segre/phys406/20S/lecture_17.pdf · Review of the integral equation solution Start with the...

Today’s Outline - March 12, 2020

• Greens function method

• Born approximation

• The impulse approximation

• Time dependent perturbation theory

Homework Assignment #08:Chapter 9:7,9,12,15,17,20due Tuesday, March 24, 2020


C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18

















































Review of the integral equation solution

Start with the time-independentSchrodinger equation

we rewrite itusing the substitutions

k ≡√

2mE

~, Q ≡ 2m

~2Vψ

this can be solved for an abitrary driv-ing function (Q = Vψ) by solving for theGreen’s function, G (~r), which is the solu-tion for a delta function source then thesolution to the actual source, Q, becomes

the Green’s function solution can be ex-pressed as a Fourier transfom

Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2

∫e i~s·~rg(~s) d3~s

the solution is obtained by using Cauchy’s fomula and contour integration

G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr



Start with the time-independentSchrodinger equation

we rewrite itusing the substitutions

k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr



Start with the time-independentSchrodinger equation we rewrite itusing the substitutions

k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~,

Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ

this can be solved for an abitrary driv-ing function (Q = Vψ) by solving for theGreen’s function, G (~r), which is the solu-tion for a delta function source

then thesolution to the actual source, Q, becomes


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ

this can be solved for an abitrary driv-ing function (Q = Vψ) by solving for theGreen’s function, G (~r), which is the solu-tion for a delta function source

then thesolution to the actual source, Q, becomes


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

= − e ikr

4πr




k ≡√

2mE

~, Q ≡ 2m

~2Vψ



Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

G (~r) =1

(2π)3/2



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s = − e ikr

4πr


General solution

The solution just obtained is not unique as it is possible to prove that thegeneral solution is G ′(~r) = G (~r) + G0(~r), where G0(~r) is any solution tothe homogeneous Helmholtz equation

(∇2 + k2)G0(~r) = 0

the general solution to the Schrodinger equation is thus

ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0

with ψ0(~r) being the solution of the free particle Schrodinger equation(∇2 + k2

)ψ0(~r) = 0

note that the wave function appears on both sides of this equation,requiring prior knowledge of the wave function to obtain an exact solution!

however, it is possible to use the integral equation for an approximationmethod called the Born approximation


General solution


(∇2 + k2)G0(~r) = 0


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


)ψ0(~r) = 0




General solution


(∇2 + k2)G0(~r) = 0


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


)ψ0(~r) = 0




General solution


(∇2 + k2)G0(~r) = 0


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


)ψ0(~r) = 0




General solution


(∇2 + k2)G0(~r) = 0


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


)ψ0(~r) = 0




General solution


(∇2 + k2)G0(~r) = 0


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


)ψ0(~r) = 0




General solution


(∇2 + k2)G0(~r) = 0


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


)ψ0(~r) = 0




Problem 10.8

Show that G (~r) = −e ikr/4πr satisfies(∇2 + k2

)G (~r) = δ3(~r) by direct

substitution

∇G = − 1

4π

(1

r∇e ikr + e ikr∇1

r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1

r∇2e ikr + e ikr∇2 1

r

]∇1

r= − 1

r2r , ∇e ikr = ike ikr r , ∇2 1

r= −4πδ3(r)

∇2e ikr = ik∇ · (e ikr r) = ik1

r2d

dr(r2e ikr ) = ike ikr

(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik

)− 4πe ikrδ3(~r)

]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)

∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]

∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1

r2r ,

∇e ikr = ike ikr r , ∇2 1

r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1

r2r , ∇e ikr = ike ikr r ,

∇2 1

r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)

∇2e ikr = ik∇ · (e ikr r)

= ik1

r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d

dr(r2e ikr )

= ike ikr(

2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)

∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]

= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]

= δ3(~r) + k2e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


Problem 10.8



substitution

∇G = − 1

4π

(1


r

)∇2G = − 1

4π

[2

(∇1

r

)·(∇e ikr

)+

1


r

]∇1

r= − 1


r= −4πδ3(r)


r2d


(2

r+ ik

)∇2G = −1

4

[2

(− 1

r2r

)·(ike ikr r

)+

1

rike ikgr

(2

r+ ik


]= δ3(~r)− 1

4πe ikr

[−2ik

r2+

2ik

r2− k2

r

]= δ3(~r) + k2

e ikr

4πr

(∇2 + k2)G = δ3(~r)


First Born approximation

Suppose we have a potential localized about ~r0 = 0

we want to solve the integral Schrodinger equation

ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0

at points far away from the scattering center, we have |~r | |~r0| for thepoints where the scattering potential is significant

|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr

e ik|~r−~r0| ∼= e ikre−i~k·~r0 −→ e ik|~r−~r0|

|~r −~r0|∼=

e ikr

re−i

~k·~r0

ψ0(~r) = Ae ikz , ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0

∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)

−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr

e ik|~r−~r0| ∼= e ikre−i~k·~r0

−→ e ik|~r−~r0|

|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0

ψ0(~r) = Ae ikz ,

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0




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 assumes that the scattering is weak and thereforethe incoming wave is not significantly 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




