Chapter 10: QUANTUM SCATTERING
Scattering is an extremely important tool to investigate particle structures and the interaction
between the target particle and the scattering particle. For example, Rutherford scattering finds
that atoms have small and massive nuclei, surrounded by tiny and light electrons. High-energy
photons can be scattered by a neutron, revealing its inner structure formed by quarks. Quantum
scattering is a very extensive subject, and we will study the basic ideas. We have briefly
discussed photon scattering, now our focus is on matter particles, such as electrons and neutrons,
within the formal and exact framework, though approximations will be taken as well in practice.
10.1 Cross Sections
Goal: finding the differential cross section dσ / dΩ of an incident particle beam with definite
momentum scattered by an object described by a (localized) potential.
Simplification: A steady particle beam with definite momentum can be described by a
planewave, and we can only consider one particle. The results for a current density of a particle
beam can be obtained readily.
Let’s begin with one dimension (1D).
⎧⎪
ikx
−ikx
⎪ Ae + Be
ψk → ⎪⎨
⎪⎪ Ce ikx + De −ikx
⎪⎩
(x → −∞)
(x → +∞)
(10.1)
(1) D = 0 .
(2) The reflection rate R =| B |2 / | A |2 .
Fig. 10.1: One-dimensional scattering.
(3) The transition rate T =|C |2 / | A |2 .
Both R and T are a function of k, therefore
(4) Theoretically, if we know V(x) , we can calculate
R(k) and T(k) .
(5) Experimentally, we can measure R(k) and T(k)
to gain knowledge of V(x) .
Now let’s look at the 3D, which shares many
features with the 1D case. However, it is much more
Fig. 10.2: 3D scattering.
complicated due to angle dependence. The differential scattering cross section is
167
dσ
number of particles scattered in dΩ/sec
.
=
dΩ number of incident particles/sec/area ×dΩ
(10.2)
The following is the derivation of dσ / dΩ based on definition of Eq. (10.2). Neglecting the
trivial normalization factor, the total wave function is
ψk = e ikz + ψsc (r, θ,φ) ,
(10.3)
where the incident wave is e ikz and the scattered wave is ψsc (r, θ,φ) . When r → ∞ , assume
rV(r) → 0 to guarantee a non-divergent solution of ψsc :
(∇2 + k 2 )ψsc = 0,
for r → ∞ .
(10.4)
The general solution for the free-particle equation in spherical coordinates:
ψsc (r, θ,φ)r→∞ = ∑ ∑ [Al jl (kr) + Blnl (kr)]Yl m (θ,φ) ,
l
(10.5)
m
where Yl m is the spherical harmonics, jl and nl the spherical Bessel and spherical Neumann
functions, respectively. The asymptotic behaviors of jl and nl :
jl (kr) →r→∞
sin(kr −lπ / 2) / kr ,
(10.6)
nl (kr) →r→∞ −cos(kr −lπ / 2) / kr .
(10.7)
To get a purely outgoing wave for ψsc = e ikr / r , Al and Bl must satisfy Al / Bl = −i , then
ψsc (r, θ,φ)r→∞ =
e ikr
kr
∑ ∑ (−i)l (−Bl )Yl m (θ,φ) =
l
m
ψk →r→∞ e ikz + f (θ,φ)
where the scattering amplitude
e ikr
r
e ikr
f (θ,φ)
r
(10.8)
(10.9)
1
(10.10)
∑ ∑ (−i)l (−Bl )Yl m (θ,φ) .
k l m
The probability flux (current density) for a particle with mass m and wave function ψ :
f (θ,φ) =
j(r,t) =

ψ*∇ψ − ψ∇ψ* ,
2im
(
)
(10.11)
which satisfies the continuity equation
∂ρ
+ ∇ ⋅ j = 0.
∂t
168
(10.12)
2
Here ρ(r,t) = ψ(r,t) is probability density, and Eq. (10.12) is derived from Schrödinger
equation.
In polar coordinate systems,
̂ ∂
̂ 1
∂
∂
+θ
+φ
.
∂r
r ∂θ
r sin θ ∂φ
(10.13)

k
.
e −ikz ∇e ikr cos θ −e ikz ∇e −ikr cos θ =
2im
m
(10.14)

ψsc*∇ψsc − ψsc∇ψsc* .
2im
(10.15)
∇ = r̂
The magnitude of the incident flux:
j inc =
(
)
The scattered flux:
jsc =
(
)
̂
̂
1
At the limit of r → ∞ , the derivatives along θ and φ directions behave as ~ 2 (Q: why?),
r
while the derivative along r̂ is
∂
∂r
Dropping the ~
ikr ⎤
ikr
ikr
⎡
⎢ f (θ,φ) e ⎥ = f (θ,φ)ik e + f (θ,φ)(−1) e .
