N

Derivation BEvery pair in the N-qubit W state has concurrence two over N

The goal is to generalize the $N=3$ result to arbitrary $N$ : establish that all $\binom N2$ pairs share the identical concurrence $C_{ij}=2/N$ , and then show these pairwise concurrences add up in squares to exactly the Coffman–Kundu–Wootters monogamy ceiling, so the W state is not merely robust but the extremal maximizer of distributed two-party entanglement. We reduce to two qubits by symmetry, apply Wootters' formula, and check the bound against the single-qubit determinant.

Step 1. Reduce to a pair. By permutation symmetry every pair is equivalent, so trace out the other $N-2$ qubits: with probability $2/N$ the excitation lands on the chosen pair as $|\Psi^+\rangle$ , and with probability $(N-2)/N$ it is elsewhere, leaving $|00\rangle$ . Hence $\rho_{ij}=\tfrac{2}{N}|\Psi^+\rangle\langle\Psi^+|+\tfrac{N-2}{N}|00\rangle\langle00|$ . Why: the excitation is on exactly one of the $N$ qubits with equal probability, so the chance it is on a given pair is $2/N$ — that combinatorics is the whole reduced state.

Step 2. Apply Wootters' formula $C=\max\{0,\sqrt{\lambda_1}-\sqrt{\lambda_2}-\sqrt{\lambda_3}-\sqrt{\lambda_4}\}$ , with $\lambda_i$ the decreasing eigenvalues of $\rho\tilde\rho$ . As at $N=3$ the flipped $|00\rangle$ block contributes nothing and only the Bell weight survives, so $C_{ij}=\tfrac{2}{N}$ . Why: this is the same X-state family with $p=2/N$ , whose concurrence is exactly the Bell weight $p$ .

Step 3. Assemble the monogamy sum for qubit $1$ . Its $N-1$ partners each contribute $(2/N)^2$ , so $\sum_{j\neq1}C_{1j}^2=(N-1)\tfrac{4}{N^2}=\tfrac{4(N-1)}{N^2}$ . Why: uniformity turns the sum into a single multiplication — every partner is on equal footing.

Step 4. Compare with the CKW ceiling. The marginal is $\rho_1=\mathrm{diag}(\tfrac{N-1}{N},\tfrac1N)$ , so $4\det\rho_1=4\cdot\tfrac1N\cdot\tfrac{N-1}{N}=\tfrac{4(N-1)}{N^2}$ , identical to the sum. The inequality holds with equality, so the residual three-tangle-type quantity vanishes. Why: every scrap of qubit $1$ 's entanglement with the rest is realized as pairwise concurrence, leaving nothing collective — the sense in which W is "all pairs, no whole." *Verified: $N=3$ gives $C=2/3$ and $\sum C_{1j}^2=8/9=4\det\rho_1$ ; $N=4$ gives $C=1/2$ and $\sum=3/4=4\det\rho_1$ . Consistent with Coffman–Kundu–Wootters PRA 61, 052306.*

Go deeper · Derivation C · 1 minWhy W and GHZ are SLOCC-inequivalent

Derivation CWhy W and GHZ are SLOCC-inequivalent

The goal is to prove rigorously that no invertible local operation — hence no stochastic LOCC protocol, however lucky — can convert $|W_3\rangle$ into $|\mathrm{GHZ}\rangle$ or back, by exhibiting an invariant that separates them: the three-tangle, which transforms under an invertible local map by a strictly positive determinant factor and so cannot turn a zero value into a nonzero one. We state the transformation law, evaluate the tangle on each state, and derive the contradiction.

Step 1. State the transformation law. Under $A\otimes B\otimes C$ the three-tangle scales as $\tau_{123}\big((A\otimes B\otimes C)|\psi\rangle\big)=|\det A|^2|\det B|^2|\det C|^2\,\tau_{123}(|\psi\rangle)$ , because $\tau$ is built from Cayley's hyperdeterminant, a relative invariant of the local $GL(2,\mathbb C)$ actions. Why: such polynomials pick up exactly the determinant of each local transformation, making $\tau$ an SLOCC monotone up to a positive factor.

Step 2. Evaluate on W. Using $\mathcal{C}_{1(23)}^2=4\det\rho_1=8/9$ and $C_{12}^2+C_{13}^2=(2/3)^2+(2/3)^2=8/9$ , the tangle is $\tau_{123}(|W_3\rangle)=8/9-8/9=0$ . Why: W saturates monogamy, so its residual is zero — all entanglement is booked as pairwise.

