# Degeneracy of the energy level

hidemi
Homework Statement:
The energy for one-dimensional particle-in-a-box is En = (n^2*h^2) / (8mL^2). For a particle in a three-dimensional cubic box (Lx=Ly =Lz), if an energy level has twice the energy of the ground state, what is the degeneracy of this energy level?
Relevant Equations:
En = (n^2*h^2) / (8mL^2)

En = [(nx/Lx)^2 + (ny/Ly)^2 + (nz/Lz)^2] *h^2 / (8mL^2)
Energy of the One-dimensional box:
ground state: En = (n^2*h^2) / (8mL^2), where n=1
twice the ground state: 2* En = 2 [(1^2*h^2) / (8mL^2)]

Energy of the Three-dimensional box:
En = (nx^2 + ny^2 + nz^2) *h^2 / (8mL^2) = 2 (1^2*h^2) / (8mL^2)
As stated, twice the ground state energy of one dimensional box is equal to that of the three -dimensional box. So, (nx^2 + ny^2 + nz^2) = 2, how would the degeneracy be 3? How should I continue?
Can someone help because I am confused, please? Thank you.

Staff Emeritus
Homework Helper
As stated, twice the ground state energy of one dimensional box is equal to that of the three -dimensional box.
The problem statement doesn't in fact say this. It's referring to the ground state of the 3-d box.

BTW, what's up with the thread title?

hidemi
The problem statement doesn't in fact say this. It's referring to the ground state of the 3-d box.
If the energy level is twice the energy of the ground state in three dimensional cubic box, then the energy would be
2*[(nx^2 + ny^2 + nz^2) *h^2 / (8mL^2)]
= 2*[(1^2 + 1^2 + 1^2) *h^2 / (8mL^2)]
= 6*h^2 / (8mL^2)

Is this correct?

Staff Emeritus
Homework Helper
Yes.

hidemi
Yes.
The degeneracy would be a total of three because for the energy to be 6*h^2/(8mL^2), there are three possibilities: (2,1,1,), (1,2,1) and (1,1,2). Is it the correct reasoning?

Staff Emeritus