⎢
r ⎥⎥⎦
r
r2
⎢⎣
(10.16)
1
terms at the limit of r → ∞ , we obtain
r2
jsc =
k | f (θ,φ) |2
r̂ .
m
r2
(10.17)
The probability (proportional to number of particles) flows into solid angle dΩ at the rate
R(dΩ) = j sc ⋅r 2dΩ ,
(10.18)
The incident flux j inc is proportional to the number of particles/sec/area, thus
dσ R(dΩ)
=
dΩ
j incdΩ
⇒
dσ
=| f (θ,φ) |2
dΩ
(10.19)
Notes:
(1) f (θ,φ) is also written as f (k, k′) , with k and k′ the incident and scattered wave vectors.
(2) We have considered elastic scattering, i.e., k ′ = k ; in general,
169
dσ k ′
= | f (k, k′) |2 .
dΩ k
(3) We only study the elastic scattering in this class.
Methods to compute the scattering amplitude f (θ,φ) :
(1) Time-dependent perturbation theory
(2) Time-independent perturbation theory
(3) Exact theory
10.2 Formal Theory and the Born Series
Up to a normalization parameter, the wave function of a free particle is φk = e ik⋅r , which satisfies
(∇2 + k 2 )φk = 0 .
(10.20)
Under full potential V(r) a particle’s wave function ψk can be obtained by solving
(∇2 + k 2 )ψk (r) =
2m
V(r)ψk (r) .
2
We write the total wave function in the form: ψk = e ik⋅r + ψsc , with eigenenergy E k =
(10.21)
 2k 2
,
2m
but momentum is not well defined. Here k is the wave vector of the incident wave ψinc = e ik⋅r ,
while we have derived that the scattered wave ψsc →r→∞ f (θ,φ)
e ikr
.
r
We need to solve the full potential Schrodinger equation [Eq. (10.21)] to obtain f (θ,φ) . We
use Green’s function G0(k; r, r′) to find the formally exact solution, which satisfies
(∇2 + k 2 )G0(k; r, r′) =
2m
δ(r − r′) .
2
(10.22)
The exact solution to Eq. (10.21) can be obtained by solving the Lippman-Schwinger equation:
ψk (r) = e ik⋅r + ∫ G0(k; r, r′)V(r′)ψk (r′)dr′ .
(10.23)
As usual, we can solve it iteratively starting from the zeroth order:
Zero-th order (for zero potential V): ψk(0) = e ik⋅r .
First-order Born approximation:
′
ψk(1)(r) = e ik⋅r + ∫ G0(k; r, r′)V(r′)e ik⋅r dr′ .
170
(10.24)
You might see the above expression in symbolic form in literature:
ψ(1) = ψ(0) +G0V ψ(0) .
(10.25)
ψ = ψ(0) +G0V ψ(0) +G0VG0V ψ(0) +  .
(10.26)
The Born Series for exact solution:
In this class, we focus on the first-order Born approximation, which requires an explicit form
to Green’s function G0(k; r − r′) , which can be obtained using contour integral:
′
e ik|r−r | m
.
| r − r′ | 2π 2
(10.27)
e ik|r−r |
V(r′)ψk (r′)dr′ ,
| r − r′ |
(10.28)
G0(k; r, r′) = G0(k; r − r′) = −
Then
ψk (r) = e
ik⋅r
m
−
2π 2
∫
′
and ψk (r) = e ik⋅r + ψsc (r) .
Now we derive the formally exact f (θ,φ) for ψsc →r→∞
e ikr
f (θ,φ)
. When r → ∞ ,
r
| r − r′ |→ r(1 − r ⋅ r′ / r 2 ),
(10.29)
1
1⎛
r ⋅ r′ ⎞
→ ⎜⎜⎜1 + 2 ⎟⎟⎟ ,
| r − r′ |
r⎝
r ⎟⎠
(10.30)
⎛
r ⋅ r′ ⎞
k | r − r′ |→ kr ⎜⎜⎜1 − 2 ⎟⎟⎟ = kr − k′ ⋅ r′ ,
r ⎟⎠
⎝
(10.31)
then
where the scattered wave vector k′ = kr̂ for elastic scattering. Thus,
′
e ik|r−r |
e ikr −ik′⋅r′
,
→
e
| r − r′ |
r
ψsc (r) →r→∞ −
e ikr m
r 2π 2
∫e
−ik′⋅r′
V(r′)ψk (r′)d 3r′ ,
(10.32)
(10.33)
and finally we obtain the exact expression for scattering magnitude:
f (θ,φ) = f (k, k′) = −
m
2π 2
171
∫e
−ik′⋅r′
V(r′)ψk (r′)d 3r′
(10.34)
Note that
∫e
−ik′⋅r′
V(r′)ψk (r′)d 3r′ = (2π)3 k′ V ψk
(10.35)
10.3 T-Matrix
Define the T-matrix as
! k′) ≡
T(k,
1
(2π)3
∫e
−ik′⋅r
V(r)ψk (r)d 3r ,
(10.36)
so that
2
2
dσ ⎛⎜ 4π 2m ⎞⎟⎟ "
= ⎜⎜ 2 ⎟ T(k, k′) .
dΩ ⎜⎝ ! ⎟⎠
(10.37)
The Born series ψ = ψ(0) +G0V ψ(0) +G0VG0V ψ(0) +  can be employed to obtain the T-matrix.