Step 3. Evaluate on GHZ. Here the marginal is maximally mixed, $\mathcal{C}_{1(23)}=1$ , while $C_{12}=C_{13}=0$ ; hence $\tau_{123}(|\mathrm{GHZ}\rangle)=1$ . Why: GHZ's entanglement is purely collective — no two qubits are entangled alone — so all of it shows up as residual three-tangle.

Step 4. Derive the contradiction. SLOCC requires $A,B,C$ invertible, so the prefactor $>0$ strictly. If W and GHZ were SLOCC-equivalent, $0=\tau(W)=(\text{positive})\times1>0$ — impossible. The two states lie in distinct SLOCC classes. Why: a positive number times $1$ can never be $0$ , so the invariant certifies the classes never meet, even probabilistically. *Verified: $\tau(W)=0$ , $\tau(\mathrm{GHZ})=1$ , matching Dür–Vidal–Cirac PRA 62, 062314 and Coffman–Kundu–Wootters PRA 61, 052306.*

Limiting cases

The W state is most revealing at the edges — losing a party, growing $N$ , cutting it in two, comparing with GHZ.

Case	Result	Meaning
Lose one qubit ( $N=3$ )	$\rho_{AB}=\tfrac13\big(2\vert\Psi^+\rangle\langle\Psi^+\vert+\vert00\rangle\langle00\vert\big)$	Reduced pair is entangled, concurrence $2/3$ — the defining robustness under loss.
Large $N$	$C_{ij}=\tfrac{2}{N}\to0$	Pairwise concurrence vanishes as $1/N$ ; monogamy spreads a finite total ever more thinly.
Bipartite cut $1$ vs $N-1$	$S(\rho_1)=H_2\!\big(\tfrac1N\big)$	Single-qubit entropy is sub-maximal and $\to0$ as $N\to\infty$ ; the Schmidt rank across any cut is only $2$ .
Three-tangle	$\tau(\vert W\rangle)=0,\ \ \tau(\vert\mathrm{GHZ}\rangle)=1$	W carries no genuine residual three-party entanglement; it is entirely pairwise.
$N=2$	$\vert W_2\rangle=\vert\Psi^+\rangle$ , $C=1$	For two qubits the W state is just a Bell state — the class distinction only appears at $N\ge3$ .

Worked examples

1 · The three-qubit W state, fully characterized (fully worked)

Take $|W_3\rangle=\tfrac{1}{\sqrt3}(|001\rangle+|010\rangle+|100\rangle)$ . Compute (a) the single-qubit reduced state and its entropy, (b) the two-qubit reduced state and its concurrence, and (c) the three-tangle — and read off what each says about the entanglement.

Step 1. Single-qubit marginal. By symmetry each qubit is excited with probability $1/3$ , so $\rho_A=\mathrm{diag}(\tfrac23,\tfrac13)$ . Its entropy measures $A$ 's entanglement with the rest: $S(\rho_A)=H_2(\tfrac13)=-\tfrac13\log_2\tfrac13-\tfrac23\log_2\tfrac23=0.528+0.390=\mathbf{0.918}$ bits — sub-maximal (a Bell pair gives $1$ bit), reflecting Schmidt rank $2$ with unequal weights.

Step 2. Two-qubit marginal. From Derivation A, tracing out $C$ gives $\rho_{AB}=\tfrac23|\Psi^+\rangle\langle\Psi^+|+\tfrac13|00\rangle\langle00|=\tfrac13\!\left(\begin{smallmatrix}1&0&0&0\\0&1&1&0\\0&1&1&0\\0&0&0&0\end{smallmatrix}\right)$ in the $\{00,01,10,11\}$ basis. Its concurrence is the Bell weight, $C(\rho_{AB})=\mathbf{2/3}$ — the pair survives the loss of $C$ still entangled.

Step 3. Three-tangle. Assemble the CKW pieces: $\mathcal{C}_{1(23)}^2=4\det\rho_A=4\cdot\tfrac23\cdot\tfrac13=\tfrac89$ and $C_{12}^2+C_{13}^2=(\tfrac23)^2+(\tfrac23)^2=\tfrac89$ , so $\tau_{123}=\tfrac89-\tfrac89=\mathbf{0}$ . Every bit of qubit $A$ 's entanglement with the rest is realized pairwise; none is genuinely three-party.

Step 4. Interpret. $S=0.918$ says $A$ is substantially but not maximally entangled with $BC$ ; $C=2/3$ says that entanglement is distributed as robust pairwise links; $\tau=0$ says there is no fragile collective part. W₃ has $S=0.918$ bit, $C=2/3$ , $\tau=0$ — the exact opposite profile to GHZ ( $S=1$ , $C=0$ , $\tau=1$ ).

