Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Today’s Outline - April 07, 2014
• Classical scattering
• Quantum scattering
• Partial wave analysis
• Phase shifts
Homework Assignment #08:Chapter 10:5,6,7,8,9,10due Wednesday, April 09, 2014
Midterm Exam #2:Monday, April 14, 2014
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 1 / 13
Generalized classical scattering problem
z
θ
b
initial direction is conventionallyalong the z-axis
b is the impact parameter, the per-pendicular (to the z-axis distancefrom the initial trajectory to thescattering center
θ is the scattering angle, the angu-lar deviation of the final trajectoryfrom the initial
assume that the scattering center is fixed (no recoil)
given the initial energy E , impact parameter b, and interaction forces,calculate the scattering angle θ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 2 / 13
Generalized classical scattering problem
z
θ
b
initial direction is conventionallyalong the z-axis
b is the impact parameter, the per-pendicular (to the z-axis distancefrom the initial trajectory to thescattering center
θ is the scattering angle, the angu-lar deviation of the final trajectoryfrom the initial
assume that the scattering center is fixed (no recoil)
given the initial energy E , impact parameter b, and interaction forces,calculate the scattering angle θ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 2 / 13
Generalized classical scattering problem
z
θ
b
initial direction is conventionallyalong the z-axis
b is the impact parameter, the per-pendicular (to the z-axis distancefrom the initial trajectory to thescattering center
θ is the scattering angle, the angu-lar deviation of the final trajectoryfrom the initial
assume that the scattering center is fixed (no recoil)
given the initial energy E , impact parameter b, and interaction forces,calculate the scattering angle θ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 2 / 13
Generalized classical scattering problem
z
θ
b
initial direction is conventionallyalong the z-axis
b is the impact parameter, the per-pendicular (to the z-axis distancefrom the initial trajectory to thescattering center
θ is the scattering angle, the angu-lar deviation of the final trajectoryfrom the initial
assume that the scattering center is fixed (no recoil)
given the initial energy E , impact parameter b, and interaction forces,calculate the scattering angle θ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 2 / 13
Generalized classical scattering problem
z
θ
b
initial direction is conventionallyalong the z-axis
b is the impact parameter, the per-pendicular (to the z-axis distancefrom the initial trajectory to thescattering center
θ is the scattering angle, the angu-lar deviation of the final trajectoryfrom the initial
assume that the scattering center is fixed (no recoil)
given the initial energy E , impact parameter b, and interaction forces,calculate the scattering angle θ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 2 / 13
Generalized classical scattering problem
z
θ
b
initial direction is conventionallyalong the z-axis
b is the impact parameter, the per-pendicular (to the z-axis distancefrom the initial trajectory to thescattering center
θ is the scattering angle, the angu-lar deviation of the final trajectoryfrom the initial
assume that the scattering center is fixed (no recoil)
given the initial energy E , impact parameter b, and interaction forces,calculate the scattering angle θ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 2 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ
will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ
=b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time
the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ
= LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ
−→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Differential cross-section
particles incident through the in-finitesmal area dσ will scatterthrough the solid angle dΩ
dσ = b dφ db
dΩ =r dθ R sin θ dφ
R2
= sin θ dθ dφ
D(θ) =dσ
dΩ=
b
sin θ
∣∣∣∣dbdθ∣∣∣∣
z
Rdθ
db
R
bdφ
Rsinθdφ
RsinθdΩ
dσ
the total cross section is
σ =
∫D(θ) dΩ
if the Luminosity, L, is defined as the number of incident particles per unitarea per unit time the number of particles incident through dσ is
dN = L dσ = LD(θ) dΩ −→ D(θ) =1
LdN
dΩ
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 3 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
)
z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα,
θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
)
z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
)
z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)
= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
)
z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)
θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
)
z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
)
z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
) z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
) z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)
=R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
) z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4
−→ σ =R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
) z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ
= πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Hard sphere scattering
A small particle strikes a large hard sphere of radius R with an impactparameter b elastically
b = R sinα, θ = π − 2α
b = R sin
(π
2− θ
2
)= R cos
(θ
2
)θ =
2 cos−1(b/R), b ≤ R
0, b ≥ R
db
dθ= −1
2R sin
(θ
2
) z
θ
bα
α
α
R
D(θ) =R cos
(θ2
)sin θ
(R sin
(θ2
)2
)=
R2
4−→ σ =
R2
4
∫dΩ = πR2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 4 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves
and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves
and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ
= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Quantum scattering
z
θ
eikz
re
ikr
treat incident particles as planewaves and scattered particles asspherical waves, about scatteringcenter
at far points from the scatteringcenter, the solutions must be
ψi (z) = Ae ikz
ψs(r) = Be ikr
k =
√2mE
~
ψ(r , θ) ≈ A
e ikz + f (θ)
e ikr
r
a superposition of the incoming andscattered waves
the differential cross-section is
