# Clifford group without the phase gate

The Clifford group is generated by the Hadamard gate $$H$$, the phase gate $$S=\sqrt{Z}$$, and the $$\text{CNOT}$$ gate. I was wondering what happens if we dropped $$S$$, so that all matrices are real.

I found Does the real Clifford group contain all real diagonal gates? all permutation gates? this article, but the "real Clifford group" this question was addressing was something different. There, they were talking about the subset of Clifford gates that leaves the set of real Paulis like $$X_iZ_j$$, and it was unclear, but the way it was written seemed like it did not leave things like $$Y_iY_j$$ which is also technically real.

I think the Clifford group without the $$S$$ should leave the "actually real" Paulis as an invariant set. Does anyone know a good reference for this group?

• You can express any quantum gate with real numbers only but at cost of one ancilla qubit. See here for details: arxiv.org/abs/quant-ph/0301040 Commented Apr 9 at 12:15
• Thanks, but that's not about Cliffords, am I correct? Commented Apr 9 at 23:59

## 1 Answer

The real Pauli group is defined as the subgroup of the Pauli matrices with only real entries. In particular, it contains $$Y_i Y_j = - X_i Z_i X_j Z_j$$. One can show that it is generated by $$X_i$$ and $$Z_j$$. The real Clifford group is the orthogonal normalizer of the Pauli group, or equivalently, the subgroup of the Clifford group given by real matrices.

The real Clifford group and its generators are first described in Calderbank, Rains, Shor, and Sloane (1997). These are

1. Hadamard gate on the first qubit, $$H_1$$,
2. Linear reversible circuits $$|x\rangle \mapsto |Ax\rangle$$ for $$A\in\mathrm{GL}(n,\mathbb{F}_2)$$,
3. Diagonal gates $$|x\rangle \mapsto (-1)^{q(x)}|x\rangle$$ where $$q$$ is a binary quadratic form.

Finally, to adress your question: Assume $$n \geq 2$$ qubits and let us consider the generators $$H_i$$ and $$CX_{ij}$$ for all $$i,j$$. Thus by assumption we have (1.) and moreover the CNOT gates $$CX_{ij}$$ generate the linear reversible circuits (2.). To obtain the diagonal gates, note that every binary quadratic form can be written as $$q(x) = \sum_{i. To realize such a gate, apply $$CZ_{ij}$$ to every qubit pair for which $$R_{ij}=1$$. Clearly, $$CZ_{ij}= H_j CX_{ij} H_j$$ and thus we have shown that we can generate the entire real Clifford group.

If you wonder how we generate Paulis, note that we have $$Z_1 = (CX_{12}\, H_2)^4$$.

EDIT: I noticed that for $$n=1$$, this argument does not work. In this case, generators are $$H$$ and $$Z$$. Formally, the quadratic form in point 3 needs to be supplemented by a linear form (not sure whether Calderbank et al. meant that in the first place).

• Thank you so much! This was very helpful. I actually have one follow up question: Do you know if there are expressions for the size of the group? I imagine it will be roughly 1/2^n smaller than the usual Clifford group (a random Clifford transformation will map X1 to a random Pauli string like XZZXYYZYX... and with approx. prob. 1/2 it will contain odd number of Ys), but I couldn't find that kind of exact expression in the reference you gave. Commented Apr 9 at 23:57
• @Jun_Gitef17 The oder of the real Clifford group follows from the fact that the quotient by the real Pauli group is the orthogonal group $O^+(2n,\mathbb{F}_2)$ (see the Calderbank et al. paper). Explicitly we have that the order of the real Clifford group is $2^{n^2+n+2} (2^n-1) \prod_{j=1}^{n-1}(4^j - 1)$ (see also Nebe et al. arxiv.org/abs/math/0001038 on page 4). Commented Apr 10 at 7:34
• Oh, I was missing that part from the other thread.. Thank you!! Commented Apr 10 at 7:52
• @Jun_Gitef17 For comparison, the order of the (complex) Clifford group is $2^{n^2 + 2n + 2} \prod_{j=1}^n (4^j -1)$. Commented Apr 10 at 7:53
• Thanks! So the real Clifford formula gives 16 for the $n=1$ case, but I am confused because I can only see 8. X can only map to X or Z, and Z needs to map to the other one: this gives us two choices. Then we have $\pm 1$ phase that we can add to them, which gives $2^2=4$ choices. This only gives 8 choices in total.. what am I missing? Commented Apr 10 at 17:11