Under the (1st-order) Born approximation:
T!(1)(k, k′) =
1
(2π)3
∫e
−ik′⋅r
V(r)e ik⋅rd 3r ,
(10.38)
which is essentially the Fourier transformation of potential:
! − k′) =
V(k
1
(2π)3
∫e
−ik′⋅r
V(r)e ik⋅rd 3r .
(10.39)
Let
 k′) = T(k
 − k′) = e −ik′⋅rT(r)e ik⋅rd 3r ,
T(k,
∫
(10.40)
combining with the definition of T-matrix [Eq. (10.36)], we obtain
T(r)φk (r) =V(r)ψk (r) .
(10.41)
 − k′) . The expansion (Born
Obviously, T (1)(r) =V(r) and then T(1)(k, k′) = T(1)(k − k′) = V(k
series) of T is similar to that of ψ , as we expect.
Recast this problem more formally: H = H 0 +V , and
H 0φk = εkφk
(10.42)
with H 0 = p 2 / 2m , εk =  2k 2 / 2m , and φk (r) = e ik⋅r . The full-potential Schrödinger equation:
H ψk = E k ψk ,
172
(10.43)
and ψk = φk + φsc . For elastic scattering, E k = εk =
(E k − H 0 )ψk =V ψk
 2k 2
. Eq. (10.43) ⇒
2m
⇒
(E k − H 0 )ψsc =V ψk ,
(10.44)
because (E k − H 0 )φk = 0 . Operating (E k − H 0 )−1 on both sides of Eq. (10.44):
ψsc = (E k − H 0 )−1V ψk ,
(10.45)
we obtain the Lippman-Schwinger equation:
ψk = φk + (E k − H 0 )−1V ψk .
(10.46)
We introduce a Green’s function defined as
G0 = GH (E) = lim η→0(E − H 0 + iη)−1 .
(10.47)
0
Here H 0 is an operator, and positive η ensures that the scatted wave ψsc is outgoing. Thus
ψk (r) = φk (r) +G0V(r)ψk (r) .
(10.48)
Using Eq. (10.41), i.e., T(r)φk (r) =V(r)ψk (r) , we derive the T-matrix equation (Dyson’s Eq.):
T(r) =V(r) +V(r)G0T(r) ,
(10.49)
T =V +VG0V +VG0VG0V + 
(10.50)
which can be solved iteratively:
Define
G = GH (E) = lim η→0
G0
1
1
=
=
E − H + iη [E − H 0 + iη](1 −G0V ) 1 −G0V
(10.51)
= G0 +G0VG0 +G0VG0VG0 + 
Then
T =V +VGV .
(10.52)
We have derived the Born series for T and ψ using two slightly different approaches, but they
essentially the same. One can verify that
G0(k; r, r′) = r
Alternatively,
′
1
1
m e ik|r−r |
r′ = r 2 2
r′ = −
. (10.53)
E − H 0 + iη
k
2π 2 | r − r′ |
− H 0 + iη
2m
173
G0(k; r, r′) = ∫
φk′(r)φk*′(r′)
εk − εk′ + iη
d 3k ′ =
1
′
′
e ik ⋅(r−r )d 3k′
∫
εk − H 0 + iη
1
=
δ(r − r′).
εk − H 0 + iη
(10.54)
10.4 Applications and Validity of the First-order Born Approximation
With the dramatic progress in computational power, now we can calculate exact differential
cross sections for any potential. But analytically, first-order Born approximation is still widely
used to reveal the basic physics and estimate the magnitude. In the following, we discuss two
examples using the first-order Born approximation.
To the first order, T (1)(r) =V(r) and then
! − k′)
T!(1)(k, k′) = T!(1)(k − k′) = V(k
=
1
(2π)3
∫e
i(k−k′ )⋅r
V(r)d 3r.
(10.55)
Fig. 10.3: Scattering with
angle .
The scattering magnitude
f (k, k′) = −
(1)
4π 2m "(1)
T (k, k′) ,
!2
(10.56)
and the differential cross section
2
2
dσ ⎛⎜ 4π 2m ⎞⎟⎟ "(1)
= ⎜⎜ 2 ⎟ T (k, k′) .
dΩ ⎜⎝ ! ⎟⎠
(10.57)
As shown in Fig. 9.3, for elastic scattering, T (1)(k, k′) = T (1)(k − k′) = T (1)(q) , and here
q =| k − k′ |= 2k sin(θ / 2) .
(10.58)
For the special case of spherical potential, V(r) =V(r) , T (1) depends only on polar angle θ :
∞
2π
π
1
′
V(r)e iqr cos(θ )r 2 sin(θ′)d θ′ dφdr
3 ∫0 ∫0 ∫0
8π
∞
1
=
∫ rV(r)sin(qr)dr
2π 2q 0
T (1)(θ) =
(10.59)
Example 1: finite square well, a simple model for nuclei.
174
Fig. 10.4: differential cross section.
