I am a undergrad and currently trying to understand and use the following paper:

Th. Busch u. a. Two Cold Atoms in a Harmonic Trap. 1997. URL: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=

There, the authors derive an analytical solution to $$ \left(H_\text{osc} - E + \sqrt 2 \pi a_0 \delta^{(3)}(\vec r) \frac{\partial}{\partial r} r \right) \Psi(\vec r) = 0, $$ which models the center-of-mass part of two interacting atoms in a harmonic trapping potential. Their interaction is the regularized $\delta$-function $$ \delta^{(3)}(\vec r) \frac{\partial}{\partial r} r. $$

The solution they give is $$ \Psi(\vec r) = \frac 12 \pi^{-3/2} A \exp\left(-\frac{r^2}2\right) \Gamma(-\nu) U\left(-\nu, \frac 32, r^2 \right),$$ where $A$ is for normalization and $U$ is the confluent hypergeometric function.

With Mathematica, I have verified that this $\Psi(\vec r)$ is indeed a solution of the Schrödinger equation.

In section 3 of the paper, they sketch how this can be done in one dimension as well. I have not succeeded in deriving $\Psi(x)$ for that case yet, but I hope that this will come eventually.

My goal is to find an analytical solution to a one dimensional anharmonic oscillator that is pertubed with a gaussian potential.

Since I am currently lacking $\Psi(x)$ for the one dimensional $\delta$-pertubed oscillator, I wanted to try the three dimensional case first. In electrodynamics we had solved potential problems with a Green's function and used convolution to extend it. I think that the same might be possible here. Convolving the Gauss potential with the unregularized $\delta^{(3)}$-function gave me $\sqrt{2} \pi \mathrm e^{-r^2} (2 r^2-1)$.

Then I did a convolution of the 3D-$\Psi(\vec r)$, but that does not solve the adapted Schrödinger equation, as far as I can tell.

Is that even a legitimate way to extend the solution in the paper to other potentials? If so, what would be the right way to do this?

  • $\begingroup$ There's one issue that will have to be considered: in the paper, the fact that the trap potential is harmonic is essential to the separability of $H=H_\text{CM}+H_\text{rel}$, which relies on the identity $\mathbf{r}_1^2+\mathbf{r}_2^2=\mathbf{R}^2+\mathbf{r}^2$ where $\mathbf{R}=2^{-1/2}(\mathbf{r}_1+\mathbf{r_2})$ and $\mathbf{r}=2^{-1/2}(\mathbf{r}_1-\mathbf{r_2})$. Because of this, the eigenstates become tensor-products of the eigenstates of $H_\text{CM}$ and $H_\text{rel}$. If you want to modify it to use an anharmonic trap, this separability is probably going to be spoiled. $\endgroup$ Commented Apr 29, 2014 at 21:36
  • $\begingroup$ The problem I have to solve is a one-particle-problem. It is a single anharmonic oscillator. Therefore, separability is not a problem. I think I only need the relative part anyway, since that is a one particle problem. $\endgroup$ Commented Apr 30, 2014 at 9:08

1 Answer 1


Right now you have the equation $L\Psi = 0$ where $L$ is the linear operator in parentheses in your first equation.

The situation to use green's functions is when you have a solution for $\Psi$ of $L\Psi = \delta(x-x')$, and you want to find a solution for $\Psi$ of $L\Psi = f(x)$ for some arbitrary function $f(x)$.

The situation you have is different. Here it is not the right hand side that is changing$-$your right hand side is always zero$-$but it is the operator that is changing. So the method of green's functions is not applicable here.

As far as ways to solve the schroedinger equation, I think it is a hard problem. I want to say that most people just do it numerically. Perhaps some one else will have a better answer.

  • $\begingroup$ I will do this numerically, but I was given this paper for my work. So I do think that there is some connection to it. And I think that I should be somehow able to come up with an analytical solution so that I can compare. $\endgroup$ Commented Apr 29, 2014 at 18:37

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.