Classically, angular momentum is only conserved in a central potential by considering the torque (correct me if I am wrong). In quantum mechanics, it is also true, isn't it?
If this is the case, then does it mean that in an infinite spherical well ($V(r)=0$ for $r<a$, and $V(r)= \text{inf}$ for $r>a$) the angular momentum is also conserved because $V$ depends on $r$ only? Also does it mean that in a 2D/3D square well it is not conserved because the potential is not radial?
Besides, usually we consider $L^2$ instead of $\vec L$. But if angular momentum is conserved and $[L^2,H]=0$, then does it automatically imply $[\vec L,H]=0$?
Best Answer
In QM, an operator is conserved iff it commutes with $H$, because $$ i\dot {\mathcal O}=[H,\mathcal O] $$
Therefore, the angular momentum is conserved iff it commutes with $H$. As the kinetic term is rotationally invariant, you only need to consider the potential: $$ \frac{\mathrm d}{\mathrm dt} L^2=0\quad\Leftrightarrow\quad [V(R),L^2]=0 $$ that is, $L^2$ is conserved iff the potential is rotationally invariant, the same as in CM!
A scalar operator $A$ satisfies, by definition, $$ R^\dagger A R=A $$ where $R$ is any rotation. As $R= \exp[i \vec\theta\cdot\vec L]$, you can expand the relation above to first order in $\theta$ to check that it is equivalent to $$ R^\dagger A R=A \quad\Leftrightarrow\quad [A,\vec L]=0 $$ so yes: a rotationally invariant operator commutes with the three components of the angular momentum operator.