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The Hamiltonian operator y for the harmonic oscillator is given by Ĥ = - a h? d? + suw? â?, where l is the reduced mass and w = 27V is the angular 2µ dæ? pwx frequency. Is the function f(x) = x exp an eigenfunction of that Hamiltonian? 2h O No. Yes, and the eigenvalue is ħw. 2 O Yes, and the eigenvalue is hw. 3hw Yes, and the eigenvalue is 4

The Hamiltonian operator y for the harmonic oscillator is given by Ĥ = - a h? d? + suw? â?, where l is the reduced mass and w = 27V is the angular 2µ dæ? pwx frequency. Is the function f(x) = x exp an eigenfunction of that Hamiltonian? 2h O No. Yes, and the eigenvalue is ħw. 2 O Yes, and the eigenvalue is hw. 3hw Yes, and the eigenvalue is 4

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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The Hamiltonian operator y for the harmonic oscillator is given by  = h? d? + suw? â?, where l is the reduced mass and w = 27V is the angular 2µ dæ? P[-] pwx? frequency. Is the function f(x) = x exp an eigenfunction of that Hamiltonian? 2h O No. Yes, and the eigenvalue is ħw. 2 Yes, and the eigenvalue is hw. 3ħw Yes, and the eigenvalue is 4
Transcribed Image Text:The Hamiltonian operator y for the harmonic oscillator is given by  = h? d? + suw? â?, where l is the reduced mass and w = 27V is the angular 2µ dæ? P[-] pwx? frequency. Is the function f(x) = x exp an eigenfunction of that Hamiltonian? 2h O No. Yes, and the eigenvalue is ħw. 2 Yes, and the eigenvalue is hw. 3ħw Yes, and the eigenvalue is 4
Expert Solution
Step 1

Given:

Hamiltonian operator H^=-ℏ22μd2 dx2+12μω2x^2

μ is the reduced mass

ω=2πν is the angular frequency

To find: if f(x)=xexp-μωx22his an eigenfunction of that Hamiltonian

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