Lec+6+Finish+P+in+B+2015+158a

Lec+6+Finish+P+in+B+2015+158a - where n = 1 2 3 … ψ n...

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Lecture 6 Translation via Particle in a Box (2) Summary of 1d results Wave function solution Normalization Energy Plot of Wave Functions/Probability Translational Spectroscopy? Observables: position and momentum 2 Dimensional Particle in a Box F. Grieman
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Translation Particle in a Box model V = 0 0<x<a Solution: ψ(x) = A’ sin (nπx/a) where n = 1, 2, 3, … ; found from ψ(x) = 0 outside box A’ via normalization < ψ | ψ > = 1 = = Look up integral!: A’ = (2/a) 1/2 so ψ n (x) = (2/a) 1/2 sin(nπx/a)
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Unformatted text preview: where n = 1, 2, 3, … ψ n are orthogonal (<ψ i |ψ j > = 0). Why? Energy via Sch. Eq. Ĥ (x) = E (x): So, (-ħ 2 /2m) d 2 /dx 2 (ψ n ) = (-ħ 2 /2m) d 2 /dx 2 (2/a) 1/2 sin (nπx/a) = (-ħ 2 /2m) (nπ/a) d/dx (2/a) 1/2 cos (nπx/a) = (-ħ 2 /2m) (nπ/a) 2 [-(2/a) 1/2 sin (nπx/a)] = (n 2 π 2 ħ 2 /2ma) 2 ψ n So, = ; n = 1,2,3, … ; Energy is Quantized !!! We then have: Ĥ ψ = (-ħ 2 /2m) (d 2 /dx 2 ) ψ = E ψ Plots of E n , ψ, and ψ*ψ...
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