Can we abstractly deduce the $L^2$ is conserved assuming only rotational symmetry of Hamiltonian?

Question

Here $L^2$ is defined as $$ L^2=L_x^2+L_y^2+L_z^2 $$ representing the observable of the magnitude of the angular momentum.

There are a lot of proofs showing the $z$-projection of the angular momentum, $L_z$, is conserved upon rotating the $xy$-plane. The proof typically uses the technic to rotate the coordinate by a tiny angle $\epsilon$, and Taylor expand the wave function to the first order, and then require the time independent Schrodinger equation to remain the same, e.g. this proof.

Intuitively, $L_z$ is not conserved under arbitrary rotation, but $L^2$ should.

Although rare, I can find some proof on the Internet, using spherical coordinate and concrete Hamiltonian for hydrogen atom, e.g. proof on page 2 of this node. However, this proof depends on the specific form of the Hamiltonian under certain choice of coordinate.

My question is, assuming rotational symmetry only, without assuming any particular form of the Hamiltonian, can we prove that $L^2$ is conserved?

I tried to mimic the proof of the conservation of $L_z$ to Taylor expand the 3D tiny rotated wave function to second order, since $L^2$ involves terms like $\partial_x^2$, but the computation is too complicated and looks not correct.

Answer 1

There is some confusion between spatial rotational invariance and rotational invariance of the dynamical evolution in the question. I assume below the rotational invariance refers to dynamics and conservation of $L^2$ means conservation along dynamical evolution.

Dynamical-Rotational symmetry means that the evolution unitary operator $U_t= e^{-itH}$ commutes with the unitary representations of the rotations around the axes. Assuming that $L$ is the total angular momentum (there are no spin contributions) the rotations around the axes ($k=x,y,z$) are represented this way in the Hilbert space $V_k(\theta)= e^{i\theta L_k}$.

Rotational invariance of the dynamics reads $$U_t e^{i\theta L_k} = e^{i\theta L_k} U_t $$ for every $k=x,y,z$ and $t,\theta \in \mathbb R$. This is equivalent to saying that $$ U_t e^{i\theta L_k}U_t^* = e^{i\theta L_k} $$ that is $$ e^{i\theta U_t L_k U_t^*} = e^{i\theta L_k} \:.$$ Taking the derivative at $\theta=0$ $$L_k = U_t L_k U_t^*$$ In other words, each $L_k$ is a constant of motion separately. Finally, ignoring some subtleties realted with to the domains of the operators (which can be easily fixed working on the Garding space for instance) $$ U_t L^2 U_t^*= U_t \sum_{k=1}^3L^2_k U_t^* = \sum_{k=1}^3 U_t L^2_k U_t^* = \sum_{k=1}^3 U_t L_k L_k U_t^*=\sum_{k=1}^3 U_t L_k U_t^* U_t L_k U_t^*= \sum_{k=1}^3 L_kL^k =L^2 \:.$$ The found identity $$ U_t L^2 U_t^*= L^2 \:.$$ just states that $L^2$ is a constant of motion.