Today’s Outline - April 11, 2017

• The Born Series

• Review problems

Midterm Exam #2:Thursday, April 13, 2017bring 1 sided page of notes

Homework Assignment #11:Chapter 11:7,9,12,14,18,20due Tuesday, April 25, 2017

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 1 / 21

Today’s Outline - April 11, 2017

• The Born Series

• Review problems

Midterm Exam #2:Thursday, April 13, 2017bring 1 sided page of notes

Homework Assignment #11:Chapter 11:7,9,12,14,18,20due Tuesday, April 25, 2017

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 1 / 21

Today’s Outline - April 11, 2017

• The Born Series

• Review problems

Midterm Exam #2:Thursday, April 13, 2017bring 1 sided page of notes

Homework Assignment #11:Chapter 11:7,9,12,14,18,20due Tuesday, April 25, 2017

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 1 / 21

Today’s Outline - April 11, 2017

• The Born Series

• Review problems

Midterm Exam #2:Thursday, April 13, 2017bring 1 sided page of notes

Homework Assignment #11:Chapter 11:7,9,12,14,18,20due Tuesday, April 25, 2017

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 1 / 21

Today’s Outline - April 11, 2017

• The Born Series

• Review problems

Midterm Exam #2:Thursday, April 13, 2017bring 1 sided page of notes

Homework Assignment #11:Chapter 11:7,9,12,14,18,20due Tuesday, April 25, 2017

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 1 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0)

= Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

r

by inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0)

= Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0)

= Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0)

= Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0)

= Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0)

= Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0

= Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0 = Ae i~k ′·~r0 ,

~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0

k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

First Born approximation review

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

= Ae ikz + Af (θ)e ikr

rby inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0

The Born approximation is that the scattering is weak and therefore theincoming wave is not much altered by the potential

ψ(~r0) ≈ ψ0(~r0) = Ae ikz0 = Ae i~k ′·~r0 , ~k ′ = kz

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ

For low energy (long wavelength)scattering, the exponential can betaken out of the integral and

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 2 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle:

κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Spherical potential

If the scattering is not at low energy but the potential is sphericallysymmetric, V (~r) ≡ V (r)

f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0

define the scattering vector

letting the polar axis for the ~r0 coordinatesystem lie along ~κ

~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2

∫e iκr0 cos θ0V (r0)r20 sin θ0 dr0 dθ0 dφ0

∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

the angular dependence of the scattering is hidden inside κ which dependson the angle: κ = 2k sin(θ/2)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 3 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

∼= −m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2

∼=(

2mV0a3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

∼= −m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2

∼=(

2mV0a3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r

∼= −m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2

∼=(

2mV0a3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r ∼= −

m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2

∼=(

2mV0a3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r ∼= −

m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2

∼=(

2mV0a3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r ∼= −

m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2 ∼=

(2mV0a

3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r ∼= −

m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2 ∼=

(2mV0a

3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ

= 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Example 11.4

Low-energy, soft-sphere scattering

V (~r) =

V0, r ≤ a

0, r > a

f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r ∼= −

m

2π~2V0

(4

3πa3)

dσ

dΩ= |f |2 ∼=

(2mV0a

3

3~2

)2

σ ∼=∫ (

2mV0a3

3~2

)2

dΩ = 4π

(2mV0a

3

3~2

)2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 4 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κ

∫ ∞0e−µr sin(κr) dr = − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κ

∫ ∞0e−µr sin(κr) dr = − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κ

∫ ∞0e−µr sin(κr) dr = − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr

= −2mβ

~2κ

∫ ∞0e−µr sin(κr) dr = − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κ

∫ ∞0e−µr sin(κr) dr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr

=1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]

=1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)

=κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κIr

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κκ

µ2 + κ2

= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Problem 11.11

The Yukawa potential is a simple spherically symmetric model for thestrong force which binds nucleons.

The model has two constant parameters, β (strength of force), and µ(range of force).

