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# hwsol6 - Physics 31 Spring 2008 Solution to HW#6 Problem A...

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Unformatted text preview: Physics 31 Spring, 2008 Solution to HW #6 Problem A The wave function of a particle is given by Ο ( x ) = N exp " β Β΅ x 3 . 1915 Λ A ΒΆ 2 # (When you evaluate this function, take all distances x in Λ A.) (e) Use the exact formula to evaluate β x for this wave func- tion: β x = p x 2 β x 2 . Compare your answer to your estimate in part (d). Hint: You need the integral R β x 2 exp( β ax 2 ) dx = 1 4 p Ο/a 3 , but you do not need the integral R β x exp( β ax 2 ) dx = 1 / (2 a ). Why not? We can evaluate β x rigorously using the expression β x = p x 2 β x 2 . Note that the average value x = 0. This result follows from the integral expression x = Z β ββ x Β― Β― Ο ( x ) Β― Β― 2 dx = N 2 Z β ββ x exp Β· β 2 x b Β΄ 2 ΒΈ dx which is zero because the integrand is an odd function [that is, f ( x ) = β f ( β x ), so the contributions to the integral from x < 0 and x > 0 exactly cancel]....
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