Greenberger–Horne–Zeilinger state

Generalization

The generalized GHZ state is an entangled quantum state of M > 2 subsystems. If each system has dimension d, i.e., the local Hilbert space is isomorphic to ${\displaystyle \mathbb {C} ^{d}}$, then the total Hilbert space of an M-partite system is ${\displaystyle {\mathcal {H}}_{\rm {tot}}=(\mathbb {C} ^{d})^{\otimes M}}$. This GHZ state is also called an M-partite qudit GHZ state. Its formula as a tensor product is^{[8]: Eqn 82}

${\displaystyle |\mathrm {GHZ} \rangle ={\frac {1}{\sqrt {d}}}\sum _{i=0}^{d-1}|i\rangle \otimes \cdots \otimes |i\rangle ={\frac {1}{\sqrt {d}}}(|0\rangle \otimes \cdots \otimes |0\rangle +\cdots +|d-1\rangle \otimes \cdots \otimes |d-1\rangle )}$.

In the case of each of the subsystems being two-dimensional, that is for a collection of M qubits, it reads

${\displaystyle |\mathrm {GHZ} \rangle ={\frac {|0\rangle ^{\otimes M}+|1\rangle ^{\otimes M}}{\sqrt {2}}}.}$

Photon implementation

The GHZ experiment was performed using photon polarization qubits,^[5] for which the basis states can be chosen to be horizontal and vertical polarization under a given coordinate system:

${\displaystyle |0\rangle \equiv |\mathrm {H} \rangle ,\qquad |1\rangle \equiv |\mathrm {V} \rangle .}$

With appropriately chosen phase factors for ${\displaystyle |\mathrm {H} \rangle }$ and ${\displaystyle |\mathrm {V} \rangle }$, both Pauli ${\displaystyle X}$ and ${\displaystyle Y}$ measurements can be implemented by two-channel polarizers:

• A linear polarizer rotated by 45° from the axes of the coordinate system implements a Pauli ${\displaystyle X}$ measurement, distinguishing between the eigenstates

${\displaystyle |{+X}\rangle \equiv |+\rangle ={\frac {1}{\sqrt {2}}}(|\mathrm {H} \rangle +|\mathrm {V} \rangle ),\qquad |{-X}\rangle \equiv |-\rangle ={\frac {1}{\sqrt {2}}}(|\mathrm {H} \rangle -|\mathrm {V} \rangle ).}$

• A circular polarizer implements a Pauli ${\displaystyle Y}$ measurement, distinguishing between the eigenstates

${\displaystyle |{+Y}\rangle \equiv |R\rangle ={\frac {1}{\sqrt {2}}}(|\mathrm {H} \rangle +i|\mathrm {V} \rangle ),\qquad |{-Y}\rangle \equiv |L\rangle ={\frac {1}{\sqrt {2}}}(|\mathrm {H} \rangle -i|\mathrm {V} \rangle ).}$

For example, to implement a ${\displaystyle Y_{1}Y_{2}X_{3}}$ measurement, one would apply circular polarizers on photons 1 and 2, and a 45° linear polarizer on photon 3. Quantum mechanics predict that the product of the results (each written as ±1) should be −1. In terms of photon polarization, only four result combinations are possible: {RL+, LR+, RR−, LL−}.

Within the experimental error, the results of the photon experiment agree with the predictions of quantum mechanics. Furthermore, the error rate is low enough (~15% for each measurement configuration) that the results are statistically in conflict with local realism.

Pairwise entanglement

An important property of the GHZ state is that taking the partial trace over one of the three qubits yields

${\displaystyle \operatorname {Tr} _{3}\left[\left({\frac {|000\rangle +|111\rangle }{\sqrt {2}}}\right)\left({\frac {\langle 000|+\langle 111|}{\sqrt {2}}}\right)\right]={\frac {(|00\rangle \langle 00|+|11\rangle \langle 11|)}{2}},}$

which is an unentangled mixed state. It has certain two-particle (qubit) correlations, but these are of a classical nature. This is also evidenced by the fact that measuring the third qubit in the Pauli Z basis will leave a product state ${\displaystyle |00\rangle }$ or ${\displaystyle |11\rangle }$ on the first two qubits, which are unentangled pure states.

