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The generators for single qubit Clifford are phase $ P $ and Hadamard $ H $.

The generators for single qutrit Clifford can be found for example here

What is the set of generators for the qutrit Clifford group?

What is a small/minimal generating set of matrices for the single qudit modular Clifford group $ \mathrm{Cl}(1,\mathbb{Z}_4) $?

For a discussion of the modular Clifford group $ \mathrm{Cl}(1,\mathbb{Z}_4) $ and a comparison with $ \mathrm{Cl}(1,\mathbb{F}_4) $ see Which Clifford groups are 2-designs?

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    $\begingroup$ The result in the answer of the post you link to extends to all integers (not just primes). So $H$ (Fourier) + $P$ (Phase or shear) are enough to generate the group. Also see arxiv.org/abs/1307.5087 $\endgroup$
    – unknown
    Commented Jul 11, 2022 at 2:47
  • $\begingroup$ @unknown You claim that the linked post extends to all integers. That appears to be false. For example plugging $ p=2 $ into that definition of $ S $ gives the matrix $ Z $ which, together with $ H $ gate, does not generate the single qubit Clifford group. Similarly, plugging in $ p=4 $, the matrices $ S $ and $ H $ constructed that way do not generate a finite group which strongly suggests that they do not generate the single 4-dit Clifford group. Indeed this fact can verified by checking that the determinant 1 versions of these matrices do not generate a finite group in $ SU_4 $. $\endgroup$ Commented Jul 11, 2022 at 21:21
  • $\begingroup$ For $p=2$, did you use equations 31 and 33?...these should work I think (there's a slight adjustment for p odd or even, eqn 32 vs 33...) $\endgroup$
    – unknown
    Commented Jul 11, 2022 at 21:57
  • $\begingroup$ for $p=2,3,4,5$ I get group size = $192,2592,1536,30000$ $\endgroup$
    – unknown
    Commented Jul 11, 2022 at 22:27
  • $\begingroup$ Nice! Looks like there was some confusion here. I thought you were saying that the result in the answer to the qutrit post I linked extended to all integers. That would be wrong. But if you were instead saying that the definition in equations 31 & 33 of the arxiv paper you mention extends to all integers then that sounds very promising, I'll definitely take a look! $\endgroup$ Commented Jul 13, 2022 at 14:57

2 Answers 2

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The generators for qudit clifford group are give here https://arxiv.org/abs/1911.08162 This is more concise than the paper in the comment and takes care of subtleties better.

Here is a short GAP program that defines these generators for any dimension:

TestA:=function(p)local Ig,Zg,Xg,Hg,Pg,grp,cen,sgrp,scen,C;
 Ig:=IdentityMat(p);
 Zg:=DiagonalMat(List([0..p-1],x->E(p)^x));
 Xg:=IdentityMat(p){\mod([0..p-1]+1,p)+1};
 Hg:=List([0..p-1],x->List([0..p-1],y->E(p)^(x*y)))/ER(p);
 Pg:=DiagonalMat(List([0..p-1],x->E(2*p)^(x*(x-\mod(p,2)))));
 grp:=Group([Hg,Pg,Zg]);cen:=Center(grp);sgrp:=Size(grp);scen:=Size(cen);
 Print(" |G| = ",String(sgrp,-5));Print(" |Cen(G)| = ",String(scen,-5));Print(" |G/Cen(G)| = ",String(sgrp/scen,-5));Print("\n");
 if(p=4)then
 Print("Z=\n");PrintArray(Zg);
 Print("X=\n");PrintArray(Xg);
 Print("H=\n");PrintArray(Hg);
 Print("P=\n");PrintArray(Pg);
 C:=KroneckerProduct([[1,0],[0,1]],[[1,1],[1,-1]]/ER(2));C:=C{[1,3,2,4]}{[1,3,2,4]};
 Zg:=C^-1*Zg*C;
 Xg:=C^-1*Xg*C;
 Hg:=C^-1*Hg*C;
 Pg:=C^-1*Pg*C;
 grp:=Group([Hg,Pg,Zg]);
 Print(" all elements are monomial? ",ForAll(Elements(grp),IsMonomialMatrix),"\n");
 fi;
return grp;end;