⎧⎪ V
⎪
V(r) = ⎪⎨ 0
⎪⎪ 0
⎪⎩
(r ≤ a)
(10.60)
(r > a)
We use Eq. (10.59) to find that
(1)
T (θ) =
⎡ sin(qa)
⎤
⎢
⎥,
−
cos(qa)
⎥
2π 2(qa)2 ⎢⎣ qa
⎦
V0a 3
(10.61)
4m 2V02a 2
dσ
=
dΩ
! 4q 4
2
⎡ sin(qa)
⎤
⎢
− cos(qa)⎥ .
⎢ qa
⎥
⎣
⎦
(10.62)
Question: in Fig. 9.4 there are many minima in differential cross section, why?
Example 2: Yukawa potential (Screened Coulomb potential), for massive scalar fields.
V(r) =
V0e −µr
µr
.
(10.63)
Using Eq. (10.59) it is straightforward to obtain
T (1)(θ) =
1 V0
1
,
2
2
16π µ q + µ 2
(10.64)
with (10.61)
q 2 = 4k 2 sin 2(θ / 2) = 2k 2(1 − cos θ) .
(10.65)
In the following we discuss the physics involved:
(1) Let µ → 0 but keep V0 / µ a constant, Yukawa potential reduces to Coulomb potential.
If V0 / µ = ZZ ′e 2 ,
2
2
dσ ⎛⎜ 2mZZ ′e 2 ⎞⎟⎟
1
1 ⎛⎜ ZZ ′e 2 ⎞⎟⎟
1
⎜⎜
.
= ⎜⎜
=
⎟⎟
⎟⎟
2
4
2
4
dΩ ⎜⎝ 
16 ⎜⎝ E K ⎟⎠ sin (θ / 2)
⎠ 4k (1 − cos θ)
(10.66)
It recovers the classical Rutherford scattering! The first-Born result is valid in the high-energy
limit, for which classical and quantum results typically agree.
(2) The total cross section:
2
⎛ 2mV ⎞⎟
4π
0⎟
.
σ = ⎜⎜⎜
2 ⎟
2
2
2
⎟
⎝⎜ µ ⎠⎟ µ (µ + 4k )
175
(10.67)
For Coulomb potential, it becomes infinity, which means that all incident probability current will
be scattered, no matter how large the incident cross section is. Therefore, the Coulomb interaction
is an infinite-range interaction, while the Yukawa potential is a finite-range interaction.
Validity: we replace ψk = e ik⋅r + ψsc by just e ik⋅r so that we need to ensure that
ψsc  e ik⋅r = 1 .
(10.68)
The scattered wave has the maximum amplitude at scattering center r = 0 , then ψsc (0)  1 .
m
ψsc (r) = −
2π 2
∫
′
e ik|r−r |
′
V(r′)e ik⋅r dr′ ,
| r − r′ |
(10.69)
Then ψsc (0)  1 ⇒
m
2π 2
∫
e ikr
V(r)e −ik⋅rd 3r  1 .
r
(10.70)
For spherical potential V(r) =V(r) , the validity condition is
2m
 2k
∫e
ikr
sin(kr)V(r)dr  1 .
(10.71)
In the following we consider r within a small scatting region of r0 .
(1) For low energy, kr → 0 , e ikr → 1 , sin(kr) → kr . ψsc (0)  1 ⇒
2m
2
∫ rV(r)dr
 1.
⎧⎪ V
⎪
Using the simplest finite square well model: V(r) = ⎪⎨ 0
⎪⎪ 0
⎪⎩
(10.72)
(r ≤ r0 )
(r > r0 )
,
we obtain the condition for the first-order Born Approximation:
mV0r02
2
 1 ⇔ V0 
(2) For high energy, kr0  1 . e ikr sin(kr) =
2
.
mr02
(10.73)
1 2ikr
(e −1) ; since kr0  1 , the e 2ikr term oscillates
2i
fast in the scattering range, which averages to zero. Therefore, ψsc (0)  1 ⇒
176
m
 2k
∫ V(r)dr
1
(10.74)
we obtain the condition for the first-order Born Approximation:
mV0r0
 2k
 1 ⇒ V0 
 2k
.
mr0
(10.75)
Conclusions: (i) If the Born approximation is good at low energy, it is also good for high energy;
(ii) The Born approximation is normally good for high-energy limit, but it might fail miserably at
the low-energy limit.
10.5 Partial Wave and Phase Shift; Optical Theorem
Motivations: (1) At low energy the first-Born approximation breaks down except for very weak
scattering potential; (2) For the scattered spherical-like wave (especially at low energy), it is of
great interest to study the angular momentum dependency on scattering amplitude.
The incident planewave can be decomposed of angular-momentum states
∞
e ikz = e ikr cos θ = ∑ (2l + 1)i l jl (kr)Pl (cos θ) ,
(10.76)
l=0
where jl is the spherical Bessel function, and Pl the Legendre polynomials,
1/2
⎛ 4π ⎞⎟
⎟ Y 0(θ) .
