My QM book states that it is possible to define ladder operators for eigenstates of Coulomb potential hamiltonian. I am unable to prove it.
$$H = \frac{\vec{p}^2}{2}-\frac{1}{r} = \frac{p^{2}_{r}}{2}-\frac{\hbar^{2}l(l+1)}{2r^{2}}-\frac{1}{r}.$$
The eigenvalues of this hamiltonian are:
$$E_{l, n_{r}}=-\frac{1}{2(l+n_{r}+1)^{2}}.$$
The eigenvectors of this hamiltonian are of the type $|n\;l\;m\rangle$ where $n = n_{r}+l+1$, $l$ is the orbital angular number and $m$ is the magnetic quantum number.
Then the operator: $d_{l}=i p_{r}+\frac{\hbar l}{r}-\frac{1}{\hbar l}$
Acts as ladder operators:
$d^{\dagger}_{l+1}|n\;l\;m\rangle = K_{+}|n\;l+1\;m\rangle$
$d_{l}|n\;l\;m\rangle = K_{-}|n\;l-1\;m\rangle$
How do I prove that?
I have tried to prove it by:
- Computing $d^{\dagger}_{l}d_{l}$ and $d_{l}d^{\dagger}_{l}$ and writing them in terms of $H$ (in terms of number l) plus a function of l.
- Trying to compute $(d_{l}d^{\dagger}_{l})d_{l}|n\;l\;m\rangle = d_{l}(d^{\dagger}_{l}d_{l})|n\;l\;m\rangle$ and $(d^{\dagger}_{l+1}d_{l+1})d^{\dagger}_{l+1}|n\;l\;m\rangle = d^{\dagger}_{l+1}(d_{l+1}d^{\dagger}_{l+1})|n\;l\;m\rangle$ with the goal to show that $d^{\dagger}_{l+1}|n\;l\;m\rangle$ and $d_{l}|n\;l\;m\rangle$ are eigenstates of $H$.
I am unable to proceed. Thank you!
