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In the context of stabilizer QECCs, the interesting/useful logical operations live in the set of valid fault-tolerant encoded operations. The automorphism group of $S$, $\text{Aut}(S)$, determines this set of valid fault-tolerant encoded operations. $S$ is necessarily Abelian, and Abelian groups have no non-trivial inner-automorphisms. So more precisely, ...


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For a stabilizer code $S$ (i.e. an abelian subgroup of the $n$-qubit Pauli group $G$), the normalizer of $S$ in $G$ is denoted by $N_G(S)$. Following Preskill's notes, the logical operators are the elements in $N_G(S)\setminus S$, where "$\setminus$" is the subtraction of sets. It follows that the set of logical operators $N_G(S)\setminus S$ cannot ...


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What you want to prove is that $P^A_{\pm}|\Psi\rangle$ is a $\pm 1$ eigenvalue of $A$. Once stated like that, hopefully it's clear how to proceed: $$ AP^A_{\pm}|\Psi\rangle=(P^A_+-P^A_-)P^A_{\pm}|\Psi\rangle $$ Now from the properties of projectors, $(P^A_+-P^A_-)P^A_{\pm}=\pm P^A_{\pm}$ ($P_+^2=P_+$ and $P_+P_-=0$). Hence, we see that $$ AP^A_{\pm}|\Psi\...


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If the operators had only the eigenvalue $1$, then they would be the identity. In the stabilizer formalism you define a state (or space) by a set of operators which all leave the state (or space) invariant. This can be done if there are other eigenvalues as well. In fact you need some other eigenvalues to rule out other states and end up with a unique one ...


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