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„ 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 (‰Â 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 ‰ Â 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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 Fall '13
 AkimasaMiyake

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