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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