2
$\begingroup$

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

GHZ

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?

Improve this question
$\endgroup$
1
  • 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
    Commented Sep 26, 2021 at 0:15

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Oh right, that's the whole point! Thanks so much. $\endgroup$
    – M. L.
    Commented Sep 25, 2021 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.