Has anyone published the procedure to generalize ladder operators for any potential in Schrodinger's equation?

I know that the ladder operator for the quantum harmonic oscillator \begin{align} H\psi_m = \left(\dfrac{p^2}{2m}+\dfrac{1}{2}m\omega^2x^2\right)\psi_m=E_m\psi_m \end{align} is \begin{align} A = \sqrt{\dfrac{m\omega}{2\hbar}}\left(x + \dfrac{i}{m\omega}p \right) \end{align} which lowers any state into its previous state, however I am at a loss on how to generalize the concept of ladders for any potential.

Has anyone published the procedure to generalize ladder operators for any potential in Schrodinger's equation?

• They are very helpful when a quantum number has integer value, or equivalently the values are equally spaced. Energy corresponding to a general potential does not have that pattern. – Xiaolei Zhu Apr 14 '15 at 0:10

A creation ladder operator $\hat{a}^\dagger$ for arbitrary states would have to be of the form $$\sum_{n=0}^\infty c_n \left| n+1 \right\rangle \left\langle n \right|$$ where you might allow a freedom to choose coefficients $c_n$ to try fulfilling whatever useful features of $\hat{a}^\dagger$ and $\hat{a}$ you may desire, say some generalization of the number operator $\hat{a}^\dagger \hat{a}$. You might choose to have its eigenvalues be the quantum number $n$ or the energy divided by the energy difference between ground state and first excited state—either, and probably some other choices are sensible generalizations of the situation for the harmonic oscillator. In fact, by choosing $c_n = \sqrt{n+1}$ and a harmonic oscillator Hamiltonian or, then, $\hat{H}=\hbar \, \omega \, \big( \hat{a}^\dagger \hat{a} + \frac{1}{2} \big)$, you reproduce the equations you gave in the question in a simpler notation abstracting away from position and momentum representation.
In general, $\hat{a}^\dagger$ (possibly through the $c_n$ and certainly through the energy eigenstates $\left| n \right\rangle$) will depend not just on your choice of what you want from the number operator, but crucially also on your system's Hamiltonian. The straight-forward way to derive $\hat{a}^\dagger$ is to solve the Hamiltonian for the energy eigenstates—which is the way you would have to go to get ladder operators from first principles if it were not for textbooks simply dropping a definition on you, (possibly) like the one you reproduced. You would simply insert the energy eigenstates $\left| n \right\rangle$ and formulate the constraint for the $c_n$ and solve for it. The simplest example, because the $c_n$ then do not depend on the Hamiltonian, is the constraint that the number operator has the quantum number as eigenvalues: $$\left\langle n \right| \hat{a}^\dagger \hat{a} \left| n \right\rangle = n \qquad \Rightarrow \qquad c_n = \sqrt{n+1}$$