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Consider a simple generalization of the Hadamard gate to qutrits, defined as follows.
\begin{equation} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} &0 \\ 0 &0&1 \end{pmatrix} \end{equation}
It is well known that the two qubit Hadamard gate is a Clifford gate. Is the above gate also a Clifford gate? I could not verify if so.
The answer is no. Define
X=[[0,1,0],[0,0,1],[1,0,0]]
Z=[[1,0,0],[0,w,0],[0,0,w^2]], w^3=1
Then the Pauli group is generated by X and Z and is of order 27. With H being your matrix, you can check that H'XH and H'ZH are not in the group.
Calculations like this are easy to do in gap
The dim=3 counterpart of the Hadamard gate is the 3 dimensional Fourier transform matrix.
