Problem 2.53
The Scattering Matrix. The theory of scattering generalizes in a pretty obvious way to arbitrary localized potentials (Figure 2.21). To the left (Region I), V(x) = 0, so
ψ(x) =Aeikx+Be−ikx, wherek≡
√ 2mE
ℏ . (2.178)
To the right (Region III),V(x) is again zero, so
ψ(x) =F eikx+Ge−ikx. (2.179) In between (Region II), of course, I can’t tell you whatψ is until you specify the potential, but because the Schr¨odinger equation is a linear, second-order differential equation, the general solution has got to be of the form
ψ(x) =Cf(x) +Dg(x),
wheref(x) andg(x) are two linearly independent particular solutions.63 There will be four boundary conditions (two joining Regions I and II, and two joining Regions II and III). Two of these can be used to eliminateC and D, and the other two can be “solved” forB and F in terms of Aand G:
B =S11A+S12G, F =S21A+S22G.
The four coefficientsSij, which depend on k(and hence onE), constitute a 2×2 matrixS, called thescattering matrix (orS-matrix, for short). The S-matrix tells you the outgoing
amplitudes (B and F) in terms of the incoming amplitudes (A andG):
B F
=
S11 S12
S21 S22 A G
. (2.180)
In the typical case of scattering from the left,G= 0, so the reflection and transmission coefficients are
Rl = |B|2
|A|2 G=0
=|S11|2, Tl= |F|2
|A|2 G=0
=|S21|2. (2.181) For scattering from the right,A= 0, and
Rr= |F|2
|G|2 A=0
=|S22|2, Tr = |B|2
|G|2 A=0
=|S12|2. (2.182)
63See any book on differential equations—for example, John L. Van Iwaarden,Ordinary Differential Equations with Numerical Techniques, Harcourt Brace Jovanovich, San Diego, 1985, Chapter 3.
(a) Construct the S-matrix for scattering from a delta-function well (Equation 2.117).
(b) Construct the S-matrix for the finite square well (Equation 2.148). Hint: This requires no new work, if you carefully exploit the symmetry of the problem.
Solution Part (a)
With
V(x) =−αδ(x), (2.117)
the Schr¨odinger equation becomes iℏ∂Ψ
∂t =−ℏ2 2m
∂2Ψ
∂x2 −αδ(x)Ψ(x, t), −∞< x <∞, t >0.
Applying the method of separation of variables [Ψ(x, t) =ψ(x)ϕ(t)] results in two ODEs, one inx and one in t.
iℏϕ′(t) ϕ(t) =E
−ℏ2 2m
ψ′′(x)
ψ(x) −αδ(x) =E
Solve the TISE for the second derivative.
d2ψ
dx2 =−2m
ℏ2 [αδ(x) +E]ψ(x) (1)
The delta function is zero everywhere exceptx= 0.
d2ψ
dx2 =−2mE
ℏ2 ψ(x), x̸= 0 For scattering states,E >0, which means the general solution is
ψ(x) =
(Aeikx+Be−ikx ifx <0 F eikx+Ge−ikx ifx >0, wherek=√
2mE/ℏ. The wave function [and consequently ψ(x)] is required to be continuous at x= 0.
x→0lim−ψ(x) = lim
x→0+ψ(x) : A+B =F +G (2)
Integrate both sides of equation (1) with respect tox from −ϵtoϵ, whereϵis a really small positive number.
ϵ
−ϵ
d2ψ
dx2 dx=−2m ℏ2
ϵ
−ϵ
[αδ(x) +E]ψ(x)dx dψ
dx
ϵ
−ϵ
=−2m ℏ2
α
ϵ
−ϵ
δ(x)ψ(x)dx+E
ϵ
−ϵ
ψ(x)dx
=−2m ℏ2
αψ(0) +Eψ(0)
ϵ
−ϵ
dx
=−2m
2 [αψ(0) +Eψ(0)(2ϵ)]
Take the limit as ϵ→0.
dψ dx
0+ 0−
=−2mα ℏ2 ψ(0)
The spatial derivative of the wave function, on the other hand, is discontinuous at x= 0.
x→0lim+ dψ
dx − lim
x→0−
dψ
dx =−2mα
ℏ2 ψ(0) : ik(F−G)−ik(A−B) =−2mα
ℏ2 (A+B) (3) Solve equations (2) and (3) for B and F.
B = imα
kℏ2−imαA+ kℏ2
kℏ2−imαG=S11A+S12G F = kℏ2
kℏ2−imαA+ imα
kℏ2−imαG=S21A+S22G Therefore, the scattering matrix for the delta-function well is
S=
imα kℏ2−imα
kℏ2 kℏ2−imα kℏ2
kℏ2−imα
imα kℏ2−imα
.
