Solution of Radial Function (Hydrogen Atom) | Quantum Science Philippines

## Solution of Radial Function (Hydrogen Atom)

Derive $\Big[\frac{d^{2}}{d\rho^{2}} - 2\lambda_{k,l} \frac{d}{d\rho} + \Big(\frac{2}{\rho} - \frac{l(l+1)}{\rho^{2}}\Big)\Big]Y_{k,l}(\rho)=0$.

Solution:

To solve the equation derived from problem (1), we look at the asymptotic behavior. First, let us assume a series of solution. But we must take into account as our $\rho\rightarrow\infty$.

A solution of the form: $U_{k,l}\sim e^{\lambda_{k,l}\rho}$ and $e^{-\lambda_{k,l}\rho}$

But $e^{\lambda_{k,l}\rho}$ can be discarded since if $\rho\rightarrow\infty$, it blows up. Then, in general, the solution for the equation at any distance is: $U_{k,l}\sim e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)$.

Substituting gives $\frac{d^{2}}{d\rho^{2}}\Big(e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)\Big) + \Big(\frac{2}{\rho}-\frac{l(l+1)}{\rho^{2}}\Big)e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho) - \lambda_{k,l}^{2}e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho) = 0$.

Taking the first term: $\frac{d^{2}}{d\rho^{2}}\Big(e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)\Big) = \lambda_{k,l}e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)+e^-\rho{\lambda_{k,l}}\frac{d}{d\rho}\big[Y_{k,l}(\rho)\big]$ $\frac{d^{2}}{d\rho^{2}}\Big(e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho)\Big) = \lambda_{k,l}e^-\rho{\lambda_{k,l}}Y_{k,l}(\rho) - 2\lambda_{k,l}e^-\rho{\lambda_{k,l}}\frac{dY_{k,l}(\rho)}{d\rho} + e^-\rho{\lambda_{k,l}}\frac{d^{2}Y_{k,l}(\rho)}{d\rho^{2}}$.

Therefore, $\Big[\frac{d^{2}}{d\rho^{2}} - 2\lambda_{k,l}\frac{d}{d\rho} + \Big(\frac{2}{\rho}-\frac{l(l+1)}{\rho^{2}}\Big)\Big] Y_{k,l}(\rho)=0$.