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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?

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. – AHusain – 2019-07-15T14:00:25.753

@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? – Niel de Beaudrap – 2019-07-15T16:39:28.013

(Hint: does being an eigenstate of $\pm i Y$ characterise a unique single-qubit state?) – Niel de Beaudrap – 2019-07-16T15:10:32.347

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. – AHusain – 2019-07-16T18:06:05.437

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. – AHusain – 2019-07-16T18:08:19.357

@AHusain: that's all well and good. It also doesn't matter for my question, which is about unique solutions. The ability to specify individual states is lost if you pass to equivalence classes of matrices. I am indicating that this is a relevant application, which requires matrices rather than equivalence classes of them, and which motivates having the Hermitian matrix $Y$ in particular. (Deleted and revised for improved concision and clarity) – Niel de Beaudrap – 2019-07-16T18:33:07.143