# SU(N) symmetry in harmonic oscillator

Starting with the D-dim. harmonic oscillator and generators of SU(D)

$$T^a;\quad [T^a,T^b] = if^{abc}T^c$$

one can construct conserved charges

$$Q^a = a^\dagger_i\,(T^a)_{ik}\,a_k;\quad [Q^a,Q^b] = if^{abc}Q^c$$

satisfying the same algebra and commuting with the Hamiltonian

$$H = a^\dagger_i\,1_{ik}\,a_k + \frac{N}{2} = \sum_{i=1}^D a^\dagger_i a_i + \frac{N}{2};\qquad [H,Q^a] = 0$$

where I introcuced a new generator for U(1).

That means that the states

$$|n_1, n_2, \ldots n_N\rangle;\quad \sum_{i=1}^D n_i = N$$

are related to integer- or vector-representations of SU(D).

The degeneracy for fixed N is calculated as

$$\text{dim}\,N_D = \left(\array{N+D-1 \\ N}\right)$$

Questions:
- how are SU(D) vector representations labelled (compared to |lm> for SU(2))
- what are their dimensions (compared to 2l+1 for l-rep. in SU(2))
- and how do the dimensions sum up to dim ND

Thanks

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Checking this explicitly for the simplest cases SU(2) and SU(3) it becomes strange already:

SU(2)

N=0: |00> is the trivial 1-rep. of SU(2)

N=1: |10>, |01> is the 2-rep. of SU(2); but this is a spinor rep. with l = 1/2 and 2l+1 = 2

N=2: |20>, |11>, |02> is the 3-rep. of SU(2)

SU(3)

N=0: |000> is the trivial 1-rep. of SU(3)

N=1: |100>, |010>, |001> is a 3-rep. of SU(3); again this is not a vector rep.; and I know there are two 3-reps, 3 and 3*

N=2: |200>, ... |110>, ... is a 6-dim. rep. of SU(3) which seems to be a sum of the two 3-reps 6 = 3 + 3*, correct?

N=3: here I find dim 33 = 10; now this can either be the 10-dim. irreducible rep., or just 1+3+6

So my first conclusion from SU(2) is that I have to include half-integer reps as well; this is strange, but OK. But for SU(3) it's unclear how to arrange rep's and how to count them.

dextercioby