SCATTERING. V 0 x a. V V x a. d Asin kx 2m dx. 2mE k. Acos x or Asin. A cosh x or Asinh

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John Hodge
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1 SCATTERING Consider a particle of mass m moving along the x axis with energy E. It encounters a potential given by V0 x a VV x a 0 In the region x > a we have with d E Asinkx m dx In the region x < a we have If E > V 0 the solutions are If E < V 0 the solutions are me k d d m V 0E EV m dx dx 1/ Acosx or Asin x A coshx or Asinh x

2 with m E V 1/ 0 The even parity solutions are xa Asin kx x a Asin kx The odd parity solutions are x a Bcos x or Bcosh x xa A'sin kx ' x a A'sin kx ' x a B'sin x or B'sinh x For the region x > a we have both even and odd parity solutions with the same energy. Hence x a Asinkx A'sinkx ' ikx ikx ikx ' ikx ' e e e e A A' i i i kx i kx ' i kx i kx ' Ae A'e Ae A'e i i Now e +ikx corresponds to a particle moving right. In the region x > a we want only particles moving right. Hence i i ' Ae A'e 0 Then A' i ' Ae

3 A ikx i i ' i ' A ikx i i ' xa e e e e e e 1 e i i For x < -a the solution is then e o i ' x x Asin kx Ae sin kx ' A e e e e e i i kx i kx i ' i kx ' i kx ' A ikx i ' i ' A ikx i i ' i ' e e e e e e e e i i A A i i ikx i ikx i i ' e e e e 1 e Ae iae e e i Hence the incoming amplitude is i ikx i i kx i ' 1 e i iae The reflection amplitude is The transmission amplitude is A e i 1 e i i with ϕ = ϕ - ϕ. Thus A e i 1 e i i

4 ref i i i i amp 1 1 e 1e 1e 1e amp 4 in For transmission we have 1 i i 1 e e cos cos sin cos cos trans i i i i amp 1 1 e 1e 1e 1e amp 4 IN 1 i i e e cos 1 cos sin 4 4 sin prob of transmission Note that these results did not depend on the form of the solution for x < a. Hence they are valid for any potential of even parity which vanishes outside a finite region. Of course, the value of does depend on the region x < a. Case 1 V 0 > E Classically this would result in all particles being reflected. However, in quantum mechanics we find otherwise. Requiring continuity of and d/dx at x = a gives even odd Asin ka Bcosh a ka cos ka Bsinh a k tan ka coth a

5 Asin ka ' B'sinh a ka cos ka ' B'cosh a k tan ka ' tanh a Since tan x tan x the equations have infinitely many solutions for and that differ by n where n is an integer. Since this leave ϕ unchanged we need only consider the lowest solutions. Thus tan cotha ka 1 k ' tan tanha ka 1 k 1k 1k tan tanha tan cotha We now consider this result in two limiting cases: E = V 0, E << V 0. If E = V 0 we have = 0 Now tanh a ka e e a a k k e e a a coth a e e a a a k k e e ka tan 1 a a 1 x x x

6 1/ 1 1 mea tan ka tan For ka =.5 this gives P trans = 0.8. For ka = 1.0 this gives P trans = 0.3. Hence particle is reflected a significant fraction of the time. Case V 0 < E Now the continuity equations at x = a are even Asin ka Bcos a kacos ka Bsina odd k tan ka ctan a A'sin ka ' B'sin a k A'cos ka ' k B'cos a k tan ka ' tan a Where m E 1 k V 1/ 0 tan ctana ka ' tan tana ka 1 k

7 1k 1 k tan tan a tan c tan a Consider the case E = V 0 so that classically there would be complete transmission. Instead we find 1 ka 1/ / 0 mv me 1 k 1 ka 1 ka tan tan tan ctan 1 ka 1 tan tan tan tan ka tan tan tan tan 1.86 Prob of reflection = cos =.0813 Hence even when E is much greater V 0 we still have a significant probability of reflection. Scattering Cross Section for V = -V 0 There are now two ways the particle can get to the region x > a. The first is to not interact at all. The second is to be trapped in the well for a while and then escape to the right. We represent these possibilities by writing the amplitude to end on the right as IN A 1 T where A IN is the amplitude IN. Then 1 represents the case of no interaction. Thus the amplitude for reflection is given by For transmission we have A 1 iae R e 1e R e e i i i i i i '

8 A iae 1 T e 1 e i i i i 1 i i ' 1 i i ' 1T e e T e e 1 We define the scattering cross section as R T Now R prob of reflection 1T prob of transmission 1 T T T But prob of reflection prob of transmission 1 Hence R 1 T T T 1 T T 1 i i ' 1 i i ' e e 1 e e 1 1 i i 1 i ' i ' e e e e coscos ' cos sin cos ' sin ' cos 1 sin ' 1 cos sin '

MATH Non-Euclidean Geometry Exercise Set #9 Solutions

MATH Non-Euclidean Geometry Exercise Set #9 Solutions MATH 6118-090 Non-Euclidean Geometry Exercise Set #9 Solutions 1. Consider the doubly asymptotic triangle AMN in H where What is the image of AMN under the isometry γ 1? Use this to find the hyperbolic