So, the energy eigenvalues are E n x , n y , n z = ( n x + n y + n z + 3 2 ) ℏ ℏ ω ω or, E n = ( n + 3 2 ) ℏ ℏ ω ω where n is a non-negative integer. So, the energy levels are degenerate and the degree of degeneracy is equal to the number of different sets { n x , n y , n z } satisfying n x + n y + n z = n The degeneracy of the n -th state can be found by considering the distribution of n quanta across n x , n y and n z . Having 0 in n x gives n + 1 possibilities for distribution across n y and n z . Having 1 quanta in n x gives n possibilities across n y and n z and so on. This leads to the general result of n − − n x + 1 and summing over all n leads to the degeneracy of the n -th state, ∑ ∑ n x = 0 n ( n − − n x + 1 ) = ( n + 1 ) ( n + 2 ) 2 For the ground state n = 0 , the degeneracy is 1 so the state is non-degenerate. For all higher states, the degeneracy is greater than 1 so the state is degenerate.