Making a Controlled-Z from a CNOT

Question

How can we draw a circuit that is based on the gates $H, CZ$ that implements $CNOT$.
I know that the $H$ gate is like that:

And also the $CZ$ gate is:

But I'm not sure how to draw this with the implementation of $CNOT$.

Durd3nT · Accepted Answer

You can check that the following is equal to a CNOT gate (the mid-part being the controlled $Z$-gate)

The first Hadamard gate rotates $q_1$ to the $X$-basis. In that basis, the $Z$-gate acts like a bit flip (the same way the $X$-gate acts in the $Z$-basis). The second Hadamard rotates $q_1$ back to $Z$-basis.

KAJ226 · Answer

@Durd3nT answered the question nicely. But here is another way to see it, and hopefully it will be useful for future purposes...
All you need to know is the identity $X = HZH$. Then now you can see that $CNOT (CX)$ can be rewritten as
$$ CX = big( I otimes H big) CZ big( I otimes H big)  $$
This is because when the controlled-qubit is in the state $|0rangle$, you are not doing anything, so the two Hadamard ($H$) gates cancel each other out. And when the controlled-qubit is in the state $|1rangle$, the combination $HZH$ will acts as an $X$ gate.

Making a Controlled-Z from a CNOT

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