{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter6_37

# Chapter6_37 - PHY3063 R D Field Quantum Mechanical...

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: PHY3063 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 Chapter6_37.doc University of Florida ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online