In this article authors define quantum states $\left|\psi\right\rangle$ (elements of some Hilbert space H - in most simple case a $2\times 1$ complex vector such that $|\psi| = 1$) by a set of operators such that $s \left|\psi\right\rangle = \left|\psi\right\rangle$. See attached.
While the concept seems clear to me, and it sounds right, I can't produce such set to identify $$\left|\psi\right\rangle = \alpha [1,0]^T + \beta[0,1]^T$$ (so, not even in the simplest case.)
I am looking for some help in understanding how, given a state, one can generate such set of operators. Can it always be done? If yes, can one always define the smallest set of operators (in finite dimensional Hilbert space). Is there a sense of orthogonality, in similiar way elements of Hilbert space are linear combinations of orthogonal basis?
Basically $S$ is a set of commuting operators and $\vert s\rangle$ is a common eigenvector of elements of this set.
To generate the two vectors you have, one might as well assume that elements of the set are diagonal, so pick them to be $$ s_1=\hat 1\, , \qquad s_2=\sigma_z $$ with $\sigma_z$ the Pauli matrix. The common eigenvectors are $[1,0]^T$ and $[0,1]^T$.
To get $\psi=(\alpha,\beta)^T$ note first that $\vert\alpha\vert^2 +\vert \beta\vert^2=1$, construct the matrix $$ U=\left(\begin{array}{cc} \alpha&-\beta^* \\ \beta&\alpha^*\end{array}\right)\, . $$ and note that $U[1,0]^T=(\alpha,\beta)$ and $U$ is unitary. Thus to get the set that will have $\psi$ as eigenvectors you need to conjugate $\hat 1$ and $\sigma_z$ by $U$, i.e. your new set is $\{\hat 1,U\sigma_zU^{-1}\}$.
If you need elements in $S$ to be unitary you can just use $e^{-i\alpha \sigma_z}$ instead of $\sigma_z$. The key point is that the eigenvalues of your second operator should be distinct.