Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

SU(N) symmetry in harmonic oscillator

  1. Nov 9, 2011 #1

    tom.stoer

    User Avatar
    Science Advisor

    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 ND

    Thanks
     
    Last edited: Nov 9, 2011
  2. jcsd
  3. Nov 9, 2011 #2

    tom.stoer

    User Avatar
    Science Advisor

    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.
     
  4. Nov 9, 2011 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Tom, the general theory of unitary representations of SU(n) is developed in Chapter 13 of Wu Ki Tung's Group Theory book and is derivable from the general theory of representations of GL(n) as it's done in the book.

    For the 3-dim isotropic oscillator a thorough investigation from a group theoretical perspective is made by Wybourne in his book in Chapter 20.
     
    Last edited: Nov 9, 2011
  5. Nov 9, 2011 #4

    Bill_K

    User Avatar
    Science Advisor

    This is equivalent to the problem of populating N cells with bosons. The states are labeled by the occupation numbers n1, n2,... nN where n = ∑ ni. The representation of SU(N) is the symmetric part of the n-fold tensor product (N ⊗ N ⊗ ... ⊗ N)sym, a reducible representation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: SU(N) symmetry in harmonic oscillator
Loading...