GHZ States

A GHZ state is a multipartite entangled state of the form

$$\lvert\mathrm{GHZ}_n\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\cdots0\rangle + \lvert11\cdots1\rangle \bigr)$$

for $n\ge2$ qubits. The name comes from Greenberger, Horne, and Zeilinger.

For $n=2$, this is the Bell state $\lvert\Phi^+\rangle$. For $n\ge3$, it is a canonical example of genuinely multipartite entanglement: the coherence is shared among all parties, but tracing out one party destroys the pure GHZ coherence among the rest.

Helpful background. Multipartite partitions explains the different cuts; conditional states distinguishes keeping a measurement outcome from discarding a subsystem.

The GHZ superposition

For $n$ qubits with product basis

$$\lvert x_1x_2\cdots x_n\rangle, \qquad x_k\in\{0,1\},$$

the standard $n$-qubit GHZ state is

$$\lvert\mathrm{GHZ}_n^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle^{\otimes n} + \lvert1\rangle^{\otimes n} \bigr).$$

The minus-phase version is

$$\lvert\mathrm{GHZ}_n^-\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle^{\otimes n} - \lvert1\rangle^{\otimes n} \bigr).$$

More generally, one may write

$$\lvert\mathrm{GHZ}_n(\phi)\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle^{\otimes n} + e^{i\phi}\lvert1\rangle^{\otimes n} \bigr).$$

A local phase rotation on one qubit can change $\phi$, so the plus state is the standard representative unless the phase is being used as part of an interferometric or stabilizer calculation.

Multipartite Entanglement

For any nontrivial bipartition of the $n$ qubits into a subset $A$ and its complement $\bar A$, the GHZ state has Schmidt form

$$\lvert\mathrm{GHZ}_n^+\rangle = \frac{1}{\sqrt2} \lvert0_A\rangle\lvert0_{\bar A}\rangle + \frac{1}{\sqrt2} \lvert1_A\rangle\lvert1_{\bar A}\rangle,$$

where

$$\lvert0_A\rangle = \lvert0\rangle^{\otimes |A|}, \qquad \lvert1_A\rangle = \lvert1\rangle^{\otimes |A|}.$$

Thus every nontrivial bipartition has Schmidt rank

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

and entanglement entropy

$$S_A=1$$

bit.

This does not mean the state is just a collection of Bell pairs. GHZ entanglement is shared globally. For $n\ge3$, its two-qubit reduced states are not Bell states. Across a cut containing at least two qubits on both sides, the rank-two state is not maximally entangled on the full subsystem spaces; a larger Schmidt rank would be possible.

Reduced States of Proper Subsets

Let $R$ be a nonempty proper subset of the $n$ qubits. Tracing out at least one qubit gives

$$\rho_R = \frac12 \lvert0_R\rangle\langle0_R\rvert + \frac12 \lvert1_R\rangle\langle1_R\rvert,$$

where

$$\lvert0_R\rangle=\lvert0\rangle^{\otimes |R|}, \qquad \lvert1_R\rangle=\lvert1\rangle^{\otimes |R|}.$$

The off-diagonal terms disappear because the traced-out states

$$\lvert0_{\bar R}\rangle \quad \text{and} \quad \lvert1_{\bar R}\rangle$$

are orthogonal.

For three qubits, tracing out qubit $C$ gives

$$\rho_{AB} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert.$$

This is a classically correlated separable state. It has perfect $Z$-basis correlation, but no two-qubit entanglement.

For $|R|=1$, this reduces to $I/2$. The same formula holds for every phase $\phi$: all proper marginals are insensitive to that phase.

Fragility Under Tracing Out

GHZ entanglement is fragile under loss of a subsystem. If one qubit is discarded, the remaining state is not a smaller pure GHZ state. It is the separable mixture

$$\frac12 \lvert00\cdots0\rangle\langle00\cdots0\rvert + \frac12 \lvert11\cdots1\rangle\langle11\cdots1\rvert.$$

This mixture can have strong classical correlations but no entanglement across a partition of the remaining individual qubits.

