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.

  • 4
    $\begingroup$ So do you consider a general Pauli group here or something? Since, normally, isn't Clifford group define to be the normalizer of the Pauli group? which is in dim of power of 2. $\endgroup$
    – KAJ226
    Aug 24 '21 at 20:09
  • 1
    $\begingroup$ The Pauli group can be defined for any dimension; same for the Clifford group. Since the OP mentions "qutrit" in the title (not qubit) I think the setting is meant to be dim=3. $\endgroup$
    – unknown
    Aug 25 '21 at 0:48
  • $\begingroup$ Yes, the setting is dimension 3. $\endgroup$
    – BlackHat18
    Aug 25 '21 at 10:34
  • $\begingroup$ why do you call this a generalization of the Hadamard gate? It doesn't really have many (if any) properties in common with it, aside from it having the form $H\oplus (1)$ $\endgroup$
    – glS
    Aug 25 '21 at 12:06
  • $\begingroup$ It behaves like the Hadamard gate for $|0\rangle$ and $|1\rangle$. The terminology is borrowed from this paper: arxiv.org/pdf/1105.5485.pdf (Section II). $\endgroup$
    – BlackHat18
    Aug 25 '21 at 12:08

The answer is no. Define

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.


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