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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.

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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.

enter image description here

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.

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  • $\begingroup$ 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. $\endgroup$
    – BlackHat18
    Aug 30, 2021 at 15:28
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    $\begingroup$ 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. $\endgroup$
    – unknown
    Aug 30, 2021 at 16:08
  • $\begingroup$ @unknown what do you mean by $Z$ above are not of order 3? I've expanded the matrix representations in my answer for clarity. $\endgroup$
    – epelaez
    Aug 30, 2021 at 16:30
  • $\begingroup$ @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. $\endgroup$
    – epelaez
    Aug 30, 2021 at 16:30
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    $\begingroup$ @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 $X^3=I, Z^3=I, X'Z'XZ=wI, w^3=I$ $\endgroup$
    – unknown
    Aug 30, 2021 at 16:53

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