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As stated in the Schuch's answer: the W state is not a stabilizer state. We can also find some operators which can "stabilizer" $W_{3}$. You can refer to Entanglement detection in the stabilizer formalism Page 9 Eq. 59 for $W_{3}$, and Efficient estimation of multipartite quantum coherence Eq.A.8 - A.11 in Appendix A for $W_{3}$ and $W_{4}$. You ...


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TL/DR The dimension of the stabilizer $\mathcal{S}$ is $2^{n}$ because there are exactly $n$ generators to make the dimension of the stabilizer's eigenspace exactly $1$. Then, each element in $\mathcal{S}$ is a unique combination of the generators for a total of $2^{n}$ combinations. Long version It's easy to show that the stabilizer $\mathcal{S}$ forms a ...


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Every $n$-qubit pure quantum state has at most $2^n$ stabilizers. There are at least two approaches to proving this bound. One makes explicit use of the language of symplectic bilinear forms and the other eschews it while using essentially the same ideas. Below, I write out the former including a short summary of the relevant definitions and the key lemma ...


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No, it's not possible. For example, being able to directly measure $X+Y$ would allow you to prepare T states and thereby perform T gates, which are not stabilizer operations. If the fact that $X$ and $Y$ don't commute bothers you, then note that preparing $|+\rangle^{\otimes n}$ and then measuring $\sum_{k=0}^{n-1} Z_k$ and getting a result of $n-2$ would ...


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