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It says in my lecture notes, that it depends on the hilbertspace in question weather the ground state $|0\rangle$ of an harmonic oscillator is degenerate or not. The ground state fullfills

$$N|0\rangle=0|0\rangle \quad \textrm{with} \quad N=a^\dagger a \quad \wedge \quad[a,a^\dagger]=1$$ Assuming that the ground state is non-degenerate one can proof that all states are non-degnerate and the spectrum is equal to $\mathbb{N_0}$

Can someone give an example for a space with a degenerate ground state? I've never encountered such an oscillator in my QM course.

In Proof that the One-Dimensional Simple Harmonic Oscillator is Non-Degenerate? the user Qmechanic relates this to Fock-Spaces. But I would appreciate a more concrete example, as this goes beyond my mathematical scope.

Qmechanic
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Solving the equation for a one-dimensional Harmonic oscillator is both a good exercise and a way of convincing oneself that it has only one ground state.

An important example of a Harmonic oscillator with degenerate ground state is the Landau levels of a particle in a magnetic field, where the $x$-oscillator Hamiltonian is given by $$\hat{H}=\frac{\hat{p}_x^2}{2m} + \frac{m\omega_c^2}{2}\left(\hat{x} - \frac{\hbar k_y}{m\omega_c}\right)^2,$$ and every state $$E_n = \hbar\omega_c(n+\frac{1}{2})$$ is inifinitely degenerate in respect to the position of the center of the oscillator: $$x_c = \frac{\hbar k_y}{m\omega_c}.$$ This example involves Hilbert space where the wave functions are plane waves in $y$-direction, and solutions to the Harmonic oscillator equation in $x$-direction. $$\Psi(x,y) = e^{ik_y y}\phi_n(x-x_c).$$

Roger V.
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