Post on 07-Aug-2020
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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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18
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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18
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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18
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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18
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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18
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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 1 / 18
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
Homework Assignment #09:Chapter 9:7,10,12,15,16,18due Tuesday, March 31, 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 2 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 3 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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) + k2e ikr
4πr
(∇2 + k2)G = δ3(~r)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 4 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 5 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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~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 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
First Born approximation
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 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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 6 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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 2020 March 12, 2020 7 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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 ∼=
(2mV0a
3
3~2
)2
σ ∼=∫ (
2mV0a3
3~2
)2
dΩ
= 4π
(2mV0a
3
3~2
)2
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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 ∼=
(2mV0a
3
3~2
)2
σ ∼=∫ (
2mV0a3
3~2
)2
dΩ
= 4π
(2mV0a
3
3~2
)2
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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 ∼=
(2mV0a
3
3~2
)2
σ ∼=∫ (
2mV0a3
3~2
)2
dΩ = 4π
(2mV0a
3
3~2
)2
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 8 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 9 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 9 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 9 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 9 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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
~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 2020 March 12, 2020 9 / 18
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!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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)
= −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 2020 March 12, 2020 10 / 18
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!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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)
]2
Classical physics, the Born approximation,and quantum field theory all givethe same result for the Coulomb potential!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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!
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 10 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 11 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 11 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 11 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 11 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 11 / 18
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 11 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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 2020 March 12, 2020 12 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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/~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 13 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
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)
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 14 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Time-dependent Schrodinger equation
HΨ = i~∂Ψ
∂t, H = H0 + H ′(t)
Ψ(t) = ca(t)ψae−iEat/~ + cb(t)ψbe
−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/~
]C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 15 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]
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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Defining equations
ca = − i
~
[caH
′aa + cbH
′abe−i(Eb−Ea)t/~
]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
~
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 16 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
Iterative corrections
ca = − i
~H ′abe
−iω0tcb, cb = − i
~H ′bae
iω0tca, ω0 ≡Eb − Ea
~
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
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 17 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18
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
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
∣∣∣∣2 6= 1
C. Segre (IIT) PHYS 406 - Spring 2020 March 12, 2020 18 / 18