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Chapter2_24

# Chapter2_24 - must be continuous and have a continuous 1 st...

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PHY4604 R. D. Field Department of Physics Chapter2_24.doc University of Florida Finite Step Potential (V 0 < E) Consider particles with energy E entering from the left encountering a potential barrier V 0 < E . Classically the particles slow down at x 0 and exit to the right with the same energy E but with less kinetic energy. We must solve ) ( )) ( ( ) ( 2 2 2 2 x x V E dx x d m ψ ψ = h , where V(x) = 0 for x < 0 and V(x) = V 0 for x 0 . Left Region (x < 0): In this region ) ( ) ( 2 2 2 x k dx x d L L ψ ψ = with 2 2 h mE k = and the most general solution is ikx L ikx L L e B e A x + + = ) ( ψ Right Region (x 0): In this region ) ( ) ( 2 2 2 x q dx x d R R ψ ψ = with E V k V E m q 0 2 0 1 ) ( 2 = = h and the most general solution is iqx R iqx R R e B e A x + + = ) ( ψ Initial Conditions: We are sending in particles from the left ( initial condition ). The second term in ψ R corresponds to particles coming in from the right and thus we set B R = 0 . Boundary Conditions: The wavefunction ( i.e. probability amplitude)
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Unformatted text preview: must be continuous and have a continuous 1 st derivative at the interface between the left and right region. ) ( ) ( = = = x x R L implies R L L A B A = + ) ( ) ( = = = x dx d x dx d R L implies R L L iqA ikB ikA = − We can express B L and A R in terms of A L with L L A r r B + − = 1 1 and L R A r A + = 1 2 where E V k q r 1 − = = . Reflection and Transmission Probability: 2 2 2 2 ) 1 ( ) 1 ( | | | | r r A B j j P L m k L m k L L R + − = = = h h r s and 2 2 2 ) 1 ( 4 | | | | r r A A j j P L m k R m q L R T + = = = h h r r It is easy to show that P R + P T = 1 . x x 0 = 0 E Step Potential V “Left Region” “Right Region” Note that T < 1 and some particles are reflected back at x !...
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