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Let's have potential $$ U(r) = -U_{0}e^{-\frac{r}{a}}. $$ I need to find energy levels for particles moving in this field (for an arbitrary values of orbital number $l$). This task isn't exactly solvable, so I need some method which can help to find approximate energy levels.

What to do?

I reduced the Schrodinger equation to the form (with normalized $r \to \frac{r}{a}$ and $\Psi (r, \varphi , \theta ) = \kappa e^{-\beta r}r^{l}\kappa (r) Y_{lm}(\theta , \varphi )$) $$ r\kappa '' + \kappa {'}(2l + 2 - 2\beta r) + \kappa (\alpha^{2}r e^{-r} - 2\beta (l + 1)) = 0, $$ where $$ \alpha^{2} = \frac{2mU_{0}a^{2}}{\hbar^{2}}, \quad \beta^{2} = \frac{2m |E|a^{2}}{\hbar^{2}}. $$ It would be tempting to use the approximation $$ r \approx \frac{1 - e^{-r}}{e^{-r}}, $$ but $r e^{-r}$ didn't reduce to the normal form.

The exact solution for $l = 0$ is existed.

Andrew McAddams
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  • Related: http://physics.stackexchange.com/questions/44942/exponential-potential-solution, http://physics.stackexchange.com/questions/47128/exponential-potential-expx – Kyle Kanos Dec 17 '13 at 20:34
  • @KyleKanos : unfortunately, it is harder, because there is $r, e^{-r}$ both in the equation. I tried to express $r$ as $r \approx \frac{1 - e^{-r}}{e^{-r}}$ near zero, but it didn't help to solve equation exactly. Also I can use quasiclassical approximation, but I believe that there is some other method. – Andrew McAddams Dec 17 '13 at 20:41
  • Well, there's always the variational method – Kyle Kanos Dec 17 '13 at 20:52
  • @KyleKanos : But it may help only for the ground state, doesn't it? – Andrew McAddams Dec 17 '13 at 21:01
  • Correct. Guess I didn't read the "arbitrary values of $\ell$" bit. If you need an approximation, might $\exp(-x)\simeq1-x$ be too approximate an approximation? – Kyle Kanos Dec 17 '13 at 21:06
  • @KyleKanos : I tried to compute the equation with this approximation. Unfortunately, even after approximation it didn't reduced to hypergeometric form. Or did you ask about the accuracy of following approximation? – Andrew McAddams Dec 17 '13 at 21:11
  • I had meant that the accuracy of the approximation might not be good enough. But I guess it doesn't matter since it didn't work. I'm not sure what the next step is, I'll have to think more about this. – Kyle Kanos Dec 17 '13 at 21:17
  • How do you know that you have bound states for arbitrary values of $\ell$? How do you know that there is a single bound eigenstate of that potential, for that matter? – Zo the Relativist Dec 17 '13 at 21:18
  • @JerrySchirmer : unfortunately, I don't know. But I was given the task, and I don't know what method to use. – Andrew McAddams Dec 17 '13 at 21:27
  • @AndrewMcAdams: well, I can certainly tell you that you wont' find any states with $\ell > U_{0}$, of course. – Zo the Relativist Dec 17 '13 at 21:31
  • @JerrySchirmer : but does it help to say anything definitely about states? In my opinion, it is useful if at least condition that allows to determine the spectrum is founded. – Andrew McAddams Dec 17 '13 at 21:35

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