# Rotational Energy and Degeneracy

1. Nov 13, 2015

### pyroknife

I have a question regarding the rotational energy equation, derived based on the rigid rotor assumption:

$\epsilon = k\theta_r J (J+1)$
where k = boltzmann constant, and J is the rotational quantum number.

The degeneracy is 2J+1.

Let's assume the constant quantity $k\theta_r$ = 1 and that the energy level is 2Joules.
Thus 2 = J(J+1) => J = +/- 1. The only physical solution is J = 1.
Thus degeneracy is 3 for J = 1.

Here is where I am confused. I thought degeneracy is when you have an energy level that DOES NOT consist of a unique set of quantum numbers. So if the degeneracy is 3, doesn't that mean I should have 3 different J values that can give me an energy level of 2 Joules?
I think I am understanding this incorrectly, because for 2 joules, the only solution is J = 1. Doesn't that mean the degenearcy is 1?

2. Nov 13, 2015

### Staff: Mentor

There are three different and linearly independent solutions to the Schrodinger equation with $J=1$. That's where the three-fold degeneracy comes from.

Google for "spherical harmonics" to see how this can be; the three solutions are multiples of the three functions $Y^1_m$ with $m$ equal to -1, 0, or 1.