This is a very basic question about the GHZ state. I know the standard construction:


A Hadamard on one qubit, and then CNOT gates with targets on all the other ones.

However, why can't I just have $n$ Hadamard gates for $n$ qubits? Why would this not be equivalent--what am I missing?

  • 1
    $\begingroup$ Applying Hadmards to $n$ qubits, each one starting in the state $|0\rangle$, produces an equal superposition over all bit strings of length $n$. $\endgroup$
    – Condo
    Sep 26, 2021 at 0:15

1 Answer 1


If you initialize three qubits to $|0\rangle$, apply a Hadamard gate to each, then measure each in the computational basis, the result will be an independent coin flip for each bit: that is, any of 000, 001, 010, ..., 111, each with probability 1/8.

If you measure all three bits of the GHZ state in the computational basis, you'll get either 000 or 111, each with probability 1/2.

  • $\begingroup$ Oh right, that's the whole point! Thanks so much. $\endgroup$
    – M. L.
    Sep 25, 2021 at 23:48

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