PHY4605–Introduction to Quantum Mechanics II
Spring 2004
Test 3 Solutions
April 16, 2004
1. Short Answer. Must attempt (only) 4 of 6. Circle answers to be graded.
(a) Using the figure sketching the wave
function of a particle with the scattering potential a) turned off and
b) turned on, state whether the potential was most likely attractive or
repulsive, and estimate (quantitatively!) the cross-section of the scattering process in the s-wave approximation in terms of the wave vector
k of the scattering particle.
Potential must be attractive in lower figure since wave function has been
“pulled in” to well (δ0 > 0). From the figure, δ0 ' π/4, so
dσ
sin2 δ0
4π sin2 δ0
2π
=
→
σ
=
=
dΩ
k2
k2
k2
(b) Explain qualitatively the difference between classical hard sphere cross
section and quantum mechanical low-energy cross section for hard spheres.
σcl = πa2 , σQM = 4πa2 . The quantum result is identical to the result
obtained for classical waves, i.e. it accounts for diffraction of waves around
the sphere into the shadow region behind the sphere. If you prefer to think
of a beam of electrons hitting the hard sphere, then quantum mechanically
the wave packets can scatter from the sphere when their impact parameter
is greater than a.
(c) Optical theorem. State the optical theorem and discuss.
σ = 4π
Imf (0). Incident plane wave brings in probability current density
k
in z direction. Some of it gets scattered into various directions. Must give
rise to decrease in current density behind target (θ = 0) due to destructive
interference of incident plane wave and scattered wave in forward direction.
So total flux scattered (vσ) related to forward scattering amplitude f (0).
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(d) Fermi Golden rule. A beam of protons is scatters off a target containing
neutrons. The scattering amplitude for p − n scattering is modelled in a
certain energy range by
f (θ) = χ†f (A + Bσp · σn )χi
(1)
where χf and χi are the final and initial spin states of the n − p system.
Assume the initial (probe) proton spin is up, the inital (target) neutron
spin is unknown, and only the final proton spin is measured. Use the
Fermi Golden rule to find the ratio of the rates of spin up and spin down
measurements.
First note that from the definition of a cross section, the rate of scattering
is τ1 = σv. What we want is
σ↑
f↑
1/τ↑
=
= | |2 ,
1/τ↓
σ↓
f↓
where 1/τ↑ is the rate of detecting scattered ↑ particles, etc. Now the
initial state is | ↑ mn i (proton first), and final state is hm0p m0n |, where
1/2 + mn = m0p + m0n by a.m. conservation.
So for final proton spin up we have
|f↑ |2 ∝ |h↑↑ |V | ↑↑i|2 + |h↑↓ |V | ↑↓i|2
= |A + B|2 + |A − B|2 ,
whereas for final proton spin down we get
|f↓ |2 ∝ |h↓↑ |V | ↑↓i|2
= |2B|2 ,
where matrix elements are easy to calculate using Sp · Sn = (Spz Snz +
2[Sp+ Sn− + Sp− Sn+ ]). So ratio of transition rates is (|A + B|2 + |A −
B|2 )/4|B|2 .
(e) Spontaneous emission. A hydrogen atom in an excited state in free space
decays after a certain time of order 10−9 sec. typically. How can this be?
After all, the atom is in an eigenstate, which is a stationary state of the
Hamiltonian, i.e. the time evolution of the state does not mix in any other
eigenstates. Discuss this paradox.
The paradox arises because the excited atom is in free space which, if it
were truly empty, would mean that the atom would be in an eigenstate of
the full Hamiltonian, i.e. it would have an infinite lifetime. Electromagnetic field turns out to have zero-point oscillations, however, constantly
interacting with atom even in ground state. Proper answer requires quantum electrodynamics, which calls these processes vacuum polarizations.
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(f) Electric dipole radiation selection rule. State the selection rule for ` derived
in the homework for an electric dipole transition in a spherically symmetric
atom, and give a hand-waving explanation.
Under the influence of electromagnetic radiation coupling to the dipole
moment of the atom, the atom can only change its state by one angular
momentum unit, ∆` = ±1. This corresponds to absorption or emission of
one photon, which carries angular momentum 1h̄.
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2. Spin exchange. The Hamiltonian which describes the interaction of a proton
and a neutron interacting by their dipolar moments (assume equal), is (for this
problem!)
H0 = J S~p · S~n
where J is a constant.
(a) What are the eigenfunctions and eigenvalues of H0 ? Give your answer for
the eigenfunctions explicitly in terms of single particle Ŝz eigenstates like
| ↑↓i, where the 1st spin is the proton and the 2nd the neutron.
~=S
~p + S
~n , just the usual
Want eigenfunctions of total spin S
|S = 1, ms i =








| ↑↑i
√
| ↑↓ + ↓↑i/ 2







| ↓↓i
√
|S = 0, 0i = | ↑↓ − ↓↑i/ 2
Now use favorite identity
~p S
~n = 1 (S
~2 − S
~2 − S
~ 2)
S
p
n
2
1 1
1 1
= (S(S + 1) − ( + 1) − ( + 1))
2 2
2 2



