In classical mechanics it is known that a certain model is solvable exactly (integrable) if it posses a sufficient amount of "well behaved" conserved charges.
On the other hand in quantum mechanics we say that a model described by Hamiltonian $H$ is integrable if it can be solved using the algebraic Bethe ansatz (look at this related question for a more clear explanation and references). If that is the case we can construct a transfer matrix which is able to generate a (possibly infinite) set of conserved charges $\{ \hat{Q}_i \}$ such that
\begin{equation} [H, \hat{Q}_i]=0 \quad \text{and} \quad [\hat{Q}_i,\hat{Q_j}]=0 \quad \forall i,j. \tag{1} \end{equation}
At this point it seems natural for me to make a parallel with classical mechanics and state that a quantum system is integrable if it posses a set of conserved charges as the one defined in $(1)$.
However given $H$ we can always construct a set of "well behaved" conserved charges whose dimension is the same as the dimension of the Hilbert space by simple taking the eigenstates $|\psi_i \rangle$ of $H$ and defining \begin{equation} \hat{Q}_i = |\psi_i\rangle \langle \psi_i |. \tag{2} \end{equation} To me it is obvious that this cannot imply that $H$ describes an integrable model because if that were the case then any model should be integrable. My immediate solution to this problem is to say that the charges $\hat{Q}_i$ constructed in $(2)$ are "functionally dependent" on $H$, in the sense that they were build using $H$ and they do not actually contribute any new information to solving the problem.
My questions are: Is my line of logic correct or am I making a huge mistake? And if I am correct is there a way to express the idea of "functionally dependent" in a more rigorous mathematical way?