V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ

∫ ∞0rV (r) sin(κr) dr = −2mβ

~2κκ

µ2 + κ2= − 2mβ

~2(µ2 + κ2)

replacing sin(κr) = (e iκr − e−iκr )/2i

Ir =1

2i

∫ ∞0

[e−(µ−iκ)r − e−(µ+iκ)r

]dr =

1

2i

[e−(µ−iκ)r

−(µ− iκ)− e−(µ+iκ)r

−(µ+ iκ)

∣∣∣∣∣∞

0

=1

2i

[1

µ− iκ− 1

µ+ iκ

]=

1

2i

(µ+ iκ− µ+ iκ

µ2 + κ2

)=

κ

µ2 + κ2

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 5 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)

= −2mq1q24πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2

Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

Example 11.6

Rutherford scattering (the Coulomb potential) can be obtained from theYukawa potential by setting: β = q1q2/4πε0 and µ = 0.

The scattering factor then becomes

f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2

since κ = 2k sin(θ/2), and k =√

2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)

]2Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 6 / 21

The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 7 / 21

The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 7 / 21

The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt

−→ θ ∼= tan−1(I

p

)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 7 / 21

The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)

this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 7 / 21

The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 7 / 21

The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 7 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0,

g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

r

this can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 8 / 21

Problem 8.3

Use the equation below to calculate the approximate transmissionprobability for a particle of energy E that encounters a finite square barrierof height V0 > E and width 2a. Compare your answer with the exactresult to which it should reduce in the WKB regime, T 1.

T ∼= e−2γ , γ ≡ 1

~

∫|p(x)| dx

Since the energy of the particle isless than the barrier height, the mo-mentum is given by

|p(x)| =√

2m(V0 − E )

γ =1

~

∫ 2a

0

√2m(V0 − E ) dx =

[1

~√

2m(V0 − E )x

∣∣∣∣2a0

=2a

~√

2m(V0 − E )

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 9 / 21

Problem 8.3

Use the equation below to calculate the approximate transmissionprobability for a particle of energy E that encounters a finite square barrierof height V0 > E and width 2a. Compare your answer with the exactresult to which it should reduce in the WKB regime, T 1.

T ∼= e−2γ , γ ≡ 1

~

∫|p(x)| dx

Since the energy of the particle isless than the barrier height, the mo-mentum is given by

|p(x)| =√

2m(V0 − E )

γ =1

~

∫ 2a

0

√2m(V0 − E ) dx =

[1

~√

2m(V0 − E )x

∣∣∣∣2a0

=2a

~√

2m(V0 − E )

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 9 / 21

Problem 8.3

Use the equation below to calculate the approximate transmissionprobability for a particle of energy E that encounters a finite square barrierof height V0 > E and width 2a. Compare your answer with the exactresult to which it should reduce in the WKB regime, T 1.

T ∼= e−2γ , γ ≡ 1

~

∫|p(x)| dx

Since the energy of the particle isless than the barrier height, the mo-mentum is given by

|p(x)| =√

2m(V0 − E )

γ =1

~

∫ 2a

0

√2m(V0 − E ) dx =

[1

~√

2m(V0 − E )x

∣∣∣∣2a0

=2a

~√

2m(V0 − E )

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 9 / 21

Problem 8.3

Use the equation below to calculate the approximate transmissionprobability for a particle of energy E that encounters a finite square barrierof height V0 > E and width 2a. Compare your answer with the exactresult to which it should reduce in the WKB regime, T 1.

T ∼= e−2γ , γ ≡ 1

~

∫|p(x)| dx

Since the energy of the particle isless than the barrier height, the mo-mentum is given by

|p(x)| =√

2m(V0 − E )

γ =1

~

∫ 2a

0

√2m(V0 − E ) dx

=

[1

~√

2m(V0 − E )x

∣∣∣∣2a0

=2a

~√

2m(V0 − E )

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 9 / 21

Problem 8.3

Use the equation below to calculate the approximate transmissionprobability for a particle of energy E that encounters a finite square barrierof height V0 > E and width 2a. Compare your answer with the exactresult to which it should reduce in the WKB regime, T 1.

T ∼= e−2γ , γ ≡ 1

~

∫|p(x)| dx

Since the energy of the particle isless than the barrier height, the mo-mentum is given by

|p(x)| =√

2m(V0 − E )

γ =1

~

∫ 2a

0

√2m(V0 − E ) dx =

[1

~√

2m(V0 − E )x

∣∣∣∣2a0

=2a

~√

2m(V0 − E )

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 9 / 21

Problem 8.3

Use the equation below to calculate the approximate transmissionprobability for a particle of energy E that encounters a finite square barrierof height V0 > E and width 2a. Compare your answer with the exactresult to which it should reduce in the WKB regime, T 1.