Do not confuse tracing out with measuring and conditioning. If the last qubit of $\lvert\mathrm{GHZ}_3^+\rangle$ is measured in the $X$ basis and the outcome is kept, the remaining two qubits are conditionally projected into a Bell state:

$$\lvert+\rangle_C \quad \Longrightarrow \quad \frac{1}{\sqrt2} \bigl( \lvert00\rangle_{AB} + \lvert11\rangle_{AB} \bigr),$$

while

$$\lvert-\rangle_C \quad \Longrightarrow \quad \frac{1}{\sqrt2} \bigl( \lvert00\rangle_{AB} - \lvert11\rangle_{AB} \bigr).$$

Each outcome has probability $1/2$. The nonselective average over both outcomes returns the separable mixture above. If the outcome is communicated, a local $Z$ correction in the minus branch makes the final pair $|\Phi^+\rangle$ in every branch. This uses existing three-party entanglement and a classical message; it does not create entanglement from a separable input.

Correlations

GHZ states have perfect agreement in the computational basis. For any pair $i,j$,

$$\langle Z_i Z_j\rangle_{\mathrm{GHZ}} = 1.$$

Each single-qubit $Z$ expectation vanishes:

$$\langle Z_i\rangle_{\mathrm{GHZ}} = 0.$$

The global phase coherence appears in an $n$-body $X$ correlation:

$$\langle X_1X_2\cdots X_n\rangle_{\mathrm{GHZ}^+} = 1,$$

and

$$\langle X_1X_2\cdots X_n\rangle_{\mathrm{GHZ}^-} = -1.$$

For an arbitrary phase,

$$\langle X_1\cdots X_n\rangle=\cos\phi, \qquad \langle Y_1X_2\cdots X_n\rangle=\sin\phi.$$

Together these correlations distinguish the relative phase that all proper marginals miss. For the plus state, a compact stabilizer description is:

$$X_1X_2\cdots X_n\lvert\mathrm{GHZ}_n^+\rangle = \lvert\mathrm{GHZ}_n^+\rangle,$$

and

$$Z_1Z_k\lvert\mathrm{GHZ}_n^+\rangle = \lvert\mathrm{GHZ}_n^+\rangle, \qquad k=2,\ldots,n.$$

The $Z_1Z_k$ operators encode the fact that all computational-basis bits agree. The global $X_1X_2\cdots X_n$ operator encodes the coherence between the all-zero and all-one branches.

Three-party correlations and a GHZ contradiction

For $|\mathrm{GHZ}_3^+\rangle$, the exact Pauli correlations are

$$\langle XXX\rangle=+1,\qquad \langle XYY\rangle=\langle YXY\rangle=\langle YYX\rangle=-1.$$

A local model assigning a fixed $\pm1$ value to each party’s $X$ and $Y$ outcomes cannot satisfy all four. Multiplying the last three predicted products cancels each local $Y$ value twice and requires the product of the three $X$ values to be $-1$, contradicting the first prediction. The quantum observables at a given site do not have simultaneous preassigned values. The ideal argument assumes the perfect correlations and locality/measurement-choice assumptions; experiments with finite noise use statistical inequalities. See entanglement and foundations.

Quantum Information Preview

GHZ states appear in quantum information as examples of multipartite resources:

• controlled secret-sharing and multiparty correlation protocols;
• cat-state and stabilizer-state constructions;
• Heisenberg-scaling sensitivity to collective phase shifts in idealized metrology settings;
• distributed sensing of a declared weighted spatial phase, with matched node-separable bounds and loss accounting;
• tests of noise, loss, and decoherence in multipartite devices.

These applications are context-dependent. A GHZ state is not automatically the best resource for every multipartite task. Its loss sensitivity is part of why W states and Graph States are studied separately.

