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Quantum Physics III (8.06) Spring 2008Solution Set 10

May 10, 2008

1. Scattering from a Reflectionless Potential (10 points)

(a)(2 points)Plugging0(x) =Asech(ax) into the Schrodinger equation, one finds that it is

an eigenstate with energyE0= h2a2

2m . Since E


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2. Simple Properties of Cross Sections (15 points)

(a)(2 points)The incident flux is

Si = hk

mz,

while the scattered flux is (to leading order in 1/r)

Ss = hk

m

|f|2

r2 r.

(b)(2 points)Using part (a),

d

dd = lim

r

Ss r

|Si| dA=

|f|2

r2 r2d,

and therefore dd = |f|2.

(c)(11 points)From conservation of probability, we must have

Stotal= 0, and thereforeStotal dA = 0, with the integral is taken over the boundary of space. Now, Stotal =

Si+ Ss+ Sint, where

Sint = h

2mi

eikz

f

reikr + eikr

f

r eikz c.c.

= hk

m

1

rRe

f eik(rz)

(z+ r).

Lets consider

Stotal dA= 0 term by term. The first term we consider is

Sint dA. The

easiest coordinate system to use is cylindrical coordinates, in which we take the boundary of

space to be two planes at z = . (You can check that the contribution to the integral from

is sub leading.) We have, then,

Sint dA =

hk

m

dd

1

rRe

f eik(rz)

(z+ r) z

hk

m

dd

1

zRe

f(= 0)eik

2/2z

(z z) z,

where in the second line I have used the approximations

r z,

r = |z| + 2

2|z| + . . . ,

0.

Because at z = , z = r, there is no contribution from the plane at z = . The plane

at z = contributes Sint dA =

4hk

mz

d Re

f(= 0)eik

2/2z

= 4hk

m

da Re

f(= 0)eika

=

4hk

m lim0

da ea [Ref(= 0) cos(ka) Imf(= 0) sin(ka)]

= 4hm

Imf(= 0).

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The other two terms are simpler, as

Si = 0 and Ss dA=

hk

m

d|f|2.

Conservation of flux then tells us |f|2d =

4

k Imf(0).

3. Born Approximations for Scattering from Yukawa and Coulomb Potentials (15

points)

(a) (6 points) The Yukawa potential is spherically symmetric, so f() = 2mh20 drrV(r)sin r,

where = |k k|. The integral we need to perform is dr er sin r =

1

2i

e(+i)r

+ i

e(i)r

i

=

2 + 2.

Using this,

f() = 2m

h2(2 + 2).

From this we find

dd

=

2mh2(2 + 2)

2.

The total cross-section is

=

|f|2d =

2m

h2

2 sin dd

1

(2 + 4k2 sin2 2)2

.

To evaluate the integral, make the variable change a = 4k2 sin2(/2)/2. Then the integral

becomes 122k24k2/20

da(1+a)2 =

22(2+4k2) , and

= 4

2 + 4k2

2m

h2

2=

4

h22 + 8mE

2m

h

2.

(b)(2 points)Putting = Q1Q2 and = 0, we find

d

d =

2mQ1Q2

2h2k2(1 cos )

2=

Q1Q2

4E

21

sin4 /2,

which is precisely the Rutherford cross-section.

(c)(2 points)To prevent confusion, I will use the symbolTfor thickness. Since the number of

particles scattered per unit solid angle per unit time per scatterer is ddd2NdtdA , and the number

density of scatterers isn, the number of particles scattered into unit solid angle per unit time

is

dA T n dd

d2NdtdA =

dd

dNdtnT. This is independent of beam area and beam uniformity, as

desired. The key point here is that the factors of area cancel independently for each areaelement, so non uniformity of the beam is not an issue.

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(d)(3 points)With the numbers

Q1 = 2e Q2= 79e =

2

T = 106m d = 104rads2 E = 8 MeV

n = Au/mAu= 5.9 1028atoms/m3 dN/dt= 3.13 1027s1,

we find that the number of scattered alpha particles seen in the detector per second is 3.7.

Note that in order to obtain the right units in SI we need to include a factor of 1 /40 in .

(e)(2 points)The number of particles detected in the detector depends on as

1

sin4 2sin (4)

where the second factor sin comes from the solid angle. The quantity in (4) takes on the

following values:

3009 = 10o

33 = 45o

0.97 = 135o

0.18 = 170o.

When we took = 90

o

and therefore

sin

sin4

/2 = 4, we found that there were 3.7 particles persecond scattered into the detector. Using this, the number of particles observed per second in

the detector at the various angular locations are

2784 = 10o

30.5 = 45o

0.9 = 135o

0.16 = 170o.