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


r

by inspection, we have

f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0


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


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0

k’

kκ


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r




2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2A

∫e−i

~k·~r0V (~r0)ψ(~r0) d~r0



f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (~r0) d~r0k’

kκ


f (θ, φ) ∼= −m

2π~2

∫V (~r) d~r


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)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ



κ = 2k sin(θ/2)


Spherical potential


f (θ, φ) ∼= −m

2π~2

∫e i(

~k ′−~k)·~r0V (r0) d~r0



~κ = ~k ′ − ~k

(~k ′ − ~k) ·~r0 = κr0 cos θ0

f (θ) ∼= −m

2π~2


∼= −2m

~2κ


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


Example 10.4

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


Example 10.4


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


Example 10.4


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


Example 10.4


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


Example 10.4


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


Example 10.4


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


Example 10.4


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


Example 10.4


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


Problem 10.11

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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κ


~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κ


~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


= −2mβ

~2κ


~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κ

∫ ∞0e−µr sin(κr) dr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr

=1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κIr

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κκ

µ2 + κ2

= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Problem 10.11



V (r) = βe−µr

r

f (θ) ∼= −2m

~2κ


~2κκ

µ2 + κ2= − 2mβ

~2(µ2 + κ2)


Ir =1

2i

∫ ∞0


]dr =

1

2i

[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


Example 10.6

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!


Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2


2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)



Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)

= −2mq1q24πε0κ2


2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)



Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2


2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)



Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2


2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)



Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2


2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)



Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2


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!


Example 10.6



f (θ) ∼= −2mβ

~2(µ2 + κ2)= −2mq1q2

4πε0κ2


2mE/~

f (θ) ∼= −q1q2

16πε0E sin2(θ/2)

dσ

dΩ=

[q1q2

16πε0E sin2(θ/2)



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 analogous to the first Born approximationwhere the incident particle is unaffected

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



θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p





θF

b


I =

∫F⊥ dt

−→ θ ∼= tan−1(I

p





θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)

this first-order correction is analogous to the first Born approximationwhere the incident particle is unaffected




θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p





θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p




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


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0,

g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

r

this can be written as


ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


Time dependent phenomena

Up to now, we have only solved problems which can be called “quantumstatics”

Most interesting systems involve transitions in energy as a function of time

The simplest model of this kind is a two level system in the presence of anelectromagnetic field. Suppose we have a two level unperturbed systemwith eigenfunctions ψa and ψb

Any possible state of the systemcan be expressed as a linear com-bination

The state of the system at any timet can be obtained by including thetime dependence of the stationarystates

H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0

Ψ(0) = caψa + cbψb

1 = |ca|2 + |cb|2

Ψ(t) = caψae−iEat/~ + cbψbe

−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~





The simplest model of this kind is a two level system in the presence of anelectromagnetic field.

Suppose we have a two level unperturbed systemwith eigenfunctions ψa and ψb



H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉

〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~








H0ψa = Eaψa

H0ψb = Ebψb

〈ψa |ψa 〉 = 1 = 〈ψb |ψb 〉〈ψa |ψb 〉 = 0


1 = |ca|2 + |cb|2


−iEbt/~


The perturbed system

When a time-dependent perturbation H ′(t) is turned on, the expression forthe system wavefunction is still the same with the exception that thecoefficients, ca and cb become functions of time as well.

Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe

−iEbt/~

A “transition” between the the two states would be described

ca(0) = 1, cb(0) = 0

−→ ca(t) = 0, cb(t) = 1

This wavefunction must satisfy the time-dependent Schrodinger equation

HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)





−iEbt/~


ca(0) = 1, cb(0) = 0

−→ ca(t) = 0, cb(t) = 1


HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)





−iEbt/~


ca(0) = 1, cb(0) = 0

−→ ca(t) = 0, cb(t) = 1


HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)





−iEbt/~


ca(0) = 1, cb(0) = 0

−→ ca(t) = 0, cb(t) = 1


HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)





−iEbt/~


ca(0) = 1, cb(0) = 0 −→ ca(t) = 0, cb(t) = 1


HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)





−iEbt/~


ca(0) = 1, cb(0) = 0 −→ ca(t) = 0, cb(t) = 1


HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)





−iEbt/~


ca(0) = 1, cb(0) = 0 −→ ca(t) = 0, cb(t) = 1


HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


Time-dependent Schrodinger equation

HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~

+((((((((

(cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~

= i~[caψae

−iEat/~

+ cbψbe−iEbt/~

+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~

]taking the inner product with ψa

ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb

⟩e−iEbt/~ = i~cae−iEat/~

ca = − i

~

[caH

′aa + cbH

′abe−i(Eb−Ea)t/~

]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~

+ cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae

−iEat/~

+ cbψbe−iEbt/~

+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~

+ ca[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae

−iEat/~

+ cbψbe−iEbt/~

+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae

−iEat/~

+ cbψbe−iEbt/~

+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~

= i~[caψae

−iEat/~

+ cbψbe−iEbt/~

+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae

−iEat/~

+ cbψbe−iEbt/~

+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~

]

taking the inner product with ψa

ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae

−iEat/~ + cbψbe−iEbt/~

+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~

]


