PHY4605–Introduction to Quantum Mechanics II
Spring 2005
Test 3 SOLUTIONS
April 15, 2005
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
=
= 2
2
2
dΩ
k
k
k
(b) State the electric dipole selection
rules. What are the most likely decay paths to the ground state for a)
an H atom starting in a 3d state?
Draw them on the picture, and explain why the other possibilities are
“forbidden”. Do the same starting
from b) the 2s state.
All electric dipole transitions should have ∆` = ±1, and ∆m = ±1, 0.
Therefore the only state to which the 322 state can decay is the 211. The
200 state is infinitely long lived in this approximation since it has nowhere
to decay to which satisfies the selection rule.
(c) Describe the three kinds of radiation processes related statistically by Einstein’s detailed balance argument. Discuss the conclusions of this argument, or alternatively state how any two of the rates in the argument are
related.
1
Einstein’s argument relates the rates of spontaneous emission, stimulated
emission, and absorption of radiation of a population of atoms in equilibrium with a bath of photons with frequency ω0 . You were supposed to
say what each process is. The rate of spontaneous emission is N2 A, rate
of stimulated emission is BnN2 , and absorption is nN1 C, where N2 , 1 is
population of upper, lower state and n the number of photons in the cavity.
Einstein showed that A = B = C.
(d) State the physical meaning of the optical theorem.
Optical theorem: σ = Im(4πf (0))/k.
The amount of flux scattered out of the incoming beam is σvcl . The forward part of the outgoing wave must be diminished by this amount, by
conservation of probability.
(e) State the Fermi Golden rule. Be sure to identify each quantity occurring
in the expression you give.
Rf =
dPf
dt
'
Pf
t
=
π|hf |V̂0 |ii|2
ρ(ω0 )
2h̄2
(1)
In other words, the rate of making transitions from state i → f is proportional to the matrix element of the perturbing potential V squared, times
the density of states at an energy h̄ω0 equal to the difference between the
two levels Ef − Ei . This form of the Fermi Golden Rule is for a bunch
of two-level atoms exposed to incoherent radiation characterized by some
distribution of energies ρ(ω), e.g. the distribution of radiation in a cavity.
(f) Explain why the quantum total cross-section σ for hard-sphere scattering
is 4 times the geometrical cross-section.
Like the classical cross section for low-energy classical waves, the QM crosssection for particles scattering from a hard sphere is a factor of 4 larger
than the classical cross-section for pariticles πa2 , since in QM the particles
can “diffract” around the sphere, filling in the classical shadow zone and
leading to enhanced scattering.
2
2. Precessing spin. The Hamiltonian for a spin 1/2 particle in a uniform mag~ where B = B0 ẑ and ~σ is
netic field B = B0 ẑ may be written H = −µ~σ · B
the Pauli spin vector. At time t = 0 the spin is along the x̂ direction (an
eigenfunction of σx with eigenvalue +1).
(a) Solve the time dependent Schrödinger equation to find the spin wave function at time t.
 
√ 1
State at t = 0 is | ↑x i ≡ (1/ 2)  . Time evolution operator is
1
e−iHt/h̄ = eiµσz B/h̄ = cos ωt + iσz sin ωt,
where ω = µB/h̄. So at time t state is

 


1 
1
 1 
|χi = √ cos ωt   + i sin ωt 
 .
2
1
−1
(b) Find the expectation value of σx at time t. What is the spin precession
frequency?
Hint: The Pauli matrices are:




