HW7_Solutions

# HW7_Solutions - PHYS851 Quantum Mechanics I, Fall 2008...

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PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 7: 1D Scattering: 1. [10pts] Step-down: An incident wave with wave-vector k 1 approaching from the left encoun- ters a step-down potential, V ( x ) = V 0 u ( x ) where u ( x ) is the unit step function. Calculate the quantum reFection probability, R . Under what conditions does it agree to a good approximation with the classical reFection probability? Answer: Let region 1 correspond to x < 0, and region 2, to x > 0. Our Ansatz is: ψ 1 ( x ) = e ikx + re ikx ψ 2 ( x ) = te iKx where K = r k 2 + 2 mV 0 / p 2 . The boundary conditions are ψ 1 (0) = ψ 2 (0) and ψ 1 (0) = ψ 2 (0), which lead to 1 + r = t k (1 r ) = Kt eliminating t gives k (1 r ) = K (1 + r ) solving for r gives r = k K k + K The reFection probability is therefore R = | r | 2 = k 2 2 kK + K 2 k 2 + 2 kK + K 2 The classical reFection probability is R = 0, which is a good approximation when K k , which requires p 2 k 2 2 m V 0 , i.e. the kinetic energy should be very large compared to the step height. 1

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2. [15pts] Square barrier plus step potential Consider a plane wave incident from the left onto the following potential V ( x ) = 0 , x < 0 V 0 , 0 < x < a V 1 , x > a , where V 0 > V 1 > 0. Using the transfer-matrix approach, Fnd the re±ection and transmission probabilities R and T as a function of the incident wave-vector, k 1 > 0. Show explicitly that T = | t | 2 k 3 k 1 , where k 3 is the wave-vector in the region x > a . Answer: Using the transfer matrix approach gives M = M [3 , 2] G [2] M [2 , 1] where M [2 , 1] = 1 2 k 2 p k 1 + k 2 k 2 k 1 k 2 k 1 k 1 + k 2 P G [2] = p e ik 2 a 0 0 e ik 2 a P M [3 , 2] = 1 2 k 3 p k 2 + k 3 k 3 k 2 k 3 k 2 k 2 + k 3 P and k 2 = r k 2 1 2 mV 0 / p 2 k 3 = r k 2 1 2 mV 1 / p 2 With r = M 21 M 22 t = M 11 M 12 M 21 M 22 we can use mathematica to Fnd r = e ik 2 a ( k 3 k 2 )( k 2 + k 1 ) + e ik 2 a ( k 3 + k 2 )( k 2 k 1 ) e ik 2 a ( k 3 k 2 )( k 2 k 1 ) + e ik 2 a ( k 3 + k 2 )( k 2 + k 1 ) t = 4 k 1 k 2 e ik 2 a ( k 3 k 2 )( k 2 k 1 ) + e ik 2 a ( k 3 + k 2 )( k 2 + k 1 ) Case I: p 2 k 2 1 2 m > V 0 In this case all of the k ’s are real, which gives r = k 2 ( k 3 k 1 ) cos( k 2 a ) i ( k 2 2 k 1 k 3 ) sin( k 2 a ) k 2 ( k 3 + k 1
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## This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.

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HW7_Solutions - PHYS851 Quantum Mechanics I, Fall 2008...

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