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



From this link:

enter image description here

Where equation 1 is: enter image description here

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 I don't understand what they mean.

Steven Sagona

Posted 2020-09-17T06:52:49.807

Reputation: 723

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 – 2020-09-17T07:21:44.523



Here are three equivalent circuits:

enter image description here

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 $HZH = X$.

Davit Khachatryan

Posted 2020-09-17T06:52:49.807

Reputation: 3 583

1Very clear, thanks for your help. Also I saw your profile and will also be looking into your algorithm tutorials - they look very nice. – Steven Sagona – 2020-09-17T08:04:06.863

@StevenSagona, Happy to help. Thanks :) – Davit Khachatryan – 2020-09-17T08:06:08.937