Composition of angular momentum (quantum): how do we know that finding common eigenspace of $J^2$ and $J_z$ is enough for degeneracy? I have some basic question on composition of angular momentum (actually spin in my case), I forgot some basis.
The fundamental commutation relations between $J_x,J_y,J_z$ (the three components of the angular momentum) imply that they do not commute. For this reason we cannot find a common eigenspace.
However, we know that $[J^2,J_i]=0$ for $i = x,y$ or $z$. Then, we can find a common eigenspace for $J^2$ and any of the $J_i$ (usually we take $i=z$).
I agree with the argument. But how do we know that we cannot reduce the degeneracy even more? I know that actually in some systems we will have extra degree of freedom such that the quantum state will be defined by $|n,j,m \rangle$ where $j$ is the eigenvalue of $J^2$, and $m$ the eigenvalue of $J_z$ (and $n$ describe some remaining degeneracy).
But for instance what tells me that I couldn't find some "smart" combination of the $J_i$ that will make this combination commute with $J^2$ and $J_z$ at the same time, removing further the degeneracy "in general" (i.e without looking at some extra degrees of freedom outside of the angular momentum).
I guess some "brute force" approach by trying to find any polynome in the $J_i$ would show that it is not possible to find such operator. But I would like some "nice" argument to show it if possible.
 A: We don't know without further information. The $su(2)$ representation $\rho: su(2) \to {\rm End}(V)$ of the angular momentum $\vec{J}$ could be (completely) reducible
$$ V~=~\oplus_i V_i. $$
Here $V$ denotes the Hilbert space of the system. E.g. the projection operator $P_i$ for each irreducible subspace $V_i$ would commute with $\rho(J^2)$ and $\rho(J_z)$.
A: We cannot reduce the degeneracy even more using only tools from angular momentum theory.  You need to find other symmetries, and even then there's no guarantee you will remove all degeneracies.
Consider the situation where you have three particles with angular momentum $\ell=1$. In the decomposition of $1\otimes 1\otimes 1$, you will find three sets of states with total angular momentum $L=1$.  One set can be made to be fully symmetric under particle permutations, and the remaining two will have mixed symmetries and transform so as to mix $\vert 1,m\rangle_1$ and $\vert 1,m\rangle_2$ is a specific way (so they will carry the irrep {2,1} of the symmetric group $S_3$).  There is no unique way of choosing the last two sets other than some constraints given by the permutations, but even then you can always take linear combinations are redefine states.
There are many instances where sets of states are mathematically equivalent; in such cases you cannot rely on math to say anything more than "the sets are equivalents".  You must instead rely on the physics of the problem to help out.  After all, this is what you do when you carefully choose the orientation of your axes: all orientations are mathematically equivalent but physically you might as well take - say - the $\hat x$ axis to be along the direction of motion of an unaccelerated particle.