Pl (cos θ) = ⎜⎜
⎜⎝ 2l + 1 ⎟⎟⎠ l
(10.77)
For spherical potential V(r) =V(r) , f (θ,φ) = f (θ) , and it also depends on k :
∞
f (θ,k) = ∑ (2l + 1)al (k)Pl (cos θ) ,
(10.78)
l=0
where al (k) is the l-th partial wave amplitude, which is the measure of the scattering strength
for the state with the (orbital) angular momentum l . For the low-energy case, l max ≈ kr0 (Q:
why?), so you don’t need to compute many al (k) .
Now the task is to find al (k) for a given potential V(r) . Let’s begin with a free particle,
jl (kr) →r→∞
sin(kr −lπ / 2)
,
kr
Then,
177
(10.79)
1 ⎡ i(kr−lπ/2)
−e −i(kr−lπ/2) ⎤⎥ Pl (cos θ)
⎢⎣e
⎦
2ikr
l=0
⎡e ikr e −i(kr−lπ) ⎤
1 ∞
⎥ P (cos θ) .
=
(2l + 1) ⎢⎢
−
∑
⎥ l
2ik l=0
r
r
⎢⎣
⎥⎦
∞
∑ (2l + 1)i
e ikz →r→∞
l
(10.80)
Here there are one incoming and one outgoing wave for each value of l, and their probability
current densities cancel out, i.e., no net probability flux flows into or comes out of the origin.
When potential V(r) is turned on, the wave function is not e ikz , instead,
∞
ψk (r) = ∑ (2l + 1)i l Rl (kr)Pl (cos θ) ,
(10.81)
l=0
where the radial wave function Rl (kr) should reduce to that of a free-particle at r → ∞ for each
value of l:
Rl (kr) →r→∞
Al sin ⎡⎢kr −lπ / 2 + δl (k)⎤⎥
⎣
⎦ = Al
kr
2ikr
⎡e iδle i(kr−lπ/2) −e −iδle −i(kr−lπ/2) ⎤ ,
⎢⎣
⎥⎦
(10.82)
where δl (k) is the phase shift.
iδ
To keep the incoming wave unchanged, we set Al = e l . Then
ψk (r) →r→∞
= e ikz
⎡ 2iδ e ikr e −i(kr−lπ) ⎤
1 ∞
⎢e l
⎥ P (cos θ)
(2l
+
1)
−
∑
⎢
⎥ l
2ik l=0
r
r
⎢⎣
⎥⎦
⎡∞
⎤ ikr
⎛e 2iδl −1 ⎞⎟
⎟⎟ P (cos θ)⎥ e
+ ⎢⎢ ∑ (2l + 1)⎜⎜⎜
⎥ r
⎜⎝ 2ik ⎟⎟⎠ l
⎢⎣ l=0
⎥⎦
(10.83)
Comparing with the expression for f (θ,k) , i.e., Eq. (10.78), we obtain
al (k) =
e
2iδl (k )
−1 1 iδl
= e sin(δl ) .
2ik
k
(10.84)
The scattering amplitude
f (θ) =
1 ∞
iδ
(2l + 1)e l sin(δl )Pl (cos θ) .
∑
k l=0
The total cross section
σ=∫
dσ
4π ∞
dΩ = ∫ | f (θ) |2 dΩ = 2 ∑ (2l + 1)sin 2(δl ) ,
dΩ
k l=0
using the orthogonality of Legendre polynomials Pl (x) :
178
(10.85)
∫
π
0
Pl (cos θ)Pl ′(cos θ) sin(θ)dθ = δll ′
2
.
2l + 1
(10.86)
On the other hand, Pl (cos 0) = Pl (1) = 1 , then
Im[f (0)] =
1 ∞
∑ (2l + 1)sin2(δl )
k l=0
(10.87)
Thus we have proved the famous optical theorem:
σ=
4π
Im[f (0)] .
k
(10.88)
An alternative proof is based on Lippman-Schwinger equation:
ψk = k +G0V ψk .
(10.89)
Operating k V from the left side on Eq. (10.89):
k V ψk = ⎡⎢ ψk − ψk V †G0† ⎤⎥V ψk = ψk V ψk − ψk V †G0†V ψk .
⎣
⎦
(10.90)
Here the Green function (operator) is
G0† = lim η→0(E − Ĥ 0 −iη)−1
= lim η→0 ∫
+∞
−∞
δ(E − E ′)
dE ′ = iπδ(E − Ĥ 0 ).
E ′ − Ĥ 0 −iη
(10.91)
Since G0† has purely imaginary part, we look at the imaginary parts on both sides of Eq. (10.90):
(
Im k V ψk
) = −π
ψk V †δ(E − Ĥ 0 )V ψk = −π k T †δ(E − Ĥ 0 )T k .