Loss and dephasing are different changes

Losing one qubit completely removes the coherence from the remaining marginal. By contrast, independent phase-flip noise $\mathcal E(\rho)=(1-q)\rho+qZ\rho Z$ on each qubit multiplies the full state’s off-diagonal term by $(1-2q)^n$. The qubits remain present, but the $X_1\cdots X_n$ contrast decays rapidly with $n$ for fixed $0<q<1/2$. Thus a metrological advantage for an ideal GHZ state does not by itself establish an advantage under a specified loss or noise model.

References

• D. M. Greenberger, M. A. Horne, A. Shimony, and A. Zeilinger, “Bell’s theorem without inequalities”, American Journal of Physics 58, 1131–1143 (1990).
• O. Gühne and G. Tóth, “Entanglement detection”, Physics Reports 474, 1–75 (2009).
• N. D. Mermin, “Extreme quantum entanglement in a superposition of macroscopically distinct states”, Physical Review Letters 65, 1838–1840 (1990).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010).

Exercises

1. Starting from $|0\rangle^{\otimes n}$, apply a Hadamard gate to the first qubit and then controlled-NOT gates from it to every other qubit. Show that the result is $|\mathrm{GHZ}_n^+\rangle$.

Solution

The Hadamard creates $(|0\rangle+|1\rangle)|0\cdots0\rangle/\sqrt2$. Each controlled-NOT leaves the zero-control branch unchanged and flips one target in the one-control branch. After all $n-1$ gates the two terms are $|0\cdots0\rangle$ and $|1\cdots1\rangle$, with equal amplitudes. Their orthogonality confirms normalization.

2. Trace out qubit $C$ from

$$\lvert\mathrm{GHZ}_3^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert000\rangle+\lvert111\rangle \bigr).$$

Solution

The projector contains diagonal terms and cross terms:

$$\frac12 \Bigl( \lvert000\rangle\langle000\rvert + \lvert000\rangle\langle111\rvert + \lvert111\rangle\langle000\rvert + \lvert111\rangle\langle111\rvert \Bigr).$$

Tracing over $C$ removes the cross terms because $\langle0\vert1\rangle=0$. Thus

$$\rho_{AB} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert.$$

3. For $|\mathrm{GHZ}_n(\phi)\rangle$, compute the parity distribution when all qubits are measured in $X$. What happens at $\phi=\pi/2$?

Solution

The product $P$ of all $\pm1$ outcomes has mean $\langle P\rangle=\cos\phi$. Thus $\Pr(P=\pm1)=(1\pm\cos\phi)/2$. At $\phi=\pi/2$, the two parities are equally probable, even though the state remains maximally entangled across every one-versus-rest cut. The $Y_1X_2\cdots X_n$ correlator is then one and reveals the phase.

4. Find the Schmidt rank across any nontrivial bipartition of $\lvert\mathrm{GHZ}_n^+\rangle$.

Solution

For a subset $A$ and complement $\bar A$,

$$\lvert\mathrm{GHZ}_n^+\rangle = \frac{1}{\sqrt2} \lvert0_A\rangle\lvert0_{\bar A}\rangle + \frac{1}{\sqrt2} \lvert1_A\rangle\lvert1_{\bar A}\rangle.$$

The two terms use orthonormal vectors on each side, so this is a Schmidt decomposition with two nonzero coefficients. The Schmidt rank is $2$.

5. Measure qubit $C$ of $\lvert\mathrm{GHZ}_3^+\rangle$ in the $X$ basis. What state remains on $AB$ after the $+$ outcome?

Solution

Use

$$\lvert0\rangle = \frac{1}{\sqrt2} \bigl( \lvert+\rangle+\lvert-\rangle \bigr), \qquad \lvert1\rangle = \frac{1}{\sqrt2} \bigl( \lvert+\rangle-\lvert-\rangle \bigr).$$

Substituting for qubit $C$ and keeping the $\lvert+\rangle_C$ component gives an unnormalized state proportional to

$$\lvert00\rangle_{AB} + \lvert11\rangle_{AB}.$$

After normalization, the conditional state is

$$\frac{1}{\sqrt2} \bigl( \lvert00\rangle_{AB} + \lvert11\rangle_{AB} \bigr).$$