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Unformatted text preview: Today’s class • The finite square well • Quantum tunneling ) ( ) ( 2 2 2 2 x E x m ψ ψ = ∂ ∂ − h Last class: “infinite square well” Energy V(x) V(x) = ∞, x<0 0, 0<x<L ∞, x>L n=2 n=3 ψ (x) 2 x m ∂ With B.C.: ψ (0)= ψ (L) =0 0 x L Solution: 2 2 2 2 2 2 2 mL n m p E n h π = = k n = n π / L , , ( n =1,2,3,4 ¡) n=1 h / ) sin( 2 ) ( ) ( ) , ( t iE n e L x n L t x t x − = = Ψ π φ ψ Interpretation Energy Classically const. KE Quantum mechanically const. KE n=2 0 x L const. KE V=0 Probability of finding e- in [x,x+dx] x 0 x L const. KE V=0 Probability of finding e- in [x,x+dx] x Electron does not bounce from one wall to the other! (time average) That’s all there is to know about the infinite square well potential. But wait! We wanted so solve this: ) ) ) ∞ ∞ Or that: But we just did this instead: 0 V(x) x 0 V(x) x V(x) x ‘finite square well’ ‘infinite square well’ ‘Finite square well’ is much closer to real physical situation. Important for thinking about “Quantum tunneling”: Radioactive decay Scanning tunneling microscope to study surfaces Need to solve for exact potential energy curve:...
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