5
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 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
    Commented Aug 24, 2021 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
    Commented Aug 25, 2021 at 0:48
  • $\begingroup$ Yes, the setting is dimension 3. $\endgroup$
    – BlackHat18
    Commented Aug 25, 2021 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 ♦
    Commented Aug 25, 2021 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
    Commented Aug 25, 2021 at 12:08

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.