# 3-d harmonic oscillator and SU(3) - how to imagine it?

1. Apr 17, 2009

### franoisbelfor

The 3-dimensional harmonic oscillator has SU(3) symmetry. This is stated in many papers. It seems to be due to the spherical symmetry of the system. (After all, the idea of a 3d harmonic oscillator is that a mass is attached to the origin with a spring, and that the mass can move in 3 dimensions, with no gravity involved.)

But I have an issue. If this system has SU(3) symmetry, the 8 generators of SU(3) must somehow act on the various eigenstates. How exactly does this happen? Is there a book or a paper on this?

For example, the eight Gell-Mann generators of SU(3) must transform the states of the harmonic oscillator. What do they do? Do they rotate the state? How? Why are there 8 of them?

Thank you for any advice in this matter.

Cheers

François

2. Apr 17, 2009

### Spinnor

3. Apr 17, 2009

### ExactlySolved

The generators are elements of the lie algebra su(3), not the group SU(3). If we take a step away from matrices and imagine the actions of the generators as abstract operators then the generators correspond to infinitesimal symmetry transformations of the Hamiltonian, therefore they do not necessarily transform eigenstates to eigenstates.

I just want to point out that the Gell-Mann matrices are one of many equivalent representations for the generators of SU(3).

Why 8 generators? Determine the dimension of SU(3), and determine the dimension of the tangent space at the identity, and tell me how many vectors are needed to span this tangent space (the answer is 8).

4. Apr 18, 2009

### franoisbelfor

D M Fradkin discussed the physics very clearly in the 1964 American Journal of Physics, I found out. The symmetries are the angular momentum and the Runge-Lentz tensor (for the harmonic oscillator in 3d, it is a tensor, not a vector). That makes 3+5=8 generators

The generators shift energy among the 3 linearly independent oscillators, keeping the total energy constant.

François

5. Apr 21, 2009

### lbrits

Thanks, François. That's insightful.

In any case, it's easy to see where the SU(N) comes from, but not terribly illuminating. If you work in the holomorphic representation, then the Hamiltonian is $$H = a^\dagger a$$ which clearly has an SU(N) symmetry. This is also a very elaborate way of determining the degeneracy of such a harmonic oscillator.

Of course, there's the unrelated issue that su(3) contains a bunch of su(2) sub-algebras.