Running in GAP gives : (takes seconds)

gap> g2:=TestA(2);;
 |G| = 192   |Cen(G)| = 8     |G/Cen(G)| = 24   
gap> g3:=TestA(3);;
 |G| = 2592  |Cen(G)| = 12    |G/Cen(G)| = 216  
gap> g4:=TestA(4);;
 |G| = 6144  |Cen(G)| = 8     |G/Cen(G)| = 768  
Z=
[ [      1,      0,      0,      0 ],
  [      0,   E(4),      0,      0 ],
  [      0,      0,     -1,      0 ],
  [      0,      0,      0,  -E(4) ] ]
X=
[ [  0,  1,  0,  0 ],
  [  0,  0,  1,  0 ],
  [  0,  0,  0,  1 ],
  [  1,  0,  0,  0 ] ]
H=
[ [        1/2,        1/2,        1/2,        1/2 ],
  [        1/2,   1/2*E(4),       -1/2,  -1/2*E(4) ],
  [        1/2,       -1/2,        1/2,       -1/2 ],
  [        1/2,  -1/2*E(4),       -1/2,   1/2*E(4) ] ]
P=
[ [     1,     0,     0,     0 ],
  [     0,  E(8),     0,     0 ],
  [     0,     0,    -1,     0 ],
  [     0,     0,     0,  E(8) ] ]
 all elements are monomial? true
gap> g5:=TestA(5);;
 |G| = 30000 |Cen(G)| = 10    |G/Cen(G)| = 3000 

As a check, the calculated $G/Cen(G)$ match the results in this paper https://arxiv.org/abs/1810.10259

The GAP code also calculates the generators in an alternate basis. This is based on this https://arxiv.org/abs/1202.3559 where the entire clifford group consists of monomial matrices (normally only the pauli's are). This can be done when $p$ is a square.

As a final note, when $p$ is prime $H$ and $P$ seem enough (so $Z$ is redundant in the generators).

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  • $\begingroup$ Wow your code is beautiful! The single qubit and single qutrit case definitely look good, I recognize those numbers for $ G/Cen(G) $. $\endgroup$ Commented Jul 13, 2022 at 15:01
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    $\begingroup$ Great answer +1 very interesting paper about basis where the entire Clifford group is monomial matrices if $ p $ is square. Interesting to note that Markus Heinrich here quantumcomputing.stackexchange.com/questions/13643/… and especially in chapter 4 of his dissertation uses a different definition of Clifford which differs from the standard one if the qudits are not of prime dimension. For example his single qudit Clifford group for d=4 has order $ 4^3*(4^2-1)= 960 $. I'll probably ask a new question about the reason for this alternative definition. $\endgroup$ Commented Jul 13, 2022 at 15:50
  • $\begingroup$ Very interesting that we need to include one of the 4dit Pauli operators (X or Z ) in addition to the $ H $ and $ P $ generators to get the whole Clifford group. The paper you got this from arxiv.org/abs/1911.08162 does a great job. Looking back at the Farinholt paper you mentioned earlier arxiv.org/abs/1307.5087 it seems that the author has totally overlooked the need to include a Pauli operator in the generating set. The group constructed from his equations 31 & 33 does not contain 4dit X or 4dit Z. Indeed the group is not even irreducible. Do you think its worth emailing the guy? $\endgroup$ Commented Jul 13, 2022 at 16:17
  • $\begingroup$ @IanGershonTeixeira It's an old paper and it looks like this was caught by others. There are many slightly different ways to define the Pauli and Clifford groups (real/complex, global phases, ...) and subtle differences are harder to resolve. That's why I like to calculate things directly when possible...I'll take a look at the other post a little later. $\endgroup$
    – unknown
    Commented Jul 13, 2022 at 16:25
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    $\begingroup$ @unknown Good point! It took a year and a half but I finally got around to asking about $ Z(p^a) $ versus $ GF(p^a) $: quantumcomputing.stackexchange.com/questions/35297/… and I got a wonderful answer from Markus Heinrich! $\endgroup$ Commented Jan 3 at 17:18
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Just wanted to give a supplementary answer providing generators for the other Clifford group $ \mathrm{Cl}(1,\mathbb{F}_4) $, taken from section 4.1.3 of the dissertation of Markus Heinrich. Note that the group $ \mathrm{Cl}(1,\mathbb{F}_4) $ discussed in this answer is a 2 design and differs from modular Clifford group $ \mathrm{Cl}(1,\mathbb{Z}_4) $ discussed in the other answer, which is not a 2-design. This difference is already mentioned in the comments below the first answer.

The generators he gives are $ X:=X(1)=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} $

$ Z:=Z(1)=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} $

$ S=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & 0 & 0 & i \end{bmatrix} $

$ H= \frac{1}{2} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \\ 1 & -1 & 1 & -1 \end{bmatrix} $

and finally a matrix that permutes the qudit basis by multiplying the $ \mathbb{F}_4 $ index by the field element $ a \in \mathbb{F}_4 $ indexing the third basis vector $ M(a)= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} $

Of course you don't really need all these generators since $ S^2=Z $ and then conjugating $ Z $ by $ H $ you can obtain $ X $. So $ H,S,M(a) $ is a generating set for $ \mathrm{Cl}(1,\mathbb{F}_4) $. A two element generating set is probably possible but I'm just not sure what it is.

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