Which Clifford groups are 2-designs?

Question

Let $ X $ be the $ q \times q $ shift matrix sending $ |y \rangle \mapsto |y+1 \rangle $ where the ket index $ y=0,\dots, q-1 $ is taken mod $ q $. Let $ Z $ be the diagonal $ q \times q $ clock matrix sending $ |y \rangle \mapsto (e^{2 \pi i /q})^m|y \rangle $. Then $ \mathrm{P}(1,\mathbb{Z}_q)=<X,Z> $ (traditionally a global phase is also added as a generator but global phases will be irrelevant throughout this question) is the modular Pauli group for one qudit of dimension $ q $. Similarly the modular Pauli group on $ n $ qudits, each of dimension $ q $, is generated by $$ \mathrm{P}(n,\mathbb{Z}_q):=<\{X_i,Z_i:i=1,\dots n \}> $$ where $ X_i $ is a tensor product with $ X $ acting on the $ i $th qudit and identity on all the other qudits. Then we can define the $ n $ qudit modular Clifford group as $$ \mathrm{Cl}(n,\mathbb{Z}_q):=\{g \in \mathrm{SU}(q^n): g h g^{-1} \in \mathrm{P}(n,\mathbb{Z}_q) \text{ for all } h \in \mathrm{P}(n,\mathbb{Z}_q) \} $$ In other words, the normalizer in the unitary group of the modular Pauli group. In this definition I use $ \mathrm{SU}(q^n) $ instead of $ \mathrm{U}(q^n) $ so that the modular Clifford group will be finite (since the question is about unitary designs, which are by definition finite sets of unitary matrices). There is no loss of generality because $ \mathrm{U}(q^n)=e^{i \theta} \mathrm{SU}(q^n) $ and global phase is irrelevant.

It is well known that $ \mathrm{Cl}(n,\mathbb{Z}_q) $ is always a 2-design if $ q $ is prime. In fact the qubit Clifford group $ \mathrm{Cl}(n,\mathbb{Z}_2) $ is actually a 3-design. It is even known that $ \mathrm{Cl}(n,\mathbb{Z}_2) $ is almost a 4-design in the sense that the frame potential is one away from the minimum that would make it a 4-design.

But what happens when $ q $ is not prime?

In Theorem III.1 of Clifford groups are not always 2-designs it is claimed that $ \mathrm{Cl}(n,\mathbb{Z}_q) $ is a 2-design if and only if $ q $ is prime. In other words, $ \mathrm{Cl}(n,\mathbb{Z}_q) $ for $ q $ composite is never a 2-design. A similar definition seems to be used in Generators for single qudit Clifford, d=4 where explicit generators of $ \mathrm{Cl}(1,\mathbb{Z}_4) $ show that $ \mathrm{Cl}(1,\mathbb{Z}_4) $ is not a 2-design

However in section 12.2.1 "The Clifford group is a unitary 2-design" of the dissertation of Markus Heinrich it is claimed that $ \mathrm{Cl}(n,\mathbb{F}_q) $ is a 2-design for any prime power $ q $. This paper by Heinrich's advisor David Gross also seems to claim that $ \mathrm{Cl}(n,\mathbb{F}_q) $ is a 2-design for any prime power $ q $.

I assume this difference is just arising from definitions. Can someone explain the difference in definition? Why do the generators in Generators for single qudit Clifford, d=4 seem to show that $ \mathrm{Cl}(1,\mathbb{Z}_4) $ is not a 2-design but the proof in Heinrich/Gross claims that $ \mathrm{Cl}(1,\mathbb{F}_4) $ is a 2-design?

1/3/2023: This is probably terrible manners but I love the answer from Markus Heinrich so much that I took his new notation $ \mathrm{Cl}(n,\mathbb{Z}_q) $ and $ \mathrm{Cl}(n,\mathbb{F}_q) $ and I went back into the original question and applied it in all the appropriate places. Same thing with the use of the term modular that he suggests.

Answer 1

The confusion stems from the existence of incompatible definitions of the "Clifford group" in dimensions which are not prime. With your definition, the Clifford group is indeed a unitary 2-design if and only if $q$ is prime.

In this definition, one chooses to label a basis of $\mathbb C^q$ by elements in the residue ring $\mathbb Z_q$. Based on modular arithmetic, the Weyl operators are then defined as in your question. I sometimes call this the "modular" definition.

This definition works for any $q\in\mathbb N$, however it is by no means canonical. For instance, if $q=p^k$ is the power of a prime, one can alternatively label the basis by elements of the finite field $\mathbb F_q$ with $q$ elements. One can then proceed to define $$ X(x) |y\rangle = |x+y\rangle, \qquad Z(z)|y\rangle = \chi(z\cdot y) |y\rangle\,, $$ where $\chi$ is an (additive) character of $\mathbb F_q$, i.e. $\chi(x) = \xi^{\mathrm{tr}(x)}$ where $\xi$ is a primitive $p$-th root of unity and $\mathrm{tr}$ is the field trace. This generally yields a different set of Weyl operators and consequently a different Clifford group as in the modular definition. Both definitions agree if $q$ is prime (as then $\mathbb{Z}_q\simeq \mathbb{F}_q$ is a field).

It would thus be best practice to explicitly denote the ring or field over which the Clifford group is defined, instead of its order, e.g. $\mathrm{Cl}(n,\mathbb F_q)$, $\mathrm{Cl}(n,\mathbb Z_q)$ or similar.

Finally, one can show that the Clifford group in the "finite field" definition is always a unitary 2-design (see e.g. Prop. 12.2 in my thesis), and it is not hard to see that the modular versions fail to do so (see Note 1 below). One should also be a bit careful when using the stabilizer formalism in non-prime dimensions as not all technical results carry over.

Note 1: The reason why the unitary 2-design property singles out the "finite field" definition is the following. In the modular definition, the Weyl operators can have many different orders, at least as many as you have distinct prime divisors of $q$. As conjugation is order-preserving, Weyl operators with given order form an invariant subspace under $\mathrm{Cl}(n,\mathbb Z_q)$. If $q$ is not prime, then there are at least three different orders, hence at least three invariant subspaces and $\mathrm{Cl}(n,\mathbb Z_q)$ cannot be a unitary 2-design. Over finite fields, the Weyl operators have either order $p$ or 1, corresponding to the two irreps of the unitary group.

Note 2: Galois extensions of $\mathbb Z_r$ can also be used to give yet another definition for $q = r^k$ where $r$ is not prime. Common to all extensions is that there are embeddings $\mathrm{Cl}(n,\mathbb F_q) \hookrightarrow \mathrm{Cl}(nk,\mathbb F_p)$, and similar for Galois rings. This can be understood as "grouping of $k$ qudits", and gives subgroups of $\mathrm{Cl}(n,\mathbb F_p)$ that are unitary 2-designs (but not 3-designs of course).