T ∼= e−2γ , γ ≡ 1

~

∫|p(x)| dx

Since the energy of the particle isless than the barrier height, the mo-mentum is given by

|p(x)| =√

2m(V0 − E )

γ =1

~

∫ 2a

0

√2m(V0 − E ) dx =

[1

~√

2m(V0 − E )x

∣∣∣∣2a0

=2a

~√

2m(V0 − E )

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 9 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ

= e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ

≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.3 (cont.)

The transmission probability isthus

while the exact answer fromChapter 2 is

in the WKB approximation thetunneling probability is assumedto be small, so γ 1 and

T ∼= e−2γ = e−4a√

2m(V0−E)/~

T =

[1 +

V 20

4E (V0 − E )sinh2 γ

]−1

sinh γ =1

2(eγ − e−γ) ≈ 1

2eγ

sinh2 ≈ 1

4e2γ

T =

[1 +

V 20

16E (V0 − E )e2γ]−1

=

16E (V0 − E )

V 20

e−2γ ≈ e−2γ

since the pre-factor is of order unity

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 10 / 21

Problem 8.5

Consider the quantum mechanical analog to the classical problem of a ballof mass m bouncing elastically on the floor

(a) What is the potential energy, as a function of height x above the floor?

(b) Solve the Schrodinger equation for this potential, expressing theanswer in terms of the appropriate Airy function. Don’t normalize thewave function

(c) Using m = 0.100 kg, find the first four allowed energies, in joules.

(d) What is the ground state energy in eV, of an electron in thisgravitational field? How high off the ground, on average, is thiselectron?

(a) The potential energy is simply V (x) = mgx

(b) the Schrodinger equation is

− ~2

2m

d2ψ

dx2+ mgxψ = Eψ −→ d2ψ

dx2=

2m2g

~2

(x − E

mg

)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 11 / 21

Problem 8.5

Consider the quantum mechanical analog to the classical problem of a ballof mass m bouncing elastically on the floor

(a) What is the potential energy, as a function of height x above the floor?

(b) Solve the Schrodinger equation for this potential, expressing theanswer in terms of the appropriate Airy function. Don’t normalize thewave function

(c) Using m = 0.100 kg, find the first four allowed energies, in joules.

(d) What is the ground state energy in eV, of an electron in thisgravitational field? How high off the ground, on average, is thiselectron?

(a) The potential energy is simply V (x) = mgx

(b) the Schrodinger equation is

− ~2

2m

d2ψ

dx2+ mgxψ = Eψ −→ d2ψ

dx2=

2m2g

~2

(x − E

mg

)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 11 / 21

Problem 8.5

Consider the quantum mechanical analog to the classical problem of a ballof mass m bouncing elastically on the floor

(a) What is the potential energy, as a function of height x above the floor?

(b) Solve the Schrodinger equation for this potential, expressing theanswer in terms of the appropriate Airy function. Don’t normalize thewave function

(c) Using m = 0.100 kg, find the first four allowed energies, in joules.

(d) What is the ground state energy in eV, of an electron in thisgravitational field? How high off the ground, on average, is thiselectron?

(a) The potential energy is simply V (x) = mgx

(b) the Schrodinger equation is

− ~2

2m

d2ψ

dx2+ mgxψ = Eψ

−→ d2ψ

dx2=

2m2g

~2

(x − E

mg

)

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 11 / 21

Problem 8.5

Consider the quantum mechanical analog to the classical problem of a ballof mass m bouncing elastically on the floor

(a) What is the potential energy, as a function of height x above the floor?

(b) Solve the Schrodinger equation for this potential, expressing theanswer in terms of the appropriate Airy function. Don’t normalize thewave function

(c) Using m = 0.100 kg, find the first four allowed energies, in joules.

(d) What is the ground state energy in eV, of an electron in thisgravitational field? How high off the ground, on average, is thiselectron?

