# Hydrogen atom – a sample wave function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.3.

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The wave function for the hydrogen atom can be obtained by a series solution of the differential equation, leading to the result (which I’ve rewritten in Shankar’s notation, although my original post used Griffiths’s notation):

$\displaystyle \psi_{nlm}\left(r,\theta,\phi\right)=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

Here, we have

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (2)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (3)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r \ \ \ \ \ (4)$

The energy levels of the hydrogen atom are

$\displaystyle E=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (5)$

where ${n=1,2,3,\ldots}$. The coefficients ${C_{k}}$ in 3 are given by a recursion relation

 $\displaystyle C_{k+1}$ $\displaystyle =$ $\displaystyle \frac{-e^{2}\lambda+2\left(k+l+1\right)}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}\ \ \ \ \ (6)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \sqrt{-\frac{2m}{\hbar^{2}E}} \ \ \ \ \ (7)$

Combining ${\lambda}$ and ${E}$, the formula becomes, for a given ${n}$

$\displaystyle C_{k+1}=\frac{2\left(k+l+1\right)-2n}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}$

The coefficient ${C_{0}}$ which starts everything off is determined by normalization.

As an example, we can find the wave function ${\psi_{210}}$. In this case ${n=2}$ and ${l=1}$ so the first term in the recursion, with ${k=0}$ gives ${k+l+1=2}$ and ${C_{1}=0}$. The full wave function is then

 $\displaystyle \psi_{210}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\rho^{2}e^{-\rho}C_{0}Y_{1}^{0} \ \ \ \ \ (8)$

To evaluate ${\rho}$ we use the energy for ${n=2}$:

$\displaystyle E_{2}=-\frac{me^{4}}{8\hbar^{2}} \ \ \ \ \ (9)$

This gives

$\displaystyle \rho=\sqrt{\frac{2m^{2}e^{4}}{8\hbar^{4}}}r=\frac{me^{2}}{2\hbar^{2}}r=\frac{r}{2a_{0}} \ \ \ \ \ (10)$

where ${a_{0}}$ is the Bohr radius

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (11)$

Plugging everything into 8, using ${Y_{1}^{0}=\sqrt{\frac{3}{4\pi}}\cos\theta}$, we have

$\displaystyle \psi_{210}=\sqrt{\frac{3}{4\pi}}\frac{C_{0}}{4a_{0}^{2}}re^{-r/2a_{0}}\cos\theta \ \ \ \ \ (12)$

Normalizing gives the condition

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{210}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi=1 \ \ \ \ \ (13)$

Working out the integral (using software or tables) gives

 $\displaystyle \frac{3}{2}a_{0}C_{0}^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{3a_{0}}} \ \ \ \ \ (15)$

So the final wave function is

$\displaystyle \psi_{210}=\frac{1}{\sqrt{32\pi a_{0}^{3}}}\frac{r}{a_{0}}e^{-r/2a_{0}}\cos\theta \ \ \ \ \ (16)$

which agrees with Shankar’s equation 13.1.27.