Chapter 6 Problem Solutions

# Chapter 6 Problem Solutions

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Unformatted text preview: is odd p2 —2 = a 2— p p H— ê aL2 H— ê aL2 Therefore D px = Printed by Mathematica for Students 2a ` The momentum uncertainty does not depend on time because px commutes with the Hamiltonian, so momentum is a constant of ` ¶ X px \ = Ÿ -¶ p y H p, tL 4 `2 ¶ X px \ = Ÿ -¶ p2 Ch6ProblemSetKey.nb 2 2 yH p, tL = = ¶ a p — p 2 Ÿ-¶ „ p p ‰ ¶ a — - Ÿ-¶ „ p p ‰ ap —2 - =0 ¬༼ integrand is odd a2 p2 —2 = a 2— p H— ê aL2 H— ê aL2 p Therefore D px = — 2a ` The momentum uncertainty does not depend on time because px commutes with the Hamiltonian, so momentum is a constant of the motion. We see that a DxD p = B 1+J 2 ht m a2 2 N FB — 2a F= — 2 1+J ht m a2 2 N ¥ — 2 for all times t so the uncertainty principle is always satisfied by this state. We also see that the minimum uncertainty state occurs at time t = 0, when the state is a Gaussian wavepacket. Problem 6.15 Given a particle of mass m in the one-dimensional potential energy well 0 0<x<L ¶ elsewhere V HxL = in the state I yHxL = 1+Â M 2 2 L sinI px M L + 1 2 2 L sinI 0 2px M L 0< x < L elsewhere at time t = 0. ü Part a The energy eigenstates yn \ of the system are those that satisfy the time-independent Schrödinger equat...
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## This note was uploaded on 02/01/2014 for the course PHYC 491/496 taught by Professor Akimasamiyake during the Fall '13 term at New Mexico.

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