 1
=
2
S=1

 − 3 S = 0,
2
so eigenvalues are J/2 (triplet), -3J/2 (singlet).
At t = 0 the system is in its ground state, and an external magnetic field in the
ẑ direction, B(t) = (B0 cos ωt)ẑ is turned on.
(b) Write a Hamiltonian which includes the magnetic field coupling to the
spins. Assume the magnetic moment operators for the particles are ~µˆp =
gp S and ~µˆn = gn S, where gp and gn are unequal constants.
~ − ~µn · B
~
H = H0 − ~µp · B
~p · B
~ − gn S
~n · B
~
= H0 − gp S
= H0 − B0 cos ωt(gp Spz + gn Snz )
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(c) Calculate perturbatively the probability for the system to make a transition
to the upper state after a time T .
For sinusoidal perturbations,
Pi→f '
|Vf i |2 sin2 (ω − ω0 )T /2
,
(ω − ω0 )2
h̄2
where Vf i = hf |V0 |ii, and V0 is the t-independent part of the potential, i.e.
B0 (gp Spz + gn Snz ). Here h̄ω0 = Ef − Ei = J/4 − (−3J/4) = J. We need
(1st spin is proton, 2nd neutron):
√
hS = 1, ms |gp Spz + gn Snz |S = 0, 0i = hS = 1, ms |gp Spz + gn Snz | ↑↓ − ↓↑i/ 2
#
"
h̄
h̄
= hS = 1, ms | gp √ − gn √ | ↑↓ + ↓↑i
2 2
2 2
#
"
h̄
h̄
= hS = 1, ms | gp − gn |S = 1, ms = 0i
2
2
h̄
= (gp − gn ) δms =0
2
⇒ Pi→f =
(gp − gn )2 2 sin2 (ω − ω0 )T /2
B0
4
(ω − ω0 )2
(d) Sketch the probability of a transition as a function of time for fixed ω.
Discuss with this picture the criterion for the breakdown of perturbation
theory.
Central peak has height |Vf i t/2h̄|2 and width 4π/t, getting higher and
narrower as time goes on (see fig.) Recall this is perturbative treatment,
however: can’t get bigger than 1, so perturbation theory breaks down
eventually.
3. Hard sphere. Particles are scattered by a potential



 V0 r < r0
V (r) 

