phys.iit.eduphys.iit.edu/~segre/phys406/19s/lecture_25.pdf · problem 11.7 (cont.) since the exact...

Today’s Outline - April 25, 2019

• Exam #2 solutions

• Chapter 11 problems

Homework Assignment #12:Chapter 11:9,11,15,20,22,26due Tuesday, April 30, 2019

Final Exam: Tuesday, May 7, 201910:30-12:30, Stuart Building 212

C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 1 / 14

Problem 11.3

Solve

ca = − i

~H ′abe

−iω0tcb cb = − i

~H ′bae

iω0tca

exactly for the case of a time-independent perturbation, assuming thatca(0) = 1 and cb(0) = 0. Check that

|ca(t)|2 + |cb(t)|2 = 1

Since the perturbation is time-independent, we can start by taking thetime derivative of the expression for cb to be able to eliminate ca.

cb = − i

~H ′ba[iω0e

iω0tca

+ e iω0t ca

]

= iω0

[− i

~H ′bae

iω0tca

]

− i

~H ′bae

iω0t

[− i

~H ′abe

−iω0tcb

]

= iω0cb

− 1

~2|H ′ab|2cb


Problem 11.3

Solve

ca = − i

~H ′abe


~H ′bae

iω0tca


|ca(t)|2 + |cb(t)|2 = 1


cb = − i

~H ′ba[iω0e

iω0tca + e iω0t ca]

= iω0

[− i

~H ′bae

iω0tca

]

− i

~H ′bae

iω0t

[− i

~H ′abe

−iω0tcb

]

= iω0cb

− 1

~2|H ′ab|2cb


Problem 11.3

Solve

ca = − i

~H ′abe


~H ′bae

iω0tca


|ca(t)|2 + |cb(t)|2 = 1


cb = − i

~H ′ba[iω0e


= iω0

[− i

~H ′bae

iω0tca

]

− i

~H ′bae

iω0t

[− i

~H ′abe

−iω0tcb

]= iω0cb

− 1

~2|H ′ab|2cb


Problem 11.3

Solve

ca = − i

~H ′abe


~H ′bae

iω0tca


|ca(t)|2 + |cb(t)|2 = 1


cb = − i

~H ′ba[iω0e


= iω0

[− i

~H ′bae

iω0tca

]− i

~H ′bae

iω0t

[− i

~H ′abe

−iω0tcb

]

= iω0cb

− 1

~2|H ′ab|2cb


Problem 11.3

Solve

ca = − i

~H ′abe


~H ′bae

iω0tca


|ca(t)|2 + |cb(t)|2 = 1


cb = − i

~H ′ba[iω0e


= iω0

[− i

~H ′bae

iω0tca

]− i

~H ′bae

iω0t

[− i

~H ′abe

−iω0tcb

]= iω0cb

− 1

~2|H ′ab|2cb


Problem 11.3

Solve

ca = − i

~H ′abe


~H ′bae

iω0tca


|ca(t)|2 + |cb(t)|2 = 1


cb = − i

~H ′ba[iω0e


= iω0

[− i

~H ′bae

iω0tca

]− i

~H ′bae

iω0t

[− i

~H ′abe

−iω0tcb

]= iω0cb −

1

~2|H ′ab|2cb


Problem 11.3 (cont.)

We now have a differential equationin cb alone

substituting α2 ≡ |H ′ab|2/~2 gives

this is a linear differential equationwith constant coefficients whichhas a solution of the form

cb = iω0cb −1

~2|H ′ab|2cb

cb − iω0cb + α2cb = 0

cb = eλt

0 = λ2eλt − iω0λeλt + α2eλt −→ λ2 − iω0λ+ α2 = 0

λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2

the general solution is thus

cb(t) = Ae i(ω0+ω)t/2 + Be i(ω0−ω)t/2 = e iω0t/2(Ae iωt/2 + Be−iωt/2

)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt

0 = λ2eλt − iω0λeλt + α2eλt

−→ λ2 − iω0λ+ α2 = 0

λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]