(10.92)
2
2
dσ ⎛⎜ 4π 2m ⎞⎟⎟
We have used Eq. (10.41), i.e., T ψk =V k . Using
= ⎜⎜ 2 ⎟ T(k, k′) , we can derive
dΩ ⎜⎝ ! ⎟⎠
k T †δ(E − Ĥ 0 )T k = k T †δ(E − Ĥ 0 ) ⎡⎢ ∫ k′ k′ d 3k′⎤⎥ T k
⎣
⎦
†
= ∫ k T δ(E − E ′) k′ k′ T k d 3k′
2
= ∫ T(k, k′) δ(E − E ′)d 3k′
2
⎛ !2 ⎞
= ⎜⎜⎜ 2 ⎟⎟⎟
⎜⎝ 4π m ⎟⎠
∫
2
dσ ⎛⎜ ! 2k 2 ! 2k ′2 ⎞⎟⎟ 2
δ⎜
−
⎟k ′ dk ′ dΩ
dΩ ⎜⎜⎝ 2m
2m ⎟⎠
⎛ !2 ⎞
k2
dσ
k! 2
= ⎜⎜⎜ 2 ⎟⎟⎟
dΩ
=
σ.
∫
2
(2π)4 m
⎝⎜ 4π m ⎟⎠ 2! k / 2m dΩ
179
(10.93)
Thus we have proved the optical theorem:
⎛ 4π 2m ⎞⎟
! 2k
k
⎜
⎡
⎤
⎟
Im ⎢ f (θ = 0)⎥ = ⎜⎜− 2 ⎟(−π)
σ=
σ
4
⎟
⎣
⎦ ⎜⎝
4π
! ⎠
(2π) m
Another interesting result is
iδ
e l sin(δl ) = −
2mk
2
∫
∞
0
jl (kr)V(r)Rl (kr)r 2 dr ,
(10.94)
(10.95)
and one can use this to determine phase shift δl .
10.6 Determination of Phase Shifts
Let’s begin with a brief review of radial wave functions for central potential
2
 2 d χl
−
+Veff (r)χl = Eχl ,
2m dr 2
where χl (r) = rRl (r) , and the effective potential Veff (r) =V(r) +
(10.96)
l(l + 1) 2
. For free particles,
2mr 2
Rl (r) = c1 jl (kr) +c2nl (kr) .
(10.97)
Asymptotically, when kr → 0 ,
⎧
⎪
⎪ 0 (l ≠ 0)
jl (kr) → ⎨
; nl (kr) → ∞.
⎪
1 (l = 0)
⎪
⎪
⎩
Thus Rl (kr) = c1 jl (kr).
(10.98)
We want to determine phase shifts δl for a scattering potential V(r) that vanishes for r > r0 .
Outside the scattering center, it is a free particle, but Rl (r) = c1 jl (kr) +c2nl (kr) , since the origin
is excluded. The wave function after scattering:
∞
ψk (r) = ∑ (2l + 1)i l Rl (kr)Pl (cos θ) .
(10.99)
l=0
We use the asymptotic behaviors of ψk , jl and nl to determine c1 and c2 :
ψk (r) →r→∞
⎡ 2iδ e ikr e −i(kr−lπ) ⎤
1 ∞
⎢e l
⎥ P (cos θ) ,
(2l
+
1)
−
∑
⎢
2ik l=0
r
r ⎥⎥⎦ l
⎢⎣
180
(10.100)
sin(kr −lπ / 2)
,
kr
cos(kr −lπ / 2)
nl (kr) →r→∞ −
,
kr
jl (kr) →r→∞
iδ
(10.101)
iδ
we obtain c1 = e l cos(δl ) and c2 = −e l sin(δl ) , so that
iδ
Rl (r) = e l ⎡⎢cos(δl )jl (kr)− sin(δl )nl (kr)⎤⎥ , for r > r0 .
⎣
⎦
(10.102)
We can match the radial wave function Rl (r) at r = r0 to find the phase shift δl .
⎧⎪ ∞,
⎪
Example 1: The Hard Sphere V(r) = ⎪⎨
⎪⎪ 0,
⎪⎩
r ≤ r0
r > r0
.
Boundary condition Rl (r0 ) = 0 ⇒
tan(δl ) =
jl (kr0 )
nl (kr0 )
⎡ j (kr ) ⎤
⇒ δl = tan−1 ⎢⎢ l 0 ⎥⎥ . (10.103)
⎢⎣ nl (kr0 ) ⎥⎦
(1) Low energy limit kr0  1 , the s-wave ( l = 0 )
scattering dominates.
⎡ sin(kr ) / kr ⎤
0
0 ⎥
δ0 = tan−1 ⎢⎢
⎥
⎢⎣ −cos(kr0 ) / kr0 ⎥⎦
= tan−1 ⎡⎢−tan(kr0 )⎤⎥
⎣
⎦
(10.104)
⇒ δ0 = −kr0 . (This is actually exact!)
1 iδ
f0(θ) = e 0 sin(δ0 )P0(cos θ) ,
k
(10.105)
Fig. 10.5: Phase shifts for attractive and
repulsive scattering potentials.
and P0(x) = 1 ⇒
2
1
1 tan (−kr0 )
2
.
= f0(θ) = 2 sin (δl ) = 2
dΩ
k
k 1 + tan 2(−kr0 )
dσ0
At the low-energy limit,
dσ0
dΩ
2
→kr
0
1
(10.106)
r02 , and we obtain a surprising result:
σ0 =
4π 2
sin (δ0 ) →kr 1 4πr02 ,
2
0
k
181
(10.107)
the total cross section is four times the classical results. For other l-values,
jl (x) →x→0
nl (x) →x→0
xl
,
(2l + 1)!!
