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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