Physics Asked by Sakh10 on December 31, 2020

I was following sidney coleman videos and lecture notes where he claimed that unique ground state requires spinless harmonic oscillator while spin introduces degeneracy in the ground state corresponding to the various values of the z component of the spin.

Can anyone explain how the degeneracy appear for ground state for non zero spin quantum 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.

Correct answer by Philip on December 31, 2020

The 1D harmonic oscillator Hamilton is given by $$H=left(n+frac{1}{2}right)hbaromega$$ where $omega$ is the frequency and $n$ is a natural number.

As you can see the Hamiltonian is independent of spin $s$. Which means it is possible to simultaneously observe/define spin and energy. Furthermore it means that the each energy level $n$ can have $2s+1$ possible spin values. Thus each level is degenerate including the ground state (except when $s=0$).

Answered by Superfast Jellyfish on December 31, 2020

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