segre/phys406/15s/lecture_20.pdfquantum scattering z q eikz r eikr treat incident particles as plane...

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

• Example 11.3

• Example 11.1

• Review problems

• Example 11.3

• Example 11.1

• Review problems

• Example 11.3

• Example 11.1

• Review problems

• Example 11.3

• Example 11.1

• Review problems

• Example 11.3

• Example 11.1

• Review problems

• Example 11.3

• Example 11.1

• Review problems

Quantum scattering

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

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

a superposition of the incoming andscattered waves

the differential cross-section is

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

treat incident particles as planewaves

and scattered particles asspherical waves, about scatteringcenter

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

treat incident particles as planewaves

and scattered particles asspherical waves, about scatteringcenter

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

dΩ= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

D(θ) =dσ

= |f (θ)|2

Quantum scattering

ψi (z) = Ae ikz

ψs(r) = Be ikr

√2mE

ψ(r , θ) ≈ A

e ikz + f (θ)

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

[V (r) +

l(l + 1)

dr2= −k2u

u(r) = Ce ikr + De−ikr

R(r) ∼ e ikr

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

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

ψ(r , θ, φ) = R(r)Yml (θ, φ)

Eu = − ~2

[V (r) +

l(l + 1)

dr2= −k2u

R(r) ∼ e ikr

Localized potential

in the Intermediate zonethe Schrodinger equation be-comes

keep only the outgoing wave,

Radiation zone - simple spherical wave so-lution

Intermediate region - only include cen-trifugal term

Scattering region - no approximations ap-plied

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

h(2)l (x) ≡ jl(x)− inl(x)

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

r→∞−−−→ e−ikr

rC. Segre (IIT) PHYS 406 - Spring 2015 April 09, 2015 4 / 12

Localized potential

dr2− l(l + 1)

r2u = −k2u

h(1)l (x) ≡ jl(x) + inl(x)

r→∞−−−→ e ikr

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

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

the real and imaginary partsof the Hankel functions are

0 π 2π 3π

l(1) ]

l=0l=1

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

l(1) ]

l=0l=1

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

l(1) ]

l=0l=1

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

l(1) ]

l=0l=1

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

l(1) ]

l=0l=1

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

l(1) ]

l=0l=1

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

h(1)0 = −i e

h(1)1 =

(− i

x2− 1

h(1)2 =

(− 3i

x3− 3

x1−−−→ (−i)l+1 eix

0 π 2π 3π

l(1) ]

l=0l=1

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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)

e ikz +

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

e ikz +

∞∑l=0

(2l + 1)alPl(cos θ)e ikr

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)

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)

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

−1Pl(x)Pl ′(x) dx =

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

and we obtain the radiation zoneform

with scattering factor

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

and we obtain the radiation zoneform

with scattering factor

ψ ≈ A

e ikz + f (θ)

f (θ) =∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2∫ 1

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ

= 4π∞∑l=0

(2l + 1)|al |2

2l + 1δll ′

ψ = A

e ikz + k

∞∑l=0

e ikz +

∞∑l=0

ψ ≈ A

e ikz + f (θ)

f (θ) =

∞∑l=0

D(θ) = |f (θ)|2 =∑l

∑l ′

∫D(θ) dΩ = 4π

∞∑l=0

(2l + 1)|al |2∫ 1

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

2l + 1δll ′

(1)l (ka)

]−→ al = − jl(ka)

ikh(1)l (ka)

Example 11.3

The total cross-section isthus

σ =4π

∞∑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

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 + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]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

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

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 + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

≈ −i(

2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

(−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1

−→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1 −→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1 −→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Example 11.3

σ =4π

∞∑l=0

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

h(1)l (z)

=jl(z)

nl(z)≈ −i

(2l l!z l

(2l + 1)!

) (−2l l!z l+1

2l + 1

[2l l!

]2z2l+1 −→ σ ≈ 4π

∞∑l=0

(2l + 1)

[2l l!

]4(ka)4l+2

Problem 9.7

The first term in the equation

cb ∼= −Vba

[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 =

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.

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?

segre/phys406/15s/lecture_20.pdfquantum scattering z q eikz r eikr treat incident particles as plane...

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