# Quantum Chemistry

Chapter07 - 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

Exercise. This is the solution to exercise 7.1 in the book.

Solution. The general equation is:

$-\frac{\hslash }{2m}{\nabla }^{2}\Psi +U\left(x\right)\Psi \left(x\right)=E\Psi \left(x\right)$

(a) In the case of the harmonic oscillator

${Ĥ}^{\left(0\right)}=-\frac{{\hslash }^{2}}{2m}\frac{{d}^{2}}{d{x}^{2}}+\frac{k}{2}{x}^{2}$

since $U\left(x\right)=\frac{k}{2}{x}^{2}$. For an oscillator governed by:

$Ĥ=-\frac{{\hslash }^{2}}{2m}\frac{{d}^{2}}{d{x}^{2}}+\frac{k}{2}{x}^{2}+\frac{\gamma }{6}{x}^{3}+\frac{b}{24}{x}^{4}$

$Ĥ={Ĥ}^{\left(0\right)}+{Ĥ}^{\left(1\right)}$

Therefore, in this case

${Ĥ}^{\left(0\right)}=-\frac{{\hslash }^{2}}{2m}{\nabla }^{2}+\frac{k}{2}{x}^{2}$

${Ĥ}^{\left(1\right)}=\frac{\gamma }{6}+\frac{b}{24}{x}^{4}$

${\Psi }^{\left(0\right)}$ and ${E}^{\left(0\right)}$ are the wavefunction/energy pair for harmonic oscillator.

${E}^{\left(0\right)}=\hslash \sqrt{\frac{k}{\mu }}\left(n+\frac{1}{2}\right)$

${\Psi }^{\left(0\right)}=\frac{1}{\sqrt{{2}^{n}n!}}{\left(\frac{\alpha }{\pi }\right)}^{1∕4}{H}_{n}\left(\sqrt{\alpha }x\right){e}^{-\alpha {x}^{2}∕2}$

with

$\alpha =\sqrt{\frac{k\mu }{{\hslash }^{2}}}$

(b) In this case the potential is that of a particle in a box. Hence,

${Ĥ}^{\left(1\right)}=\left\{\begin{array}{cc}0\phantom{\rule{1em}{0ex}}\hfill & \phantom{\rule{1em}{0ex}}if\phantom{\rule{1em}{0ex}}0\le x\le \frac{a}{2}\hfill \\ b\phantom{\rule{1em}{0ex}}\hfill & \phantom{\rule{1em}{0ex}}if\phantom{\rule{1em}{0ex}}\frac{a}{2}\le x\le a\hfill \end{array}\right\$

${Ĥ}^{\left(0\right)}$, ${\Psi }^{\left(0\right)}$ and ${E}^{\left(0\right)}$ are the Hamiltonian, wavefunctions, and energies of the particle in a box.

$\begin{array}{rcll}{Ĥ}^{\left(0\right)}& =& -\frac{{\hslash }^{2}}{2m}\frac{{d}^{2}}{d{x}^{2}}& \text{}\\ {\Psi }^{\left(0\right)}& =& \sqrt{\frac{2}{a}}sin\frac{n\pi x}{a}& \text{}\\ {E}_{n}^{\left(0\right)}& =& \frac{{n}^{2}{h}^{2}}{8m{a}^{2}}& \text{}\end{array}$

(c) For a He atom:

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

But

$\begin{array}{rcll}-\frac{{\hslash }^{2}}{2{m}_{e}}{\nabla }_{1}^{2}-\frac{2{e}^{2}}{4\pi {ϵ}_{0}}\frac{1}{{r}_{1}}& & & \text{}\\ -\frac{{\hslash }^{2}}{2{m}_{e}}{\nabla }_{2}^{2}-\frac{2{e}^{2}}{4\pi {ϵ}_{0}}\frac{1}{{r}_{2}}& & & \text{}\end{array}$

are Hamiltonians for hydrogen-like systems. Therefore, we can take:

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

and

${Ĥ}^{\left(1\right)}=\frac{{e}^{2}}{4\pi {ϵ}_{0}}\frac{1}{{r}_{12}}$

Then, the wavefunction ${\Psi }^{\left(0\right)}$ is a product of two hydrogen-like wavefunctions (since ${Ĥ}^{\left(0\right)}$ is decomposable as sum).

${\Psi }^{\left(0\right)}\left({r}_{1},{r}_{2}\right)={\Psi }_{1}\left({r}_{1}\right){\Psi }_{2}\left({r}_{2}\right)$

and energy is

${E}^{\left(0\right)}={E}_{1}^{\left(0\right)}+{E}_{2}^{\left(0\right)}$

the sum of energies of hydrogen-like atoms.

(d) In this case

$\begin{array}{rcll}{Ĥ}^{\left(0\right)}& =& -\frac{{\hslash }^{2}}{2\mu }{\nabla }^{2}-\frac{{e}^{2}}{e\pi {ϵ}_{0}}\frac{1}{r}& \text{}\\ {Ĥ}^{\left(1\right)}& =& e\zeta rcos\theta & \text{}\end{array}$

and ${\Psi }^{\left(0\right)}$ and ${E}^{\left(0\right)}$ are the wavefunction/energy of the hydrogen atom.

(e) In this case:

$\begin{array}{rcll}{Ĥ}^{\left(0\right)}& =& -\frac{{\hslash }^{2}}{2I}{\nabla }^{2}& \text{}\\ {Ĥ}^{\left(1\right)}& =& \mu \zeta cos\theta & \text{}\end{array}$

and ${\Psi }^{\left(0\right)}$ and ${E}^{\left(0\right)}$ are the wavefunction/energy of the rigid rotator.