{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 6 Problem Solutions

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

This preview shows pages 1–4. Sign up to view the full content.

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 = Â ¶∂ ¶∂ p B 1 2 p Ÿ x p x y H x LF = Â ¶∂ ¶∂ p A Ÿ x X p x \ X x y \E therefore Y p x ` y ] = Â ¶∂ ¶∂ 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 \ * Â ¶∂ ¶∂ p X p y \ This suggests we should represent the position operator in momentum space as x ` Ø momentum space Â ¶∂ ¶∂ 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 fl A p ` x , x ` E = H L Ø x ` Ì - H L p = Â 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 - Â H ` t y \ = Ÿ p X x - Â p ` x 2 2 m t p \ X p y \ y H x , t L = Ÿ p - Â p 2 2 m t X x p \ y è H p L = 1 2 p ÿ a p Ÿ p - Â p 2 2 m t Â p x - a 2 p 2 2 2 y H x , t L = 1 a 2 p p Ÿ p Â p x - 1 2 K a 2 2 + Â 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 Â t m a NF - x 2 2 a 2 1 + Â t m a 2 and y H x , t L = a - J Â t m a N p B a 2 + J t m a N 2 F - x 2 1 - Â 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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern