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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?

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  • $\begingroup$ For $V=0$ on $S^1$, you need operators that take $e^{2\pi i n x}$ to $e^{2\pi i (n\pm 1)x}$. In general, it seems that the ladder operators exist, but their explicit form seems to need knowledge of the solutions $\psi_n$. They are generally manifestations of the representation theory of $\mathfrak{su}_2$ or the Heisenberg Lie algebra. $\endgroup$
    – S. Carnahan
    May 18, 2014 at 2:20
  • $\begingroup$ It may be a stupid question, but I don't understand why your operator has discrete spectrum. What happens if $V\equiv 0$? The spectrum of the second derivative on $L^2(I)$ is absolutely continuous if $I=\mathbb R$. Or are you assuming $I$ to be bounded? $\endgroup$ May 18, 2014 at 14:53

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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 (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.

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    $\begingroup$ great answer, thank you. indeed, such a solution to $W(x)^2-W'(x) = V(x)-E_0$ exists. at least, in some relevant cases. $\endgroup$
    – user37929
    May 18, 2014 at 12:06
  • $\begingroup$ $W^2-W'=V-E_0$ has a global solution precisely if $H$ has no spectrum below $E_0$. $\endgroup$ May 19, 2014 at 0:02
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Schrödinger's equations in periodic potentials are usually analyzed using Bloch theorem.

Because your potential is periodic it has a Fourier series, $V = \sum V_i \, e^{inx}$ and the wave function must also have a Fourier series, $\phi(x) = \sum \psi_n(x) e^{i n x}$.

Bloch's theorem says you can write the wavefunctions as $\phi(x) = u(x) e^{i\mathbf{k} x}$ where $u(x)$ is also periodic and $\mathbf{k}$ lives in the 1st Broullin zone, the fundamental domain dual to the lattice.

The proof says that any translation of the lattice should multiply the wavefunction by a phase. This phase determines the value of $\mathbf{k}$.


One example is the Kronig-Penny potential in solid state physics:

$$ V(x) = V_0 \sum_{n \in \mathbb{Z}} \delta(x - an) $$

This potential is invariant under translations $x \mapsto x + a$ and therefore we guess $\psi(x+a) = \psi(x) e^{ika}$ for the wavefunction. Then we get a relationship between the lattice momentum and the energy:

$$ \cos \lambda = \frac{v}{2\beta} \sin \eta + \cos \beta $$

with $\lambda = ka$ and $\beta = a \sqrt{\tfrac{2mE}{\hbar}}$ and $V = \tfrac{2mV_0 a}{\hbar^2}$ (these are just formulas taken from the link).


The general framework for this type of equation is Floquet-theory which deals with equations of the type $\dot{\psi} = A(x) \psi$ with $A(x)$ periodic in $x$.

There may be integrable systems which exhibit this type of monodromy behavior.

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  • $\begingroup$ mhmm... I don't really see through your answer. My first concern is, that your theorem applies only to translations, which is slightly different from my problem which is more like rotation. Even if it applies to rotation (as I restrict myself to $[0,2\pi]$ with $\psi(0)=\psi(2\pi)$ and the same for the derivative), then I don't see how it helps me getting all the energy eigenvalues if I just know the ground state ? Actually, your theorem does not give much more insight into the problem, as I already assumed my wavefunction to be periodic. So sorry, I don't see how this helps. $\endgroup$
    – user37929
    May 24, 2014 at 20:22
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    $\begingroup$ @Lipschitz Your Hamiltonian is invariant under translations $x \mapsto x + 2\pi$. Therefore, the translation group $\mathbb{Z}^2$ induces a representation on the space of wave-functions $\psi_k(x)$. $\endgroup$ May 24, 2014 at 20:36
  • $\begingroup$ so you say: $\psi(x+2 \pi) = \psi(x) e^{i 2\pi x}$? I mean, this is clear, from $\psi(0) = \psi(2\pi)$. I agree, Floquet-theory and Bloch theorem were the first classical results in some sense to the theory of periodic Schrödinger operators, but I don't see how all this helps me with my actual problem: finding the ladder operators for higher eigenvalues if the ground state is known. $\endgroup$
    – user37929
    May 24, 2014 at 21:07
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    $\begingroup$ @Lipschitz I am suggesting a ladder structure may not exist... the Harmonic oscillator in phase space is a circle $H = p^2 + x^2$ which has a rotational symmetry, but your hamiltonain is $H = p^2 + \sin(x)$ which does not have the same type of symmetry. We have $[p,x] = i$ but $[p, \sin x] = i \cos x$. $\endgroup$ May 24, 2014 at 21:09
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    $\begingroup$ @Lipschitz I do know a way to generalize the ladder operators... and I would love to write about them. As your question discusses periodic potentials I had left it out... may I put a second answer here? $\endgroup$ May 24, 2014 at 21:17

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