(a) The potential energy is simply V (x) = mgx

(b) the Schrodinger equation is

− ~2

2m

d2ψ

dx2+ mgxψ = Eψ −→ d2ψ

dx2=

2m2g

~2

(x − E

mg

)C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 11 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)

d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3

finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)

d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3

finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)]

(c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

Making the substitutions

y ≡ x − E

mg, α ≡

(2m2g

~2

)1/3finally, substituting z ≡ αy

this is simply the Airy equationwhich has solutions

but the Bi(z) solution is un-bounded at ∞

the un-normalized solution is thus

d2ψ

dx2=

2m2g

~2

(x − E

mg

)d2ψ

dy2= α3yψ

d2ψ

dz2= zψ

ψ = aAi(z) + bBi(z)

ψ = aAi

[α

(x − E

mg

)](c) Since the ball cannot drop below the floor, there is a boundarycondition ψ(0) = 0 which means that we need to find the zeroes of theAiry function Ai [α(−E/mg)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 12 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan = −

(1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan = −

(1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan = −

(1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3

En = −mg

αan = −

(1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan

= −(

1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan = −

(1

2mg2~2

)1/3

in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan = −

(1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

The first four zeros of Ai [α(−E/mg)] can be computed athttp://keisan.casio.com/exec/system/1180573400 and are

a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787

the nth root defines the correspond-ing eigenvalue of the system

an = −αEn

mg, α =

(~2

2m2g

)1/3En = −mg

αan = −

(1

2mg2~2

)1/3in Joules, these energies become

E1 = 0.88× 10−22 J E2 = 1.54× 10−22 J

E3 = 2.08× 10−22 J E4 = 2.56× 10−22 J

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 13 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩

=1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩

=1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉

=1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉

=3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉

=3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

Problem 8.5 (cont.)

(d) According to the virial theorem,for stationary states

but dV /dx = mg

the total energy of the state is thus

thus

〈x〉 =2En

3mg

〈T 〉 =1

2

⟨xdV

dx

⟩=

1

2〈mgx〉 =

1

2〈V 〉

En = 〈T 〉+ 〈V 〉 =3

2〈V 〉

=3

2〈mgx〉 =

3

2mg〈x〉

for an electron E1 = 1.84× 10−32 J = 1.15× 10−13 eV

and 〈x〉 = 1.37× 10−3 m = 1.37 mm

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 14 / 21

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 dropping the 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 2017 April 11, 2017 15 / 21

Problem 9.7 (cont.)

(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 the perturbation theory result whenthe perturbation is “small,” and state precisely what small means inthis context, as a constraint on V .

(d) At what time does the system first return to its initial state?

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 16 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb

cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]

= i(ω0 − ω)cb − iVba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb

− iVba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]

= i(ω0 − ω)cb −|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

(a) Applying the rotating wave approximation to the time-dependentperturbation equations, we have

ca = − i

2~Vabe

iωte−iω0tcb cb = − i

2~Vbae

−iωte iω0tca

taking the derivative of the equation on the right

cb = −i Vba

2~

[i(ω0 − ω)e i(ω0−ω)tca + e i(ω0−ω)t ca

]= i(ω0 − ω)cb − i

Vba

2~e i(ω0−ω)t

[− i

2~Vabe

iωte−iω0tcb

]= i(ω0 − ω)cb −

|Vab|2

(2~)2cb

this is a second order differential equation for cb which can be solvedexactly

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 17 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλt

λ2 + i(ω − ω0)λ+|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]

= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]

= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

0 =d2cbdt2

+ i(ω0 − ω)dcbdt

+|Vab|2

4~2cb

the solution is of the form cb = eλtλ2 + i(ω − ω0)λ+

|Vab|2

4~2= 0

λ =1

2

[−i(ω − ω0)±

√−(ω − ω0)2 − |Vab|2

4~2

]= i

[−(ω − ω0)

2± ωr

]

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

the general solution is thus

cb(t) = Ae i [−(ω−ω0)/2+ωr ]t + Be i [−(ω−ω0)/2−ωr ]t

= e−i(ω−ω0)t/2[Ae iωr t + Be−iωr t

]= e−i(ω−ω0)t/2 [C cos(ωr t) + D sin(ωr t)]

