Degeneracy of the ground state of harmonic oscillator

Degeneracy occurs when a system has more than one state for a particular energy level. Considering the three dimensional harmonic oscillator, the energy is given by

$$E_n = (n_x + n_y + n_z) ,hbar omega + frac{3}{2},$$

where $n_x, n_y$, and $n_z$ are integers, and a state can be represented by $|n_x, n_y, n_zrangle$. It can be easily seen that all states except the ground state are degenerate.

Now suppose that the particle has a spin (say, spin-$1/2$). In this case, the total state of the system needs four quantum numbers to describe it, $n_x, n_y, n_z,$ and $s$, the spin of the particle and can take (in this case) two values $|+rangle$ or $|-rangle$. However, the spin does not appear anywhere in the Hamiltonian and thus in the expression for energy, and therefore both states

$$|n_x, n_y, n_z, +rangle quad quadtext{and} quad quad |n_x, n_y, n_z, -rangle$$

are distinct, but nevertheless have the same energy. Thus, if we have non-zero spin, the ground state can no longer be non-degenerate.