=i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2


cb(t) = Ae i(ω0+ω)t/2 + Be i(ω0−ω)t/2

= e iω0t/2(Ae iωt/2 + Be−iωt/2

)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)

= e iω0t/2[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)= e iω0t/2

[C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0






cb = iω0cb −1

~2|H ′ab|2cb


cb = eλt


λ =1

2

[iω0 ±

√−ω2

0 − 4α2

]=

i

2(ω0 ± ω), ω ≡

√ω20 + 4α2



)= e iω0t/2

[��C cos

(ω2 t)

+ D sin(ω2 t)]

cb(0) = 0→ C ≡ 0



cb(t) = De iω0t/2 sin(ω2 t)

cb(t) = D

[iω0

2e iω0t/2 sin

(ω2 t)

+ω

2e iω0t/2 cos

(ω2 t)

]=ω

2De iω0t/2

[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

= − i

~H ′bae

iω0tca

ca(t) =i~ω

2H ′bae−iω0t/2D

[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

since we know that ca(0) = 1, then D = 2H ′ba/(i~ω) and

ca(t) = e−iω0t/2[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω

e iω0t/2 sin(ω2 t), ω ≡

√ω20 +

4

~2|H ′ab|2




cb(t) = D

[iω0

2e iω0t/2 sin

(ω2 t)

+ω

2e iω0t/2 cos

(ω2 t)

]

=ω

2De iω0t/2

[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

= − i

~H ′bae

iω0tca

ca(t) =i~ω


[cos(ω2 t)

+ iω0

ωsin(ω2 t)]



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2




cb(t) = D

[iω0

2e iω0t/2 sin

(ω2 t)

+ω

2e iω0t/2 cos

(ω2 t) ]

=ω

2De iω0t/2

[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

= − i

~H ′bae

iω0tca

ca(t) =i~ω


[cos(ω2 t)

+ iω0

ωsin(ω2 t)]



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2




cb(t) = D

[iω0

2e iω0t/2 sin

(ω2 t)

+ω

2e iω0t/2 cos

(ω2 t) ]

=ω

2De iω0t/2

[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

= − i

~H ′bae

iω0tca

ca(t) =i~ω


[cos(ω2 t)

+ iω0

ωsin(ω2 t)]



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω

e iω0t/2 sin(ω2 t),

ω ≡√ω20 +

4

~2|H ′ab|2




cb(t) = D

[iω0

2e iω0t/2 sin

(ω2 t)

+ω

2e iω0t/2 cos

(ω2 t) ]

=ω

2De iω0t/2

[cos(ω2 t)

+ iω0

ωsin(ω2 t)]

= − i

~H ′bae

iω0tca

ca(t) =i~ω


[cos(ω2 t)

+ iω0

ωsin(ω2 t)]



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2




+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2

Now check for the normalization condition to be maintained over all times

|ca|2 + |cb|2 = cos2(ω2 t)

+ω20

ω2sin2

(ω2 t)

+4|H ′ab|2

~2ω2sin2

(ω2 t)

= cos2(ω2 t)

+��1

ω2��(ω20 + 4

|H ′ab|2

~2ω2

)sin2

(ω2 t)

= cos2(ω2 t)

+ sin2(ω2 t)

= 1




+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2


|ca|2 + |cb|2 = cos2(ω2 t)

+ω20

ω2sin2

(ω2 t)

+4|H ′ab|2

~2ω2sin2

(ω2 t)

= cos2(ω2 t)

+1

ω2

(ω20 + 4

|H ′ab|2

~2ω2

)sin2

(ω2 t)

= cos2(ω2 t)

+ sin2(ω2 t)

= 1




+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2


|ca|2 + |cb|2 = cos2(ω2 t)

+ω20

ω2sin2

(ω2 t)

+4|H ′ab|2

~2ω2sin2

(ω2 t)

= cos2(ω2 t)

+��1

ω2��(ω20 + 4

|H ′ab|2

~2ω2

)sin2

(ω2 t)

= cos2(ω2 t)

+ sin2(ω2 t)

