# Creation and annihilation operators in the hydrogen atom

In my quantum chemistry course, we have been discussing the wavefunctions of the hydrogen atom, further, I am familiar with the idea of ladder operators from the quantum harmonic oscillator. Can we use ladder operators to jump between orbitals in a hydrogen atom? It appears from a cursory check that $$\hat a^+ \psi_{1s} \neq \psi_{2s}$$ such that $$\hat a^+ = \frac{-i}{\sqrt{2w \hbar}}(\hat P+iw \hat X)$$ Is there any such set of ladder operators such that the above relation would be true? Thanks, and I apologize in advance if this question is naive. (Further, I am asking about ONLY wavefunctions of the solved hydrogen atom)

Yes, there is set of ladder operators for the discrete spectrum of the Hydrogen atom, too. Following Fitts, we can set the variables of the radial equation so that this is exactly as:

$$\left(\rho^2 \frac{d^2}{d\rho^2} + 2\rho\frac{d}{d\rho}+\lambda\rho-\frac{\rho^2}{4}\right)S_{\lambda l}=l(l+1) S_{\lambda l} \tag{1}.$$

Then the sought ladder operators are defined as:

$$\hat{A}_{\lambda} := \rho\frac{d}{d\rho}+\frac{\rho}{2}-\lambda \tag{2},$$ $$\hat{B}_{\lambda} := \rho\frac{d}{d\rho}-\frac{\rho}{2}+\lambda \tag{3}.$$

By using the operators in (2) and (3) to rewrite (1), one can show that $$\hat{A}_{\lambda}+\hat{1}$$ is a lowering ladder operator (in $$\lambda$$), and $$\hat{B}_{\lambda}+\hat{1}$$ is a raising ladder operator (also in $$\lambda$$).

For the rest of the algebraic treatment, see the quoted article by Fitts or his textbook, which is based on his article, just with minor changes.

Just in case the article and book are out of reach, here are the definitions:

The radial equation from textbooks is:

$$\frac{1}{r^2} \frac{d}{dr}\left(r^2\frac{d R(r)}{dr}\right) +\left[\frac{2\mu}{\hbar^2}\left(E+\frac{Ze^2}{r}\right)-\frac{l(l+1)}{r}\right]R(r) =0 \tag{4}.$$

To get (1), we need to define as follows (remember, $$E<0$$):

$$\lambda :=\frac{\mu Z e^2}{\hbar \sqrt{-2\mu E}} \tag{5},$$

$$\rho := 2\frac{\sqrt{-2\mu E}}{\hbar} r = \frac{2\mu Z e^2}{\lambda \hbar^2} r = \frac{2Z}{\lambda a_{\text{Bohr}}} \tag{6},$$

$$a_{\text{Bohr}} := \frac{\hbar^2}{\mu e^2} \tag{7},$$

$$R_{El} (r) =: \left(\frac{2Z}{\lambda a_{\text{Bohr}}} \right)^{3/2} S_{\lambda l}(\rho) \tag{8}.$$