# Is the CNOT in the standard three-qubit circuit for the GHZ state necessary?

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?

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

## 1 Answer

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.

• Oh right, that's the whole point! Thanks so much. Sep 25 at 23:48