A powerful method in theoretical physics are ladder operators. They are used in QM to solve problems like the harmonic oscillator and the hydrogen atom. The idea is to solve with their help the groundstate problem in order to get the full spectrum and eigenfunctions afterwards by successively applying them to the preceding states. Now the thing is, that most oftens these problems were solved somehow earlier so that somebody came up with a good way to choose the operators so that it works out. I have never heard of an analytical way to choose them. 


My question is: Do we know how to construct them for periodic potentials on $[0,2\pi]$?


So, if I have a Hamiltonian $H \psi = (- \frac{d^2}{dx^2} + V)\psi $, where $V$ is a smooth($\in C^{\infty}$) $2 \pi $ periodic function. If it helps, we could assume that it has a finite Fourier series expansion. In that case spectral theory (see for example Simon/Reed Analysis of Operators) tells us that $H$ has a completely discrete spectrum. Hence, in principle the chances should not be that bad that such operators exist.
Furthermore, we are able to choose eigenvectors $(\psi\_n)$ so that they are analytic in their argument for $x \in (0,\pi) \cup (\pi,2\pi)$ and continuous at $\{0,\pi,2\pi\}$ with $\psi\_n(0) = \psi\_n(2\pi)$. 


Now the question is:
Can we find operators $A$ and an adjoint version $A^\*$ such that $A^\* \psi\_n = \lambda\_n \psi\_{n+1}$ and $A \psi\_{n+1} = \mu\_n \psi\_n$ in this general setting without solving the problem completely? 



Mathematical equivalent to ladder operators?
An explicit construction of generalized ladder operators $A^\pm=\mp d/dx+W(x)$ exists if the Hamiltonian can be factorized as
$$H=-\frac{d^2}{dx^2}+V(x)=A^+ A^- +E\_0,$$
with $E\_0$ the lowest eigenvalues of $H$. The function $W(x)$ satisfies the Ricatti equation,
$$W(x)^2-W'(x)=V(x)-E\_0.$$
A class of "shape-invariant" potentials that can be treated in this way is discussed in [Generalized Ladder Operators for Shape-invariant Potentials](http://arxiv.org/abs/nucl-th/0108073) (2001).


Note that typically one also wants the ladder operators to satisfy a commutation relation of the form
$$B^+ (x)B^- (x)-B^- (x)B^+ (x)=B\_0$$
with $B\_0$ independent of $x$. How to transform $A^\pm$ into $B^\pm$ satisfying this property is also discussed in this paper.


