4-Fold Degeneracy of 14th Energy Level in Cubic Well

[YoungEverest]

Homework Statement

In reference to the particle in an infinite, cubic well I have been asked to calculate the degeneracy of the 14th energy level and comment on its special nature.

2. The attempt at a solution
Notation: E_n = E (n_x , n_y , n_z)

E_14 = E(3,3,3)=E(5,1,1)=E(1,5,1)=E(1,1,5) and so is 4-fold degenerate.

Now, I know that a double degeneracy leads to a symmetry with respect to a 90 degree rotation. Am I along the right lines that symmetry is what is special about this case?

Thanks!

 

[JaWiB]

Shouldn't the 14th energy level be the one where n_x^2+n_y^2+n_z^2 = 14?

 

[YoungEverest]

The 14th energy level is the 14th highest energy value, if the energy level number was the product of 3 integers squared then you couldn't have a 1st, 2nd 4th etc. energy level. n_x^2+n_y^2+n_z^2 = 14 is the 6th energy level. :)