Note that the observed number of particles shoots up very sharply near = 0.

4. The Size of Nuclei (10 points)

(a)(4 points)Using the charge distribution (r) = 3Z4R3

for r Rand 0 otherwise, we find

r2dr sin ddeiqr cos (r) =

3Z

R3q2

1

qsin qR R cos qR

,

and therefore

f= 6mZe2

h2q2(qR)3(sin qR qR cos qR).

The scattering cross-section is

d

d = |f|2 = 6mZe2

h2q2(qR)32

(sin qR qR cos qR)2.

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5. The Born Approximation in One Dimension (15 points)

(a)(6 points)We need to show that (x) = 0(x) imh2k

e

ik|xx0|V(x0)(x0)dx0satisfies

the Schrodinger equation with eigenvalue h2k2

2m . Now,

2

x2eik|xy| =

2ik(x y) k2

eik|xy|,

so

h2

2m

2

x2=

h2k2

2m

0

im

h2k2

dy eik|xy|V(y)(y)

(x)V(x).

It is therefore easy to see that

h2

2m

2

x2 + V(x)

(x) = h2k2

2m (x).

(b)(5 points)Using 0(x) Aeikx, the first Born approximation becomes

(x) Aeikx Aim

h2k

dy eik|xy|V(y)eiky .

We want to find the reflection coefficient, which as Griffiths defines it is R= |reflected wave|2/|incident wave|2

(since this problem comes from Griffiths, we will use his definitions for R and T in this prob-

lem). The reflected wave is the component of that behaves like eikx atx . Now, as

x , |x y| =y x, and thus the coefficient of the reflected wave is Aimh2k

dy e2ikyV(y).

The incident wave is the component of that behaves likeeikx atx , and the coefficient

of this piece of isA. Therefore,

R=

mh2k

dy e2ikyV(y)

2

. (5)

(c)(4 points)First, we setV(x) = (x). In this case, the integral in equation (5) becomes

dy e2iky(y) = . The reflection coefficient is thereforeR= m22

h4k2 = m

2

2h2E. From this,

we find the first Born approximation to the transmission coefficient,

T = 1 R= 1 m2

2h2E.

The exact answers are, defining w = m2

2h2E,

Rexact= w

w+ 1, Texact =

1

1 + w,

so the first Born approximation has found the first term in an expansion ofRexact, Texact in

smallw. Smallw means that the strength of the potential (measured by ) is small compared

to the energy of the particle, and so this means that the Born approximation is good in the

weak scattering regime.

Second, we set V(x) =V0 when |x| < a and 0 otherwise. In this case, the integral of (5) is

V0aa

dy e2iky = V0sin(2ka)k

. The reflection coefficient is thenR= mh2k2

V0sin(2ka)2

, and

the transmission coefficient is

T = 1 R= 1

V02E

sin(2ka)2

.

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The exact answer is

Texact=

1 +

V204E(E+ V0)

sin2(2ka)

1 1

V02E

sin(2ka)

2

where the approximation holds when V0/E 1, that is, in the weak scattering regime. Again,

we have found the first term in the weak scattering expansion of the exact answer.

6. Scattering from a Small Crystal (8 points)

(a) (4 points) In the first Born approximation, the scattering amplitude is given by the

formula f(, ) = m2h2 i d

3r eiqrv(r Xi). In order to proceed further, we Fourier

expand the scattering potential of the single atom, v(x) =

(2)3d3k eikxvk. Plugging thisin, we find

f(, ) = m

2h2

i

(2)3d3k vke

ikXi

d3r ei(qk)r.

The integral over r yields a delta function,

d3r ei(qk)r = (2)33(q k), which makes the

integration overk trivial, so that

f(, ) = m

2h2vq

i

eiqXi

.

The differential cross-section is given by

d

d=

m

2h2

2|vq|

2

i

eiqXi

2

.

Note that the last factor contains all the information about the crystal structure, in the form

of the sum over atom positionsXi, while the factor |vq|2contains all the information about the

individual atom potential.

(b) (4 points) Consider scattering from two atoms, separated by a distance d. Let the

incoming momentum bek, while the outgoing momentum is k. Constructive interference will

occur whend (k k) = d q= 2n, for some integern. Taking the entire crystal into account,

constructive interference will occur when Xi q= 2n, for allXi. (This is simply a fancy way

of defining a Bragg plane.) Now, if the scattered wave satisfies the conditionXi q = 2n,

theneiqXi = 1, so that dd N2, whereNis the number of atoms in the crystal; otherwise,

generically, the interference between scattered wave will lead to dd 0. Thus, scattering

amplitudes are only large when the Bragg condition is satisfied.

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