D(θ) =dσ
dΩ= |f (θ)|2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 5 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Partial wave analysis
For a spherically symmetric potential, the scattered wave can be separatedinto a radial portion and spherical harmonics
the modified radial func-tion, u(r) = rR(r), mustsatisfy
for very large r , V (r)→ 0and we neglect the cen-trifugal term
the second term, the in-coming wave can be ig-nored (D = 0)
ψ(r , θ, φ) = R(r)Yml (θ, φ)
Eu = − ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u
d2u
dr2= −k2u
u(r) = Ce ikr + De−ikr
R(r) ∼ e ikr
r
this is precisely the form expected from the physical argument in thekr 1 regime (the radiation zone)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 6 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
r
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
rC. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Localized potential
V 0
V 0
kr 1
in the Intermediate zonethe Schrodinger equation be-comes
keep only the outgoing wave,
h(1)l
Radiation zone - simple spherical wave so-lution
Intermediate region - only include cen-trifugal term
Scattering region - no approximations ap-plied
d2u
dr2− l(l + 1)
r2u = −k2u
the solutions are linear combinations ofthe spherical Hankel functions
h(1)l (x) ≡ jl(x) + inl(x)
r→∞−−−→ e ikr
r
h(2)l (x) ≡ jl(x)− inl(x)
r→∞−−−→ e−ikr
rC. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 π 2π 3π
Re
[hl(1
) ]
x
l=0
l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Hankel function of the first kind
Hankel functions are combi-nations of Bessel functionsand the Hankel function ofthe first kind is
h(1)0 = −i e
ix
x
h(1)1 =
(− i
x2− 1
x
)e ix
h(1)2 =
(− 3i
x3− 3
x2+
i
x
)e ix
h(1)l
x1−−−→ (−i)l+1 eix
x
the real and imaginary partsof the Hankel functions are
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0 π 2π 3π
Im[h
l(1) ]
x
l=0l=1
l=2
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 8 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen
and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen
and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients
giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients
giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Solution in intermediate region
since u(r) ∼ h(1)l (kr), then
R(kr) ∼ h(1)l (kr) and
because the potential isspherically symmetric, therecan be no φ dependence andonly the m = 0 terms are al-lowed
the spherical harmonics arethen and redefining the co-efficients giving
ψ = A
e ikz +∑l ,m
Cl ,mh(1)l (kr)Ym
l (θ, φ)
= A
e ikz +
∑l
Cl ,0h(1)l (kr)Y 0
l (θ, φ)
Y 0l =
√2l + 1
4/πPl(cos θ)
Cl ,0 ≡ i l+1k√
4π(2l + 1)al
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 9 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform
with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform
with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ
= 4π∞∑l=0
(2l + 1)|al |2
∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Connection to the radiation zone
ψ = A
e ikz + k
∞∑l=0
i l+1(2l + 1)alh(1)l (kr)Pl(cos θ)
≈ A
e ikz +
∞∑l=0
(2l + 1)alPl(cos θ)e ikr
r
at large r , h(1)l → (−i)l+1e ikr/kr
and we obtain the radiation zoneform with scattering factor
the differential and total cross-sections then become
ψ ≈ A
e ikz + f (θ)
e ikr
r
f (θ) =
∞∑l=0
(2l + 1)alPl(cos θ)
D(θ) = |f (θ)|2 =∑l
∑l ′
(2l + 1)(2l ′ + 1)a∗l al ′Pl(cos θ)Pl ′(cos θ)
σ =
∫D(θ) dΩ = 4π
∞∑l=0
(2l + 1)|al |2∫ 1
−1Pl(x)Pl ′(x) dx =
2
2l + 1δll ′
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 10 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Solving the interaction zone
The partial wave amplitudes are obtained by solving the Schrodingerequation in the scattering region, where V 6= 0
we know that for V = 0 the general solution is
ψ =∑l ,m
[Al ,mjl(kr) + Bl ,mnl(kr)]Yml (θ, φ)
The incident particle is expressedin Cartesian coordinate and mustbe changed to spherical usingRayleigh’s formula
the full solution can now be written
e ikz =∞∑l=0
i l(2l + 1)jl(kr)Pl(cos θ)
noting that the Neumann functionscannot be used and m ≡ 0 for ourgeometry
ψ(θ, φ) = A∞∑l=0
i l(2l + 1)[jl(kr) + ikalh
(1)l (kr)
]Pl(cos θ)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 11 / 13
Example 11.3
The potential for quantum hardsphere scattering is
, with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is
, with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ 1
−1Pl(x)Pl ′(x) dx
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l′[jl ′(ka) + ikal ′h
(1)l ′ (ka)
]
−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]
−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The potential for quantum hardsphere scattering is , with boundarycondition
applying the boundary condition
V (r) =
∞, r ≤ a
0, r > a
ψ(a, θ) = 0
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
]Pl(cos θ)
multiplying by Pl ′(cos θ) sin θ dθ and integrating from 0→ π
0 =∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] ∫ π
0Pl(cos θ)Pl ′(cos θ) sin θ dθ
=∞∑l=0
i l(2l + 1)[jl(ka) + ikalh
(1)l (ka)
] 2
2l + 1δll ′
= 2i l[jl (ka) + ikal h
(1)l (ka)
]−→ al = − jl(ka)
ikh(1)l (ka)
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 12 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)
≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)
≈ −i(
2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
)
(−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1
−→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1 −→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1 −→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13
Example 11.3
The total cross-section isthus
σ =4π
k2
∞∑l=0
(2l + 1)
∣∣∣∣∣ jl(ka)
h(1)l (ka)
∣∣∣∣∣2
the limiting case is more instructive, take low energy scattering(z = ka 1), that is long wavelength
jl(z)
h(1)l (z)
=jl(z)
jl(z) + inl(z)≈ −i jl(z)
nl(z)≈ −i
(2l l!z l
(2l + 1)!
) (−2l l!z l+1
(2l)!
)
=i
2l + 1
[2l l!
(2l)!
]2z2l+1 −→ σ ≈ 4π
k2
∞∑l=0
1
(2l + 1)
[2l l!
(2l)!
]4(ka)4l+2
thus the cross-section becomes a sum dominated by the l = 0 term andconsequently independent of θ: σ ≈ 4πa2
this is 4 times the geometrical cross-section and equal to the total surfacearea of the sphere so the particles “see” the entire sphere when scattering
C. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 13 / 13