Does energy level determine degeneracy in 3D systems?

[vabite]

Hi everyone.

I can not remember if, in 3D, the higher it is the energy level, the higher it is its degeneracy. With a cubic well and with a 3D harmonic oscillator it holds... Does anyone know if it is a general rule or not (and in the case it is, where does this rule come from)?

 

[WannabeNewton]

No that's not true, not even for a particle in a box of equal sides.

Consider the energy eigenvalues of this system: $E_{\vec{n}} = \frac{\hbar^2 \pi^2}{2m L^2}(n_x^2 + n_y^2 + n_z^2)$. The first excited state has degeneracy 3 and so do the second and third excited states but the fourth excited state has no degeneracy just like the ground state.

 

[Telemachus]

You must look at the energy, and see how different eigenvalues generates the same value of energy (because that means that for the same value of the energy, you have more than one eigenfunction, and that's the definition of degeneracy). For example, for the particle in a cubical box of side a, the energies are:

$E=\frac{h^2}{8ma^2}(n^2_x+n^2_y+n^2_z)$

Look at what happens for different eigenvalues, for example for $n_x=1,n_y=2,n_z=1$ you have the same value of the energy than for: $n_x=2,n_y=1,n_z=1$ and $n_x=1,n_y=1,n_z=2$, so that energy level is degenerate. Now, if the box weren't cubical, this wouldn't hold. Now, for higher energies, in general you will have more combinations for the triad of n's that gives the same value of energy, and that's why the degeneracy grows with the energy (not always as WannaBeNewton said).

 

[vabite]

Ok. You have been both very clear. Thanks.