= 1


Problem 11.5

Suppose you don’t assume that H ′aa = H ′bb = 0. Find ca(t) and cb(t) infirst-order perturbation theory, for the case ca(0) = 1, cb(0) = 0, showthat the time-dependent coefficients are normalized to first order in H ′.

ca = − i

~[caH

′aa + cbH

′abe−iω0t

], cb = − i

~[cbH

′bb + caH

′bae

+iω0t]

these are exact solutions and with the initial conditions that ca(0) = 1,cb(0) = 0, to first order, we can write

c(1)a = − i

~H ′aa −→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t −→ cb(t) = − i

~

∫ t

0H ′ba(t ′)e iω0t′ dt ′


Problem 11.5


ca = − i

~[caH

′aa + cbH

′abe−iω0t

],

cb = − i

~[cbH

′bb + caH

′bae

+iω0t]


c(1)a = − i

~H ′aa −→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t −→ cb(t) = − i

~

∫ t



Problem 11.5


ca = − i

~[caH

′aa + cbH

′abe−iω0t

], cb = − i

~[cbH

′bb + caH

′bae

+iω0t]


c(1)a = − i

~H ′aa −→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t −→ cb(t) = − i

~

∫ t



Problem 11.5


ca = − i

~[caH

′aa + cbH

′abe−iω0t

], cb = − i

~[cbH

′bb + caH

′bae

+iω0t]


c(1)a = − i

~H ′aa

−→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t −→ cb(t) = − i

~

∫ t



Problem 11.5


ca = − i

~[caH

′aa + cbH

′abe−iω0t

], cb = − i

~[cbH

′bb + caH

′bae

+iω0t]


c(1)a = − i

~H ′aa −→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t −→ cb(t) = − i

~

∫ t



Problem 11.5


ca = − i

~[caH

′aa + cbH

′abe−iω0t

], cb = − i

~[cbH

′bb + caH

′bae

+iω0t]


c(1)a = − i

~H ′aa −→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t

−→ cb(t) = − i

~

∫ t



Problem 11.5


ca = − i

~[caH

′aa + cbH

′abe−iω0t

], cb = − i

~[cbH

′bb + caH

′bae

+iω0t]


c(1)a = − i

~H ′aa −→ ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′

c(1)b = − i

~H ′bae

iω0t −→ cb(t) = − i

~

∫ t




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t


Now it is possible to check the normalization

|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t

0H ′ba(t ′)e−iω0t′ dt ′

]≈ 0

thus to first order, |ca|2 + |cb|2 ≈ 1



ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

]

[1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t


]≈ 0




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]

= 1−[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t


]≈ 0




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2

≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t


]≈ 0




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t


]≈ 0




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


]

[i

~

∫ t


]≈ 0




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t


]

