# Ground state degeneracy [closed]

Today, I have learned from Prof. Cramer's (univ. of minnesota) lecture that the ground state can have degeneracy.

and he showed entropy $S = k_{b} ~ ln_{}~ \bf{n}$ , if ground state is $\bf{n}$ - fold degenerate.

Since I started learning this newly, I couldn't resist myself from asking,

Do ground state degenerate? If yes, could you explain through simple example. (pls. dont confuse me with lengthy equations)

## closed as unclear what you're asking by Jon Custer, John Rennie, Asher, Raskolnikov, MikeNov 30 '17 at 17:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• -1. Not clear what you are asking. Why do you think degeneracy should not be possible for ground states? What did Prof. Cramer have to say about this in his lecture? – sammy gerbil Nov 15 '17 at 14:18
• If some particles in the excited state, I could imagine, they can have a different combination of translation, vibration, rotational and electronic energies to represent the same energy. If it is in the ground state assuming zero energy(at absolute zero temperature), What combination will give me zero/ground energy? – mustang Nov 16 '17 at 8:50
• I am learning through offline videos. no communication between us. – mustang Nov 16 '17 at 8:51
• Offline videos still contain explanations. – sammy gerbil Nov 17 '17 at 5:52
• Your problem is that you assume the ground state has zero energy. This is not always true for quantum systems; a very simple example is the quantum simple harmonic oscillator, which has a ground state energy of $\hbar\omega/2$. – probably_someone Nov 17 '17 at 6:14

Ground states can indeed be degenerate. A simple example is a half-spin particle experiencing a quantum harmonic oscillator potential. Assuming there's no magnetic field here, then the ground state will be degenerate, namely $|E=\frac{1}{2}\hbar\omega, S_z = \frac{1}{2}\hbar\rangle$ and $|E=\frac{1}{2}\hbar\omega, S_z = -\frac{1}{2}\hbar\rangle$. These two states are distinguished by their spin state, nevertheless, they are ground states and have the same energy. In general, if an observable (i.e. an operator associated to a measurement, for instance, $\hat S_z$ in this particular case) commutes with the Hamiltonian, then there would be energy degeneracy whose magnitude (the dimension of the vector subspace of states of the same energy eigenvalue) is determined by the number of possible distinct eigenvalues of the observable commuting with the Hamiltonian.