Problem Set 4 Solution

Problem Set 4 Solution

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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 ȹ1 / 2 * ψ 0 ( x ) ψ1( x ) dx = ȹ ȹ π ȹ ȹ −∞ ȹ Ⱥ ∞ ∞ ∫−∞ x e−α x ȹ 2α 2 ȹ1 / 2 Ⱥ = ȹ ȹ π ȹ Ⱥ ȹ Ⱥ ȹ Ⱥ 0 ∞ dx ∫−∞ x e−α x ȹ 2α 2 ȹ1 / 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 ȹ1 / 2 ( x) ψ1( x...
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This document was uploaded on 12/05/2013.

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