However, measuring the third qubit in the Pauli X or Y basis can leave the first two qubits in a maximally entangled Bell state.^[9] The 3-qubit GHZ state written with the third qubit expressed in a Pauli X basis is:

${\displaystyle {\begin{aligned}|\mathrm {GHZ} \rangle &={\frac {1}{\sqrt {2}}}\left(|000\rangle +|111\rangle \right)\\&={\frac {1}{\sqrt {2}}}\left(|00\rangle \otimes {\frac {|+\rangle +|-\rangle }{\sqrt {2}}}+|11\rangle \otimes {\frac {|+\rangle -|-\rangle }{\sqrt {2}}}\right)\\&={\frac {1}{2}}\left(|00\rangle +|11\rangle \right)\otimes |+\rangle +{\frac {1}{2}}\left(|00\rangle -|11\rangle \right)\otimes |-\rangle \\&={\frac {1}{\sqrt {2}}}\left(|\Phi ^{+}\rangle \otimes |+\rangle +|\Phi ^{-}\rangle \otimes |-\rangle \right).\end{aligned}}}$

An X basis measurement of the third qubit that yields ${\displaystyle |+\rangle }$ selects ${\displaystyle |\Phi ^{+}\rangle =(|00\rangle +|11\rangle )/{\sqrt {2}}}$; similarly, if ${\displaystyle |-\rangle }$ results then ${\displaystyle |\Phi ^{-}\rangle =(|00\rangle -|11\rangle )/{\sqrt {2}}}$ is selected. Either case is a maximally entangled Bell state.

As a corollary, measurement statistics of the first two qubits will exhibit quantum nonlocality. However, the exact nature of this correlation depends on the measurement result of the third qubit, and thus it does not manifest when the latter measurement result is unknown. For example, measuring both qubits of ${\displaystyle |\Phi ^{+}\rangle }$ in the X basis will always yield the same result, but the same measurement on ${\displaystyle |\Phi ^{-}\rangle }$ will always yield opposite results. Therefore, when the measurement result of the third qubit is unknown, the X measurement results of the first two qubits appear uncorrelated. This phenomenon, where entanglement of two quantum systems are dependent on the measurement of a third system, was named "entangled entanglement" by Krenn and Zeilinger.^[10]

The correlation between the first two qubits can also be made definite by applying a local quantum gate depending on the measurement result of the third qubit. For example, when the result is ${\displaystyle |-\rangle }$, one can apply a Z gate to either of the first two qubit to transform ${\displaystyle |\Phi ^{-}\rangle }$ into ${\displaystyle |\Phi ^{+}\rangle }$. Therefore the 3-qubit GHZ state can be converted into a 2-qubit Bell state via local operations and classical communication.

Multi-partite entanglement

There is no standard measure of multi-partite entanglement because different, not mutually convertible, types of multi-partite entanglement exist. Nonetheless, many measures define the GHZ state to be a maximally entangled state.^{[citation needed]}

A pure state ${\displaystyle |\psi \rangle }$ of ${\displaystyle N}$ parties is called biseparable, if one can find a partition of the parties in two nonempty disjoint subsets ${\displaystyle A}$ and ${\displaystyle B}$ with ${\displaystyle A\cup B=\{1,\dots ,N\}}$ such that ${\displaystyle |\psi \rangle =|\phi \rangle _{A}\otimes |\gamma \rangle _{B}}$, i.e. ${\displaystyle |\psi \rangle }$ is a product state with respect to the partition ${\displaystyle A|B}$. The GHZ state is non-biseparable and is the representative of one of the two non-biseparable classes of 3-qubit states which cannot be transformed (not even probabilistically) into each other by local quantum operations, the other being the W state, ${\displaystyle |\mathrm {W} \rangle =(|001\rangle +|010\rangle +|100\rangle )/{\sqrt {3}}}$.^[8]: 903 Thus ${\displaystyle |\mathrm {GHZ} \rangle }$ and ${\displaystyle |\mathrm {W} \rangle }$ represent two very different kinds of entanglement for three or more particles.^[11]

The W state is, in a certain sense, "less entangled" than the GHZ state; however, that entanglement is more robust against qubit loss, in the sense that a 3-qubit W state remains a bipartite entangled state when one of the qubits is discarded. If one qubit is measured in the Z basis, the entanglement between the other two qubits is lost with probability 1/3, but is promoted to maximal entanglement with probability 2/3. By contrast, for the 3-qubit GHZ state, discarding a qubit results in a separable mixed state, and a Z measurement on a qubit will always leave behind a pure product state. Both the GHZ state and the W state can be generalized to systems with more than 3 qubits, and both retain their own characteristics: The N-qubit GHZ state becomes unentangled if even one qubit is lost, and the N-qubit W state keeps some entanglement even if only two qubits remain.