I learned that the symmetry group of the quantum isotropic harmonic oscillator in n-dimensions is $SU(n)$. Specifically, in two-dimensions, it is $SU(2)$ and hence the eigenstates are given by the irreducible representations of $SU(2)$. Sticking to the usual notation for angular momentum in physics, we can label the eigenstates by $|l,m>$, where $m=-l, -l+1,…, l$ and $l$ runs over non-negative half integers. Here each $l$ gives a different irreducible representation with dimension $2l+1$. This is precisely the degeneracy of the $l$-th harmonic oscillator shell, with energy $\hbar\omega(2l+1)$. Furthermore, the angular momentum in z-direction is given by $2\hbar m$.
So far, so good. Now I could form tensor products of the different representations if I'm interested in describing multiple particles inside the harmonic oscillator. For non-interacting particles the tensor product $V_{l_1}\otimes V_{l_2}$ is a subspace of the eigenspace of the combined Hamiltonian with energy $2\hbar\omega(l_1+l_2+1)$.
On the other hand, I can decompose this thensor product representation into irreducible reps in the standard way,
$V_{l_1}\otimes V_{l_2} = \bigoplus_{L=|l_1-l_2|}^{l_1+l_2} V_L.$
However, these $V_L$ now seem to play a very different role, as the energy of the system is apparently not given by $\hbar\omega(2L+1)$. So what is the proper interpretation of the $|L,M>$?
Best Answer
You are trying to understand the language of the Jordan-Schwinger realization of angular momentum. You must understand that throughout there are only two different oscillators, not more, even when you "add angular momenta" by tensoring states. If, somehow, bizarrely, you changed your Hilbert space, you'd have more oscillators, instead, and su(n) instead of su(2)! But you don't. Call the oscillators a and b. $$ H/\hbar\omega= 1 + a^\dagger a + b^\dagger b ,\\ L_z=(a^\dagger a - b^\dagger b)/2 , ~~L_+ = a^\dagger b , ~~~L_-= b^\dagger a,\\ a^{\dagger ~k} b^{\dagger ~n}|0\rangle \equiv |k;n\rangle . $$ You then have $$ E=1+ k+n, ~~~L_z |k;n\rangle={k-n\over 2}|k;n\rangle\equiv m |k;n\rangle, \\ L^2|k;n\rangle = {k+n\over 2} \left ({k+n\over 2} +1\right )|k;n\rangle \equiv l(l+1) |k;n\rangle \leadsto\\ k=l+m, ~~~ n= l-m. $$
So the doublet representation is $$ |\uparrow\rangle =|1;0\rangle, ~~~(m=1/2, l=1/2)\\ |\downarrow\rangle =|0;1\rangle, ~~~(m=-1/2, l=1/2)\\ E=1+1, $$ the first excited state.
Tensor two ups. This is the state $$ |\uparrow \uparrow\rangle= |2;0\rangle ~~\implies l=1, m=1, E=1+2. $$ Apply the lowering operator to it, to get $$ {|\uparrow\downarrow\rangle +|\downarrow\uparrow\rangle \over \sqrt{2}}=|1;1\rangle ~~\implies l=1, m=0, E=1+2, $$ and further $$ |\downarrow \downarrow\rangle= |2;0\rangle ~~\implies l=1, m=-1, E=1+2, $$ the three second excited states.
It is evident that, for any $l$, the $m=0$ states require $k=n$. So the state orthogonal to the middle state above must be the $$ {|\uparrow\downarrow\rangle -|\downarrow\uparrow\rangle \over \sqrt{2}}=|0;0\rangle=|0\rangle ~~\implies l=0, m=0, E=1, $$ the ground state! But it's inaccessible to tensoring: you cannot combine two excitations to drop to the ground state,
The reason is the oscillators commute among themselves, and so you only get the symmetric representation in the composition, and you project out all others. (To get the above singlet, you'd need fermion oscillators, antisymmetrizing the two tensor factors.) So the only surviving factor in your (last) "Clebsch reduction" direct sum formula is the symmetric rep, $L=l_1+l_2$. There are no other Ls to consider.
Edit post comments
It appears the question was about the 4-dimensional isotropic oscillator, after all, whose symmetry group is SU(4) with 15 symmetry generators, not just 3! The energy/$\hbar \omega$ is then $4/2 + n$, where $n\equiv n_1+n_2+n_3+n_4$, the sum of the occupation numbers of all 4 uncoupled oscillators.
The degeneracy for a given energy shell characterized by n, above, is the dimensionality of the symmetric representation of SU(4) with n boxes in the Young tableau (so, arrayed next to each other),
$$ {4+n-1 \choose n}= \frac{(n+3)(n+2) (n+1)}{3!} ~~, $$ hence 4,10,20,... for n=1,2,3,... Check this.