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## New answers tagged qutrit

4

From the paper Normal form for single-qutrit Clifford+T operators and synthesis of single-qutrit gates, the Clifford group in $p>2$ dimensions acting on a sigle qudit is generated by $S$ and $H$ given by: $$\begin{gather} S=\sum_{j=0}^{p-1}\omega^{j(j+1)2^{-1}}|j\rangle\langle j| \\ H = \frac{1}{\sqrt{p}}\sum_{j=0}^{p-1}\sum_{k=0}^{p-1}\omega^{jk}|j\... 4 Based on the paper Elementary gates of ternary quantum logic circuit, the extensions of the Z gate are as follows$$ \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 &...

5

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 ...

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