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.

Helpful background. Multipartite partitions specifies the cut; partial trace and PPT tests distinguish a surviving entangled state from merely an entangled component in one mixture. We take $n\ge2$, with $|W_1\rangle=|1\rangle$ when it occurs in a reduction.

A symmetric single-excitation state

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

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

where $H_2(p)=-p\log_2p-(1-p)\log_2(1-p)$, in bits.

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

$$S_A = - \frac{m}{n}\log_2\frac{m}{n} - \frac{n-m}{n}\log_2\frac{n-m}{n}.$$

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.

Any retained pair from $|W_n\rangle$ has the state

$$\rho_{ij}=\frac{n-2}{n}|00\rangle\langle00|+\frac2n|\Psi^+\rangle\langle\Psi^+|.$$

An entangled component in a mixture is not by itself proof of entanglement. Here a direct test supplies that proof: partial transpose has two eigenvalues $1/n$ and a remaining block

$$\frac1n\begin{pmatrix}n-2&1\\1&0\end{pmatrix}.$$

Its smaller eigenvalue is $[n-2-\sqrt{(n-2)^2+4}]/(2n)<0$ for every finite $n\ge2$. Thus every retained pair is entangled. Its concurrence is $2/n$, so pairwise entanglement becomes weaker as more parties share the excitation.

Robustness Under Loss

For $n\ge3$, discarding a named qubit leaves a mixed state, but every pair of the remaining qubits still has the NPT marginal just computed. This is the precise loss robustness. At least two qubits must remain to discuss entanglement among the survivors, and this calculation does not imply protection against arbitrary noise.

If an accessible 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 $(n-m)/n$;
• for $n\ge3$, tracing out one qubit leaves a mixed state with entangled pair marginals;
• the defining coherence is between different locations of a single excitation.

For pure three-qubit states, GHZ and W represent inequivalent classes under invertible local filters, equivalently reversible conversion with nonzero success probability by stochastic LOCC (Dür, Vidal, and Cirac). This is stronger than a local-unitary distinction; it is not a classification of arbitrary mixed or larger multipartite states.

Correlations

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

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

These events are mutually exclusive: seeing one excitation rules out excitations elsewhere. For $n>2$, two selected qubits can both be zero, so their binary outcomes are not perfectly anticorrelated. 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.

For distinct qubits $i,j$, with $Z|0\rangle=|0\rangle$,

$$\langle Z_i\rangle=1-\frac2n,\quad \langle Z_iZ_j\rangle=1-\frac4n,\quad \langle X_iX_j\rangle=\langle Y_iY_j\rangle=\frac2n.$$

The connected $ZZ$ correlation is $-4/n^2$. A classical mixture of the excitation locations has the same $Z$ statistics but zero $XX$ and $YY$ correlations; these latter observables expose the surviving coherence.

References

• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed entanglement”, Physical Review A 61, 052306 (2000).
• W. Dür, G. Vidal, and J. I. Cirac, “Three qubits can be entangled in two inequivalent ways”, Physical Review A 62, 062314 (2000).
• O. Gühne and G. Tóth, “Entanglement detection”, Physics Reports 474, 1–75 (2009).

Exercises

1. Compare the connected $ZZ$ correlation of two qubits in $|W_n\rangle$ with their Pearson correlation coefficient. When is the anticorrelation perfect?

Solution

The connected correlation is $(1-4/n)-(1-2/n)^2=-4/n^2$. Each $Z$ variance is $1-(1-2/n)^2=4(n-1)/n^2$. The Pearson coefficient is therefore $-1/(n-1)$, which equals $-1$ only for $n=2$. A single excitation excludes two ones without excluding two zeros when $n>2$.

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. Discard two qubits from $|W_4\rangle$. Compute the negativity of the remaining pair.

Solution

The state is $\rho=\tfrac12|00\rangle\langle00|+\tfrac12|\Psi^+\rangle\langle\Psi^+|$. Its partial transpose has eigenvalues $1/4,1/4,(1+\sqrt2)/4,(1-\sqrt2)/4$. Only the last is negative, so $\mathcal N=(\sqrt2-1)/4\simeq0.104$. The pair remains entangled even though the two discarded parties were not measured or used for postselection.