Griffiths Quantum Mechanics 3e: Problem 2.53 Page 1 of 5
Problem 2.53
The Scattering Matrix. The theory of scattering generalizes in a pretty obvious way toarbitrary localized potentials (Figure 2.21). To the left (Region I), V (x) = 0, so
ψ(x) = Aeikx +Be−ikx, where k ≡√
2mE
~. (2.178)
To the right (Region III), V (x) is again zero, so
ψ(x) = Feikx +Ge−ikx. (2.179)
In between (Region II), of course, I can’t tell you what ψ is until you specify the potential, butbecause the Schrödinger equation is a linear, second-order differential equation, the generalsolution has got to be of the form
ψ(x) = Cf(x) +Dg(x),
where f(x) and g(x) are two linearly independent particular solutions.63 There will be fourboundary conditions (two joining Regions I and II, and two joining Regions II and III). Two ofthese can be used to eliminate C and D, and the other two can be “solved” for B and F in termsof A and G:
B = S11A+ S12G, F = S21A+ S22G.
The four coefficients Sij , which depend on k (and hence on E), constitute a 2× 2 matrix S, calledthe scattering matrix (or S-matrix, for short). The S-matrix tells you the outgoingamplitudes (B and F ) in terms of the incoming amplitudes (A and G):(
BF
)=
(S11 S12S21 S22
)(AG
). (2.180)
In the typical case of scattering from the left, G = 0, so the reflection and transmissioncoefficients 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 withNumerical Techniques, Harcourt Brace Jovanovich, San Diego, 1985, Chapter 3.
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Griffiths Quantum Mechanics 3e: Problem 2.53 Page 2 of 5
(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 nonew work, if you carefully exploit the symmetry of the problem.
Solution
Part (a)
WithV (x) = −αδ(x), (2.117)
the Schrödinger equation becomes
i~∂Ψ
∂t= − ~
2
2m
∂2Ψ
∂x2− αδ(x)Ψ(x, t), −∞ < x 0.
Applying the method of separation of variables [Ψ(x, t) = ψ(x)φ(t)] results in two ODEs, one in xand one in t.
i~φ′(t)
φ(t)= E
− ~2
2m
ψ′′(x)
ψ(x)− αδ(x) = E
Solve the TISE for the second derivative.
d2ψ
dx2= −2m
~2[αδ(x) + E]ψ(x) (1)
The delta function is zero everywhere except x = 0.
d2ψ
dx2= −2mE
~2ψ(x), x 6= 0
For scattering states, E > 0, which means the general solution is
ψ(x) =
{Aeikx +Be−ikx if x < 0
Feikx +Ge−ikx if x > 0,
where k =√
2mE/~. The wave function [and consequently ψ(x)] is required to be continuous atx = 0.
limx→0−
ψ(x) = limx→0+
ψ(x) : A+B = F +G (2)
Integrate both sides of equation (1) with respect to x from −� to �, where � is a really smallpositive number.
ˆ �−�
d2ψ
dx2dx = −2m
~2
ˆ �−�
[αδ(x) + E]ψ(x) dx
dψ
dx
∣∣∣∣�−�
= −2m~2
[α
ˆ �−�δ(x)ψ(x) dx+ E
ˆ �−�ψ(x) dx
]= −2m
~2
[αψ(0) + Eψ(0)
ˆ �−�dx
]= −2m
~2[αψ(0) + Eψ(0)(2�)]
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Griffiths Quantum Mechanics 3e: Problem 2.53 Page 3 of 5
Take the limit as �→ 0.dψ
dx
∣∣∣∣0+0−
= −2mα~2
ψ(0)
The spatial derivative of the wave function, on the other hand, is discontinuous at x = 0.
limx→0+
dψ
dx− limx→0−
dψ
dx= −2mα
~2ψ(0) : ik(F −G)− ik(A−B) = −2mα
~2(A+B) (3)
Solve equations (2) and (3) for B and F .
