Generators for single qudit Clifford, $d=4$

Question

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 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 this post.

Answer 1

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).

Answer 2

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.