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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).

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  • $\begingroup$ Find the probability of each state, then Use those probabilities as distribution coefficients for the degeneracy. $\endgroup$ – Bill N Jun 13 '18 at 19:35
  • $\begingroup$ 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. $\endgroup$ – chandrasekhar17 Jun 14 '18 at 21:16
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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}$

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