Skip to main content

# 444: Hydrogen Atom Calculation Assuming the Electron is a Particle in a Sphere of Radius R

Trial wave function:

$\Phi (r, R) := \frac{1}{ \sqrt{2 \pi R}} \frac{ \sin ( \frac{ \pi r}{R})}{r}$

Integral:

$\int_{0}^{ \infty} \blacksquare 4 \pi r^2 dr$

Kinetic energy operator:

$T = \frac{1}{2r} \frac{d^2}{dr^2} (r \blacksquare )$

Potential energy operator:

$V = \frac{1}{r}$

Demonstrate the wave function is normalized.

Set up the variational energy integral.

$E(R) := \int_{0}^{R} \Phi (r, R) [ \frac{-1}{2r} \frac{d^2}{dr^2} (r \Phi (r, R))] 4 \pi r^2 dr + \int_{0}^{R} \Phi (r, R) \frac{-1}{r} \Phi (r, R) 4 \pi r^2 dr]$

Minimize the energy with respect to the variational parameter R.

R := 1 R := Minimize(E, R) R = 4.049 E(R) = -0.301

The exact ground state energy for the hydrogen atom is -.5 Eh. Calculate the percent error.

$\frac{-.5 - E(R)}{-.5} = 39.793$

Compare optimized trial wave function with the exact solution by plotting the radial distribution functions.

$S(r) := \frac{1}{ \sqrt{ \pi}} exp(-r)~~~r := 0,.02..4.2$ • Was this article helpful?