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+ caψa

(− iEa

~

)e−iEat/~

+ cbψb

(− iEb

~

)e−iEbt/~

]


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

ca[H0ψa

]e−iEat/~ + cb

[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+ caψa

(− iEa

~

)e−iEat/~ + cbψb

(− iEb

~

)e−iEbt/~

]


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~ +((((

(((((

cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~

]


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~ +((((

(((((

cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~ +((((

(((((

cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~ +((((

(((((

cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψa

∣∣H ′∣∣ψb

⟩e−iEbt/~

= i~cae−iEat/~

ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~ +((((

(((((

cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]



HΨ = i~∂Ψ

∂t, H = H0 + H ′(t)


−iEbt/~

((((((((

(ca[H0ψa

]e−iEat/~ +((((

(((((

cb[H0ψb

]e−iEbt/~ + ca

[H ′ψa

]e−iEat/~

+ cb[H ′ψb

]e−iEbt/~ = i~

[caψae


+

caψa

(− iEa

~

)e−iEat/~ +

cbψb

(− iEb

~

)e−iEbt/~


ca⟨ψa

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψa

∣∣H ′∣∣ψb


ca = − i

~

[caH

′aa + cbH


]C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18

Defining equations

ca = − i

~

[caH

′aa + cbH


]

Similarly, taking the inner product with ψb

ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψb

∣∣H ′∣∣ψb

⟩e−iEbt/~ = i~cbe−iEbt/~

cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]

for most cases, the diagonal terms will vanish (H ′aa = H ′bb = 0) and wehave equations for the coefficients which are just a recasting of theSchrodinger equation

ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae

iω0tca, ω0 ≡Eb − Ea

~


Defining equations

ca = − i

~

[caH

′aa + cbH


]Similarly, taking the inner product with ψb

ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~

+ cb⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb

⟩e−iEbt/~

= i~cbe−iEbt/~

cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb,

cb = − i

~H ′bae


~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe

−iω0tcb, cb = − i

~H ′bae

iω0tca,

ω0 ≡Eb − Ea

~


Defining equations

ca = − i

~

[caH

′aa + cbH



ca⟨ψb

∣∣H ′∣∣ψa

⟩e−iEat/~ + cb

⟨ψb

∣∣H ′∣∣ψb


cb = − i

~

[cbH

′bb + caH

′bae

i(Eb−Ea)t/~]


ca = − i

~H ′abe


~H ′bae


~


Iterative corrections

ca = − i

~H ′abe


~H ′bae


~

Let us now assume that H ′ is smalland that the system starts out inthe lower energy state

in Zeroth order, the coefficients willremain the same forever

for the First order correction, applythe zeroth order coefficients to thetime dependent equations

and the Second order correction issimilarly obtained

c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t

0H ′ba(t ′)e iω0t′ dt ′

c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t

0H ′ba(t ′)e iω0t′ dt ′ c

(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1

c(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0

→ c(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


c(2)b ≡ c

(1)b



ca = − i

~H ′abe


~H ′bae


~





c(0)a (t) = 1 c

(0)b (t) = 0

c(1)a = 0 → c

(1)a = 1

c(1)b = − i

~H ′bae

iω0t

c(1)b = − i

~

∫ t


c(2)a = − i

~H ′bae

−iω0t

(− i

~

)∫ t


(2)b ≡ c

(1)b


The full second order solution

The full second order solution includes the zeroth order terms as well

ca(t) ≈ 1− 1

~2

∫ t

0H ′ba(t ′)e−iω0t′

[∫ t′

0H ′ba(t ′′)e iω0t′′dt ′′

]dt ′

cb(t) ≈ − i

~

∫ t


This iterative solution can be continued for higher precision and in general,the Nth order solution has N factors of H ′

note that while the exact coefficients, must be normalized, thisapproximate solution is only normalized to order N in H ′

for example to first order, normalization becomes

|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1




ca(t) ≈ 1− 1

~2

∫ t


[∫ t′


]dt ′

cb(t) ≈ − i

~

∫ t





|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1




ca(t) ≈ 1− 1

~2

∫ t


[∫ t′


]dt ′

cb(t) ≈ − i

~

∫ t





|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1




ca(t) ≈ 1− 1

~2

∫ t


[∫ t′


]dt ′

cb(t) ≈ − i

~

∫ t





|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1




ca(t) ≈ 1− 1

~2

∫ t


[∫ t′


]dt ′

cb(t) ≈ − i

~

∫ t





|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1




ca(t) ≈ 1− 1

~2

∫ t


[∫ t′


]dt ′

cb(t) ≈ − i

~

∫ t





|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1




ca(t) ≈ 1− 1

~2

∫ t


[∫ t′


]dt ′

cb(t) ≈ − i

~

∫ t





|ca|2 + |cb|2 ≈ 12 +

∣∣∣∣− i

~

∫ t


∣∣∣∣2 6= 1


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