Timeline for Finite subgroup of $U(4)$ containing a non-Clifford gate and all local Cliffords
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 28, 2022 at 3:49 | comment | added | Jonas Anderson | You're right. I didn't realize that the real Clifford group, $C_n$, was maximal with respect to $SO(2^n, \mathbb{R})$. In regards to the paper you posted, we might be able to use some of their results, but the answer to my question isn't immediate (at least to me). They allow arbitrary one-qubit unitaries with a fixed entangling gate. Here the one-qubit unitaries are restricted to Clifford unitaries. | |
| Jun 27, 2022 at 20:18 | comment | added | unknown | @JonasAnderson, I think real $C_1$ is probably maximal in orthogonal group. The complex case looks a lot different. It doesn't look possible; this paper shows that "any entangling gate is universal"...arxiv.org/abs/quant-ph/0207072 ...it might give clues for a formal proof | |
| Jun 25, 2022 at 19:55 | comment | added | Jonas Anderson | If you have time, definitely take a look at the complex case. I'm very interested in your findings. | |
| Jun 25, 2022 at 19:19 | comment | added | Jonas Anderson | I believe $U$ is non-Clifford in both settings. It's that $C_1$ isn't a maximal finite subgroup to begin with and the question becomes less interesting. | |
| Jun 25, 2022 at 15:14 | comment | added | unknown | The example is for the real Paul and Clifford groups; $U$ is "non-clifford" in that setting. The real example was quick to find; I expect the complex to be more difficult...I'll look if I have time | |
| Jun 25, 2022 at 9:57 | comment | added | Jonas Anderson | The (complex) Clifford group I'm referring to has $C_1= \langle S, H \rangle$ where $S=diag(1,i)$. I believe this definition is standard, but I should have been more clear. The $C_1$ you've written is not a maximal finite subgroup of $U(2,\mathbb{C})$ since $S$ can be added to it. With LC operations from the complex $C_1$ we can see that $(S^3 X \otimes S^3) U^2 (X\otimes S) = C(X)$ for your $U$ above. If $U$ is non-Clifford as you claim then $U$ will violate restriction (3) above and must generate an infinite group with $C_1 \otimes C_1$ as I've defined $C_1$. | |
| Jun 25, 2022 at 2:32 | history | answered | unknown | CC BY-SA 4.0 |