# Quantum Chemistry

Chapter04 - 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

Exercise. This is the solution to exercise 4.1 in the book.

(a)

${\int }_{-\infty }^{\infty }{e}^{-{x}^{2}}dx=\sqrt{\pi }$

This is the Gaussian integral. Hence, the normalized function is $\frac{1}{\sqrt{\pi }}{e}^{-{x}^{2}}$. If we treat this as a normlizable function then we must have:

$\psi \left(x\right)=C{e}^{-{x}^{2}}$

The normalization condition requires:

${\int }_{-\infty }^{\infty }{\psi }^{\ast }\left(x\right)\psi \left(x\right)dx=1$

Upon variable change and integration we find $C={\left(\frac{2}{\pi }\right)}^{1∕4}$, thus the function is:

$\psi \left(x\right)={\left(\frac{2}{\pi }\right)}^{1∕4}{e}^{-{x}^{2}}$

(b) ${e}^{x}$ is unbounded in $\left(0,\infty \right)$, hance it cannot be normalized.

(c) A normalizable function based on ${e}^{i\theta }$ over the domain $\left(0,2\pi \right)$ is

$\psi \left(\theta \right)=C{e}^{i\theta }\phantom{\rule{1em}{0ex}}{\psi }^{\ast }\left(\theta \right)={C}^{\ast }{e}^{-i\theta }$

Integrating

${\int }_{0}^{2\pi }{\psi }^{\ast }\left(\theta \right)\psi \left(\theta \right)=2\pi |C{|}^{2}=1$

Hence, the normalized function is:

$\psi \left(\theta \right)=\frac{1}{\sqrt{2\pi }}{e}^{i\theta }$

(d) A normalized function based on $sinhx$ over the domain $\left(0,\infty \right)$ cannot be found because $sinhx$ is unbounded at $+\infty$.

(e) A normalized function based on $x{e}^{-x}$ is

$\psi \left(x\right)=Cx{e}^{-x}$

Normalization requires that

${\int }_{0}^{\infty }{C}^{2}{x}^{2}{e}^{-2x}=1$

or

${\int }_{0}^{\infty }{x}^{2}{e}^{-2x}=\frac{1}{{C}^{2}}$

Evaluating the integral we have

${\int }_{0}^{\infty }{x}^{2}{e}^{-2x}=\frac{1}{4}=\frac{1}{{C}^{2}}$

Hence

$\psi \left(x\right)=2x{e}^{-x}$

is normalized.