W States

A W state is the symmetric superposition in which exactly one qubit is excited and the excitation is coherently delocalized over all parties:

$$\lvert W_n\rangle = \frac{1}{\sqrt n} \sum_{k=1}^{n} \lvert0\cdots 010\cdots0\rangle,$$

where the $1$ is in slot $k$. For three qubits,

$$\lvert W_3\rangle = \frac{1}{\sqrt3} \bigl( \lvert100\rangle + \lvert010\rangle + \lvert001\rangle \bigr).$$

W states are canonical examples of multipartite entanglement with a very different pattern from GHZ states. GHZ entanglement is stored in coherence between two macroscopically distinct branches. W entanglement is stored in the coherent location of one excitation.

Definition and Symmetry

Let

$$\lvert e_k\rangle = \lvert0\rangle_1\cdots \lvert0\rangle_{k-1} \lvert1\rangle_k \lvert0\rangle_{k+1}\cdots \lvert0\rangle_n.$$

The $n$-qubit W state is

$$\lvert W_n\rangle = \frac{1}{\sqrt n} \sum_{k=1}^{n} \lvert e_k\rangle.$$

The basis vectors $\lvert e_k\rangle$ are orthonormal, so the normalization is immediate:

$$\langle W_n\vert W_n\rangle = \frac{1}{n}\sum_{k=1}^{n} \langle e_k\vert e_k\rangle = 1.$$

The state is invariant under any permutation of qubits. It also lies entirely in the one-excitation sector. If

$$N_1 = \sum_{k=1}^{n} \lvert1\rangle\langle1\rvert_k,$$

then

$$N_1\lvert W_n\rangle = \lvert W_n\rangle.$$

Here $N_1$ counts how many qubits are in $\lvert1\rangle$. This number operator is only a convenient finite-qubit notation; it should not be confused with the full Fock-space number operator for identical particles.

One-Qubit Reductions

Because of permutation symmetry, all one-qubit reductions are the same. Separating qubit $k$ from the other $n-1$ qubits gives

$$\lvert W_n\rangle = \frac{1}{\sqrt n} \lvert1\rangle_k \lvert0\rangle^{\otimes n-1} + \sqrt{\frac{n-1}{n}}\, \lvert0\rangle_k \lvert W_{n-1}\rangle.$$

The two states of the remaining qubits are orthonormal, so this is a Schmidt decomposition for the split

$$k\vert\{1,\ldots,n\}\setminus\{k\}.$$

The one-qubit reduced state is

$$\rho_k = \frac{n-1}{n} \lvert0\rangle\langle0\rvert + \frac{1}{n} \lvert1\rangle\langle1\rvert.$$

Thus a single qubit is not maximally mixed unless $n=2$. Its entropy is the binary entropy

$$S(\rho_k) = H_2\!\left(\frac1n\right),$$

with the logarithm base chosen consistently with the rest of the calculation.

Bipartitions

Let $A$ contain $m$ qubits and let $\bar A$ contain $n-m$ qubits, with

$$1\le m\le n-1.$$

The excitation is either inside $A$ or inside $\bar A$, so

$$\lvert W_n\rangle = \sqrt{\frac{m}{n}}\, \lvert W_m\rangle_A \lvert0\rangle_{\bar A}^{\otimes n-m} + \sqrt{\frac{n-m}{n}}\, \lvert0\rangle_A^{\otimes m} \lvert W_{n-m}\rangle_{\bar A}.$$

The two vectors on each side are orthonormal. Therefore every nontrivial bipartition has Schmidt rank

$$\operatorname{SR}=2,$$

with Schmidt probabilities

$$\frac{m}{n} \quad \text{and} \quad \frac{n-m}{n}.$$

The entanglement entropy across this cut is

$$S_A = - \frac{m}{n}\log\frac{m}{n} - \frac{n-m}{n}\log\frac{n-m}{n}.$$

Unlike a GHZ state, the amount of bipartite entanglement depends on the size of the cut.

Two-Qubit Reductions

For $\lvert W_3\rangle$, tracing out qubit $C$ gives

$$\rho_{AB} = \frac23 \lvert\Psi^+\rangle\langle\Psi^+\rvert + \frac13 \lvert00\rangle\langle00\rvert,$$

where

$$\lvert\Psi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr).$$

This is a mixed state, but it still contains coherent two-qubit entanglement. That behavior contrasts sharply with the two-qubit reduction of $\lvert\mathrm{GHZ}_3\rangle$, which is a separable classical mixture of $\lvert00\rangle$ and $\lvert11\rangle$.

