![Page 1: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/1.jpg)
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
![Page 2: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/2.jpg)
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
![Page 3: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/3.jpg)
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
![Page 4: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/4.jpg)
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
![Page 5: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/5.jpg)
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
![Page 6: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/6.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 7: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/7.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 8: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/8.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 9: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/9.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 10: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/10.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 11: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/11.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 12: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/12.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 13: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/13.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 2 / 14
![Page 14: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/14.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 15: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/15.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 16: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/16.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 17: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/17.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 18: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/18.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 19: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/19.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 20: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/20.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 21: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/21.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 22: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/22.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 23: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/23.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 24: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/24.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 25: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/25.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 26: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/26.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 27: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/27.jpg)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 3 / 14
![Page 28: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/28.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 29: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/29.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 30: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/30.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 31: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/31.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 32: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/32.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 33: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/33.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 34: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/34.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 35: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/35.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 36: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/36.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 37: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/37.jpg)
Problem 11.3 (cont.)
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 4 / 14
![Page 38: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/38.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 39: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/39.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 40: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/40.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
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Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 42: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/42.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 43: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/43.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 44: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/44.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 45: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/45.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 46: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/46.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 47: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/47.jpg)
Problem 11.3 (cont.)
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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 5 / 14
![Page 48: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/48.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 49: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/49.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 50: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/50.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 51: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/51.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 52: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/52.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 53: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/53.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 54: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/54.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 6 / 14
![Page 55: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/55.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 56: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/56.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 57: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/57.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 58: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/58.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 59: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/59.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 60: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/60.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 61: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/61.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
]
[i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 62: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/62.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]
≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 63: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/63.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 64: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/64.jpg)
Problem 11.5 (cont.)
ca(t) = 1− i
~
∫ t
0H ′aa(t ′) dt ′, cb(t) = − i
~
∫ t
0H ′ba(t ′)e iω0t′ dt ′
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
0H ′ba(t ′)e iω0t′ dt ′
] [i
~
∫ t
0H ′ba(t ′)e−iω0t′ dt ′
]≈ 0
thus to first order, |ca|2 + |cb|2 ≈ 1
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 7 / 14
![Page 65: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/65.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 66: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/66.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 67: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/67.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 68: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/68.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 69: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/69.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 70: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/70.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 71: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/71.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 72: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/72.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 73: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/73.jpg)
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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 8 / 14
![Page 74: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/74.jpg)
Problem 11.6 (cont.)
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
0H ′ab(t ′)e−iω0t′ dt ′
− 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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 9 / 14
![Page 75: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/75.jpg)
Problem 11.6 (cont.)
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
0H ′ab(t ′)e−iω0t′ dt ′
− 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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 9 / 14
![Page 76: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/76.jpg)
Problem 11.6 (cont.)
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
0H ′ab(t ′)e−iω0t′ dt ′
− 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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 9 / 14
![Page 77: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/77.jpg)
Problem 11.6 (cont.)
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
0H ′ab(t ′)e−iω0t′ dt ′
− 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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 9 / 14
![Page 78: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/78.jpg)
Problem 11.6 (cont.)
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
0H ′ab(t ′)e−iω0t′ dt ′
− 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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 9 / 14
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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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
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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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
![Page 81: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/81.jpg)
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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
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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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
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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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
![Page 84: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/84.jpg)
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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
![Page 85: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/85.jpg)
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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
![Page 86: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/86.jpg)
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
−iω0tcb cb = − i
~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 ′
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 10 / 14
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Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
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Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
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Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
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Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
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Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
![Page 92: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/92.jpg)
Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
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Problem 11.7 (cont.)
c(2)a (t) = 1−
|H ′ab|2
~2
∫ t
0e−iω0t′
[∫ t′
0e iω0t′′ dt ′′
]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
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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 11 / 14
![Page 94: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/94.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
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Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 96: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/96.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
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Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 98: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/98.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 99: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/99.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 100: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/100.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 101: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/101.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 102: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/102.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 103: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/103.jpg)
Problem 11.7 (cont.)
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
e iω0t/2 sin(ω02 t)
=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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 12 / 14
![Page 104: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/104.jpg)
Problem 11.7 (cont.)
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)]}
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 13 / 14
![Page 105: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/105.jpg)
Problem 11.7 (cont.)
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)]}
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 13 / 14
![Page 106: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/106.jpg)
Problem 11.7 (cont.)
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)]}
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 13 / 14
![Page 107: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/107.jpg)
Problem 11.7 (cont.)
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)]}
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 13 / 14
![Page 108: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/108.jpg)
Problem 11.7 (cont.)
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)]}
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 13 / 14
![Page 109: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/109.jpg)
Problem 11.7 (cont.)
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)]}
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 13 / 14
![Page 110: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/110.jpg)
Problem 11.7 (cont.)
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)]}
= 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
(e iω0t/2 − e−iω0t/2
)]}= 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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 14 / 14
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Problem 11.7 (cont.)
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)]}
= 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
(e iω0t/2 − e−iω0t/2
)]}= 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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 14 / 14
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Problem 11.7 (cont.)
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)]}
= 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
(e iω0t/2 − e−iω0t/2
)]}
= 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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 14 / 14
![Page 113: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/113.jpg)
Problem 11.7 (cont.)
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)]}
= 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
(e iω0t/2 − e−iω0t/2
)]}= 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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 14 / 14
![Page 114: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/114.jpg)
Problem 11.7 (cont.)
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)]}
= 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
(e iω0t/2 − e−iω0t/2
)]}= 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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 14 / 14
![Page 115: csrri.iit.educsrri.iit.edu/~segre/phys406/19S/lecture_25.pdf · Problem 11.7 (cont.) Since the exact result for c b is already rst order in H0, we can allow!!! 0 and drop the extra](https://reader036.vdocuments.site/reader036/viewer/2022080717/5f781c3a5a0ee376f200e469/html5/thumbnails/115.jpg)
Problem 11.7 (cont.)
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)]}
= 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
(e iω0t/2 − e−iω0t/2
)]}= 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
C. Segre (IIT) PHYS 406 - Spring 2019 April 25, 2019 14 / 14