The real Pauli group is the subgroup of $ O_{2^n}(\mathbb{R}) $ generated by products and tensor products of $ X $ and $ Z $ (this deviates from the usual Pauli group in that only real Paulis are allowed, in other words there is no global phase of $ i $).

The real Clifford group is defined to be the normalizer in $ O_{2^n}(\mathbb{R}) $ of the real Pauli group.

The real Clifford group is finite of size $$ 2^{n^2+n+2}(2^n-1)\prod_{j=1}^{n-1}(4^j-1) $$ see https://arxiv.org/abs/math/0001038

According to the same reference the real Clifford group contains all diagonal gates of the form $$ diag((−1)^{q(v)}+a) $$ where $ q $ is a binary quadratic form and $ a $ is $ 0 $ or $ 1 $. The real Clifford group also contains all permutation matrices that correspond to the action of an element of $ AGL(n,2) $ on the $ 2^n $ basis vectors.

These two types of gates are mentioned in the context of a "convenient generating set" for the real Clifford group. Are all real diagonal gates and all permutation matrices in the in the real Clifford group? (I assume not because Toffoli is a permutation matrix that is not in the Clifford group) Or are the diagonal gates in terms of quadratic forms and the permutation matrices correspond to $ AGL(n,2) $ the only diagonal and permuation gates, respectively, in the real Clifford group?

  • $\begingroup$ I might be misunderstanding the question but I dont see how the group could contain all real diagonal matrices (of which there are infinitely many) if it is a finite group. $\endgroup$
    – Rammus
    Commented Nov 30, 2022 at 7:38
  • $\begingroup$ I'm assuming everything is a gate meaning it's unitary so ya I just mean real diagonal unitary matrices of which there are $ 2^{2^n} $ for $ n $ Qubits, good question! $\endgroup$ Commented Nov 30, 2022 at 20:12
  • $\begingroup$ Ah, of course. I was being a bit slow. :/ $\endgroup$
    – Rammus
    Commented Nov 30, 2022 at 20:33

1 Answer 1


I assume by "real diagonal" you mean diagonal with $\pm1$ entries since, at least in the quantum arena, we're talking about unitaries.

I'm not familiar with the real Clifford group, so I'm going to assume that its evolutions can be efficiently simulated on a classical computer, by a version of the Gottesman-Knill theorem (real Paulis are mapped to real Paulis, so we just have to keep track of them).

The Toffoli gate is a permutation matrix that is not in the real Clifford group (which we know because Toffoli+Hadamard is universal, Hadamard is in the real Clifford group, and the real Clifford group is not universal).

By extension, if I take Toffoli and pre- and post-multiply the target qubit by a Hadamard gate, I get controlled-controlled-$Z$, which is real diagonal but not in the real Toffoli group.

  • $\begingroup$ Ok cool this is exactly the kind of argument I was looking for. $CCX=$Toffoli is a permutation that isn't Clifford and $ CCZ $ is a diagonal that isn't Clifford since its just some Hadamards times Toffoli. This all requires at least 3 qubits. For $ n=1 $ it's obvious that all permutations and all diagonals are Clifford (in fact they are Pauli). What about for $ n=2 $ qubits? Are all permutations and all diagonals in the 2 qubit Clifford group? $\endgroup$ Commented Nov 30, 2022 at 15:26
  • $\begingroup$ Ya all permutations and real diagaonal gates are in there you just need Paulis and controlled $ X $ and controlled $ Z $. $ X \otimes X $ corresponds to the permutation $ (14)(23) $, and CNOT corresponds to $ (34) $. So $ (X \otimes X)CNOT $ corresponds to $ (1423) $ and any $ n $ cycle together with any transposition (like CNOT) generates all of $ S_n $, in this case generates all of $ S_4 $. It is also true that all real diagonal gates are in the 2 qubit Clifford group. The group of real diagonal gates is $ 2^4 $ and is generated by $ -I \otimes I, I\otimes Z, Z \otimes I, CZ $ $\endgroup$ Commented Dec 1, 2022 at 19:51

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