4. The GHZ state

The Bell state correlates two qubits. The GHZ state correlates three: one Hadamard and two CNOTs produce a state that is only ever measured as 000 or 111.

1. The idea

Copy the superposition on qubit 0 onto qubit 1, then onto qubit 2. Each CNOT flips its target only in the branch where the control is 1, so the three bits stay equal.

GHZ state. One Hadamard and two CNOTs.

2. The code


import numpy as np
from fqkit import QuantumCircuit, Hadamard, CNOT, run, measure_all
 
np.set_printoptions(precision=4, suppress=True)
 
qc = QuantumCircuit(3)
qc.add_gate(Hadamard(), [0])
qc.add_gate(CNOT(), [0, 1])
qc.add_gate(CNOT(), [0, 2])
 
state = run(qc)
print("Probabilities:", np.round(np.abs(state) ** 2, 4))
print("Counts       :", measure_all(state, shots=1024))

3. The output


Probabilities: [0.5 0.  0.  0.  0.  0.  0.  0.5]
Counts       : {'000': 499, '111': 525}

Eight amplitudes, two of them nonzero. They are the first entry (000) and the last (111). The counts move around 512 / 512; the zeros do not.

4. The explanation

Hadamard on qubit 0 creates the branches 000 and 100.
CNOT from 0 to 1 turns 100 into 110.
CNOT from 0 to 2 turns 110 into 111.
The 000 branch is never flipped. No amplitude is left on 001, 010, 011, 100, 101, or 110.
Any one qubit, looked at alone, is random. Any two already determine the third.

5. The math

|\mathrm{GHZ}\rangle = \frac{|000\rangle + |111\rangle}{\sqrt{2}}

In fqkit’s big endian order the vector has length 8:

\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}

💡

Key insight: adding a qubit did not add a new independent coin. Both CNOTs copy the same superposition, so the only surviving outcomes are the two in which every bit agrees.

Try it yourself

Drop the second CNOT. Which two bitstrings remain, and which qubit is now independent?
Change the second CNOT to [1, 2]. Is the final state the same GHZ state?
The algorithms that use these states are in Algorithms, starting with Deutsch’s algorithm.