Uniquness of stabilizer generators of a state $\left| \psi \right>$?

Question

Let $G_n$ be the $n$-qubit Pauli group and $S$ an abelian subgroup which does not contain the element ($-I$). I know that if $S$ has $n$ generators then it specifies a single state $\left| \psi \right>$ which it stabilizes. Contrary to this I have seen it said that e.g. $\left| 0\right>$ has the stabilizer generator $Z$ and $\left| 0 0\right>$ the stabilizer generators $\Bbb{I}\otimes Z$ and $Z\otimes \Bbb{I}$. This hints at the following theorem which I can't find explicitly stated or proved:

The stabilizers of the state $\left| \psi \right>$ (i.e. elements of $G_n$ for which $g\left| \psi \right>=\left| \psi \right>$) form a abelian subgroup of $G_n$ which does not contain ($-I$) and has $n$ generators.

I assume the theorem to be true otherwise the association of $\left| 0 0\right>$ with the subgroup generated by $\Bbb{I}\otimes Z$ and $Z\otimes \Bbb{I}$ would not be unique. Am I correct? And how is this proved/disproved?

Answer 1

The proposition you make, as written, is false even for $n=1$ by the theorem you quote. Indeed the Pauli group on $n$ q-bits is a finite group, and so has finitely many abelian subgroups $H_i$ with $n$ generators which don't contain $-I$. Each specifies a state $| \psi_i \rangle$. $\{ |\psi_i \rangle \}$ can not be all the states because this is a finite set and there are infinitely many states. For a concrete example with $n=1$, the stabilizer of $\frac1{\sqrt3} |0 \rangle + \sqrt{\frac23} |1\rangle$ is trivial.

On the other hand, it is true that the stabilizer of any state is an abelian subgroup of $G_n$ which does not contain $-I$. That it does not contain $-I$ is clear since the stabilized state must have eigenvalue $1$ and by definition every vector is an eigenvector of $-I$ with eigenvalue $-1$. That it is abelian is not much more complicated to prove: given any two elements $g,h \in G_n$, it's easy to show that $\frac1{2} [g,h] \in \{0\} \cup G_n$ using the fact that every pair of elements in each 1 particle Pauli group either commutes or anticommutes. If an element is stabilized by both $g$ and $h$ it is in $\ker \frac1{2} [g,h]$, but all elements of $G_n$ are nondegenerate and hence $[g,h] = 0$. What fails is that the stabilizer of the state is in general not maximal among such subgroups of the Pauli group; indeed the stabilizer for almost all states (in the measure theoretic sense) is the trivial group.