Homework 02

now we can find x as x 3a 2 1 x x 4 sin 4 dx 0

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Unformatted text preview: 2π Ⱥ−π / a ȹ π Ⱥ π Ⱥ Ⱥ 0 ȹ Ⱥ 3a Ⱥ 2 ȹ πx ȹ π / a ka = Ⱥ sinȹ ȹ + 2 ∫ cos( kx ) dk Ⱥ 2π Ⱥ x ȹ a Ⱥ Ⱥ 0π = ȹ πx ȹȺ 3a 2 Ⱥ Ⱥ1 − cosȹ ȹȺ ȹ a ȺȺ π 2 x 2 Ⱥ € <k> = 0, since the momentum space wave function is symmetric about k = 0. . Now, we can find Δx as ∞ ȹ πx ȹȺ 3a 2 Ⱥ 1 x =∫ x Ⱥ 4 sin 4 ȹ ȹȺdx = 0 4 ȹ 2 a ȺȺ Ⱥ x −∞ 4 π since the integral has an odd symmetry integrand, and ∞ ȹ ȹȺ ȹ πx ȹ 3a 2 2 Ⱥ 1 3a 2 ∞ 1 2 4 πx x =∫ x Ⱥ 4 sin ȹ ȹȺdx = ∫ x 2 sin4 ȹ 2a ȹdx 4 ȹ 2 a ȺȺ ȹ Ⱥ 4 π 4 −∞ Ⱥ x −∞ 4 π € 3a 2 ȹ π ȹ π 3a...
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This note was uploaded on 10/03/2010 for the course EEE 434 taught by Professor Roedel during the Fall '08 term at ASU.

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