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?

  • 2
    $\begingroup$ 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. $\endgroup$ Apr 14, 2015 at 0:10

1 Answer 1


I know of no such publication. However, this issue may simply be on one hand too trivial and on the other hand too far removed from practical relevance. Let's derive what you are after to see:

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: $$\left\langle n \right| \hat{a}^\dagger \hat{a} \left| n \right\rangle = n \qquad \Rightarrow \qquad c_n = \sqrt{n+1}$$

That means that the ladder operators for any specific (non-harmonic oscillator) Hamiltonian is usually not helpful in solving this Hamiltonian but at best for re-expressing it. A harmonic oscillator is a rare case where this leads to a simplification, but this will not be the case in general. The fact that you obviously cannot retain all useful features e.g. of the number operator further limits the utility of specialized ladder operators.

  • $\begingroup$ Thanks for the answer. May you expand on the first paragraph. Why is this too trivial and not relevant? Should I delete this question? Thank you for your time and consideration. $\endgroup$ Apr 14, 2015 at 1:49
  • $\begingroup$ No, it's a fine question; these comments only meant to explain why such publication might(?) not exist. With some experience in quantum mechanics, you should become able to quickly derive these things (hence trivial, although perhaps only for an advanced practitioner). By not of practical relevance I meant to opinionate that it usually won't give you practical benefit (e.g. for solving a Hamiltonian) because it only seems to offer such shortcuts for harmonic oscillators after such an Hamiltonian has once been solved. In general, the shortcut does not work as such. $\endgroup$
    – user73762
    Apr 14, 2015 at 1:56
  • $\begingroup$ Thanks. I don't think I am advanced enough to see the trivial procedure. It sounds like you might be experienced enough to provide a procedure. I give you full support. $\endgroup$ Apr 14, 2015 at 2:01
  • $\begingroup$ Maybe not! I tried to explain the procedure, but if that is not understandable, then obviously my explanation is no good. Perhaps someone else can contribute a better answer? $\endgroup$
    – user73762
    Apr 14, 2015 at 2:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.