≈ 0




ca(t) = 1− i

~

∫ t

0H ′aa(t ′) dt ′, cb(t) = − i

~

∫ t



|ca|2 =

[1− i

~

∫ t

0H ′aa(t ′) dt ′

] [1 +

i

~

∫ t

0H ′aa(t ′) dt ′

]= 1−

[i

~

∫ t

0H ′aa(t ′) dt ′

]2≈ 1

|cb|2 =

[− i

~

∫ t


] [i

~

∫ t


]≈ 0



Problem 11.6

Solve

ca = − i

~H ′abe

−iω0tcb, cb = − i

~H ′bae

+iω0tca

to second order in perturbation theory, for the general case ca(0) = a,cb(0) = b

The zeroth order solution is justgiven by the initial conditions

use the defining equations to gen-erate the higher order perturbationterms

c(0)a = a

c(0)b = b

ca = − i

~H ′abe

−iω0tb −→ c(1)a (t) = a− ib

~

∫ t

0H ′ab(t ′)e−iω0t′ dt ′

cb = − i

~H ′abe

−iω0ta −→ c(1)b (t) = b − ia

~

∫ t

0H ′ab(t ′)e+iω0t′ dt ′


Problem 11.6

Solve

ca = − i

~H ′abe


~H ′bae

+iω0tca




c(0)a = a

c(0)b = b

ca = − i

~H ′abe

−iω0tb

−→ c(1)a (t) = a− ib

~

∫ t


cb = − i

~H ′abe

−iω0ta −→ c(1)b (t) = b − ia

~

∫ t



Problem 11.6

Solve

ca = − i

~H ′abe


~H ′bae

+iω0tca




c(0)a = a

c(0)b = b

ca = − i

~H ′abe

−iω0tb −→ c(1)a (t) = a− ib

~

∫ t


cb = − i

~H ′abe

−iω0ta −→ c(1)b (t) = b − ia

~

∫ t



Problem 11.6

Solve

ca = − i

~H ′abe


~H ′bae

+iω0tca




c(0)a = a

c(0)b = b

ca = − i

~H ′abe

−iω0tb −→ c(1)a (t) = a− ib

~

∫ t


cb = − i

~H ′abe

−iω0ta

−→ c(1)b (t) = b − ia

~

∫ t



Problem 11.6

Solve

ca = − i

~H ′abe


~H ′bae

+iω0tca




c(0)a = a

c(0)b = b

ca = − i

~H ′abe

−iω0tb −→ c(1)a (t) = a− ib

~

∫ t


cb = − i

~H ′abe

−iω0ta −→ c(1)b (t) = b − ia

~

∫ t




For the second order

ca =− i

~H ′abe

−iω0t

[b − ia

~

∫ t

0H ′ab(t ′)e+iω0t dt ′

]−→

c(2)a (t) = a− ib

~

∫ t


− a

~2

∫ t

0H ′ab(t ′)e−iω0t′

[∫ t′

0H ′ba(t ′′)e+iω0t′′ dt ′′

]dt ′

cb =− i

~H ′bae

+iω0t

[a− ib

~

∫ t

0H ′ba(t ′)e−iω0t dt ′

]−→

c(2)b (t) = b − ia

~

∫ t

0H ′ba(t ′)e+iω0t′ dt ′

− b

~2

∫ t

0H ′ba(t ′)e+iω0t′

[∫ t′

0H ′ab(t ′′)e−iω0t′′ dt ′′

]dt ′




ca =− i

~H ′abe

−iω0t

[b − ia

~

∫ t


]

−→

c(2)a (t) = a− ib

~

∫ t


− a

~2

∫ t


[∫ t′

0H ′ba(t ′′)e+iω0t′′ dt ′′

]dt ′

cb =− i

~H ′bae

+iω0t

[a− ib

~

∫ t


]−→

c(2)b (t) = b − ia

~

∫ t


− b

~2

∫ t


[∫ t′

0H ′ab(t ′′)e−iω0t′′ dt ′′

]dt ′




ca =− i

~H ′abe

−iω0t

[b − ia

~

∫ t


]−→

c(2)a (t) = a− ib

~

∫ t


− a

~2

∫ t


[∫ t′

0H ′ba(t ′′)e+iω0t′′ dt ′′

]dt ′

cb =− i

~H ′bae

+iω0t

[a− ib

~

∫ t


]−→

c(2)b (t) = b − ia

~

∫ t


− b

~2

∫ t


[∫ t′

0H ′ab(t ′′)e−iω0t′′ dt ′′

]dt ′




ca =− i

~H ′abe

−iω0t

[b − ia

~

∫ t


]−→

c(2)a (t) = a− ib

~

∫ t


− a

~2

∫ t


[∫ t′

0H ′ba(t ′′)e+iω0t′′ dt ′′

]dt ′

cb =− i

~H ′bae

+iω0t

[a− ib

~

∫ t


]

−→

c(2)b (t) = b − ia

~

∫ t


− b

~2

∫ t


[∫ t′

0H ′ab(t ′′)e−iω0t′′ dt ′′

]dt ′




ca =− i

~H ′abe

−iω0t

[b − ia

~

∫ t


]−→

c(2)a (t) = a− ib

~

∫ t


− a

~2

∫ t


[∫ t′

0H ′ba(t ′′)e+iω0t′′ dt ′′

]dt ′

cb =− i

~H ′bae

+iω0t

[a− ib

~

∫ t


]−→

c(2)b (t) = b − ia

~

∫ t


− b

~2

∫ t


[∫ t′

0H ′ab(t ′′)e−iω0t′′ dt ′′

]dt ′


Problem 11.7

Calculate ca(t) and cb(t) to second order, for a time-independentperturbation and compare to the exact result.

