Chapter2_27 - PHY4604 R. D. Field Quantum Mechanical...

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Unformatted text preview: PHY4604 R. D. Field Quantum Mechanical Tunneling (2) Boundary Conditions: There are four boundary conditions. At x = 0 we have ψ 1 ( 0) = ψ 2 ( 0 ) which implies A + B = C + D dψ 1 dψ 2 = which implies ikA − ikB = κC − κD dx dx x =0 and at x = L we have ψ 2 ( L) = ψ 3 ( L) dψ 2 dψ 3 = dx x=L dx x=0 which implies Ce + κL + De −κL = Fe + ikL which implies κCe + κL − κDe −κL = ikFe + ikL x=L We now must solve for B, C, D, and F in terms of A. We have four equations and four unknowns: + 1B − 1C − 1D + 0 F = − A − ikB − κC + κD + 0 F = −ikA 0 B + e + κLC + e −κL D − e + ikL F = 0 0 B + κe + κLC − κe −κL D − ike + ikL F = 0 I will use determinates to solve these simultaneous linear equations as follows: +1 Det = −1 −1 0 − ik −κ +κ 0 + κL − κL e − e + ikL 0 e 0 κe +κL − κe −κL − ike + ikL −1 = κe + ikL +1 + κL − κL e e −1 0 − 1 + ike κe +κL − κe −κL − ik ( + ikL ) −1 0 + κL − κL −1 e e κe +κL − κe −κL − ik ( (e (ik + κ ) − e (ik − κ ) ) = e (− (κ (− (κ − k ) sinh(κL) + 2ikκ cosh(κL)) = κe + ikL e −κL (ik + κ ) + e +κL (ik − κ ) + ike + ikL e −κL (ik + κ ) − e +κL (ik − κ ) = e + ikL −κL = 2e + ikL 2 2 + κL 2 + ikL 2 ) − k 2 )(e +κL − e −κL ) + 2ikκ (e +κL + e −κL ) ) 2 where I used e ±θ = cosh θ ± sinh θ . Department of Physics Chapter2_27.doc University of Florida ...
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This note was uploaded on 05/29/2011 for the course PHY 4064 taught by Professor Fields during the Spring '07 term at University of Florida.

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