# Compute degeneracy in energy levels of a wavefunction which is a sum of other wavefunctions

Given a hydrogen atom-like wavefunction at time $t=0$, say:

$\Psi(r,t=0)=A(2\psi_{100}+\psi_{210}+\sqrt{2}\psi_{211}+\sqrt{3}\psi_{21-1})$

How can I compute the degeneracy of the energy levels for the state of this wavefunction?

I know that for a given state represented by $\psi_{nlm}$ the degeneracy of the energy levels would be $n^2$ (without considering spin).

• Find the probability of each state, then Use those probabilities as distribution coefficients for the degeneracy. – Bill N Jun 13 '18 at 19:35
• I don't think that's a good idea, since as I know degeneracy must be an integer and that procedure leads to a rational number. – chandrasekhar17 Jun 14 '18 at 21:16

Neglecting spin quantum number of the number of degeneracy of $H$-Atom is given by $n^2$ where $n =$ is principal quantum number.
Thereby, $\Psi(r,t=0)=A(2\psi_{100}+\psi_{210}+\sqrt{2}\psi_{211}+\sqrt{3}\psi_{21-1}) = A(2\psi_{n_{1}l_{1}m_{1}}+\psi_{n_{2}l_{2}m_{2}}+\sqrt{2}\psi_{n_{2}l_{2}m_{3}}+\sqrt{3}\psi_{n_{2}l_{2}m_{4}})$
Which gives the total number of degeneracy $= n^{2}_{1} + n^{2}_{2} = 1^2 + 2^2 = 5$
However, $\psi_{200}$ is missing in the given wavefunction. Hence total number of degeneracy of the energy levels for the state of this wavefunction is $= 5-1 = \boxed {4}$