ca = − i

~H ′abe


~H ′bae

iω0tca

Given H ′ independent of t and the initial conditions ca(0) = 1, cb(0) = 0,the equations above give

c(2)b (t) = c

(1)b (t) = − i

~H ′ba

∫ t

0e iω0t′ dt ′ =

[− i

~H ′baiω0

e iω0t′∣∣∣∣t0

= −H ′ba~ω0

(e iω0t − 1

)c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′

0e iω0t′′ dt ′′

]dt ′


Problem 11.7


ca = − i

~H ′abe


~H ′bae

iω0tca


c(2)b (t) = c

(1)b (t)

= − i

~H ′ba

∫ t


[− i

~H ′baiω0

e iω0t′∣∣∣∣t0

= −H ′ba~ω0

(e iω0t − 1

)c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′


Problem 11.7


ca = − i

~H ′abe


~H ′bae

iω0tca


c(2)b (t) = c

(1)b (t) = − i

~H ′ba

∫ t

0e iω0t′ dt ′

=

[− i

~H ′baiω0

e iω0t′∣∣∣∣t0

= −H ′ba~ω0

(e iω0t − 1

)c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′


Problem 11.7


ca = − i

~H ′abe


~H ′bae

iω0tca


c(2)b (t) = c

(1)b (t) = − i

~H ′ba

∫ t


[− i

~H ′baiω0

e iω0t′∣∣∣∣t0

= −H ′ba~ω0

(e iω0t − 1

)c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′


Problem 11.7


ca = − i

~H ′abe


~H ′bae

iω0tca


c(2)b (t) = c

(1)b (t) = − i

~H ′ba

∫ t


[− i

~H ′baiω0

e iω0t′∣∣∣∣t0

= −H ′ba~ω0

(e iω0t − 1

)

c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′


Problem 11.7


ca = − i

~H ′abe


~H ′bae

iω0tca


c(2)b (t) = c

(1)b (t) = − i

~H ′ba

∫ t


[− i

~H ′baiω0

e iω0t′∣∣∣∣t0

= −H ′ba~ω0

(e iω0t − 1

)c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′



c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′

= 1−|H ′ab|2

i~2ω0

∫ t

0

(1− e−iω0t′

)dt ′ = 1 +

i |H ′ab|2

~2ω0

[t ′ +

1

iω0e−iω0t′

∣∣∣∣t0

= 1 +i |H ′ab|2

~2ω0

[t +

e−iω0t − 1

iω0

]c(2)b (t) = −

H ′ba~ω0

(e iω0t − 1

)Comparing with the exact result from Problem 11.3


+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2



c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′

= 1−|H ′ab|2

i~2ω0

∫ t

0

(1− e−iω0t′

)dt ′

= 1 +i |H ′ab|2

~2ω0

[t ′ +

1

iω0e−iω0t′

∣∣∣∣t0

= 1 +i |H ′ab|2

~2ω0

[t +

e−iω0t − 1

iω0

]c(2)b (t) = −

H ′ba~ω0

(e iω0t − 1



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2



c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′

= 1−|H ′ab|2

i~2ω0

∫ t

0

(1− e−iω0t′

)dt ′ = 1 +

i |H ′ab|2

~2ω0

[t ′ +

1

iω0e−iω0t′

∣∣∣∣t0

= 1 +i |H ′ab|2

~2ω0

[t +

e−iω0t − 1

iω0

]c(2)b (t) = −

H ′ba~ω0

(e iω0t − 1



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2



c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′

= 1−|H ′ab|2

i~2ω0

∫ t

0

(1− e−iω0t′

)dt ′ = 1 +

i |H ′ab|2

~2ω0

[t ′ +

1

iω0e−iω0t′

∣∣∣∣t0

= 1 +i |H ′ab|2

~2ω0

[t +

e−iω0t − 1

iω0

]

