Chapter 6 Problem Solutions

Neither of these terms is zero a priori because there

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Unformatted text preview: 2 „ x2 = E yHxL HinsideL yHxL = - 2mE yHxL = - 2 m HE - V0 L —2 HoutsideL yHxL —2 HoutsideL yHxL HinsideL Solutions to the equation outside of the barrier are complex exponentials. In the region to the left of the barrier (x < 0), the solution is a superposition of an incoming incident wave (‰Â k x ) and an outgoing reflected wave (‰- k x ). Solutions to the equation inside of the barrier (0 § x § a) for E < V0 are real exponentials and will in general be a superposition of a decay term (‰-q x ) and a growing term (‰q x ). Neither of these terms is zero a priori because there will be a reflected and transmitted contribution from each of the boundaries at x = 0 and x = a. Finally, the solution to the equation to the right of the barrier (x > a) will be an outgoing transmitted wave, as we do not expect an incident wave from the right. We thus have the ansatz yHxL = A ‰Â k x + B ‰-  k x x < 0 F ‰q x + G ‰- q x 0 § x § a C ‰Â k x k= where q= x>a 2mE — 2 m HV0 - E L — We solve for the coefficients by imposing continuity of the wavefunction and its derivative at the boundaries at x = 0 and x = a, as limx Ø 0- yHxL = limx Ø 0+ yHxL limx Ø a- y HxL = limx Ø a+ y HxL „ „x „ limx Ø a- „ x limx Ø 0- „ „x „ limx Ø a+ „ x y HxL = limx Ø...
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