Given potentials, how does one find conserved quantities using Noether's theorem? I've been asked to find the conserved quantities of the following 3D potentials: 


*

*$U(\vec{r}) = U(x^2)$, 

*$U(\vec{r}) = U(x^2 + y^2)$ and 

*$U(\vec{r}) = U(x^2 + y^2 + z^2)$. 


For the first one, there is no time dependence or dependence on the y or z coordinate therefore energy is conserved and linear momentum in the y and z direction are conserved. I'm having trouble with the angular momentum. It would seem to me that since there is only a dependence on x, that the Lagrangian would be invariant under rotations around the y and z-axis and thus angular momentum in those directions is conserved. Similar approach for the other two potentials. Can anyone give me any more depth or background on this, what would be a concrete way to approach these types of problems so I can be more confident in my answer.
 A: Noether theorem tells you that if you can find a (one parameter) group of infinitesimal transformations $\alpha$ and $\beta$ such that:
\begin{equation}
 t'=t+\alpha\epsilon
\end{equation}
\begin{equation}
q'^\mu=q^\mu+\beta^\mu\epsilon
\end{equation}
and your lagrangian is invariant under this group of transformations, then the quantity
\begin{equation}
 p_\mu\beta^\mu-H\alpha
\end{equation}
is conserved, where $p$ is the canonical momentum and $H$ is the hamiltonian.
So for example, for the potentials you mention, the lagrangian would be invariant under a rotation around the $q^2=y$ axis and the infinitesimal transformations are
\begin{equation}
 q'^1=q^1+\epsilon q^3
\end{equation}
\begin{equation}
 q'^2=q^2
\end{equation}
\begin{equation}
 q'^3=q^3-\epsilon q^1
\end{equation}
so, $\beta^1=q^3$, $\beta^2=0$, $\beta^3=-q^1$ and suppose $\alpha=0$ (no time transformation) then the conserved quantity is
\begin{equation}
 p_\mu\beta^\mu=p_1\beta^1+p_3\beta^3=p_1q^3-p_3q^1=(\boldsymbol{r}\times\boldsymbol{p})_2=L_2
\end{equation}
The same goes for the other components of the angular momentum $\boldsymbol{L}$.
