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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 Spring '07
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 Physics, Energy, Kinetic Energy, Boundary value problem, general solution

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