# Why is the $N$-qubit stabilizer group abelian?

In Devitt et al. 2013's introduction to quantum error correction, the authors mention (bottom of page 12) how the stabilizer group for $$N$$ qubits is abelian.

More specifically, here is the quote:

An $$N$$-qubit stabilizer state $$\lvert\psi\rangle_N$$ is then defined by the $$N$$ generators of an Abelian (all elements commute) subgroup $$\mathcal G$$ of the $$N$$-qubit Pauli group, $$\mathcal G=\{K^i\,|\,K^i\lvert\psi\rangle=\lvert\psi\rangle,\,[K^i,K^j]=0,\forall (i,j)\}\subset \mathcal P_N.$$

I am confused by this. Is the stabilizer subgroup $$\mathcal G$$ defined as an abelian subgroup of elements of $$\mathcal P_N$$ that stabilizes $$\lvert\psi\rangle$$, or is it instead the case that the subgroup of elements of $$\mathcal P_N$$ that stabilize $$\lvert\psi\rangle$$ is abelian?

If the latter, doesn't this introduce ambiguity in the definition? There could be other elements that stabilize $$\lvert\psi\rangle$$ but not commute with $$\mathcal G$$.

If the former, how is this shown? I can see why the action of $$K^i K^j$$ and $$K^j K^i$$ is identical on $$\lvert\psi\rangle$$, but how do you show that $$K^i=K^j$$?

• It they are not abelian, they don't stablize a state (as they don't have a common eigenvector). I seem to me missing the problem. – Norbert Schuch Oct 25 '18 at 19:07
• @NorbertSchuch mh, couldn't you have $AB\lvert\psi\rangle=BA\lvert\psi\rangle$ for the state to be stabilized, but $AB\lvert\phi\rangle\neq BA\lvert\phi\rangle$ for other states? That is, can it not be the case that they act as commuting on the specific state that defines the subgroup, but not on other states? – glS Oct 25 '18 at 19:09
• Not for Paulis. Otherwise: Sure. – Norbert Schuch Oct 25 '18 at 19:16

It is not necessary to define the group as commuting —$$\def\ket#1{\lvert#1\rangle}$$ by virtue of every element in the group stabilising the state $$\ket{\psi}$$, this property follows.
Because we are considering subgroups of the $$N$$-qubit Pauli group, any two elements either commute or anti-commute. Let $$P \in \mathcal P_N$$ be an operator which stabilises some vector $$\ket{\psi}$$, that is such that $$P \ket{\psi} = \ket{\psi}$$. Suppose that $$Q$$ anticommutes with $$P$$. It then follows that $$Q \ket{\psi} = Q P \ket{\psi} = - P Q \ket{\psi}.$$ Now, if $$Q \ket{\psi} = \lambda_Q \ket{\psi}$$ for any scalar $$\lambda_Q$$ at all, we have $$\lambda_Q \ket{\psi} = - P Q\ket{\psi} = - \lambda_Q \ket{\psi}.$$ But this implies that either $$\lambda_Q = 0$$ (which is impossible as $$Q$$ is unitary) or $$\ket{\psi} = \mathbf 0$$, the zero vector.
It follows that if $$\ket{\psi}$$ is actually a state (so that in particular it has norm 1), any operator in $$\mathcal P_N$$ which has $$\ket{\psi}$$ as a $$\pm1$$ eigenvector must commute with all operators which stabilise $$\ket{\psi}$$. Thus the subgroup of $$\mathcal P_N$$ which stabilises $$\ket{\psi}$$ is abelian.