# 439: Variational Calculation on the Two-dimensional Hydrogen Atom

- Page ID
- 136972

Normalized trial wave function:

\[ \psi ( \alpha , r) = \sqrt{ \frac{2}{ \pi}} \alpha e^{- \alpha r}\]

\[ \int_{0}^{ \infty} \psi ( \alpha , r)^2 2 \pi r dr~~assume,~ \alpha > 0 \rightarrow 1\]

Calculate electron kinetic energy:

\[ T ( \alpha ) = \int_{0}^{ \infty} \psi ( \alpha , r) \frac{-1}{2r} \frac{d}{dr} (r \frac{d}{dr} \psi ( \alpha , r)) 2 \pi r dr~~assume,~ \alpha >0 \rightarrow (-2) \alpha Z\]

Calculate electron-nucleus potential energy:

\[ V_{NE} ( \alpha , Z) = \int_{0}^{ \infty} \psi ( \alpha , r) \frac{-Z}{r} \psi ( \alpha , r) 2 \pi r dr~assume,~ \alpha > 0 \rightarrow (-2) \alpha Z\]

Calculate total electronic energy for the 2D H atom:

\( \alpha\) = 1 \( \alpha\) = Minimize (E, \( \alpha\)) \( \alpha\) = 2 E( \( \alpha\)) = -2

Demonstrate that the virial theorem is satisfied:

\[ \frac{T ( \alpha )}{E( \alpha )} = -1~~~ \frac{T( \alpha )}{V_{NE} ( \alpha , 1)} = -0.5~~~ \frac{V_{NE} ( \alpha , 1)}{E( \alpha )} = 2\]