Question

Show that most of the energies determined by Equation 5.64are doubly degenerate. What are the exceptional cases? Hint: Try it for N=1,2,3,4.... , to see how it goes. What are the possible values of cos(ka)in each case?

Short answer

There is no degeneracy at the top and the bottom of the band when$\cos \left(ka\right)=\pm 1$

Step-by-step solution

Define the Schrodinger equation

• A differential equation for the quantum mechanical description of matter in terms of the wave-like characteristics of particles in a field. The solution has to do with the probability density of a particle in space and time.

• The time-dependent Schrodinger equation is represented as

$\mathbf{iħ}\frac{\mathbf{d}}{\mathbf{dt}}\overline{)\mathbf{\psi }\left(t\right)\mathbf{>}\mathbf{=}\stackrel{\mathbf{⏜}}{\mathbf{H}}\overline{)\mathbf{\psi }\left(t\right)\mathbf{>}}}$

Analyze the condition

For $Ka=2\mathrm{\pi n}/\mathrm{N}$we have a condition on $n:n=0,12,...,N-1$. Because for larger $n$, we don't get new solutions.

We are going to write solutions for $\cos \left(Ka\right)$for couple of N

$Ka=\frac{2\mathrm{\pi n}}{N}$

Value of different $n$corresponds to a distinct state

Analyze the energies

$N=1,n=0\cos \left(ka\right)=1$Non- degenerate

$N=2,n=0,1\cos \left(ka\right)=1,1$Non- degenerate

$N=3,n=0,1,2\cos (ka)=1,-\frac{1}{2},-\frac{1}{2}$First one is non -degenerate other one is degenerate

$N=4,n=0,1,2,3\cos (ka)=1,0,-1,0$Two are non -degenerate and two are degenerate

When we have $\cos \left(ka\right)=\pm 1$, then we don't have degeneracy. This happens at the top and at the bottom of the band.

So, we don’t have degeneracy at the top and the bottom of the band when$\cos (ka)=\pm 1$