Sometimes, there is a bit of confusion around the Clifford group in the field ... and it's a matter of definition.
A lot of people define the Clifford group $\mathrm{Cl}_n(p)$ of $n$ qudits of prime dimension $p$ as the unitary normaliser of the generalised Pauli group (e.g. Gottesman, Nielsen & Chuang). As such, it is clearly not a finite group as the centre is $U(1)$.
However, this is also a bad definition, both from the mathematical and the physical point of view. Why? First, because the centre is mathematically boring and doesn't add anything to the structure of the group and second, because the global phase of a unitary is unphysical.
There are three ways of resolving this:
- Define it by its generators (see Gottesman's paper on higher-dimensonal generators). Then, it's clear that it is finite since all generators have algebraic matrix entries. Using the standard generators, you will still get a too large centre in the qubit case: $\mathbb{Z}_8$. The minimal one is $\mathbb{Z}_4$ in the qubit case an $\mathbb{Z}_p$ in the qudit case. Redefining the $H$ gate as $\frac{1+i}{\sqrt{2}} H$ makes the centre minimal for qubits.
- You can define it as the normaliser in the unitary group with entries in $\mathbb{Q}[i]$ (rational complex numbers). This will give you a group with minimal centre (see Ref. 1).
- Define the projective group $\overline{\mathrm{Cl}}_n(p):=\mathrm{Cl}_n(p)/U(1)$, or equivalently, the image under the adjoint representation, i.e.~the unitary channels associated to Cliffords. Its elements can be seen as permutations of the Pauli operators (possibly with added phases) which preserve commutation relations. Since there are $p^{2n}$ Pauli operators, the number of permutations is finite (and thus is $\overline{\mathrm{Cl}}_n(p)$).
Moreover, it holds
$$
\overline{\mathrm{Cl}}_n(p)/\overline{\mathcal{P}}_n(p) \simeq \mathrm{Sp}_{2n}(p),
$$
where $\overline{\mathcal{P}}_n(p)$ is the (generalised) Pauli group up to phases and $\mathrm{Sp}_{2n}(p)$ is the symplectic group over the finite field $\mathbb{F}_p$. Its cardinality is
$$
|\mathrm{Sp}_{2n}(p)| = p^{n^2} \prod_{i=1}^n (p^{2i} - 1).
$$
Thus, the projective Clifford group has cardinality
$$
|\overline{\mathrm{Cl}}_n(p)| = |\overline{\mathcal{P}}_n(p)| |\mathrm{Sp}_{2n}(p)| = p^{2n} p^{n^2} \prod_{i=1}^n (p^{2i} - 1)
$$
For the non-projective ones, one has to multiply with the order of the centre. For a reference, see e.g. Ref. 2
References:
- Gabriele Nebe, E. M. Rains, and N. J. A. Sloane. “The invariants of the Clifford groups”
- D. Gross, "Hudson’s theorem for finite-dimensional quantum systems"
