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 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

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