c(2)b (t) = −

H ′ba~ω0

(e iω0t − 1



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2



c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′

= 1−|H ′ab|2

i~2ω0

∫ t

0

(1− e−iω0t′

)dt ′ = 1 +

i |H ′ab|2

~2ω0

[t ′ +

1

iω0e−iω0t′

∣∣∣∣t0

= 1 +i |H ′ab|2

~2ω0

[t +

e−iω0t − 1

iω0

]c(2)b (t) = −

H ′ba~ω0

(e iω0t − 1

)

Comparing with the exact result from Problem 11.3


+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2



c(2)a (t) = 1−

|H ′ab|2

~2

∫ t

0e−iω0t′

[∫ t′


]dt ′

= 1−|H ′ab|2

i~2ω0

∫ t

0

(1− e−iω0t′

)dt ′ = 1 +

i |H ′ab|2

~2ω0

[t ′ +

1

iω0e−iω0t′

∣∣∣∣t0

= 1 +i |H ′ab|2

~2ω0

[t +

e−iω0t − 1

iω0

]c(2)b (t) = −

H ′ba~ω0

(e iω0t − 1



+ iω0

ωsin(ω2 t)]

cb(t) =2H ′bai~ω


√ω20 +

4

~2|H ′ab|2



Since the exact result for cb is already first order in H ′, we can allowω → ω0 and drop the extra correction term of the order of H ′

cb(t) =2H ′bai~ω

e iω0t/2 sin(ω2 t)

≈2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i

(e iω0t/2 − e−iω0t/2

)= −

H ′ba~ω0

(e−iω0t − 1

)this is exactly the result from the second order correction

the calculation for ca is somewhat more complicated and starts byapproximating ω to first order in the perturbation

ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)= ω0 +

2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20




cb(t) =2H ′bai~ω

e iω0t/2 sin(ω2 t)≈

2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i


)= −

H ′ba~ω0

(e−iω0t − 1



ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)= ω0 +

2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20




cb(t) =2H ′bai~ω


2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i


)

= −H ′ba~ω0

(e−iω0t − 1



ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)= ω0 +

2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20




cb(t) =2H ′bai~ω


2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i


)= −

H ′ba~ω0

(e−iω0t − 1

)

this is exactly the result from the second order correction


ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)= ω0 +

2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20




cb(t) =2H ′bai~ω


2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i


)= −

H ′ba~ω0

(e−iω0t − 1



ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)= ω0 +

2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20




cb(t) =2H ′bai~ω


2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i


)= −

H ′ba~ω0

(e−iω0t − 1



ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)

= ω0 +2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20




cb(t) =2H ′bai~ω


2H ′bai~ω0


=2H ′bai~ω0

e iω0t/21

2i


)= −

H ′ba~ω0

(e−iω0t − 1



ω = ω0

√1 +

4|H ′ab|2

~2ω20

≈ ω0

(1 +

2|H ′ab|2

~2ω20

)= ω0 +

2|H ′ab|2

~2ω0

ω0

ω≈ 1−

2|H ′ab|2

~2ω20



ca(t) = e−iω0t/2[cos(ω

2t)

+ iω0

ωsin(ω

2t)]

≈ e−iω0t/2

[cos

(ω0

2t +

2|H ′ab|2

~2ω20

)+i

(1−

2|H ′ab|2

~2ω20

)sin

(ω0

2t +

2|H ′ab|2

~2ω20

)]considering that the second term in each parenthesis is small compared tothe first, we can apply some Taylor expansions

cos(x + ε) ≈ cos x − ε sin x , sin(x + ε) ≈ sin x + ε cos x

ca(t) ≈ e−iω0t/2

{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}




2t)

+ iω0

ωsin(ω

2t)]

≈ e−iω0t/2

[cos

(ω0

2t +

2|H ′ab|2

~2ω20

)+i

(1−

2|H ′ab|2

~2ω20

)sin

(ω0

2t +

2|H ′ab|2

~2ω20

)]

considering that the second term in each parenthesis is small compared tothe first, we can apply some Taylor expansions



