Is there any way to derive the quantum mechanical operators that form the integrals of motion in quantum mechanics from the Hamiltonian for general systems?
Let's for simplicity focus on stationary problems.
For example if I have a system with a(n obvious) continuous spatial symmetry, like the electron in the potential of the proton with its apparent $SO(3)$ symmetry (neglecting now the 'hidden' Runge-Lenz vector conservation) and the Hamiltonian $\hat{H}_0$, and I perturb the potential e.g. by adding a second proton to the potential $\hat{V}'$. Since the number degrees of freedom remain unchanged, we have as many constants of motion as in the $SO(3)$ symmetric case, with the same number of integrals of motion for time independent problems. But for the $SO(3)$ case we knew that the total orbital angular momentum $\hat{L}^2$ and one of its components $\hat{L}_z$ were integrals of motion, which were induced by the Noether principle by the continuous $SO(3)$ symmetry. My question is now: What are the two new quantities that are conserved in stationary states of the system described by the Hamiltonian: $\hat{H}_0+\hat{V}'$?
Is there a way to determine any* representation of these operators (that commute with the Hamiltonian) in explicit form?
On a more concrete basis if we consider the particle in a ring of radius $R$ there we have the Schrödinger equation:
\begin{align} \hat{H}_0 \psi & = E \psi\\ \hat{H}_0 & = -\frac{\hbar^2}{2R^2m}\frac{\partial^2}{\partial \vartheta^2} \end{align}
for $0\le \vartheta < 2 \pi$ with the eigenfunctions $$ \psi_{ml}(\vartheta) = \frac{1}{\sqrt{2\pi}}e^{iml\theta} $$ for $m\in\{-1,1\}$ and $l\in \mathbb{N^+_0}$ and the corresponding energies $$ E_{l} =\frac{l^2 \hbar^2}{2R^2m}.$$
Since the system has a continuous rotational symmetry we have conservation of angular momentum $L_z$ with the corresponding angular momentum operator
$$\hat{L}_z = -i\hbar\frac{\partial}{\partial \vartheta} $$
As obviously
$$ [\hat{H}_0,\hat L_z] = 0 $$
angular momentum is conserved in all solutions of the stationary problem. Now lets suppose we break the symmetry by adding a small attractive $\delta$ function potential of "height" $\epsilon$ at $\vartheta_0 = 0$
$$ [\hat{H}_0 - \epsilon \delta(\vartheta)] \psi = E \psi $$
This system has exactly one bound state (i.e. one state with $E<0$)
$$ \psi_b = \frac{1}{\sqrt{N}} \left(e^{d \vartheta} + e^{-d(\vartheta - 2\pi)}\right) $$
With the some $d>0$ (on a side note: which is determined by $ \coth(\pi d) = \frac{2\pi \hbar^2}{ \epsilon e m R} d $)
My question instantiated for this system is basically if there exists a non-trivial operator $\hat{\Lambda}_\epsilon$ with
$$ [\hat{H}_0 - \epsilon \delta(\theta) ,\hat \Lambda_\epsilon] = 0 $$ and
$$ \lim_{\epsilon\to 0} \Lambda_\epsilon = \hat{L}_z $$
and how it can be determined.
*) I understand that the quantities are not unique because any "algebraic combination" will be another representative
