Solving the radial part of the Schrodinger equation for a central potential with two radial terms

Question

a "hydrogen-like" atom has the following modified coulomb potential $$V(r)=\frac{-Ze^2}{r}+\frac{\alpha}{r^2}$$ Where Z is the number of positive charges and $\alpha$ is a positive energy constant. Deduce the energy levels.

Here's where I'm at: Since we have a central potential, we can deduce the eigenstates only from the radial part of the schrodinger equation. We have $$\frac{1}{R}\frac{\mathrm{d} }{\mathrm{d} r}(r^2\frac{\mathrm{d} R}{\mathrm{d} r})+\frac{2\mu r^2 }{h^2}(E-V(r)-\frac{l(l+1)h^2}{2\mu r^2})=0$$

$$\frac{1}{r^2}\frac{\mathrm{d} }{\mathrm{d} r}(r^2\frac{\mathrm{d} R}{\mathrm{d} r})+\frac{2\mu }{h^2}(E-V(r)-\frac{l(l+1)h^2}{2\mu})=0$$

Solving this will give us eigenvalues. Let $$R(r)=\frac{u(r)}{r}$$ $$\frac{\mathrm{d} R}{\mathrm{d} r}=\frac{-1}{r^2}u(r)+\frac{1}{r}\frac{\mathrm{d} u(r)}{\mathrm{d} r}$$ Plugging back into the equation gives $$\frac{\mathrm{d}^2 u}{\mathrm{d} r^2} +\frac{2\mu}{h^2}(E-V(r)-\frac{l(l+1)h^2}{2\mu r^2})u(r)=0$$ Now using our potential V(r) $$\frac{\mathrm{d}^2 u}{\mathrm{d} r^2} +\frac{2\mu}{h^2}(E+\frac{Ze^2}{r}-\frac{\alpha}{r^2}-\frac{l(l+1)h^2}{2\mu r^2})u(r)=0$$ Change of variable, let $$\rho =\gamma r$$ I calculated $$\frac{\mathrm{d} u}{\mathrm{d} r}=\frac{\mathrm{d} u}{\mathrm{d} \rho}\gamma$$ $$\frac{\mathrm{d} u^2}{\mathrm{d} r^2}=\frac{\mathrm{d} u^2}{\mathrm{d} \rho^2}\gamma^2$$ plugging back in, $$\frac{\mathrm{d}^2 u}{\mathrm{d} \rho^2} +(\frac{2\mu E}{h^2 \gamma ^2}+\frac{2\mu Ze^2}{h^2 \gamma }\frac{1}{\rho }-\frac{2\mu \alpha}{h^2 \rho^2}-\frac{l(l+1)}{\gamma^2 r^2})u(\rho)=0$$ Choosing $\gamma$ such that $\frac{2\mu E}{h^2 \gamma^2}=\frac{-1}{4}$ (bound state) so $\gamma^2\equiv \frac{-8\mu E}{h^2}$ Also, let $\lambda^2 \equiv \frac{2\mu Ze^2}{h^2 \gamma}$ our equation becomes $$\frac{\mathrm{d}^2 u}{\mathrm{d} \rho^2}+(\frac{-1}{4}+\frac{\lambda}{\rho}-\frac{2\mu \alpha}{h^2 \rho^2}-\frac{l(l+1)}{\rho^2})u(\rho)=0$$

$$\frac{\mathrm{d}^2 u}{\mathrm{d} \rho^2}+(\frac{-1}{4}+\frac{\lambda}{\rho}-\frac{2\mu \alpha-h^2l(l+1)}{h^2 \rho^2})u(\rho)=0$$

I know the solution will be a combination of both as $\rho\rightarrow \infty$ and $\rho\rightarrow 0$

$\rho\rightarrow \infty$: $$\frac{\mathrm{d}^2 u}{\mathrm{d} \rho^2}-\frac{-1}{4}u(\rho)=0$$ $$\rightarrow u(\rho)=e^{-1/2 *\rho}$$ $\rho\rightarrow 0$: $$\frac{\mathrm{d}^2 u}{\mathrm{d} \rho^2}=\frac{2\alpha-h^2l(l+1)}{h^2 \rho^2} u(\rho)$$

Where do I go from here? I am not sure how to solve this last equation.

I know the entire solution will be a multiple of the solutions of the two limits times another function of $\rho$. Then I can get said solution to satisfy something like the confluent hypergeometric function, and then get the eigenstates from there...

Any help or suggestions will be very appreciated. Thanks in advance,

Ali

Answer 1

While the substitution provided in the comments works, sometimes such a substitution is not at all obvious, and I will present a more general method which does not require a specific guess. Let, $$u''(\rho) - \frac{c}{\rho^2}u(\rho) = 0$$

be your equation, with $c = \frac{1}{h^2}(2\alpha-h^2 l(l+1))$. We can make the ansatz,

$$u(\rho) = \sum_{n= 0}^\infty A_n \rho^{n+r}$$

where $r \in \mathbb C$ and $A_0 \neq 0$, which is known as the Frobenius method. Plugging in and differentiating, one finds that,

$$\sum_{n=0}^\infty \left[ (k+r-1)(k+r) - c\right] A_n \rho^{n+r-2} = 0.$$

From the $n=0$ case we find $r(r-1) -c = 0$ and thus $r = \frac{1}{2}(1\pm\sqrt{1+4c})$. Notice in this case that the equation is only satisfied for $A_n = 0$ for $n>0$; in most instances one finds a recurrence relation. Thus, the solution is simply,

$$u(\rho) = \sqrt{\rho} \left( c_1 \rho^{\frac{1}{2}\sqrt{1+4c}}+c_2 \rho^{-\frac{1}{2}\sqrt{1+4c}} \right)$$

which you would have found with the $u(\rho) = \rho^z$ guess.