# Qutrit analogues of controlled Z and cc-Z gates

I am trying to look for the qutrit analogues of a controlled-Z, and a cc-Z (Z gate with two controls) for qubits.

There is a previous answer that gives a qutrit analogue of a CNOT gate, but does not talk about the gates I mentioned.

Note that controlled-Z is a Clifford gate. So, it is expected any qutrit analogue should also be a Clifford gate.

• Aug 30 '21 at 14:55
• Look at the qudit formulation of the stabiliser formailsm. There are tons of papers, e.g. one by Gottesman from 1999 arxiv.org/abs/quant-ph/9802007 ... Sep 1 '21 at 10:59

Based on the paper Elementary gates of ternary quantum logic circuit, the extensions of the Z gate are as follows

$$\begin{gather} Z^{[0]} = \text{diag}\{-1, I_2\} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ Z^{[1]} = \text{diag}\{1, -1, 1\} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ Z^{[2]} = \text{diag}\{I_2,-1\} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \end{gather}$$

The controlled variant, denoted $$TCZ$$, is defined as the gate that implements $$Z^{[n']}$$ on the target qutrit iff the control qutrit is in the state $$|n\rangle$$ where $$n\in\{0, 1, 2\}$$. Figure 3 (a) below gives the circuit representation of the $$TCZ$$ gate.

However, as noted in the comments by @unknown, this gates are not Clifford. For a Clifford formulation of the $$Z$$ gate, we can turn to Canonical forms for single-qutrit Clifford+T operators that defines it as

$$Z = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^2 \end{bmatrix},$$

with $$\omega = e^{2 \pi i / 3}$$ which means $$Z^3=I$$. To convert this into a qutrit controlled-$$Z$$ operator, I recommend taking a look at this answer, taken from a question linked in the comment to your original question.

• Is there a way to see that this gate still remains a Clifford gate? The qutrit extension of the Hadamard gate, as mentioned in that paper, is not a Clifford gate. Aug 30 '21 at 15:28
• I doubt that they are Clifford. The process to check if $CZ$ is Clifford or not is the same as for the Hadamard gate but now the Pauli group involved is of order $3*9*9=243$ and the matrices are $9x9$. Note that $Z$ above are not of order 3; so even these don't look like natural generalization of qubit $Z$ gates. Aug 30 '21 at 16:08
• @unknown what do you mean by $Z$ above are not of order 3? I've expanded the matrix representations in my answer for clarity. Aug 30 '21 at 16:30
• @BlackHat18: as stated in the other comment, I think this generalization of $TCZ$ is not a Clifford a gate, but it is what I've found most in literature. Aug 30 '21 at 16:30
• @epelaaez the matrices are of dimension 3 ($3x3$) but order 2 not 3. So $Z^2=I$; the most "natural" generalization of qubits to qutrits defines $X^3=I, Z^3=I, X'Z'XZ=wI, w^3=I$ Aug 30 '21 at 16:53