Chapter 6 Problem Solutions

We recognize this equation as that of the simple

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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...
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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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