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1 HW.nb HW #. D scattering (a) The delta function potential requires a discontinuity of the wave function, y' H+0L - y' H-0L = ÅÅÅÅÅÅÅÅÅÅÅÅ m m yh0l, while the wavefunction itself is continuous. SolveA9 + R ã T, I k T - I k H - RL ã m m ÅÅÅÅÅÅÅÅÅÅÅ T=, R, T<E  m m 99R Ø - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ Â m m + k, T Ø k ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ Â m m + k == Simplify@%D 99R Ø m m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ -m m +  k, T Ø k ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ Â m m + k == SimplifyA i k j- m m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ y z i m m -  k { k j- m m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ y z + i k y m m +  k j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ { k  m m + k z i k y j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ { k - m m + k z E { It is worth noting that the reflection probabiliy» R» is 0% at the low momentum limit k = 0, while it monotonically decreases to zero at higher momenta. (b) We assume T and R to be approximately constant. Hence the Gaussian wave packet is given by the integrals IntegrateAE I q x E -Hq-kL êês E -I H q êêml tê, q, -, <, Assumptions Ø s œ Reals && m œ Reals && œ Reals && t œ Reals<E  k m x-m x s - k t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ è!!!!!!! m+  t s p ÅÅÅÅÅÅ s m IntegrateAE -I q x E -Hq-kL êês E -I H q êêml tê, q, -, <, Assumptions Ø s œ Reals && m œ Reals && œ Reals && t œ Reals<E  k m x+m x s + k t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ è!!!!!!! - m-  t s p ÅÅÅÅÅÅ s m
2 HW.nb y = IfAx < 0, Â k m x-m x s -Â k t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ è!!!!!!! m+ Â t s p ÅÅÅÅÅÅ + R - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Â k m x+m x s ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ +Â k ÅÅÅÅÅÅÅÅÅ t è!!!!!!! m+ Â t s p ÅÅÅÅÅÅÅÅ, T Â k m x-m x s -Â k t ÅÅÅÅ è!!!!!!! m+ Â t s p ÅÅÅÅÅÅ E #### #### #### ÅÅÅÅÅ ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ ÅÅÅÅÅÅÅÅ s m s m s m IfAx < 0, Â k m x-m x s -Â k t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ è!!!!!!! m+ Â t s p ÅÅÅÅÅÅ + R J - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Â k m x+m x s ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ +Â k ÅÅÅÅÅÅÅÅÅ t è!!!!!!! m+ Â t s p N ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ, T J Â k m x-m x s -Â k t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ è!!!!!!! m+ Â t s p N ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E s m s m s m TableA m m PlotAAbs@yD ê. 9R Ø - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ m m - Â k, T Ø k ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ = ê. k Ø 0, m Ø, m Ø 30, s Ø, Ø <, Â m m + k x, -, <, PlotRange Ø 0, <E, t, -0., 0., 0.0<E
3 HW.nb
4 HW.nb
5 HW.nb
6 HW.nb
7 HW.nb Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü, Ü Graphics Ü< Select the graphics and go to "Animate Selected Graphics." (c) Rewriting Eq. () into the form Eq. (3) gives f HpL = -i R, f H0L = i H - TL. Therefore, the unitarity relation is rewritten as =» R» +» T» =» i f HpL» +» + i f H0L» =» f HpL» + + i f H0L - i f * H0L +» f H0L» = s + - Im f H0L. Therefore, s = Im f H0L, the one-dimensional version of the optical theorem.. Probability currents The plane wave y = e i kø ÿx Ø gives simply Ø j = ÅÅÅÅÅÅÅ kø, while the spherical wave y = m ei k r ê r gives Ø j = ÅÅÅÅÅÅ k = ÅÅÅÅÅÅ k p m dixø M indeed. There- former, Ø j is constant and hence its divergence simply vanishes. For the latter, Ø ÿ Ø j =- ÅÅÅÅÅÅ k fore, the spherical wave is a wave that emanates from the origin. For fun, I plot below the probability current on the x-y plane at z=0 for both cases. D ÅÅÅÅ m r ÅÅÅÅÅ Ø x m r 3 = - ÅÅÅÅÅÅ k m Ø ÅÅÅÅ r. For the In[0]:= <<Graphics`PlotField`
8 HW.nb In[]:= 0<, x, -, <, y, -, <D Out[]= In[]:= Ü Graphics Ü x, y< PlotVectorFieldA ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ, x, -, <, y, -, <E Hx + y 3ê L Power::infy : Infinite expression ÅÅÅÅ 0 encountered. More ::indet : Indeterminate expression 0 ComplexInfinity encountered. More ::indet : Indeterminate expression 0 ComplexInfinity encountered. More Out[]= Ü Graphics Ü
9 HW.nb 9 3. Classical Hard Sphere Scattering For the impact parameter b > a, the point-like particle will not get scattered by the hard sphere and merely passes it by. On the other hand, if b < a, it gets scattered into the angle q = p - sin - ÅÅÅÅ b q. Solving it, we find b = a cos ÅÅÅÅ. Because the a original beam is assumed to have a uniform area density, the cross section is d s = b d b d f = -a cos ÅÅÅÅ q a sin ÅÅÅÅ q d q d f = - ÅÅÅÅÅ a a sin q d q d f = ÅÅÅÅÅ d cos q d f. Hence the differential cross section is d s ÅÅÅÅÅÅÅÅ d W = ÅÅÅÅÅ a, and the total cross section is s = Ÿ ÅÅÅÅÅ a d W = p a, nothing but the geometric cross section of the hard sphere.
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