B =imα
k~2 − imαA+
k~2
k~2 − imαG = S11A+ S12G
F =k~2
k~2 − imαA+
imα
k~2 − imαG = S21A+ S22G
Therefore, the scattering matrix for the delta-function well is
S =
imα
k~2 − imαk~2
k~2 − imα
k~2
k~2 − imαimα
k~2 − imα
.Part (b)
Now solve the Schrödinger equation,
i~∂Ψ
∂t= − ~
2
2m
∂2Ψ
∂x2+ V (x, t)Ψ(x, t), −∞ < x 0,
with
V (x) =
{−V0 if − a ≤ x ≤ a0 if |x| > a
, (2.148)
Applying the method of separation of variables [Ψ(x, t) = ψ(x)φ(t)] results in two ODEs, one in xand one in t.
i~φ′(t)
φ(t)= E
− ~2
2m
ψ′′(x)
ψ(x)+ V (x) = E
Solve the TISE for the second derivative.
d2ψ
dx2=
2m
~2[V (x)− E]ψ(x) (4)
For scattering states, E > 0, which means the general solution is
ψ(x) =
Aeikx +Be−ikx if x < −aC sin(lx) +D cos(lx) if − a ≤ x ≤ aFeikx +Ge−ikx if x > a
,
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Griffiths Quantum Mechanics 3e: Problem 2.53 Page 4 of 5
where
k =
√2mE
~and l =
√2m(V0 + E)
~.
Require the wave function [and consequently ψ(x)] to be continuous at x = −a.
limx→−a−
ψ(x) = limx→−a+
ψ(x) : Ae−ika +Beika = −C sin(la) +D cos(la) (5)
Require the wave function to be continuous at x = a as well.
limx→a−
ψ(x) = limx→a+
ψ(x) : C sin(la) +D cos(la) = Feika +Ge−ika (6)
Integrate both sides of equation (4) with respect to x from −a− � to −a+ �, where � is a reallysmall positive number.
ˆ −a+�−a−�
d2ψ
dx2dx =
2m
~2
ˆ −a+�−a−�
[V (x)− E]ψ(x) dx
dψ
dx
∣∣∣∣−a+�−a−�
=2m
~2
[ˆ −a−a−�
(0− E) dx+ˆ −a+�−a
(−V0 − E) dx]
=2m
~2[(−E)(�) + (−V0 − E)(�)]
Take the limit as �→ 0.dψ
dx
∣∣∣∣−a+−a−
= 0
It turns out that the spatial derivative of the wave function ∂Ψ/∂x is continuous at x = −a.
limx→−a−
dψ
dx= lim
x→−a+dψ
dx: ik(Ae−ika −Beika) = l[C cos(la) +D sin(la)] (7)
Now integrate both sides of equation (4) with respect to x from a− � to a+ �.ˆ a+�a−�
d2ψ
dx2dx =
2m
~2
ˆ a+�a−�
[V (x)− E]ψ(x) dx
dψ
dx
∣∣∣∣a+�a−�
=2m
~2
[ˆ aa−�
(−V0 − E) dx+ˆ a+�a
(0− E) dx]
=2m
~2[(−V0 − E)(�) + (−E)(�)]
Take the limit as �→ 0.dψ
dx
∣∣∣∣a+a−
= 0
It turns out that the spatial derivative of the wave function ∂Ψ/∂x is also continuous at x = a.
limx→a−
dψ
dx= lim
x→a+dψ
dx: l[C cos(la)−D sin(la)] = ik(Feika −Ge−ika) (8)
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Griffiths Quantum Mechanics 3e: Problem 2.53 Page 5 of 5
To summarize, there are four equations involving A, B, C, D, E, and F .Ae−ika +Beika = −C sin(la) +D cos(la)C sin(la) +D cos(la) = Feika +Ge−ika
ik(Ae−ika −Beika) = l[C cos(la) +D sin(la)]l[C cos(la)−D sin(la)] = ik(Feika −Ge−ika)
Solve for B and F , eliminating C and D.
B =i(l2 − k2)e−2ika sin 2la
2lk cos 2la− i(l2 + k2) sin 2laA+
2lke−2ika
2lk cos 2la− i(l2 + k2) sin 2laG
F =2lke−2ika
2lk cos 2la− i(l2 + k2) sin 2laA+
i(l2 − k2)e−2ika sin 2la2lk cos 2la− i(l2 + k2) sin 2la
G
Therefore, the scattering matrix for the finite square well is
S =
i(l2 − k2)e−2ika sin 2la
2lk cos 2la− i(l2 + k2) sin 2la2lke−2ika
2lk cos 2la− i(l2 + k2) sin 2la
2lke−2ika
2lk cos 2la− i(l2 + k2) sin 2lai(l2 − k2)e−2ika sin 2la
2lk cos 2la− i(l2 + k2) sin 2la
.
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