(2l −1)!!
− −(l+1) .
x
(10.108)
Then
tan(δl ) →kr 1 −
0
(kr0 )2l+1
⇒
(2l + 1)!!(2l −1)!!
2l+1
δl →kr 1 ∝ −(kr0 )
0
(10.109)
.
At low energies, the scattering into the high angular momentum states is negligible.
(2) High-energy limit kr0  1 ,
sin(kr0 −lπ / 2)
jl (kr0 ) →kr →∞
kr0
0
nl (kr0 ) →kr →∞ −
,
cos(kr0 −lπ / 2)
0
kr0
(10.110)
.
Then
tan(δl ) = −tan(kr0 −lπ / 2) ⇒ δl →kr 1 −kr0 + lπ / 2,
0
sin 2(δl ) = sin 2(−kr0 + lπ / 2) ⇒
sin 2(δl ) + sin 2(δl+1 ) = 1,
(10.111)
(10.112)
and we obtain
kr
σ=
4π 0
4π
1
(2l + 1)sin 2(δl ) = 2 (kr0 )2 = 2πr02 .
2 ∑
2
k l=0
k
(10.113)
The total cross section at high-energy limit is twice the classical results.
⎪⎧⎪ V ,
Example 2: The finite Square Well V(r) = ⎪⎨ 0
⎪⎪ 0,
⎪⎩
r ≤ r0
r > r0
⎪⎧⎪ V > 0
and ⎪⎨ 0
⎪⎪ V < 0
⎪⎩ 0
(repulsive)
(attactive)
.
To simplify our discussions, let’s consider only the s-wave scattering.
For r > r0 ,
iδ
R0(r) = e
iδ0
0
⎡cos(δ )j (kr)− sin(δ )n (kr)⎤ = e sin(kr + δ0 ) .
0 0
0
0
⎣⎢
⎦⎥
kr
For r < r0 ,
182
(10.114)
χ0(r) = rR0(r) = A0 sin(κr) ,
where E −V0 =
(10.115)
 2κ 2
. Note: when E <V0, κ = ik ′ with k ′ real. Then sin(κr) = sinh(k ′r) .
2m
Boundary conditions: χ0(r) and its derivative χ0′ (r) are continuous at r = r0 , then
iδ
e 0
sin(kr0 + δ0 ) = A0 sin(κr0 )
k
iδ
e 0 cos(kr0 + δ0 ) = A0κ cos(κr0 )
⇒ tan(kr0 + δ0 ) =
k − κ cot(κr0 )tan(kr0 )
k
tan(κr0 ) ⇒ tan(δ0 ) =
.
κ
κ cot(κr0 ) + k tan(kr0 )
(10.116)
(10.117)
Thus for attractive potential, V0 < 0 , κ > k , δ0 > 0 ; for repulsive potential, V0 > 0 , κ < k ,
δ0 < 0 , as illustrated in Fig. 9.5.
10.7 Ramsauer-Townsend Effect
Let’s focus on the attractive case. For a given k
(E is fixed), when V0 increases, δ0 also increases;
one can adjust the values of V0 and r0 so that
δ0 = π / 2 , resulting in the maximum cross section
for the s-wave scattering: σ0 =
4π
. If V0
k2
increases still, eventually one can make δ0 = π , so
that σ0 = 0 , i.e., the s-wave partial cross section
vanishes. Therefore, in the low-energy limit (
kr0  1 ), strong attractive potential could lead to
perfect transmission without scattering. This is
Fig. 10.6: Electron scattering from rare
gases. Rev. Mod. Phys. 5, 257 (1933).
known as the Ramsauer-Townsend effect, which was first observed for electron scattering by
rare gas atoms such as Ar, Kr and Xe in 1923.
To understand the Ramsauer-Townsend effect , we need to use Levinson’s theorem:
δl (k = 0) = N l π
183
(10.118)
where N l is the number of bound states in the potential with angular momentum l . N l is finite
for a potential of short range. Previously we have omitted N l π in δl when we discuss if the
phase shift is positive and negative.
The phase shift is usually determined by the expression like tan(δl ) = F(k) . We have
encountered the case that F(k = 0) = 0 ; since tan(N π) = 0 , how do we determine N? The
unique answer is obtained by graphing the phase shift as a function of k. For any finite attractive
scattering potential, we demand that this graph has two characteristics: (1) The phase shift
vanishes at very large values of k, and (2) The phase shift is a continuous function of k.
