Problem Set 4 Solution

# Problem Set 4 Solution

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , you can look the integral up in a table. From integral tables, € ∞ ∫ 2 x e− bx dx = 0 1 . 2b Note however that this integral is defined only from 0 to ∞ and not from −∞ to ∞ . If the integrand f ( x ) is an odd function, we know that € 0 ∫ −∞ € € f ( x ) dx = − ∞ ∫ 0 € € f ( x ) dx . € 6 3 . continued So, the integral becomes ∫ ȹ 2α 2 ȹ1 / 2 * ψ 0 ( x ) ψ1( x ) dx = ȹ ȹ π ȹ ȹ −∞ ȹ Ⱥ ∞ ∞ ∫−∞ x e−α x ȹ 2α 2 ȹ1 / 2 Ⱥ = ȹ ȹ π ȹ Ⱥ ȹ Ⱥ ȹ Ⱥ 0 ∞ dx ∫−∞ x e−α x ȹ 2α 2 ȹ1 / 2 Ⱥ 1 = ȹ ȹ π ȹ Ⱥ− 2α + ȹ Ⱥ ȹ Ⱥ * ∫−∞ ψ0 ( x) ψ1( x) dx 2 2 dx + ∞ ∫ 0 x e−α x 2 Ⱥ dx Ⱥ Ⱥ 1 Ⱥ Ⱥ 2α Ⱥ = 0. € 4 . Show b y e xplicit i ntegration t hat t he f irst e xcited s tate h armonic o scillator w avefunction, ψ1 ( x ) , i s n ormalized. For normalization, we must show that € ∞ ∫−∞ ψ1*( x) ψ1( x) dx = 1 . Substituting, ∫−∞ * ψ1 € ȹ 4α 3 ȹ1 / 2 ( x) ψ1( x...
View Full Document

## This document was uploaded on 12/05/2013.

Ask a homework question - tutors are online