# Hydrogen atom – wave function examples

Required math: calculus

Required physics: 3-d Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.11.

A few more examples of working out the hydrogen atom wave functions. Using the formulas in the last example, we can get ${R_{20}}$. The recursion formula for ${n=2}$, ${l=0}$ is

$\displaystyle c_{j+1}=\frac{2(j+1)-4}{(j+1)(j+2)}c_{j} \ \ \ \ \ (1)$

The series has 2 terms, and we get ${c_{1}=-c_{0}}$, so

 $\displaystyle R_{20}\left(r\right)$ $\displaystyle =$ $\displaystyle \frac{1}{r}u_{20}\left(r\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r}\rho e^{-\rho}v_{20}(\rho)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2a}e^{-r/2a}c_{0}\left(1-\frac{r}{2a}\right) \ \ \ \ \ (4)$

To find ${c_{0}}$ we normalize the radial function:

 $\displaystyle \int_{0}^{\infty}r^{2}|R_{20}(r)|^{2}dr$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}c_{0}^{2}r^{2}\left[\frac{1}{2a}\left(1-\frac{1}{2a}\right)\right]^{2}e^{-r/a}dr\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{2}c_{0}^{2}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (7)$

So ${c_{0}=\sqrt{2/a}}$ and ${R_{20}(r)}$ is

$\displaystyle R_{20}\left(r\right)=\frac{1}{\sqrt{2}a^{3/2}}e^{-r/2a}\left(1-\frac{r}{2a}\right) \ \ \ \ \ (8)$

The complete wave function is then

 $\displaystyle \psi_{200}$ $\displaystyle =$ $\displaystyle R_{20}(r)Y_{00}(\phi,\theta)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{8\pi}}a^{-3/2}\left(1-\frac{r}{2a}\right)e^{-r/2a} \ \ \ \ \ (10)$

For ${R_{21}\left(r\right)}$, we have

$\displaystyle c_{j+1}=\frac{2(j+2)-4}{(j+1)(j+4)}c_{j} \ \ \ \ \ (11)$

This time, there is only a single term in the series, so we have

 $\displaystyle R_{21}\left(r\right)$ $\displaystyle =$ $\displaystyle \frac{1}{r}u_{21}\left(r\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r}\rho^{2}e^{-\rho}v_{21}(\rho)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{r}{\left(2a\right)^{2}}e^{-r/2a}c_{0} \ \ \ \ \ (14)$

Doing the normalization integral for ${R_{21}(r)}$ gives ${c_{0}=\sqrt{2/3a}}$ which gives the final result

$\displaystyle R_{21}\left(r\right)=\frac{r}{2\sqrt{6}a^{5/2}}e^{-r/2a} \ \ \ \ \ (15)$

There are 3 wave functions corresponding to ${n=2,\; l=1}$, for which we need the spherical harmonics

 $\displaystyle Y_{1}^{1}$ $\displaystyle =$ $\displaystyle -\left(\frac{3}{8\pi}\right)^{1/2}\sin\theta e^{i\phi}\ \ \ \ \ (16)$ $\displaystyle Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \left(\frac{3}{8\pi}\right)^{1/2}\sin\theta e^{-i\phi}\ \ \ \ \ (17)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \left(\frac{3}{4\pi}\right)^{1/2}\cos\theta \ \ \ \ \ (18)$

The three wave functions are thus

 $\displaystyle \psi_{211}$ $\displaystyle =$ $\displaystyle -\frac{1}{8\sqrt{\pi}}\frac{r}{a^{5/2}}e^{-r/2a}\sin\theta e^{i\phi}\ \ \ \ \ (19)$ $\displaystyle \psi_{21-1}$ $\displaystyle =$ $\displaystyle \frac{1}{8\sqrt{\pi}}\frac{r}{a^{5/2}}e^{-r/2a}\sin\theta e^{-i\phi}\ \ \ \ \ (20)$ $\displaystyle \psi_{210}$ $\displaystyle =$ $\displaystyle \frac{1}{4\sqrt{2\pi}}\frac{r}{a^{5/2}}e^{-r/2a}\cos\theta \ \ \ \ \ (21)$