Clifford groups are used in at least 3 places I've encountered so far in QIP:

  1. A circuit that contains only Clifford operations, which are generated from CNOT, H and P, is sufficient for a wide variety of quantum circuits, as shown here.
  2. Clifford circuits are used in randomized benchmarking since we can get averages over the the entire unitary space by averaging only over Cliffords.
  3. They are sufficient, together with a single non-Clifford gate, to perform universal quantum computation.

However, in the literature, there are at least 2 (useful) definitions of the Clifford group I've seen:

  1. Here, as the group that preserves the set $\pm P_n^{*}=\pm P_n\backslash\{I_n\}$, where $P_n$ is the Pauli group modulo phase, $P_n/U(1)$. In this case, $C_1$ has 24 elements, $C_2$ has 11520 and so on. Also, this definition gives us the same group that is generated by P, H, CNOT modulo $U(1)$, and I've seen it used for RB as well, for example here.

  2. As the normalizer of the Pauli group, where (presumably) we're talking about the Pauli group whose elements are $P_n$ along with the phase $i^k$, and thus has a size of $4^{n+1}$. I've also see this definition used in the context of RB, here.

Clearly, these definitons do not coincide since the order of the Pauli group as defined in (2) does not divide the order of the Clifford group as defined in (1).

My question is: What is the relation between the two definitions? And, is there any practical scenario (e.g when running and analyzing an actual quantum circuit) where we have to use one of the definitions and the other would be wrong?

  • $\begingroup$ Is there any explanation why qiskit.quantum_info.Clifford only produces 11328 unitaries? $\endgroup$
    – Hadamard
    Commented Jun 16 at 18:18

1 Answer 1


Does https://quantumcomputing.stackexchange.com/a/13650/2305 answer your question?

In short: The second definition is not even a finite group because it includes $U(1)$ as a centre. The same holds for the first definition. Nevertheless, the contrary is often claimed, because what people are actually thinking about, is the projective group. This is ok because global phases don't matter and all this subtle differences of course vanish projectively.

When defining a finite Clifford group, one has some freedom in choosing the size of the centre. The minimal centre is $\mathbb Z_4$ while the (usual) definition in terms of $H$, CNOT, and $S$ gives you a centre $\mathbb Z_8$.

To answer your final question: The precise definition does usually not matter, it only influences the precise order of the group (so you have take some care when you count).

There's only a few practical scenarios where global phases actually matter, namely when they are turning into relative phases. This happens, for instance, when you control a unitary. There is usually no danger here, you just have to define what you mean with e.g. $CS$.

Sometimes, one has to take some care for stabilizer-based simulators. Everything in the stabilizer formalism works projectively, hence you have to be careful when not working with density matrices. For instance, if you expand a pure state $|\psi\rangle$ as a linear combination of stabilizer states, and you want to update this under a Clifford unitary, some stabilizer states might get an additional phase compared to others, so this would be a relative phase. That's why a "phase-sensitive" Clifford simulator is used in the stabilizer rank papers.

  • $\begingroup$ Is there an intuitive explanation for where the Z8 center is coming from? $\endgroup$
    – Lior
    Commented Mar 27, 2022 at 10:36
  • $\begingroup$ @Lior The standard generators have matrix entries $1, -1, i, 1/\sqrt{2}$, which generates matrices with coefficients in $\mathbb Z[i,1/\sqrt{2}] = \mathbb Q[\xi_8]$, where $\xi_8 = (1+i)/\sqrt{2}$ is an eigth root of unity. One can explicitly check that $(HS)^3 = \xi_8 \mathbb I$, so the centre is $\mathbb Z_8$ (and it cannot be larger!). In fact, one can show that the generated group is exactly the normalizer of the Pauli group, when the coefficients are restricted to $ \mathbb Q[\xi_8]$. $\endgroup$ Commented Mar 28, 2022 at 8:08

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