 0 r > r0
with V0 > 0.
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(2)
(a) Find the differential cross section in Born approximation.
Z r0
m Z 3 0
m Z1
0 i~
q ·~
r0
d
cos
θ
dr0 r02 V0 eiqr cos θ
d
r
V
(r
)e
=
−
0
2πh̄2
2πh̄2 −1
2mV0 Z r0 0 0
m Z r0 0 02 2 sin qr0
=−
dr r
=− 2
dr r sin qr0
qr0
2πh̄2 0
h̄ q 0
2mV0
= − 2 3 (sin qr0 − qr0 cos qr0 ),
h̄ q
f =−
and dσ/dΩ = |f |2 .
(b) What is the limit of your result when the particles are very slow?
0
Very slow means |q| → 0, so ei~q·~r ' 1. So you can take limit as q → 0 of
part a), or just say
m Z r0
2mV0 r02
02
0
f =−
4πr
dr
V
=
−
0
2πh̄2 0
3h̄2
2 2 6
dσ
4m V0 r0
=
⇒
dΩ
9h̄4
(c) For this part, take dσ/dΩ = 10−12 cm2 . If 1016 particles per cm2 per sec are
incident on this target at slow speeds, how many per second are detected
in a 1 cm2 detector located at 90◦ from the incident direction 1m (À r0 )
from the target?
dN
dσ
= (nv) ·
· δΩ
dt
dΩ
' 1016 · 10−12 · 10−4 = 1 particle/sec,
where I took δΩ ' 10−4 since the detector area 1 cm2 was approx. 10−4
of the total surface area at a distance of 100 cm. Since the particles are
slow, the scattering angle is irrelevant.
(d) Let V0 → ∞. Write down and solve the radial Schrödinger equation for
the radial wave fctn. u0 = rψ|`=0 in the s-wave scattering approximation.
Find the s-wave scattering phase shift δ0 (k), and use it to calculate the
differential scattering cross section. Compare with your answer in b) and
explain the difference.
Radial Schrödinger eqn. for radial function u = rψ (remember just like
1D eqn. for r):
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−
h̄2 ∂ 2
h̄2 k 2
u
=
u0
0
2m ∂r2
2m
Outside the scattering potential, the solutions are just sums of 1D plane
waves, e.g.
u0 = Aeikr + Be−ikr = rψ0 .
Boundary conditions u0 (r0 ) = 0 since V0 → ∞, gives you −A/B =
exp(−2ikr0 ). and now recall asymptotic ψ0 in scattering form:
1
ψ'
2i
Ã
!
−e−ikr
eikr
+ η0
,
kr
kr
and compare the 2 forms of ψ0 you have, giving
η0 = −
δ0 = −kr0 ,
A
= e−2ikr0
B
where I used the definition of η0 = e2iδ0 . The differential scattering cross
section is then (compare with form of plane wave & identify coefficients of
outgoing wave) dσ/dΩ = sin2 kr0 /k 2 .
Difference with b) has to do that, although both parts assume particle is
slow (` = 0), in part b) we had a weak potential to apply Born approx.,
whereas in part d) we assumed potential went to ∞. So we don’t expect
the same answer.
4. Scattering from molecular hydrogen. A light, neutral spinless probe particle of wave vector k scatters weakly off protons. The asymptotic (r → ∞)
wave function has the form
ψ(r) = e−k·r + a
eikr
,
r
(3)
where a is a constant.
(a) Find the differential scattering cross section and the total scattering cross
section for a single hydrogen atom in this approximation.
The coefficient of the outgoing wave eikr /r is a, but this is also the definition
of the scattering amplitude f . Therefore
|f |2 = dσ/dΩ = a2 . Since a is
R
constant the total cross section is σ = dΩa2 = 4πa2 .
(b) A hydrogen molecule now consists of two protons separated by a displacement d, which is unaltered by the scattering event. For small k, the scattering is not exactly isotropic. Find the differential cross section for an
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angle α between k and d, as a function of the scattering angle θ, assuming
that the scattering plane contains d.
Scattered ψ will be a superposition of spherical waves:
Ã
ψscatt
eikr1 eikr2
+
∼a
r1
r2
!
,
where r1 and r2 are the distances from each H atom to the detection point
r.
r
r1
d/2
d/2
α
θ
k
r2
q
Geometry: r 1 = r2 ∓ r · d + O(d2 ) ' r(1 ∓ (r · d/(2r2 )), where r · d =
2
rd cos(α − θ). So
Ã
ψscatt
!
r·d
r·d
eikr −ikr̂·d/2
e
(1 +
) + eikr̂·d/2 (1 −
)
a
2
r
2r
2r2
Ã
!
eikr
kd
id
kd
2a
cos( cos(α − θ)) −
sin( cos(α − θ))
r
2
2r
2
=
=
2a
'
eikr
kd
cos( cos(α − θ)).
r
2
r→∞
So scattering amplitude f ∝ cos( kd
cos(α − θ)), dσ/dΩ = |f |2 .
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(c) Find the ` = 0 and ` = 1 phase shifts δ0 and δ1 in the low energy limit
k → 0. If you could not do part b), discuss how you would go about this
calculation, beginning from an arbitrary scattering amplitude f (θ, φ).
Low energy limit
!
Ã
k 2 d2
cos2 (α − θ)
f 'a 1−
4
Ã
!
k 2 d2
2
=a 1−
[cos α cos θ + sin α sin θ]
4
In notes we derived
f = −iπ 1/2
X
P` (cos θ)(2` + 1)
`
8
(η` − 1)
k
Use orthonormality of Legendre polynomials:
to project out:
2iδ`
η` − 1 = e
R1
−1
P (x)P`0 (x) = δ``0
dx dx
2 `
Z 1
ik
dx
− 1 = 1/2
f (x)P` (x)
π (2` + 1) −1 2
For ` = 0, 1, P` = 1, cos θ, so I find
ika 1
(24 − 3d2 k 2 + d2 k 2 cos 2α)
π 1/2 24
−ik 3 d2 aπ 1/2
η1 − 1 =
sin 2α
64
η0 − 1 =
(d) Find the total cross section for fixed α, either in terms of α, if you did part
b) or in terms of the δ` .
Z
σ=
σ=
Z
dΩf 2 = 2πa
dθ sin θ cos2 (
4π X
(2` + 1) sin2 δ`
k2 `
9
kd
cos(α − θ)) , or
2