Correct interpretation of a Quantum State

Question

If a quantum system is described to be only in Quantum states $s_1$ and $s_2$ (described by computational basis $\vert {0} \rangle$ and $\vert {1}\rangle$) or a combination of both (for a qubit system for example), should not all the operators, which we interpret as representations of measurable properties of the system, be only diagonal matrices? What happens if we have an operator whose basis is not the computational basis?

If we measure a property of the quantum system (operator O) and measure an eigenvalue for an eigenvector which is not associated to $s_1$ nor $s_2$, but an eigenvector which is a linear combination of both, does this mean the quantum state collapsed to such eigenstate? Would not that be contradictory to the fact that the only observed states from the Quantum System are $s_1$ and $s_2$?

Answer 1

You seem to be referring specifically to measurement operators ("operator" without qualification also refers to unitary operators, which evolve the system). In the case of projective measurements, the measurement operator decomposes into a set of projection operators, say $M = \sum m \, P_m$, where $m$ is an eigenvalue of $M$ and $P_m$ is a projection operator (POVM is slightly more complicated, but conceptually similar).

The projection operator $P_m$ projects every vector in the relevant vector space into the eigenspace associated with $m$. So if the state vector, say $\vert \psi \rangle$, is an eigenvector of $M$ associated with the eigenvalue $k$, then the probabilities of the measurement will be $p(m=k)=1$ and $p(m \ne k)=0$.

For the general case when $\vert \psi \rangle$ is not necessarily an eigenvector of $M$, then $$p(m)=\langle \psi \vert P_m \vert \psi \rangle = \sum P_{m,ij} \, \psi_i^\ast \psi_j.$$ So each $P_m$ is playing the role of an Hermitian form (the complex version of a quadratic form) that gives the probability of the measurement observing $\vert \psi \rangle$ in the $P_m$ eigenspace.

The $P_m$ are always complete, $\sum P_m = I$, positive semi-definite, and, in this case (projective measurements), also orthogonal, $P_i \, P_j = \delta_{ij} P_i$. Two $M$'s with distinct eigenspaces are simply measuring $\vert \psi \rangle$ over distinct bases that are related by some coordinate transform.

If $M$ is diagonal, the $P_m$ are permutations of $\text{diag}(1,0,...,0)$. This makes the calculations easy, but it's certainly not necessary for a valid measurement operator.