For general $n$, tracing out one qubit gives

$$\rho_{\mathrm{rest}} = \frac{n-1}{n} \lvert W_{n-1}\rangle\langle W_{n-1}\rvert + \frac{1}{n} \lvert0\rangle^{\otimes n-1} \langle0\rvert^{\otimes n-1}.$$

The two terms correspond to whether the excitation was not in the lost qubit or was in the lost qubit. There is no coherence between those alternatives after tracing out the lost subsystem.

Robustness Under Loss

W states are often described as robust under particle loss. The precise statement is not that loss has no effect. Tracing out a qubit changes the state from a pure state to a mixed state. The robust feature is that the remaining parties still contain an entangled component:

$$\rho_{\mathrm{rest}} = \frac{n-1}{n} \lvert W_{n-1}\rangle\langle W_{n-1}\rvert + \frac{1}{n} \lvert0\cdots0\rangle\langle0\cdots0\rvert.$$

If the lost or measured qubit is instead measured in the computational basis and the outcome is known, then:

• outcome $0$ occurs with probability $(n-1)/n$ and leaves the remaining qubits in $\lvert W_{n-1}\rangle$;
• outcome $1$ occurs with probability $1/n$ and leaves the remaining qubits in $\lvert0\rangle^{\otimes n-1}$.

The nonselective average over these outcomes is the same mixed state obtained by tracing out the qubit.

Contrast with GHZ States

GHZ and W states are both genuinely multipartite entangled, but they organize correlations differently.

For $\lvert\mathrm{GHZ}_n\rangle$:

• every nontrivial bipartition has Schmidt probabilities $1/2,1/2$;
• tracing out one qubit leaves a separable classical mixture on the remaining qubits;
• the phase coherence is visible in an $n$-body observable such as $X_1X_2\cdots X_n$.

For $\lvert W_n\rangle$:

• every nontrivial bipartition has Schmidt rank $2$, but the probabilities are $m/n$ and $(n-m)/n$;
• tracing out one qubit leaves a mixed state with an entangled $\lvert W_{n-1}\rangle$ component;
• the defining coherence is between different locations of a single excitation.

For three qubits, GHZ-type and W-type entanglement are inequivalent under stochastic local operations and classical communication. That classification is a preview of multipartite entanglement theory, where no single scalar measure captures all operational distinctions.

Correlations

The computational-basis statistics of $\lvert W_n\rangle$ are simple: exactly one qubit is found in $\lvert1\rangle$. Therefore

$$\Pr(x_k=1) = \frac1n, \qquad \Pr(x_i=1,x_j=1) = 0 \quad (i\ne j).$$

The state has perfect anticorrelation in the sense that seeing one excitation rules out excitations elsewhere. But the state is not merely a classical mixture over the possible excitation locations. The off-diagonal terms

$$\lvert e_i\rangle\langle e_j\rvert, \qquad i\ne j,$$

carry phase coherence between locations.

This is why the reduced one-qubit density matrix alone is not enough to describe the state. The one-qubit state only says how often a chosen qubit is excited; the multipartite state also says that the alternatives are coherently superposed.

Common Mistakes

• Thinking W states are less important than GHZ states because their branches differ by only one excitation.
• Assuming a one-qubit reduction of $\lvert W_n\rangle$ is maximally mixed for all $n$.
• Saying W states are immune to decoherence; they are robust under a specific loss comparison, not protected against arbitrary noise.
• Treating GHZ and W states as related by local unitaries.
• Calling the one-excitation superposition “particle entanglement” without specifying whether the tensor factors are qubits, sites, modes, or distinguishable parties.
• Inferring genuine multipartite entanglement from pairwise entanglement alone; multipartite classification requires more structure.