 
1 0 
0 −i
0 1
σz = 
σy = 
σx =  


0 −1
i 0
1 0
 
1
0 1
hχ|σx |χi = √ (cos ωt (1 1) − i sin ωt (1 − 1))  
2
1 0



 
1 
 1 
1
× √ cos ωt   + i sin ωt 
 = cos 2ωt
2
1
−1
so precession frequency is 2ω = 2µB/h̄.
(c) The field is now changed to
Bx = B1 cos ωt
By = B1 sin ωt
Bz = const. À B1
3
What value of ω gives resonant transitions (takes an up spin and converts
it to a down spin with maximum probability)?
The large z-component of the field makes the energy levels of the spin
to zeroth order just µBz . Resonant absorption will take place when the
frequency of the time-varying magnetic field matches the energy difference
between the up and down spin configurations, i.e. h̄ω = 2µBz .
(d) At t=0 all the protons are polarized in the state | ↑i. What is the probability for a proton at time t to have spin in the −z direction?
From t-dependent perturbation theory, the amplitude for a spin to be in
the down state at some later time is
Z
1 t 0
c↓ '
dt h↓ | − µx B1 cos ωt0 − µy B1 sin ωt0 | ↑i × ei(E↓ −E↑ )t/h̄
ih̄ 0
Now use µ = (g/2mp c)S = (g/4mp c)σ, and matrix elements h↓ |σx | ↑i = 1
and h↓ |σy | ↑i = i to find
Z
−1 g B1 t 0 i(ω+ω0 )t
c↓ =
dt e
ih̄ 2mp c 2 0
−gB1 ei(ω+ω0 )t/h̄ − 1
,
=
ih̄4mp c i(ω + ω0 )
where ω0 = (E↓ − E↑ )/h̄. Probability to find proton spin down is then
|c↓ (t)|2 .
4
3. Born approximation. A particle of mass m is scattered by a potential V (r) =
V0 exp(−r/a).
(a) Find the differential scattering cross section in the first Born approximation.
Z
m
0
d3 r0 V (r0 )eiq·r
f (θ) = −
2
2πh̄
Z π
Z
mV0
0
02
−r0 /a
=−
r dr e
dθ sin θ eiqr cos θ
2 (2π)
2πh̄
0
Z
2mV0 ∞ 0 0
0
=− 2
dr r sin(qr0 )e−r /a
h̄ q 0
4mV0 a3
=− 2
,
h̄ (1 + q 2 a2 )2
where q = k|k̂ − r̂| = 2k sin(θ/2). So
dσ
16m2 V02 a6
= |f |2 = 4
dΩ
h̄ (1 + 4k 2 a2 sin2 θ/2)4
(b) Sketch the angular dependence for small and large k, where k is the wave
number of the particle being scattered. At what k value does the scattering
start to become significantly non-isotropic (estimate)?
If ka ¿ 1, the sin θ/2 in the denominator won’t matter. So roughly speaking if k ¿ a−1 , there will be no angular dependence of the cross section.
One could have guessed this without doing the calculation for part a), because as we discussed, the scattering is always isotropic when k is much
smaller than the inverse range of the scattering (only length in the problem).
5
(c) The criterion for the validity of the first Born approximation is that the
1st-order correction to the incident plane wave be small, |ψ (1) | ¿ |eik·r |.
Using the integral form of the Schrödinger equation, translate this condition into a condition on V0 , and simplify it for small k, ka ¿ 1. Hint:
you may compare the incoming wave function and the first Born scattering
correction to it at r = 0.
Integral form of the S.-eqn is (see helpful formulae)
m
ψ(r) = ψ0 (r) −
2πh̄2
Z
0
eik|r−r |
V (r0 )ψ(r0 )d3 r0 ,
|r − r0 |
and the first Born correction is obtained by inserting ψ0 = eik·r for ψ, so
¯ ¯
¯
¯
Z ikr0
¯ ∆ψ (1) (0) ¯ ¯ V0 m
¯
e
0
0
−r /a ik·r 3 0 ¯
¯ =¯
¯
e
e
d
r
¯ ψ (0) (0) ¯ ¯ 2πh̄2
¯
r0
¯
¯
Z ∞
Z π
¯ V0 m
¯
0
0
0
0 ikr 0 −r0 /a ik·r0 ¯
¯
=¯ 2
r dr
sin θ dθ e e
e ¯
h̄
¯
¯
Z0
Z0
¯ V0 m ∞ 0 0 π
¯
0
0 ikr 0 −r0 /a+ikr 0 cos θ0 ¯
¯
=¯ 2
r dr
sin θ dθ e
¯
h̄
0
0
¯
µ
¶¯
Z
2ikr 0
¯ V0 m ∞ 0 0
− 1 ¯¯
−r0 /a e
¯
=¯ 2
r dr −ie
¯
kr0
h̄
0
2m|V0 |a2
,
= 2√
h̄ 1 + 4k 2 a2
so the criterion that the Born approximation be valid is
2m|V0 |a2
√
¿1
h̄2 1 + 4k 2 a2
and in the limit ka ¿ 1,
|V0 | ¿
h̄2
2ma2
(d) What is the limit on V0 which comes from this criterion in the high-k limit,
i.e. ka >> 1?
Compare with answer to c). In limit ka À 1, find |V0 | ¿ h̄2 k/(ma) =
h̄2 (ka)/(ma2 ). So that if ka À 1, the Born approximation is actually valid
over a much larger regime.
6