# Solving the schrodinger equation in cylindrical coordinates

I'm trying to solve the Schrodinger equation in cylindrical coordinates using this potential $V=\frac{c}{\rho}$ where c is a constant.

$$\psi(\rho,\phi,z)=R(\rho)\Phi(\phi)Z(z)$$

The $\phi$ and Z part are simple to solve but in the radial part I get this:

$$\rho\frac{1}{R}\frac{\partial}{\partial\rho}\left(\frac{\partial R}{\partial\rho}\right)-\frac{2m}{\hbar^2}V\rho^2+l^2\rho^2=m^2$$

Where $l^2$ and $m^2$ are the separations constant when I separate the Z and $\phi$ part.

I think the solution must be in terms in Bessel function like the H atom but I'm not sure and I don't know how to try to solve this.

• Something is not right. You don't have the energy in the equation. Also, your equation should be independent of $m$ – Andrei Mar 5 '17 at 17:22
$$\frac{1}{r}\frac{\partial}{\partial{r}}\Big(r\frac{\partial\Psi}{\partial{r}}\Big)+\frac{1}{r^{2}}\frac{\partial^{2}\Psi}{\partial\varphi^{2}}+\frac{\partial^{2}\Psi}{\partial{z}^{2}}+\Big(E-\frac{2mc}{\hbar{r}}\Big)=0$$ Let $\Psi=e^{in\varphi}e^{iqz}\rho(r)$ with $q\in\mathbb{R}$, $n\in\mathbb{Z}$. $$\frac{1}{r}\frac{\partial}{\partial{r}}\Big(r\frac{\partial\rho}{\partial{r}}\Big)-\frac{n^{2}}{r^{2}}\rho+\Big(\frac{2mE}{\hbar}-q^{2}-\frac{2mc}{\hbar{r}}\Big)\rho=0$$ $$r^{2}\rho''+r\rho'+\Big(\Big[\frac{2mE}{\hbar}-q^{2}\Big]r^{2}-\frac{2mc}{\hbar}r-n^{2}\Big)\rho=0$$ Let $\rho(r)=r^{n}\chi(r)$, then $$r\chi''+(2n+1)\chi'+\Big(\Big[\frac{2mE}{\hbar}-q^{2}\Big]r-\frac{2mc}{\hbar}\Big)\chi=0$$ Now, by letting $$\chi=\exp\Big(\Big[\frac{2mE}{\hbar}-q^{2}\Big]r\Big)\eta(r)$$ you get the confluent hypergeometric equation $$r\eta''+\Big(2\Big[\frac{2mE}{\hbar}-q^{2}\Big]r+2n+1\Big)\eta'+\Big((2n+1)\Big[\frac{2mE}{\hbar}-q^{2}\Big]-\frac{2mc}{\hbar}\Big)\eta=0$$ The solution is $$\chi=\exp\Big(-2\Big[\frac{2mE}{\hbar}-q^{2}\Big]r\Big)\mathcal{L}^{2n}_{k}\Big(2\Big[\frac{2mE}{\hbar}-q^{2}\Big]r\Big)$$ $$k+\frac{2n+1}{2}=-\frac{2mc}{\hbar\Big[\frac{2mE}{\hbar}-q^{2}\Big]}$$ $$E_{q, n, k}=-\frac{2c}{2k+2n+1}+\frac{\hbar{q}^{2}}{2m}, \ k\in\mathbb{N}$$ Here $\mathcal{L}^{b}_{a}(z)$ is the assosiated Lagurre polynomial. This solution was chosen with this particular energy, for the wave-function not to be singular and for it to form a basis for the Hilbert space.