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I'm currently learning about spherical potentials (ex. hydrogen and hydrogen-like systems) and am trying to work through the problem of a generic spherical potential well such as:

$$V(r) = \left\{\begin{matrix} & -V_0 : r \leq a\\ & 0 : r > a \end{matrix}\right.$$

I'm comfortable with the separation of variables and how the radial equation is obtained. My issue is with actually solving the radial equation -- specifically, the transformation of the radial equation to the spherical Bessel equation.

I've searched through several QM textbooks and tutorials online. All of them suggest that once the radial equation is of the form (or similar to the form):

$$\frac{-\hbar^{2}}{2\mu} \frac{\mathrm{d}^2 u}{\mathrm{d} r^2}+\frac{\hbar^{2}}{2\mu}\frac{l(l+1)}{r^2}u=(E+V_{0})u$$

$$\frac{\mathrm{d}^2 u}{\mathrm{d} r^2}-\frac{l(l+1)}{r^2}u=-\frac{2\mu}{\hbar^{2}}(E+V_{0})u$$

$$\frac{\mathrm{d}^2 u}{\mathrm{d} r^2}-\frac{l(l+1)}{r^2}u=-k^{2}u$$

where $$u = rR(r)$$ and $$k=\frac{1}{\hbar}\sqrt{2\mu(E+V_0)}$$

Then make the transformation

$$\rho = kr$$

and the ODE transforms into the spherical Bessel equation. Trouble is, none of the examples I've seen explicitly show the steps of carrying out the transformation, and my own attempts have been fruitless. My attempt is as follows:

Multiply through by $\frac{1}{k^{2}}$

$$\frac{1}{k^{2}}{}\frac{\mathrm{d}^2 u}{\mathrm{d} r^2}-\frac{l(l+1)}{(kr)^2}u=-u$$

Rearrange and substitute $\rho=kr$

$$\frac{1}{k^{2}}{}\frac{\mathrm{d}^2 u}{\mathrm{d} r^2}-\Big(\frac{l(l+1)}{\rho^2}+1\Big)u=0$$

But it's unclear to me how to transform $\frac{\mathrm{d}^2 u}{\mathrm{d} r^2}$ to be in terms of only $\rho$. I believe the chain rule is somehow involved as the Bessel equation contains both first and second order derivatives. I know $du=(r\frac{\mathrm{dR} }{\mathrm{d} r}+R)dr$ but this doesn't seem to help anything from what I can tell.

Would someone be willing to help me fill in the gaps of how to arrive at this seemingly magical equation?

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  • $\begingroup$ Perhaps you could show us what you tried that didn't work. That will help us focus in on the specific problem you are having. $\endgroup$ – march Dec 3 '15 at 17:18
  • $\begingroup$ Sure, I've updated my post with some more details. $\endgroup$ – Anna W. Dec 3 '15 at 18:21
  • $\begingroup$ Great! Thanks. The magical "equation" is the chain rule. That is, since $kr = \rho$, $du/dr = (du/d\rho)(d\rho/dr)=kdu/d\rho$. Another application of the chain rule gets you the second derivative. $\endgroup$ – march Dec 3 '15 at 18:25
  • $\begingroup$ D'oh, how embarrassing. My calculus is a little rusty, so thank you! $\endgroup$ – Anna W. Dec 3 '15 at 18:34

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