Chapter 6 Problem Solutions

Chapter 6 Problem Solutions - Intermediate QMI Ch 6 Problem...

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Intermediate QMI Ch 6 Problem Solutions Problem 6.2 We have Y p x ` y ] = Ÿ x x £ X p x \ Y x x ` x £ ] X x £ y \ ¬ Y x x ` x £ ] = x £ X x x £ \ = x £ d H x - x £ L X p x \ = 1 2 p p x X x £ y \ = y H x £ L = Ÿ x x £ 1 2 p p x x £ d H x - x £ L y H x £ L = 1 2 p Ÿ x x p x y H x L =  p B 1 2 p Ÿ x p x y H x LF =  p A Ÿ x X p x \ X x y \E therefore Y p x ` y ] =  p X p y \ and Y j x ` y ] = Ÿ p X j p \ Y p x ` y ] Y j x ` y ] = Ÿ p X p j \ *  p X p y \ This suggests we should represent the position operator in momentum space as x ` Ø momentum space  p Note: x ` and p ` x are “symmetric” up to a sign in the commutation relation A x ` , p ` x E = Â Ø p ` x Ì x in the basis of the first observable f A p ` x , x ` E = H L Ø x ` Ì - H L p =  p in the p ` eigenbasis Problem 6.4 ü Part a Given a free particle of mass m initially in the state y H x L = X x y \ = 1 p a - x 2 ë 2 a 2 Printed by Mathematica for Students
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The free particle Hamiltonian is given by H ` = p ` x 2 2 m so its eigenstates are just the momentum eigenstates, with energy eigenvalues E p = p 2 2 m . We thus expand our wavefunction in the momentum basis as X p y \ = Ÿ x X p x \ X x y \ ¬ X p x \ = 1 2 p p x X x y \ = y H x L = 1 p a - x 2 ë 2 a 2 X p y \ = y è H p L y è H p L = 1 2 p p a Ÿ x p x - x 2 ë 2 a 2 This is the Fourier transform of a Gaussian, and thus we have y è H p L = a p - a 2 p 2 2 2 Thus, the time evolution of the particle state in the position basis is given by X x -  H ` t y \ = Ÿ p X x -  p ` x 2 2 m t p \ X p y \ y H x , t L = Ÿ p -  p 2 2 m t X x p \ y è H p L = 1 2 p ÿ a p Ÿ p -  p 2 2 m t  p x - a 2 p 2 2 2 y H x , t L = 1 a 2 p p Ÿ p  p x - 1 2 K a 2 2 +  t m O p 2 This is the inverse Fourier transform of a Gaussian. Therefore y H x , t L = 1 p B a + J  t m a NF - x 2 2 a 2 1 +  t m a 2 and y H x , t L = a - J  t m a N p B a 2 + J t m a N 2 F - x 2 1 -  t m a 2 2 a 2 B 1 + t m a 2 2 F y H x , t L 2 = a 2 + J t m a N 2 p B a 2 + J t m a N 2 F 2 - 2 x 2 2 a 2 B 1 + t m a 2 2 F 2 Ch6ProblemSetKey.nb Printed by Mathematica for Students
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y H x , t L 2 = 1 2 p a 2 2 B 1 + J t m a 2 N 2 F - x 2 2 a 2 2 B 1 + t m a 2 2 F The uncertainty in x is given by H D x L 2 = ZI x ` - X x ` \M 2 ^ = X x ` 2 \ - X x ` \ 2 so X x ` \ = 1 2 p a 2 2 B 1 + J t m a 2 N 2 F Ÿ x x - x 2 2 a 2 2 B 1 + t m a 2 2 F = 0 ¬ integrand is odd X x ` 2 \ = 1 2 p a 2 2 B 1 + J t m a
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Chapter 6 Problem Solutions - Intermediate QMI Ch 6 Problem...

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