⎪⎧⎪ V ,
For the attractive square well potential V(r) = ⎪⎨ 0
⎪⎪ 0,
⎪⎩
with V0 < 0 . Set k02 =
−2mV0
2
κ 2 = k 2 + k02 . (Note: k 2 =
r ≤ r0
r > r0
, then
2mE
)
2
We have derived
tan(δ0 ) =
k − κ cot(κr0 )tan(kr0 )
κ cot(κr0 ) + k tan(kr0 )
previously, see Eq. (10.117). As
plotted in Fig. 9.7, in case (1), there is
no bound state, N 0 = 0 . But in case
(2), there are one bound state and two
bound states in case (3). In case (3),
Fig. 10.7: The s-wave phase shift
well for three values of
: (1)
for an attractive square
, (2)
, and (3)
, as plotted in solid, long-dash and dashed lines.
the phase shift passes through π ; however, the Ramsauer-Townsend effect occurs only when
kr0 < 1 . Higher value of k0r0 (i.e. higher |V0 | ) is required. For example, kr0 ~ 0.324 for the
experiment of electron scattering by rare gas atoms.
At the low-energy limit:
184
2
⎡
tan(k0r0 ) ⎤⎥
dσ
,
= r02 ⎢⎢1 −
dΩ
k0r0 ⎥⎥⎦
⎢⎣
(10.119)
so that the differential cross section can be either smaller or larger (sometimes much larger) than r02 .
10.8 Resonance Scattering
In atomic, nuclear and particle physics, the scattering for a given partial wave often has a
pronounced peak, which is called the resonance scattering. This is because the phase shift δl
passes through π / 2 or (n + 1 / 2)π , then σl = 4π / k 2 , the maximum value of σl for a given k .
The following is an example to show that resonance condition is equivalent to the boundstate condition. Therefore if the potential is attractive enough to form a bound state, then there is
a strong possibility that there sill also be resonance scattering.
⎧⎪ V ,
⎪
Let’s consider the attractive square well potential V(r) = ⎪⎨ 0
⎪⎪ 0,
⎪⎩
r ≤ r0
r > r0
with V0 < 0 .
This time we look at the bound states for l ≠ 0 . The radial wave function
⎪⎧⎪ A j (κr),
Rl (r) = ⎪⎨ l l (1)
⎪⎪ B h (ikr),
⎪⎩ l l
r ≤ r0
r > r0
(10.120)
where κ = 2m(E −V0 ) / ! , k = −2mE / ! , and the first type spherical Hankel function
h (1)(x) = jl (x) + inl (x) .
(10.121)
Under the low-energy and the strong-attraction limits, i.e., kr0 ≪ l and κ0r0  l (where
κ0 = −2mV0 / ! ), respectively, the asymptotic behaviors are
sin(x −lπ / 2)
,
x
(2l −1)!!
hl (1)(x) →x→0 −i
.
x l+1
jl (x) →x→∞
(10.122)
Employing the boundary conditions at r = r0 , i.e., continuity of radial wavefunction and its firstorder derivative, one finds that the condition for a bound state to occur is
lπ
κ0r0 cot(κ0r0 − ) = −l .
(10.123)
2
185
Since cot(κ0r0 −
lπ
) ≈ 0 (for the strong-attraction limit), we obtain
2
κ0r0 −lπ / 2 = (n + 1 / 2)π .
(10.124)
The scattering states have E > 0 , and the general expression for the phase shift δl is:
cot(δl ) =
knl′(kr0 )jl (κ0r0 )− κ0nl (kr0 )jl′(κ0r0 )
.
kjl′(kr0 )jl (κ0r0 )− κ0 jl (kr0 )jl′(κ0r0 )
(10.125)
At a resonance, δl = π / 2 and cot(δl ) = 0 ⇒
knl′(kr0 )jl (κ0r0 )− κ0nl (kr0 )jl′(κ0r0 ) = 0 .
(10.126)
Using the asymptotic properties of jl and nl under the limits of kr0  l and κ0r0  l :
jl (x) →x→∞
nl (x) →x→0
sin(x −lπ / 2)
,
x
(2l −1)!!
−
,
x l+1
(10.127)
we obtain
−(l + 2)sin(κ0r0 −lπ / 2) + κ0r0 cos(κ0r0 −lπ / 2) = 0 ,
(10.128)
then again due to κ0r0  l ,
cos(κ0r0 −lπ / 2) ≈ 0 ⇒ κ0r0 −lπ / 2 = (n + 1 / 2)π,
(10.129)
which is exactly the same condition for a bound state.
From Eq. (10.84) one can show that the partial wave amplitude
al (k) =
1
.
k cot(δl )−ik
(10.130)
We consider the neighborhood of resonance, k = kR (or E = E R ), and δl (k = kR ) = π / 2 , by
expanding δl (E) as follows
2
2
cot[δl (E)] ≈ cot[δl (E = E R )]− (E − E R ) = − (E − E R ) .
Γ
Γ
(10.131)
Here the width Γ is defined by
d cot[δl (E)]
dE
=−
E =ER
which leads to
186
2
,
Γ
(10.132)
al (k) = −
Γ/2
,
⎡
⎤
iΓ
k ⎢(E − E R ) + ⎥
⎢
2 ⎥⎦
⎣
(10.133)
and
σl =
4π (2l + 1)(Γ / 2)2
.
k 2 (E − E R )2 + (Γ / 2)2
(10.134)
The peak cannot be infinitely sharp because
σl (k) ≤
4(2l + 1)π
.
k2
(10.135)
187
Fig. 10.8:
scattering.
versus k in a resonance