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 18 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]

ca(t) = i2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]

since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1

1 = i2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr

−→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

cb(t) = e i(ω0−ω)t/2 [C cos(ωr t) + D sin(ωr t)]

applying the initial conditions that cb(0) = 0 forces C = 0 and thesolution becomes

cb(t) = De i(ω0−ω)t/2 sin(ωr t)

cb(t) = D

[i

(ω0 − ω

2

)e i(ω0−ω)t/2 sin(ωr t) + ωre

i(ω0−ω)t/2 cos(ωr t)

]ca(t) = i

2~Vba

e i(ω−ω0)t cb

=2~DVba

e i(ω−ω0)t/2

[iωr cos(ωr t)−

(ω0 − ω

2

)sin(ωr t)

]since the inital condi-tions require ca(0) = 1 1 = i

2~DVba

ωr −→ D = − iVba

2~ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 19 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]

cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2

=

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1

and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1

and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2

≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1

and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t)

= 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

ca = e i(ω−ω0)t/2

[ωr cos(ωr t) + i

(ω0 − ω

2

)sin(ωr t)

]cb = − i

2~ωrVbae

i(ω−ω0)t/2 sin(ωr t)

(b) The probability of transi-tion, Pa→b is given by

The maximum probability iswhen the sine is 1 and is only1 when ω = ω0

Pa→b = |cb|2 =

(|Vab|2~ωr

)2

sin2(ωr t)

Pmax =|Vab|2/~2

(ω − ω0)2 + |Vab|2/~2≤ 1

|ca|2 + |cb|2 = cos2(ωr t) +

(ω0 − ω

2ωr

)2

sin2(ωr t) +

(|Vab|2~ωr

)2

sin2(ωr t)

= cos2(ωr t) +(ω − ω0)2 + |Vab|2/~2

4ω2r

sin2(ωr t) = 1

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 20 / 21

Problem 9.7 (cont.)

(c) The perturbation can be considered small when |Vab|2 ~2(ω − ω0)2

in which case

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

≈ 1

2|ω − ω0|

Pa→b ≈|Vab|2

~2sin2

(ω−ω0

2 t)

(ω − ω0)2which confirms the PT result

(d) the system will return to its initial state when ωr t = π and thust = π/ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 21 / 21

Problem 9.7 (cont.)

(c) The perturbation can be considered small when |Vab|2 ~2(ω − ω0)2

in which case

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

≈ 1

2|ω − ω0|

Pa→b ≈|Vab|2

~2sin2

(ω−ω0

2 t)

(ω − ω0)2which confirms the PT result

(d) the system will return to its initial state when ωr t = π and thust = π/ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 21 / 21

Problem 9.7 (cont.)

(c) The perturbation can be considered small when |Vab|2 ~2(ω − ω0)2

in which case

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

≈ 1

2|ω − ω0|

Pa→b ≈|Vab|2

~2sin2

(ω−ω0

2 t)

(ω − ω0)2which confirms the PT result

(d) the system will return to its initial state when ωr t = π and thust = π/ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 21 / 21

Problem 9.7 (cont.)

(c) The perturbation can be considered small when |Vab|2 ~2(ω − ω0)2

in which case

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

≈ 1

2|ω − ω0|

Pa→b ≈|Vab|2

~2sin2

(ω−ω0

2 t)

(ω − ω0)2

which confirms the PT result

(d) the system will return to its initial state when ωr t = π and thust = π/ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 21 / 21

Problem 9.7 (cont.)

(c) The perturbation can be considered small when |Vab|2 ~2(ω − ω0)2

in which case

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

≈ 1

2|ω − ω0|

Pa→b ≈|Vab|2

~2sin2

(ω−ω0

2 t)

(ω − ω0)2which confirms the PT result

(d) the system will return to its initial state when ωr t = π and thust = π/ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 21 / 21

Problem 9.7 (cont.)

(c) The perturbation can be considered small when |Vab|2 ~2(ω − ω0)2

in which case

ωr =1

2

√(ω − ω0)2 +

|Vab|2~2

≈ 1

2|ω − ω0|

Pa→b ≈|Vab|2

~2sin2

(ω−ω0

2 t)

(ω − ω0)2which confirms the PT result

(d) the system will return to its initial state when ωr t = π and thust = π/ωr

C. Segre (IIT) PHYS 406 - Spring 2017 April 11, 2017 21 / 21