- #1

tom.stoer

Science Advisor

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

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

one can construct conserved charges

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

satisfying the same algebra and commuting with the Hamiltonian

[tex]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 [/tex]

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

That means that the states

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

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

The degeneracy for fixed N is calculated as

[tex]\text{dim}\,N_D = \left(\array{N+D-1 \\ N}\right)[/tex]

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 N

Thanks

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

one can construct conserved charges

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

satisfying the same algebra and commuting with the Hamiltonian

[tex]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 [/tex]

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

That means that the states

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

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

The degeneracy for fixed N is calculated as

[tex]\text{dim}\,N_D = \left(\array{N+D-1 \\ N}\right)[/tex]

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 N

_{D}Thanks

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