# Hydrogen atom – complete wave function

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.43.

If we combine all the results from the solution of the angular and radial equations for the hydrogen atom, we get a formula for the spatial wave function, which is given in Griffiths’s book as eqn 4.89:

$\displaystyle \psi_{nlm}=\sqrt{\left(\frac{2}{na}\right)^{3}\frac{\left(n-l-1\right)!}{2n\left[\left(n+l\right)!\right]^{3}}}e^{-r/na}\left(\frac{2r}{na}\right)^{l}L_{n-l-1}^{2l+1}\left(\frac{2r}{na}\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

where ${L}$ is an associated Laguerre polynomial and ${Y}$ is a spherical harmonic.

The spherical harmonics can be calculated from a standard formula, as can the associated Laguerre polynomials. The forms of these functions vary according to the normalization. My version is

$\displaystyle L_{p}^{q}(x)=c_{0}\sum_{j=0}^{p}\frac{(-1)^{j}(p+q)!}{(p-j)!(q+j)!j!}x^{j} \ \ \ \ \ (2)$

The version in Griffiths sets ${c_{0}=\left(p+q\right)!}$ (so it is that version that is used in the above formula for ${\psi_{nlm}}$) while some other sources use ${c_{0}=1}$.

For the spherical harmonics, the formulas are

$\displaystyle Y_{l}^{m}(\theta,\phi)=\left[\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}\right]^{1/2}e^{im\phi}P_{l}^{m}(\cos\theta) \ \ \ \ \ (3)$

with the ${P_{l}^{m}}$ being the associated Legendre function:

$\displaystyle P_{l}^{m}\left(x\right)=(1-x^{2})^{m/2}\sum_{k=0}^{[(l-m)/2]}\frac{(2l-2k)!}{2^{l}(l-k)!k!(l-2k-m)!}(-1)^{k}x^{l-m-2k} \ \ \ \ \ (4)$

As an example of using this formula, we’ll construct ${\psi_{321}}$. We get ${L_{0}^{5}(x)=120}$ and we worked out ${Y_{2}^{1}}$ in the previous post as

$\displaystyle Y_{2}^{1}(\theta,\phi)=-\sqrt{\frac{15}{8\pi}}e^{i\phi}\sin\theta\cos\theta \ \ \ \ \ (5)$

Plugging these into the overall formula gives

 $\displaystyle \psi_{321}$ $\displaystyle =-$ $\displaystyle \sqrt{\left(\frac{2}{3a}\right)^{3}\frac{1}{6\times120^{3}}}e^{-r/3a}\left(\frac{2r}{3a}\right)^{2}(120)\left(\frac{15}{8\pi}\right)^{\frac{1}{2}}\sin\theta\cos\theta e^{i\phi}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{81\sqrt{\pi a^{7}}}r^{2}e^{-r/3a}\sin\theta\cos\theta e^{i\phi} \ \ \ \ \ (7)$

To check the normalization, we do the integral:

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}|\psi_{321}|^{2}r^{2}\sin\theta drd\theta d\phi$ $\displaystyle =$ $\displaystyle \frac{1}{81^{2}\pi a^{7}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}r^{6}e^{-2r/3a}\sin^{3}\theta\cos^{2}\theta drd\theta d\phi\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (9)$

using Maple for the integral.

The expectation value of ${r^{s}}$ is (using Maple)

 $\displaystyle \langle r^{s}\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}|\psi_{321}|^{2}r^{2+s}\sin\theta drd\theta d\phi\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{81^{2}\pi a^{7}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}r^{6+s}e^{-2r/3a}\sin^{3}\theta\cos^{2}\theta drd\theta d\phi\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{720}\left(\frac{3a}{2}\right)^{s}\Gamma(7+s) \ \ \ \ \ (12)$

The gamma function is infinite at all non-positive integral arguments, so this value is finite for ${s>-7}$ and all non-integer values less than ${-7}$. The smallest integer value for which this is finite is ${s=-6}$. Note in particular that if ${s=0}$, the formula reduces to ${\langle r^{0}\rangle=1}$ as it should.

The expectation value of ${r}$ itself is

$\displaystyle \left\langle r\right\rangle =\frac{21a}{2} \ \ \ \ \ (13)$

This is quite a bit bigger than the value for the ground state, which is ${\left\langle r\right\rangle =3a/2}$.