Problem Set 4 Solution

# In order to show that the function is an

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Unformatted text preview: ) dx = ȹ ȹ ȹ π ȹ ȹ Ⱥ ∫−∞ ȹ 4α 3 ȹ1 / 2 = ȹ ȹ π ȹ ȹ ȹ Ⱥ ∞ ∫−∞ ∞ x e−α x ∞ x 2 e−α x dx . 2 /2 x e−α x 2 /2 dx 2 From integral tables, € ∞ ∫ 0 2 − bx 2 xe 1 1 ȹ π ȹ 2 dx = ȹ ȹ . 4b ȹ b Ⱥ Note however that this integral is defined only from 0 to ∞ and not from −∞ to ∞ . If the integrand f ( x ) is an even function like it is in this c€ we know that ase, ∞ ∫ −∞ € € f ( x ) dx = 2 ∞ ∫ 0 € € f ( x ) dx . € 7 4 . continued So, the integral becomes ∫ ȹ 4α 3 ȹ1 / 2 * ψ1 ( x ) ψ1( x ) dx = ȹ ȹ π ȹ ȹ −∞ ȹ Ⱥ ∞ ∞ ∫−∞ x 2 e−α x ȹ 4α 3 ȹ1 / 2 Ⱥ = ȹ ȹ π ȹ Ⱥ 2 ȹ Ⱥ ȹ Ⱥ 2 dx ∞ ∫ 0 x 2 e−α x 2 Ⱥ dx Ⱥ Ⱥ ȹ 4α 3 ȹ1 / 2 Ⱥ 1 ȹ π ȹ1 / 2 Ⱥ Ⱥ Ⱥ = ȹ ȹ π ȹ 2 Ⱥ 4α ȹ α ȹ Ⱥ ȹ ȹ Ⱥ Ⱥ ȹ Ⱥ Ⱥ ȹ 4α 3π ȹ1 / 2 1 = ȹ ȹ π α ȹ ȹ 2α ȹ Ⱥ () = 4α 2 ∞ ∫−∞ ψ1*( x) ψ1( x) dx 1/ 2 1 2α = 1. Therefore, we see that the first excited state wavefunction is normalized. € 5 . Demonstrate that the ground state wavefunction ψ...
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## This document was uploaded on 12/05/2013.

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