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 $n$ 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.

Definition

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. Its two-qubit reduced states are not Bell states.

One-Qubit Reduced States

For

$$\rho_{\mathrm{GHZ}} = \lvert\mathrm{GHZ}_n^+\rangle \langle\mathrm{GHZ}_n^+\rvert,$$

the reduced state of any one qubit is maximally mixed:

$$\rho_k = \operatorname{Tr}_{\{1,\ldots,n\}\setminus\{k\}} \rho_{\mathrm{GHZ}} = \frac12 I.$$

Therefore no single qubit carries the phase-coherent information that distinguishes the GHZ superposition from a classical mixture. The information is stored in joint correlations.

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.

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

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

The nonselective average over both outcomes returns the separable mixture above.

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 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.

Foundations Preview

GHZ states are central in foundations because three or more parties allow an all-or-nothing contradiction with certain local hidden-variable assignments. The three-qubit GHZ argument does not rely on statistical inequality violation in the same way as the standard CHSH presentation; it uses perfect correlations among selected Pauli measurements.

This page does not give the full GHZ theorem. The important composite-systems point is that multipartite entanglement can have correlation patterns that are not reducible to pairwise Bell-state intuition.

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.

Common Mistakes

• Thinking the three-qubit GHZ state is just a Bell pair with an extra spectator qubit.
• Assuming two-qubit reductions of a GHZ state are entangled.
• Confusing tracing out a qubit with measuring it and conditioning on the outcome.
• Treating strong $Z$-basis agreement as proof of pairwise entanglement.
• Forgetting that GHZ entanglement is relative to the chosen qubit tensor-product structure.
• Ignoring the relative phase when discussing $X$-basis or stabilizer correlations.

Cross-Links

• Entangled States
• Bell States
• Classical Correlation versus Entanglement
• Reduced Density Operators
• Schmidt Rank
• Entanglement Entropy
• Mutual Information
• Multipartite Systems
• W States
• Graph States
• Stabilizer States Preview
• Multipartite Separability
• Monogamy of Entanglement
• Entanglement Sharing
• Heisenberg Scaling
• Distributed Quantum Sensing
• Formula Sheet

References

• D. M. Greenberger, M. A. Horne, and A. Zeilinger, “Going Beyond Bell’s Theorem,” in Bell’s Theorem, Quantum Theory, and Conceptions of the Universe, Kluwer, 1989.
• N. D. Mermin, “Extreme Quantum Entanglement in a Superposition of Macroscopically Distinct States,” Physical Review Letters 65, 1838-1840, 1990.
• D. M. Greenberger, M. A. Horne, A. Shimony, and A. Zeilinger, “Bell’s Theorem without Inequalities,” American Journal of Physics 58, 1131-1143, 1990.
• 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.
• M. Hein, W. Dur, J. Eisert, R. Raussendorf, M. Van den Nest, and H.-J. Briegel, “Entanglement in Graph States and Its Applications,” in Quantum Computers, Algorithms and Chaos, IOS Press, 2006.

Exercises

1. Verify that $\lvert\mathrm{GHZ}_n^+\rangle$ is normalized.

Solution

The two product-basis states $\lvert0\rangle^{\otimes n}$ and $\lvert1\rangle^{\otimes n}$ are orthonormal. Therefore

$$\langle\mathrm{GHZ}_n^+\vert\mathrm{GHZ}_n^+\rangle = \frac12(1+1) = 1.$$

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. Show that a one-qubit reduction of a GHZ state is $I/2$.

Solution

Trace out all qubits except qubit $k$. The diagonal branches contribute

$$\frac12\lvert0\rangle\langle0\rvert \qquad \text{and} \qquad \frac12\lvert1\rangle\langle1\rvert.$$

The cross terms vanish because the traced-out all-zero and all-one strings are orthogonal. Therefore

$$\rho_k = \frac12 \bigl( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \bigr) = \frac12 I.$$

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

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