# Quantum Chemistry

Chapter09 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44 - 45 - 46 - 47 - 48 - 49

Exercise. This is the solution to exercise 9.1 in the book.

Solution. For the hydrogen molecule the Hamiltonian in SI units is:

$Ĥ=-\frac{{\hslash }^{2}}{2m}\left({\nabla }_{1}^{2}+{\nabla }_{2}^{2}\right)-\frac{{e}^{2}}{4\pi {ϵ}_{0}{r}_{1A}}-\frac{{e}^{2}}{4\pi {ϵ}_{0}{r}_{1B}}-\frac{{e}^{2}}{4\pi {ϵ}_{0}{r}_{2A}}-\frac{{e}^{2}}{4\pi {ϵ}_{0}{r}_{2B}}+\frac{{e}^{2}}{4\pi {ϵ}_{0}{r}_{12}}+\frac{{e}^{2}}{4\pi {ϵ}_{0}R}$

or

$Ĥ=-\frac{{\hslash }^{2}}{2m{a}_{0}^{2}}\left({\stackrel{̃}{\nabla }}_{1}^{2}+{\stackrel{̃}{\nabla }}_{2}^{2}\right)-\frac{{e}^{2}}{{\kappa }_{0}{a}_{0}{\stackrel{̃}{r}}_{1A}}-\frac{{e}^{2}}{{\kappa }_{0}{a}_{0}{\stackrel{̃}{r}}_{1B}}-\frac{{e}^{2}}{{\kappa }_{0}{a}_{0}{\stackrel{̃}{r}}_{2A}}-\frac{{e}^{2}}{{\kappa }_{0}{a}_{0}{\stackrel{̃}{r}}_{2B}}+\frac{{e}^{2}}{{\kappa }_{0}{a}_{0}{\stackrel{̃}{r}}_{12}}+\frac{{e}^{2}}{{\kappa }_{0}{a}_{0}\stackrel{̃}{R}}$

or

$\frac{{\kappa }_{0}{a}_{0}}{{e}^{2}}Ĥ=-\frac{1}{2}\left(\frac{{\hslash }^{2}}{m{a}_{0}^{2}}\frac{{\kappa }_{0}{a}_{0}}{{e}^{2}}\right)\left({\stackrel{̃}{\nabla }}_{1}^{2}+{\stackrel{̃}{\nabla }}_{2}^{2}\right)-\frac{1}{{\stackrel{̃}{r}}_{1A}}-\frac{1}{{\stackrel{̃}{r}}_{1B}}-\frac{1}{{\stackrel{̃}{r}}_{2A}}-\frac{1}{{\stackrel{̃}{r}}_{2B}}+\frac{1}{{\stackrel{̃}{r}}_{12}}+\frac{1}{\stackrel{̃}{R}}$

or

$\stackrel{̂}{\stackrel{̃}{H}}=-\frac{1}{2}\left(\frac{{\hslash }^{2}}{m{a}_{0}}\frac{{\kappa }_{0}}{{e}^{2}}\right)\left({\stackrel{̃}{\nabla }}_{1}^{2}+{\stackrel{̃}{\nabla }}_{2}^{2}\right)-\frac{1}{{\stackrel{̃}{r}}_{1A}}-\frac{1}{{\stackrel{̃}{r}}_{1B}}-\frac{1}{{\stackrel{̃}{r}}_{2A}}-\frac{1}{{\stackrel{̃}{r}}_{2B}}+\frac{1}{{\stackrel{̃}{r}}_{12}}+\frac{1}{\stackrel{̃}{R}}$

but

${a}_{0}=\frac{{\kappa }_{0}{\hslash }^{2}}{m{e}^{2}}$

Hence,

$\stackrel{̂}{\stackrel{̃}{H}}=-\frac{1}{2}\left({\stackrel{̃}{\nabla }}_{1}^{2}+{\stackrel{̃}{\nabla }}_{2}^{2}\right)-\frac{1}{{\stackrel{̃}{r}}_{1A}}-\frac{1}{{\stackrel{̃}{r}}_{1B}}-\frac{1}{{\stackrel{̃}{r}}_{2A}}-\frac{1}{{\stackrel{̃}{r}}_{2B}}+\frac{1}{{\stackrel{̃}{r}}_{12}}+\frac{1}{\stackrel{̃}{R}}$