segre/phys406/14S/lecture_20.pdf · Generalized classical scattering problem z q b initial direction is conventionally along the z-axis b is the impact parameter, the per-pendicular

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





• Phase shifts








• Phase shifts








• Phase shifts








• Phase shifts








• Phase shifts








• Phase shifts




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



z

θ

b








z

θ

b








z

θ

b








z

θ

b








z

θ

b







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



particles incident through the in-finitesmal area dσ

will scatterthrough the solid angle dΩ

dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ

=b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


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




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ

= LD(θ) dΩ −→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ

−→ D(θ) =1

LdN

dΩ




dσ = b dφ db


R2

= sin θ dθ dφ

D(θ) =dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

z

Rdθ

db

R

bdφ

Rsinθdφ

RsinθdΩ

dσ


σ =

∫D(θ) dΩ


dN = L dσ = LD(θ) dΩ −→ D(θ) =1

LdN

dΩ


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




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





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





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





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





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





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





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





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





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





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


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


Quantum scattering

z

θ

eikz

re

ikr

treat incident particles as planewaves

and scattered particles asspherical waves, about scatteringcenter


ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr

treat incident particles as planewaves

and scattered particles asspherical waves, about scatteringcenter


ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ

= |f (θ)|2


Quantum scattering

z

θ

eikz

re

ikr



ψi (z) = Ae ikz

ψs(r) = Be ikr

k =

√2mE

~

ψ(r , θ) ≈ A

e ikz + f (θ)

e ikr

r



D(θ) =dσ

dΩ= |f (θ)|2


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)







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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r








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


R(r) ∼ e ikr

r



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


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

r


Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


r→∞−−−→ e−ikr

rC. Segre (IIT) PHYS 406 - Spring 2014 April 07, 2014 7 / 13

Localized potential

V 0

V 0

kr 1



h(1)l




d2u

dr2− l(l + 1)

r2u = −k2u


h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r


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




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


-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




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


-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




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


-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




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


-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




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


-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




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


-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




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


-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


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)







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








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








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







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







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







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







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








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








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



Connection to the radiation zone

ψ = A

e ikz + k

∞∑l=0


≈ 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 ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r


and we obtain the radiation zoneform

with scattering factor


ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r


and we obtain the radiation zoneform

with scattering factor


ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

∫ 1


2

2l + 1δll ′



ψ = A

e ikz + k

∞∑l=0


≈ A

e ikz +

∞∑l=0


r




ψ ≈ A

e ikz + f (θ)

e ikr

r

f (θ) =

∞∑l=0


D(θ) = |f (θ)|2 =∑l

∑l ′


σ =

∫D(θ) dΩ = 4π

∞∑l=0

(2l + 1)|al |2∫ 1


2

2l + 1δll ′


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)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)





ψ =∑l ,m




e ikz =∞∑l=0



ψ(θ, φ) = A∞∑l=0


(1)l (kr)

]Pl(cos θ)


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


(1)l (ka)

] ∫ π

0Pl(cos θ)Pl ′(cos θ) sin θ dθ

=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′

= 2i l[jl (ka) + ikal h

(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3

The potential for quantum hardsphere scattering is

, with boundarycondition


V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3

The potential for quantum hardsphere scattering is , with boundarycondition


V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] ∫ 1

−1Pl(x)Pl ′(x) dx


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′

= 2i l′[jl ′(ka) + ikal ′h

(1)l ′ (ka)

]

−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]

−→ al = − jl(ka)

ikh(1)l (ka)


Example 11.3



V (r) =

∞, r ≤ a

0, r > a

ψ(a, θ) = 0

0 =∞∑l=0


(1)l (ka)

]Pl(cos θ)


0 =∞∑l=0


(1)l (ka)

] ∫ π


=∞∑l=0


(1)l (ka)

] 2

2l + 1δll ′


(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)


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


Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




Example 11.3


σ =4π

k2

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2


jl(z)

h(1)l (z)

=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




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segre/phys406/14S/lecture_20.pdf · Generalized classical scattering problem z q b initial direction is conventionally along the z-axis b is the impact parameter, the per-pendicular