Cross-Links

• GHZ States
• Multipartite Systems
• Graph States
• Multipartite Separability
• Monogamy of Entanglement
• Entanglement Sharing
• Entangled States
• Bell States
• Classical Correlation versus Entanglement
• Reduced Density Operators
• Schmidt Rank
• Entanglement Entropy
• Mutual Information
• Formula Sheet

References

• W. Dur, G. Vidal, and J. I. Cirac, “Three Qubits Can Be Entangled in Two Inequivalent Ways,” Physical Review A 62, 062314, 2000.
• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed Entanglement,” Physical Review A 61, 052306, 2000.
• A. Acin, D. Bruss, M. Lewenstein, and A. Sanpera, “Classification of Mixed Three-Qubit States,” Physical Review Letters 87, 040401, 2001.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• O. Guhne and G. Toth, “Entanglement Detection,” Physics Reports 474, 1-75, 2009.

Exercises

1. Verify the normalization of $\lvert W_n\rangle$.

Solution

The vectors $\lvert e_k\rangle$ are orthonormal because they place the only excitation in different qubit slots. Thus

$$\langle W_n\vert W_n\rangle = \frac1n \sum_{j,k=1}^{n} \langle e_j\vert e_k\rangle = \frac1n \sum_{k=1}^{n}1 = 1.$$

2. Derive the one-qubit reduced state of $\lvert W_n\rangle$.

Solution

Separate qubit $k$ from the rest:

$$\lvert W_n\rangle = \frac{1}{\sqrt n} \lvert1\rangle_k \lvert0\rangle^{\otimes n-1} + \sqrt{\frac{n-1}{n}}\, \lvert0\rangle_k \lvert W_{n-1}\rangle.$$

The two states of the remaining system are orthonormal. Tracing out the remaining qubits leaves

$$\rho_k = \frac1n \lvert1\rangle\langle1\rvert + \frac{n-1}{n} \lvert0\rangle\langle0\rvert.$$

3. Trace out qubit $C$ from $\lvert W_3\rangle$ and express the result on $AB$ using $\lvert\Psi^+\rangle$.

Solution

Write

$$\lvert W_3\rangle = \frac{1}{\sqrt3} \bigl( \lvert10\rangle_{AB}\lvert0\rangle_C + \lvert01\rangle_{AB}\lvert0\rangle_C + \lvert00\rangle_{AB}\lvert1\rangle_C \bigr).$$

The two terms with $\lvert0\rangle_C$ remain coherent with each other after tracing out $C$, while the term with $\lvert1\rangle_C$ is orthogonal to them. Therefore

$$\rho_{AB} = \frac13 \bigl( \lvert10\rangle+\lvert01\rangle \bigr) \bigl( \langle10\rvert+\langle01\rvert \bigr) + \frac13 \lvert00\rangle\langle00\rvert.$$

Since

$$\bigl( \lvert10\rangle+\lvert01\rangle \bigr) \bigl( \langle10\rvert+\langle01\rvert \bigr) = 2\lvert\Psi^+\rangle\langle\Psi^+\rvert,$$

we get

$$\rho_{AB} = \frac23 \lvert\Psi^+\rangle\langle\Psi^+\rvert + \frac13 \lvert00\rangle\langle00\rvert.$$

4. Compute the Schmidt probabilities of $\lvert W_n\rangle$ across a split with $m$ qubits on one side.

Solution

The excitation is in the $m$-qubit side with total probability $m/n$ and in the complementary side with total probability $(n-m)/n$. The normalized branch states are $\lvert W_m\rangle_A\lvert0\rangle_{\bar A}^{\otimes n-m}$ and $\lvert0\rangle_A^{\otimes m}\lvert W_{n-m}\rangle_{\bar A}$. Hence the Schmidt coefficients are

$$\sqrt{\frac{m}{n}}, \qquad \sqrt{\frac{n-m}{n}},$$

and the Schmidt probabilities are

$$\frac{m}{n}, \qquad \frac{n-m}{n}.$$

5. Compare the result of losing one qubit from $\lvert\mathrm{GHZ}_3\rangle$ and from $\lvert W_3\rangle$.

Solution

Tracing out one qubit from $\lvert\mathrm{GHZ}_3\rangle$ gives

$$\frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert,$$

which is separable. Tracing out one qubit from $\lvert W_3\rangle$ gives

$$\frac23 \lvert\Psi^+\rangle\langle\Psi^+\rvert + \frac13 \lvert00\rangle\langle00\rvert.$$

The latter is mixed, but it still contains a coherent Bell-state component. This is the elementary sense in which W-state entanglement is more robust under loss than GHZ-state entanglement.