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Confused about the action of the $CCZ$ gate on Pauli operators:

I understand the action of the $CZ$ gate: $$CZ: XI \rightarrow XX$$ $$CZ: IX \rightarrow IX$$ $$CZ: ZI \rightarrow ZI$$ $$CZ: IZ \rightarrow IZ$$ I have verified this by means of matrix multiplication. However, I then get confused when I see: $$CCZ: XII \rightarrow X(CZ)$$ $$CCZ: ZII \rightarrow ZII$$ (similar for acting on $X_{2}$, $X_{3}$, $Z_{2}$, $Z_{3}$).

I just don't understand the action of $CCZ$ on $XII$. Specifically, I don't understand the notation $X(CZ)$. At first I thought it was matrix multiplication but $X$ is $2 \times 2$ and $CZ$ is $4 \times 4$.

If anybody could help me understand what $X(CZ)$ is, that would be very helpful!

am567
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2 Answers2

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$CCZ$ will map $X_iI_jI_k$ to $(X_iI_jI_k)(CZ_{j,k})$ by conjugation where the parenthesis indicates the regular product (i.e. gate chaining) and $CZ_{j,k}$ is the $CZ$ operator between qubit $j$ and $k$ (seen as an operator on three qubits).

This notation is confusing if you do not include indices: $I_1X_2I_3$ is mapped to $(I_1X_2I_3)(CZ_{1,3})$, which is not intuitively understood from $X(CZ)$ reading from left to right, expecting the left most operator to act on the first qubit.

The action of $CCZ$ is more complex than the one of $CZ$ because they do not belong to the same class of the Clifford hierarchy (see this question where its precise definition appears). $CZ$ is a Clifford gate (belonging to the first class) and therefore maps Pauli operators to Pauli operators. $CCZ$ belongs to the second class: it maps Pauli operators to elements of the first class of the hierarchy, here a product between a Pauli operator ($X$) and a non-Pauli operator (the $CZ$ gate).

AG47
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CCZ is defined by

$$ CCZ \vert{x_1,x_2,x_3} \rangle = \vert x_1, x_2 \rangle Z^{x_1\cdot x_2} \vert{x_3}\rangle = (-1)^{x_1 x_2 x_3} \vert{x_1,x_2,x_3} \rangle $$

Therefore, \begin{align} (CCZ)(XII)(CCZ)\vert{x_1,x_2,x_3}\rangle = (-1)^{x_1 x_2 x_3} CCZ\vert{1 \oplus x_1,x_2, x_3} \rangle = (-1)^{x_1 x_2 x_3} (-1)^{(1 \oplus x_1) x_2 x_3} \vert 1 \oplus x_1,x_2, x_3\rangle \end{align} On the other hand, \begin{align} ()\vert{x_1,x_2,x_3} \rangle = \vert{1 \oplus x_1} \rangle \vert {x_2}\rangle Z^{x_2} \vert{x_3} \rangle = (-1)^{x_2 \oplus x_3} \vert 1 \oplus x_1,x_2, x_3\rangle. \end{align} 

These to expressions are the same because exactly one of $x_1$ and $1 \oplus x_1$ equals 1.

Frederik Ravn Klausen
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