Part (b)
Now solve the Schr¨odinger equation, iℏ∂Ψ
∂t =−ℏ2 2m
∂2Ψ
∂x2 +V(x, t)Ψ(x, t), −∞< x <∞, t >0, with
V(x) =
(−V0 if −a≤x≤a
0 if|x|> a , (2.148)
Applying the method of separation of variables [Ψ(x, t) =ψ(x)ϕ(t)] results in two ODEs, one inx and one in t.
iℏϕ′(t) ϕ(t) =E
−ℏ2 2m
ψ′′(x)
ψ(x) +V(x) =E
Solve the TISE for the second derivative.
d2ψ dx2 = 2m
ℏ2 [V(x)−E]ψ(x) (4)
For scattering states,E >0, which means the general solution is
ψ(x) =
Aeikx+Be−ikx ifx <−a Csin(lx) +Dcos(lx) if −a≤x≤a F eikx+Ge−ikx ifx > a
,
where
k=
√ 2mE
ℏ and l=
p2m(V0+E)
ℏ .
Require the wave function [and consequentlyψ(x)] to be continuous at x=−a.
x→−alim−ψ(x) = lim
x→−a+ψ(x) : Ae−ika+Beika=−Csin(la) +Dcos(la) (5) Require the wave function to be continuous atx=aas well.
x→alim−ψ(x) = lim
x→a+ψ(x) : Csin(la) +Dcos(la) =F eika+Ge−ika (6) Integrate both sides of equation (4) with respect tox from −a−ϵto−a+ϵ, whereϵis a really small positive number.
−a+ϵ
−a−ϵ
d2ψ
dx2 dx= 2m ℏ2
−a+ϵ
−a−ϵ
[V(x)−E]ψ(x)dx dψ
dx
−a+ϵ
−a−ϵ
= 2m ℏ2
−a
−a−ϵ
(0−E)dx+
−a+ϵ
−a
(−V0−E)dx
= 2m ℏ2
[(−E)(ϵ) + (−V0−E)(ϵ)]
Take the limit as ϵ→0.
dψ dx
−a+
−a−
= 0
It turns out that the spatial derivative of the wave function∂Ψ/∂x is continuous atx=−a.
lim
x→−a−
dψ
dx = lim
x→−a+
dψ
dx : ik(Ae−ika−Beika) =l[Ccos(la) +Dsin(la)] (7) Now integrate both sides of equation (4) with respect tox froma−ϵtoa+ϵ.
a+ϵ
a−ϵ
d2ψ
dx2 dx= 2m ℏ2
a+ϵ
a−ϵ
[V(x)−E]ψ(x)dx dψ
dx
a+ϵ a−ϵ
= 2m ℏ2
a
a−ϵ
(−V0−E)dx+
a+ϵ
a
(0−E)dx
= 2m ℏ2
[(−V0−E)(ϵ) + (−E)(ϵ)]
Take the limit as ϵ→0.
dψ dx
a+ a−
= 0
It turns out that the spatial derivative of the wave function∂Ψ/∂x is also continuous atx=a.
x→alim− dψ
dx = lim
x→a+
dψ
dx : l[Ccos(la)−Dsin(la)] =ik(F eika−Ge−ika) (8)
To summarize, there are four equations involvingA,B,C,D,E, andF.
Ae−ika+Beika=−Csin(la) +Dcos(la) Csin(la) +Dcos(la) =F eika+Ge−ika ik(Ae−ika−Beika) =l[Ccos(la) +Dsin(la)]
l[Ccos(la)−Dsin(la)] =ik(F eika−Ge−ika) Solve for B and F, eliminating C and D.
B = i(l2−k2)e−2ikasin 2la
2lkcos 2la−i(l2+k2) sin 2laA+ 2lke−2ika
2lkcos 2la−i(l2+k2) sin 2laG F = 2lke−2ika
2lkcos 2la−i(l2+k2) sin 2laA+ i(l2−k2)e−2ikasin 2la 2lkcos 2la−i(l2+k2) sin 2laG Therefore, the scattering matrix for the finite square well is
S=
i(l2−k2)e−2ikasin 2la 2lkcos 2la−i(l2+k2) sin 2la
2lke−2ika
2lkcos 2la−i(l2+k2) sin 2la 2lke−2ika
2lkcos 2la−i(l2+k2) sin 2la
i(l2−k2)e−2ikasin 2la 2lkcos 2la−i(l2+k2) sin 2la
.
