385lec12 - PHYS 385 Lecture 12 - Tunneling in one dimension...

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Unformatted text preview: PHYS 385 Lecture 12 - Tunneling in one dimension 12 - 1 2003 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. Lecture 12 - Tunneling in one dimension What's important : tunneling in one dimension WKB approximation alpha decay Text : Gasiorowicz, Chap. 5 Square barrier Let's now take the potential well from the last lecture and invert it to form a barrier: V ( x ) 2 a E V o We take E < V and have the object approach the barrier from the left. As usual, there are the usual plane wave solutions exp( ikx ) where E < V o , and exponentially decaying solutions inside the barrier. We define the normalization coefficients as follows: For x < - a u ( x ) = exp( ikx ) + R exp(- ikx ) k 2 = 2 mE / h 2 , For - a < x < a u ( x ) = A exp(- x ) + B exp(+ x ) 2 = 2 m ( V o- E ) / h 2 , (real) (1) For x > a u ( x ) = T exp( ikx ) k 2 = 2 mE / h 2 . We can go through the same steps as before with the square well, but make the replacement q-> i . Matching u ( x ) and its derivative du / dx at the discontinuities a give the set of equations: exp(- ika ) + R exp( ika ) = A exp( a ) + B exp(- a ) ik exp(- ika ) + (- ik ) R exp( ika ) = - A exp( a ) + B exp(- a ) (2) A exp(- a ) + B exp( a ) = T exp( ika ) - A exp(- a ) + B exp( a ) = ikT exp( ika ) PHYS 385 Lecture 12 - Tunneling in one dimension 12 - 2 2003 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. Four equations and four unknowns yields the previous expressions for R and T with the substitution q-> i : T = e ! 2 ika 2 kq 2 kq cos(2 qa ) ! i ( q 2 + k 2 )sin(2 qa ) = e ! 2 ika 2 ik " 2 ik " cos(2 i " a ) !...
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This note was uploaded on 09/07/2009 for the course PHYS 385 taught by Professor Davidboal during the Spring '09 term at Simon Fraser.

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385lec12 - PHYS 385 Lecture 12 - Tunneling in one dimension...

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