# Why are so many wave functions associated with hydrogen?

According to Wikipedia, there's an infinite set of possible wavefunctions (orbitals) for the hydrogen atom: $$\psi_{n\ell m}(r,\theta,\phi) = \sqrt {{\left ( \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^{\ell} L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^{m}(\theta, \phi )$$

1. Would an unperturbed electron ever go outside the lowest orbital?
2. At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions? "Probability densities for the first few hydrogen atom orbitals" (source)

1.Would an unperturbed electron ever go outside the lowest orbital?

First, there is no such thing as "outside the lowest orbital" because the lowest orbital has probability density at all points in space (no nodes).

But yes, at a given temperature electrons will be distributed amoung the energy levels in accordance with a Boltzmann distribution.

2.At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?

There are infinite solutions none of which exceed the ionization energy (the ionization energy is only approached as n approaches infinity).

Think of a series like 0, 3/4, 8/9, 15/16, etc. where there are infinite members without reaching 1 (1 is approached in the limit of infinity).

• so if I have only one electron, which can be at all points in space via the lowest orbital, what additional information I get from all the other orbitals? especially since they completely change (hybridize etc.) when around other atoms. – Sparkler Mar 28 '15 at 18:28
• @Sparkler although in the ground state the electron can be anywhere is space, each point has a probability density. If the electron enters a higher energy state, the probability density will change. For example, if the atom is in a 2s state instead of a 1s state, the electron has a greater probability of being further from the proton. – DavePhD Mar 28 '15 at 19:13
• and is there any "use" of superposition of orbitals when discussing a single isolated atom? – Sparkler Mar 28 '15 at 22:05
• "There are infinite solutions none of which exceed the ionisation energy" is not the whole story, strictly speaking. Technically, the solutions are not limited to such bound states which give rise to discrete energy spectrum, the are also infinite number of solutions all of which exceed the ionisation energy. These solutions correspond to unbound states and give rise to continuous part of energy spectrum. – Wildcat Mar 29 '15 at 8:36
• See, for example, here. – Wildcat Mar 29 '15 at 8:38