segre/phys406/15s/lecture_20.pdfquantum scattering z q eikz r eikr treat incident particles as plane...
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Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
Today’s Outline - April 09, 2015
• Quantum scattering
• Partial wave analysis
• Example 11.3
• Example 11.1
• Review problems
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Building
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 1 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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
r
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 2015 April 09, 2015 2 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 3 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 4 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
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 2015 April 09, 2015 5 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
Solution in intermediate region
since u(r) ∼ rh(1)l (kr), then
R(r) ∼ 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 2015 April 09, 2015 6 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 7 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 8 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 9 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
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 2015 April 09, 2015 10 / 12
Problem 9.7
The first term in the equation
cb ∼= −Vba
2~
[e i(ω0+ω)t − 1
ω0 + ω+
e i(ω0−ω)t − 1
ω0 − ω
]
comes from the e iωt/2 part of cos(ωt). and the second from e−iωt/2.Thus droppingthe first term is formally equivalent to writingH ′ = (V /q)e−iωt , which is to say,
H ′ba =Vba
2e−iωt , H ′ab =
Vab
2e iωt
Rabi noticed that if you make the rotating wave approximation at thebeginning of the calculation, the time dependent coefficient equations canbe solved exactly with no need for perturbation theory, and no assumptionof field strength.
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 11 / 12
Problem 9.7
(a) Solve for the time dependent coefficients with the usual startingconditions: ca(0) = 1, cb(0) = 0. Express your results in terms of theRabi flopping frequency,
ωr ≡ 12
√(ω − ω0)2 + (|Vab|/~)2
(b) Determine the transition probability, Pa→b(t), and show that it neverexceeds 1. Confirm |ca(t)|2 + |cb(t)|2 = 1.
(c) Check that Pa→b(t) reduces to perturbation theory result when theperturbation “small,” and state precisely what small means in thiscontext, as a constraint on V .
(d) At what time does the system first return to its initial state?
C. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 12 / 12