I would like to find the eigenstates of a potential with terms like $$ \left(\frac{1}{4}A^{4}x^{4}-A^{2}x^{2}+1\right)^{2} $$ and am planning to use ladder operators to find the solution in the basis of harmonic oscillator eigenstates. In that case (absorbing new constants into $B$) I must work with terms like $$ \left(\frac{1}{4}B^{4}\left(a^{+}+a\right)^{4}-B^{2}\left(a^{+}+a\right)^{2}+1\right)^{2} $$ I can just expand this out and work from there, but that carries with it the risk of computational errors which may be difficult to track down. It would be safer to use an identity if one is available.

Is there a better method to approach this than just brute-forcing it? I thought there was an identity for the action of a function of ladder operators upon a given state $|j\rangle$, but if there is, I can't seem to remember it.

EDIT: just to clarify, I do not expect a closed-form solution. I just want to expand in the basis of states $|j\rangle$ and diagonalize in a truncated (e.g. $j=0,1,2,...100$) basis. If I can figure out the matrix elements of terms like this, i.e. $\langle k|V|j\rangle$ where $V$ is the potential term above, that's all I need.

  • 2
    $\begingroup$ Do you have any reason to suppose that there is a closed-form solution to this problem? There are only a few non-quadratic polynomial potentials that have such solutions, and this is not one that I have seen. $\endgroup$ – mike stone Sep 6 '19 at 16:48
  • $\begingroup$ No, not in the least - rather, I expect that I should be able to solve it with a superposition of not too many terms. Since it's 8th-order, each $|j\rangle$ couples only up to $|j+8\rangle$ or down to $|j-8\rangle$. That suggests to me that I can find a solution with the most weight centered on only a (relatively) small range of $j$. $\endgroup$ – BGreen Sep 6 '19 at 17:01
  • 2
    $\begingroup$ FWIW, the potential can be written as $\left(\frac{1}{2}(Ax)^2-1\right)^4$. $\endgroup$ – Qmechanic Sep 6 '19 at 17:14
  • $\begingroup$ Thank you - now that you mention that, it should have been obvious. If there's a relevant identity for evaluating matrix elements of functions of ladder operators, that could come in handy. $\endgroup$ – BGreen Sep 6 '19 at 17:17
  • 1
    $\begingroup$ This can be useful? $\endgroup$ – Sunyam Sep 6 '19 at 19:23

You need to think about the wavefunction $\psi(x) = Ae^{-x^2}$, and the momentum operator definition $p=-i\frac{d}{dx}$. If you apply the momentum operator on the wavefunction, you get Hermite polynomials (which are in the form of the potential). The question wants the expectation value of the function in x basis: $$<|V|> =A^2 \int_{-\infty }^{\infty}e^{-\alpha x^2}V(x)e^{-\alpha x^2}dx$$ Where $\alpha = \frac{m\omega}{2 \hbar}$. You will see that multiple applications of the momentum operator on the wavefunction will produce the potential in question. Off the top of my head, $H_2(x) = 4x^2-2$, which results from the momentum operator applied to the wave function twice. if raised to the 4th power will produce something similar to your problem. If you put the momentum operator in terms of $p=\frac{i\hbar \omega}{2}\sqrt{a^\dagger - a }$ then: $$A^2 \int_{-\infty }^{\infty}e^{-\alpha x^2}(a^\dagger-a )^4 e^{-\alpha x^2}dx$$ When you expand this, only the $a^\dagger a$ terms will survive for i=j, and you will get some coupling terms too.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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