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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!

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    $\begingroup$ I think the parentheses in $X(CZ)$ just tell you that this is a tensor product of two gates, where the first is the $X$ gate (2x2 matrix) and the second the $CZ$ gate (4x4 matrix) in contrast to being the tensor product of three gates ($X$, $C$ and $Z$) as in the case of $ZII$ ($Z$, $I$ and $I$). $\endgroup$
    – Mathias
    Commented Apr 4 at 12:06

2 Answers 2

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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).

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  • $\begingroup$ In terms of the matrix representation. $CCZ =diag(1,1,1,1,1,1,1,-1)$. Then $CZ_{1,2}=diag(1,1,0,1,-1,0)$. $CZ_{2,3}=diag(0,1,1,0,1,-1)$ and $CZ_{1,3}=diag(1,0,1,1,0,-1)$? Is this correct? So then CCZ will map $IXI$ to $X_{2} \otimes CZ_{1,3}$, which is a $24 \otimes 24$ matrix? Or does matrix representation not really work here?....I think I've got something wrong along the way $\endgroup$
    – am567
    Commented Apr 4 at 14:19
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    $\begingroup$ $CZ_{1,2}$ might not be as clear as a notation as I thought. By $X_2 \otimes CZ_{1,3}$ I meant the operator which acts as $X$ on qubit 2 and as $CZ$ on qubits 1 and 3 (i.e. a $8\times 8$ matrix). diag(1,1,0,1,-1,0) doesn't make sense as an operator, it is not of size $2^i \times 2^i$. It is probably more rigorous to say that $IXI$ is mapped to $X_2$ followed by a $CZ$ between qubits 1 and 3 and to not try to express it as a tensor product. I will edit my answer. $\endgroup$
    – AG47
    Commented Apr 4 at 14:36
  • $\begingroup$ Okay, thank you! Yes, I had thought that maybe $CZ1,3$ could be expressed as $I \otimes |0\rangle \langle 0| + Z \otimes |2 \rangle \langle 2|$ (hence the $diag(1,1,0,1,-1,0)$) because I had seen CZ1,2 expressed as $I \otimes |0\rangle \langle0| + Z \otimes |1 \rangle \langle 1| $ but obviously this is not the case! $\endgroup$
    – am567
    Commented Apr 4 at 14:48
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    $\begingroup$ @am567 You can have a look at this question to better grasp controlled operations in a system with more than two qubits: quantumcomputing.stackexchange.com/questions/4252/… $\endgroup$
    – AG47
    Commented Apr 4 at 15:24
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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} 𝑋(𝐶𝑍)\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.

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