2 · The four-qubit W state and monogamy

Step 1. Pairwise concurrence. From equation (5), $C_{ij}=2/N=2/4=\mathbf{1/2}$ for every one of the $\binom42=6$ pairs — half the value it had at $N=3$ , exactly the $1/N$ thinning the second figure shows.

Step 2. Single-qubit entropy. Each qubit is excited with probability $1/4$ , so $\rho_1=\mathrm{diag}(\tfrac34,\tfrac14)$ and $S(\rho_1)=H_2(\tfrac14)=\tfrac14(2)+\tfrac34(0.415)=\mathbf{0.811}$ bits — lower than the $0.918$ bits of $W_3$ , since the excitation is now less likely to be on any one qubit.

Step 3. Monogamy sum. Qubit $1$ has three partners, so $\sum_{j\neq1}C_{1j}^2=3\cdot(\tfrac12)^2=\mathbf{3/4}$ . The ceiling is $4\det\rho_1=4\cdot\tfrac14\cdot\tfrac34=\mathbf{3/4}$ . The two match, so the CKW inequality is saturated and the three-tangle-type residual vanishes.

Step 4. Read the trend. From $N=3$ to $N=4$ the per-pair concurrence fell $2/3\to1/2$ and the entropy $0.918\to0.811$ bit, yet monogamy stayed saturated. W₄: $C_{ij}=1/2$ , $S=0.811$ bit, $\sum C_{1j}^2=4\det\rho_1=3/4$ — W stays the extremal distributor of pairwise entanglement as $N$ grows.

Physics of the portrait

The W state is named for a person, not a letter: the "W" is widely taken to stand for Wolfgang — Wolfgang Dür, who with Vidal and Cirac gave three-qubit entanglement its two-class structure.

Origins · 1 minWolfgang Dür and how this was discovered

Wolfgang Dür

Austrian physicist, University of Innsbruck · the W and GHZ SLOCC classes of tripartite entanglement

Wolfgang Dür is an Austrian theoretical physicist whose name is attached, letter for letter, to one of the two fundamental classes of multiqubit entanglement. In 2000, working with Guifré Vidal and Juan Ignacio Cirac in Innsbruck, he published "Three qubits can be entangled in two inequivalent ways" (Phys. Rev. A 62, 062314), the paper that overturned the tacit assumption that entanglement is a single quantity to be maximized. Classifying pure three-qubit states under stochastic local operations and classical communication (SLOCC), they proved there are exactly two genuinely tripartite classes — the GHZ class and a second, previously unnamed class whose representative they wrote as $|W\rangle=\tfrac{1}{\sqrt3}(|001\rangle+|010\rangle+|100\rangle)$ , the "W" of the title. Using invertible local operators and invariants such as the three-tangle, they showed the classes are separated by an unbridgeable gap that no local protocol, however lucky, can cross — and that the W class is robust, keeping bipartite entanglement when a qubit is lost, exactly where GHZ is fragile. Dür has since worked broadly on entanglement theory, quantum repeaters, and physics education, but it is the 2000 classification — and the little state carrying his initial — that put his name into every quantum-information course.

History

1935

— Schrödinger names entanglement the characteristic trait of quantum mechanics

1954

— Dicke introduces the symmetric collective states now called Dicke states

1989

— Greenberger, Horne, and Zeilinger introduce the GHZ state

2000

— Dür, Vidal, and Cirac prove W and GHZ are the two SLOCC classes of three-qubit entanglement

2000

— Coffman, Kundu, and Wootters define the three-tangle and prove monogamy of entanglement

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, §12 (entanglement, distance measures) · Watrous, The Theory of Quantum Information, ch. 6 · Wilde, Quantum Information Theory, 2nd ed., ch. 11 · Preskill, Lecture Notes on Quantum Computation, ch. 4 · Horodecki, Horodecki, Horodecki, Horodecki, "Quantum entanglement," Rev. Mod. Phys. 81, 865 (2009).

Key papers. Dür, Vidal, Cirac, "Three qubits can be entangled in two inequivalent ways," Phys. Rev. A 62, 062314 (2000) · Coffman, Kundu, Wootters, "Distributed entanglement," Phys. Rev. A 61, 052306 (2000) · Wootters, "Entanglement of formation of an arbitrary state of two qubits," Phys. Rev. Lett. 80, 2245 (1998) · Dicke, "Coherence in spontaneous radiation processes," Phys. Rev. 93, 99 (1954) · Häffner et al., "Scalable multiparticle entanglement of trapped ions," Nature 438, 643 (2005).