{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}




2t)

+ iω0

ωsin(ω

2t)]

≈ e−iω0t/2

[cos

(ω0

2t +

2|H ′ab|2

~2ω20

)+i

(1−

2|H ′ab|2

~2ω20

)sin

(ω0

2t +

2|H ′ab|2

~2ω20




{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}




2t)

+ iω0

ωsin(ω

2t)]

≈ e−iω0t/2

[cos

(ω0

2t +

2|H ′ab|2

~2ω20

)+i

(1−

2|H ′ab|2

~2ω20

)sin

(ω0

2t +

2|H ′ab|2

~2ω20


cos(x + ε) ≈ cos x − ε sin x ,

sin(x + ε) ≈ sin x + ε cos x


{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}




2t)

+ iω0

ωsin(ω

2t)]

≈ e−iω0t/2

[cos

(ω0

2t +

2|H ′ab|2

~2ω20

)+i

(1−

2|H ′ab|2

~2ω20

)sin

(ω0

2t +

2|H ′ab|2

~2ω20




{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}




{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}

= e−iω0t/2{[

cos(ω0

2t)

+ i sin(ω0

2t)]

−|H ′ab|2t~2ω0

[t(

sin(ω0

2t)− i cos

(ω0

2t))

+2i

ω0sin(ω0

2t)]}

= e−iω0t/2

{e iω0t/2 −

|H ′ab|2t~2ω0

[−ite iω0t/2 +

1

ω0


)]}= 1−

|H ′ab|2t~2ω0

[−it +

(1− e−iω0t

)ω0

]= 1+

i

~2ω0|H ′ab|2

[t +

(e−iω0t − 1

)iω0

]

this matches the second order perturbation theory result




{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}

= e−iω0t/2{[

cos(ω0

2t)

+ i sin(ω0

2t)]

−|H ′ab|2t~2ω0

[t(

sin(ω0

2t)− i cos

(ω0

2t))

+2i

ω0sin(ω0

2t)]}

= e−iω0t/2

{e iω0t/2 −

|H ′ab|2t~2ω0

[−ite iω0t/2 +

1

ω0


)]}

= 1−|H ′ab|2t~2ω0

[−it +

(1− e−iω0t

)ω0

]= 1+

i

~2ω0|H ′ab|2

[t +

(e−iω0t − 1

)iω0

]





{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}

= e−iω0t/2{[

cos(ω0

2t)

+ i sin(ω0

2t)]

−|H ′ab|2t~2ω0

[t(

sin(ω0

2t)− i cos

(ω0

2t))

+2i

ω0sin(ω0

2t)]}

= e−iω0t/2

{e iω0t/2 −

|H ′ab|2t~2ω0

[−ite iω0t/2 +

1

ω0


)]}= 1−

|H ′ab|2t~2ω0

[−it +

(1− e−iω0t

)ω0

]

= 1+i

~2ω0|H ′ab|2

[t +

(e−iω0t − 1

)iω0

]





{cos(ω0

2t)−

2|H ′ab|2t~2ω2

0

sin(ω0

2t)

+i

(1−

2|H ′ab|2

~2ω20

)[sin(ω0

2t)

+2|H ′ab|2t~2ω2

0

cos(ω0

2t)]}

= e−iω0t/2{[

cos(ω0

2t)

+ i sin(ω0

2t)]

−|H ′ab|2t~2ω0

[t(

sin(ω0

2t)− i cos

(ω0

2t))

+2i

ω0sin(ω0

2t)]}

= e−iω0t/2

{e iω0t/2 −

|H ′ab|2t~2ω0

[−ite iω0t/2 +

1

ω0


)]}= 1−

|H ′ab|2t~2ω0

[−it +

(1− e−iω0t

)ω0

]= 1+

i

~2ω0|H ′ab|2

[t +

(e−iω0t − 1

)iω0

]



phys.iit.eduphys.iit.edu/~segre/phys406/19s/lecture_25.pdf · problem 11.7 (cont.) since the exact...

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