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Tuesday, October 11, 2016

The isotropic harmonic oscillator

While studying Lie-algebras I read that the three-dimensional harmonic oscillator has an SU(3) symmetry. I found this very unexpected; I thought it was ''obvious'' that the symmetry is only SO(3).
The Hamiltonian of the three-dimensional harmonic oscillator is (setting \hbar=m=\omega = 1 for simplicity) \begin{equation*} H = \frac{1}{2} \vec{p}^2 + \frac{1}{2} r^2 \end{equation*} with r^2 = x^2 + y^2 + z^2. The energy levels are of the form [1] \begin{equation*} E_n = n + 3/2 \end{equation*} with n \ge 0 an integer. The degeneracy of the n-th level d(n) is equal to the number of ways n can be written as n = n_1 + n_2 + n_3 where 0 \le n_1, n_2, n_3 are integers. For example,
  • If n=0, d(0)=1 because one only has 0 = 0+0+0 
  • If n=1, d(1)=3 because one has the three possibilities \begin{align*} 1 &= 1+0+0\\ 1 &= 0+1+0\\ 1 &= 0+0+1 \end{align*}
  • The case n=2 is already more interesting. d(2) = 6 coming from \begin{align*} 2 &= 2+0+0\\ 2 &= 0+2+0\\ 2 &= 0+0+2\\ 2 &= 1+1+0\\ 2 &= 1+0+1\\ 2 &= 0+1+1 \end{align*} Here one can calculate that the states transform under SO(3) as the representation \bf{5} + \bf{1}. There is thus a degeneracy in the energy levels that is not explained by the SO(3) symmetry.
  • At the next level, there is even more degeneracy: d(3) = 10 coming from \begin{align*} 3 &= 3+0+0,\quad\text{3 permutations in total}\\ 3 &= 2+1+0,\quad\text{6 permutations in total}\\ 3 &= 1+1+1\end{align*} Here one can calculate that the states transform under SO(3) as the representation \bf{7} + \bf{3}.
Mathematically, it is very easy to construct the SU(3) symmetry. Because the harmonic oscillator is three-dimensional, there are three independent creation- and annihilation operators with commutation relations \begin{align*} [a_i, a_j] &=0\\ [a_i^{\dagger}, a_j^{\dagger}] &=0\\ [a_i, a_j^{\dagger}] &=\delta_{ij} \end{align*} In general, if T_a are 3\times3 matrices satisfying \begin{equation*} [T_a,T_b] = \sum_c f_{ab}^c T_c \end{equation*} then the operators Q_a = \sum_{ij} a_i^{\dagger} T_{a,ij} a_j satisfy the same algebra \begin{equation*} [Q_a,Q_b] = \sum_c f_{ab}^c Q_c \end{equation*} If the matrices T_a form the 3-dimensional representation of SU(3), then the Q_a form a representation of the SU(3) algebra, and it is easy to see that they also commute with the Hamiltonian. The states for n=2 form the SU(3) representation \bf{6}. The states for n=3 form the SU(3) representation \bf{10}, and so on. All in all, the mathematics is easy, but I would not have guessed that there is this extra symmetry. All this is explained in chapter 14 of [2], and also in notes by Kirson [3].
References
[1] Griffiths, Introduction to Quantum Mechanics, exercise 4.39
[2] Georgi, Lie Algebras In Particle Physics
[3] Kirson, Isotropic harmonic oscillator

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