# Definition of the Pauli group and the Clifford group

There seem to be two definitions of the Pauli group. In Nielsen and Chuang, the Pauli group on 1 qubit is defined as \begin{align*} \mathcal{P}_1 = \{\pm I, \pm iI, \pm X, \pm iX, \pm Y, \pm iY, \pm Z, \pm iZ\} = \langle X, Y, Z\rangle. \end{align*} But in other references, such as https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.127, \begin{align*} \mathcal{P}_1 = \{\pm I, \pm X, \pm iY, \pm Z\} = \langle X, Z\rangle. \end{align*} The second definition makes more sense in the stabilizer formalism, since we usually only consider the action of Pauli strings of $$X$$ and $$Z$$. What bothers me is that by the second definition, the phase gate $$S$$, which is an element of the Clifford group, results in \begin{align*} SXS^\dagger = Y \notin \mathcal{P}_1. \end{align*} I know that the prefactor $$\pm i$$ somehow doesn't matter, but can someone justify it from a more mathematical/group-theoretical perspective?

Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $$\pm i Y$$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $$\pm i Y$$. This is only enough to simulate real stabiliser circuits, which is what you're noticing.

For that reason, I would use the first definition; and anyone whose work involves taking the Pauli operators first (which consist of $$\mathbf 1$$, $$X$$, $$Z$$, and also $$Y$$) and then generating a group from them will inevitably use the first definition, for the simple reason that this is the group $$\langle X, Y, Z\rangle$$. And because the operators $$\mathbf 1$$, $$X$$, $$Y$$, and $$Z$$ form an operator basis for the Hermitian operators on a single qubit, there are often very good theoretical reasons to include the operator $$Y$$ (in preference to $$iY$$) in one's analysis.

It must be said that [Phys. Rev. A 57 (127), 1998] is not just any old article which uses the Pauli group, nor will the point I made above have escaped its author. The reason (I believe) why Gottesman didn't make the distinction there, that I am making now, is because his main interest in that article is in describing error correcting codes — in which the scalar factors preceding the Pauli operators are in principle less important.

• To start with, many of the best liked and interesting codes — from the 9-qubit code and the Steane $$[\! [7,1,3]\! ]$$ code, to the toric code — happen to be CSS codes, whose stabiliser groups are generated by separate X and Z stabilisers; any stabiliser operators involving Y, will indeed have Y operators in pairs. So, in practise, the point I made about $$iY$$ not having a +1 eigenspace often doesn't come up.

• Secondly, if one decides to just do a hand-wave and forget all scalars, then the important thing is not whether a state is a joint +1 eigenstate of the stabilisers, but just that the state is some eigenstate of all the stabilisers. Considering a stabiliser code, the important thing is that all of the states of the code have the same eigenvalues as each other. This second point gives rise to the notion of the Pauli Frame: the eigenvalues which the codespace has for the 'stabilisers', interpreted as a reference frame. The main thing is for the Pauli frame to be well-defined, and for it to change slowly enough to track how it is changing by stabiliser measurements. And if you're not actually responsible for realising a system to do so, but just want to explore the theory of quantum error correction codes itself, specific eigenvalues are largely a distraction.

The thing is, if you want to use the stabiliser formalism for a different purpose — say, to efficiently simulate stabiliser circuits on standard basis states — then you can't be quite so carefree with scalars: a collection of $$n$$-qubit stabiliser generators define a specific basis in which the state will live, but flip some signs and you change which of the $$2^n$$ states you are talking about. To perform the correct simulation, signs do matter here. Keeping track of them may be slightly tedious using the usual convention, but in the end it's important to do.

• But $[\pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $\pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion. Jul 15, 2019 at 14:00
• @AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[\pm i Y]$. Which state am I thinking of? Jul 15, 2019 at 16:39
• (Hint: does being an eigenstate of $\pm i Y$ characterise a unique single-qubit state?) Jul 16, 2019 at 15:10
• Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^{n-1}$. In this case, the equivalence class of $(\alpha,\beta)$ goes to that of $(\beta,-\alpha)$. So $(\alpha,\beta) \equiv (\beta,-\alpha)$ is the condition for fixed points of group action. This translates into $(1,\frac{\beta}{\alpha})=(1,\frac{-\alpha}{\beta})$. Which becomes $\alpha^2=-\beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $\alpha = \pm i \beta$. Rescale solutions and you get two equivalence classes $(1,\pm i)$. I never claimed anything about unique solutions. Jul 16, 2019 at 18:06
• We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic. Jul 16, 2019 at 18:08

The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $$\pm i$$ that are missing.

In lieu of a tikz commutative diagram

\begin{align} & \hookrightarrow & U(2)\\ \downarrow & & \downarrow\\ <[X],[Z]> & \hookrightarrow & PU(2) \end{align}

For both of the inclusion arrows $$H \hookrightarrow G$$, you can form normalizers $$N_G (H)$$ which fit in between on each row.

• You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $\mathcal{P}_1/U(1)$ rather than normalizing $\mathcal{P}_1$? Jul 14, 2019 at 23:45
• It would be nicer if could do the commutative diagram with tikz here Jul 14, 2019 at 23:57