Chapter 6 Problem Solutions

# We recognize this equation as that of the simple

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: ion ` H yn \ = En yn \ position basis —2 Ø B- 2 m „2 „ x2 + V HxLF yn HxL = En yn HxL For our potential, these are states that satisfy —2 -2m 0 „2 „ x2 yn HxL = En yn HxL 0 < x < L elsewhere We see that the latter statement imposes the boundary conditions yn H0L = yn HLL = 0 on our differential equation, the former statement above. We recognize this equation as that of the simple harmonic oscillator, which has solutions yn HxL = An cosHkn xL + Bn sinHkn xL where kn = 2 m En — Imposing the boundary conditions gives Printed by Mathematica for Students yn H0L = 0 : An cosH0L + Bn sinH0L = 0 yn HLL = 0 : An cosHkn LL + Bn sinHkn LL = 0 ﬂ An = 0 sinHkn LL = 0 HBn = 0 is trivialL We see that the latter statement imposes the boundary conditions yn H0L = yn HLL = 0 on our differential equation, the former statement above. We recognize this equation as that of the simple harmonic oscillator, which has solutions Ch6ProblemSetKey.nb 5 yn HxL = An cosHkn xL + Bn sinHkn xL 2 m En — where kn = Imposing the boundary conditions gives yn H0L = 0 : An cosH0L + Bn sinH0L = 0 yn HLL = 0 : An cosHkn LL + Bn sinHkn LL = 0 An = 0 sinHkn LL = 0 ﬂ HBn = 0 is trivialL Thus kn L = n p 2 m En — n œ Zཻ+ for L = np En = 1 2m I np— 2 M L with energy eigenstates Bn sin I yn HxL = npx M L 0<x<L 0 elsewhere We calculate the Bn by requiring that the eigenstates be normalized ¶ 2 =1 ¶ 2 = Ÿ-¶ „ x yn HxL where Ÿ-¶ „ x yn HxL = ¶ Ÿ-¶ „ x 2 yn HxL...
View Full Document

## 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.

Ask a homework question - tutors are online