# CNOT expressed with CZ and H gates by taking into account HZH =X

Where equation 1 is:

I can probably brute-force this by explicitly calculating this quantum circuit's effective 4x4 matrix and seeing that its equivalent to this teleportation operation - but what does this have to do with X=HZH? Where is x used here? Is X somehow equivalent to a cNOTgate? X, Z and H are all one-qubit gates...so I don't understand what they mean.

• No, the gates are not the same, as you noticed. The claim is that Eq. (3) does the same as Eq. (1) when applied to $|0\rangle$ on the first qubit and an arbitrary state $|\psi\rangle$ on the second (i.e. it swaps the states). Just compute the action of Eq. (3) on $|0\rangle|\psi\rangle$. – Markus Heinrich Sep 17 '20 at 7:21

The second equation can be understood from this related answer about the fact that $$CZ_{1 \rightarrow 2} = CZ_{2 \rightarrow 1}$$, where first index is the control qubit and the second index is the target qubit. The first equation is similar to this related answer, but let's prove it explicitly where we will use $$HZH = X$$:
$$\begin{equation*} I \otimes H \; (CZ_{1 \rightarrow 2}) \; I \otimes H= \\ = I \otimes H \; (|0\rangle \langle 0 | \otimes I + |1\rangle \langle 1 | \otimes Z) \; I \otimes H =\\ = |0\rangle \langle 0 | \otimes H H + |1\rangle \langle 1 | \otimes HZH = \\ = |0\rangle \langle 0 | \otimes I + |1\rangle \langle 1 | \otimes X = CNOT_{1 \rightarrow 2} \end{equation*}$$
because $$HH = I$$ and $$HZH = X$$.