Provenance. A verifier should: (1) confirm normalization and $|W_N\rangle=|D^{(1)}_N\rangle$ ; (2) recompute $\rho_{AB}=\tfrac23|\Psi^+\rangle\langle\Psi^+|+\tfrac13|00\rangle\langle00|$ and its concurrence $2/3$ ; (3) verify $C_{ij}=2/N$ and the saturation $\sum C_{1j}^2=4(N-1)/N^2=4\det\rho_1$ ; (4) check $S(\rho_1)=H_2(1/N)$ , giving $0.918$ bit at $N=3$ and $0.811$ bit at $N=4$ ; (5) confirm $\tau(W)=0$ , $\tau(\mathrm{GHZ})=1$ and the $|\det|^2$ scaling under SLOCC.

Prerequisites. Separable versus Entangled States · Schmidt Decomposition · Concurrence and Wootters' Formula · Entanglement Entropy of Pure States. Leads to. Entanglement Witnesses · PPT Criterion and the Peres–Horodecki Test · CHSH Inequality and Its Quantum Violation.

Ask the tutor: Ask me to trace out a qubit from any $|W_N\rangle$ and watch the surviving pair stay entangled, or to prove for your own state that its three-tangle puts it in the W class or the GHZ class.

What does the "W" in W state actually stand for?

It is generally understood to stand for Wolfgang — Wolfgang Dür, first author of the 2000 paper (with Vidal and Cirac) that introduced and named the state. Unlike "GHZ," an acronym of three surnames, "W" is a single-letter label for the second SLOCC class of three-qubit entanglement, and the paper simply wrote its representative as $|W\rangle$ . The letter stuck, and the state is now called the W state regardless of qubit number — $|W_3\rangle$ , $|W_4\rangle$ , the general $|W_N\rangle$ .

Why is the W state called "robust" when GHZ is called "maximally entangled"?

They are extremal in different senses. GHZ maximizes genuine multipartite entanglement — three-tangle $1$ — but that entanglement is all-or-nothing, so measuring or losing one qubit leaves a fully separable remainder. W has zero three-tangle, so by that measure it is less entangled, yet its entanglement is distributed as pairwise links that partly survive loss: trace out a qubit of $|W_3\rangle$ and the survivors keep concurrence $2/3$ . "Robust" refers to this survival of bipartite entanglement under particle loss, which GHZ lacks. Which is preferable depends on the task.

If both are genuinely tripartite entangled, why can't SLOCC turn W into GHZ?

Because SLOCC preserves a separating invariant. The three-tangle $\tau$ transforms under an invertible local operation only by a strictly positive factor $|\det A|^2|\det B|^2|\det C|^2$ , so whether it is zero or nonzero is a class label no local operation can flip. W has $\tau=0$ and GHZ has $\tau=1$ ; a positive multiple of $1$ can never equal $0$ , so the two lie in distinct classes and cannot be interconverted even probabilistically. This is the technical heart of Dür–Vidal–Cirac: genuine tripartite entanglement is not one thing but two.

Why does the pairwise concurrence fall off as 1/N?

Monogamy. Qubit $1$ 's total entanglement with the rest is bounded by $\mathcal{C}_{1(2\cdots N)}^2=4\det\rho_1=4(N-1)/N^2$ , which itself shrinks with $N$ . In the W state that total is shared equally among $N-1$ partners, so each gets a $1/(N-1)$ slice of a quantity already $O(1/N)$ , giving $C_{1j}=2/N$ . This is not inefficiency — the squared concurrences add up to exactly the monogamy ceiling — but the unavoidable cost of splitting a bounded resource among more partners.

How is a W state related to a magnon or a Dicke state?

They are the same object in different languages. A Dicke state $|D^{(k)}_N\rangle$ is the symmetric state of $N$ two-level systems with $k$ excitations; the W state is the single-excitation member $|W_N\rangle=|D^{(1)}_N\rangle$ . In a spin chain, one flipped spin delocalized symmetrically over all sites — a zero-momentum spin wave, or magnon — is exactly this state, and in atomic physics it is the $J_z=-J+1$ rung above the collective ground state. This is why W states arise wherever a single quantum of collective excitation is weakly driven: superradiance, cavity QED, cold atoms, spin-wave memories.

Is the entanglement entropy of a W state maximal across a bipartition?

No — it is sub-maximal, a structural feature rather than a defect. Across a single-qubit-versus-rest cut the Schmidt rank is only $2$ (excited or not), with unequal weights $\tfrac1N$ and $\tfrac{N-1}{N}$ , so $S=H_2(1/N)$ : about $0.918$ bit at $N=3$ , $0.811$ bit at $N=4$ , tending to $0$ as $N\to\infty$ because the excitation is almost surely elsewhere. A maximally entangled cut would need equal Schmidt weights; W never reaches that, consistent with its entanglement being thinly distributed. (The clean two-class picture is itself special to three qubits — four qubits already form a continuum of SLOCC classes.)