3. Eliminate c, and c, from Eq.(3b) by using Eq.(2). You should get the following: ö + ißå, + a°cq = 0 (4) 4. Solve this ODE (Hint: assume c, = e*). Then you should get the following general solution, c = e-i/2(Act + Be-«t) (5) where A and B are integration constant, and 2 = Va² + (3/2)² . 5. Based on the initial state, what is ca(0)? Show that this value leads to B = -A in Eq.(5). %3D
3. Eliminate c, and c, from Eq.(3b) by using Eq.(2). You should get the following: ö + ißå, + a°cq = 0 (4) 4. Solve this ODE (Hint: assume c, = e*). Then you should get the following general solution, c = e-i/2(Act + Be-«t) (5) where A and B are integration constant, and 2 = Va² + (3/2)² . 5. Based on the initial state, what is ca(0)? Show that this value leads to B = -A in Eq.(5). %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please do 4, 5, and 6
![Exercise # 8
In class we introduced Rabi's model, a two level system with a time-dependent perturbation
Ha = H, = 0 and H = (H)* = aheut
where a and w are real parameters. This leads to the dynamical equation
aħe'(w-ww)t]
ih-
(1)
=
where wo
(Es – Ea)/h is a positive constant provided Ea < Eg. Then the state vector is
written in terms of the solutions, Ca(t) and c(t), as
|V(t)) = ca(t) e¯iEat/h|Ea) + c»(t) e¯-iEst/h|EL).
In what follows, you solve Eq. (1) for the initial state
|V(0)) = |Ea).
1. Let 3 = w - wo and write down Eq.(1) in components. You should get
Ca = -iaest
(2a)
Cb
C = -iae-ißt
(2b)
Ca
2. Take the time derivative of the above. You get two coupled second-order equations,
č, = aet (Bc, – ić,)
č, = -ae-ist (Bc, + ic,)
(За)
(3b)
3. Eliminate ca and ca from Eq.(3b) by using Eq.(2). You should get the following:
č, + ißc, + a²c, = 0
(4)
4. Solve this ODE (Hint: assume c, = et). Then you should get the following general
solution,
Cb = e-ißt/2(Aest + Be¬it)
(5)
where A and B are integration constant, and 2 = Va? + (B/2)2 .
5. Based on the initial state, what is c(0)? Show that this value leads to B = -A in
Еq (5).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa4426dc-92c3-4ac7-bd5a-9ba1dd10b9af%2F39ef7708-ded0-4154-b653-f292d26e7e03%2F69908tl_processed.png&w=3840&q=75)
Transcribed Image Text:Exercise # 8
In class we introduced Rabi's model, a two level system with a time-dependent perturbation
Ha = H, = 0 and H = (H)* = aheut
where a and w are real parameters. This leads to the dynamical equation
aħe'(w-ww)t]
ih-
(1)
=
where wo
(Es – Ea)/h is a positive constant provided Ea < Eg. Then the state vector is
written in terms of the solutions, Ca(t) and c(t), as
|V(t)) = ca(t) e¯iEat/h|Ea) + c»(t) e¯-iEst/h|EL).
In what follows, you solve Eq. (1) for the initial state
|V(0)) = |Ea).
1. Let 3 = w - wo and write down Eq.(1) in components. You should get
Ca = -iaest
(2a)
Cb
C = -iae-ißt
(2b)
Ca
2. Take the time derivative of the above. You get two coupled second-order equations,
č, = aet (Bc, – ić,)
č, = -ae-ist (Bc, + ic,)
(За)
(3b)
3. Eliminate ca and ca from Eq.(3b) by using Eq.(2). You should get the following:
č, + ißc, + a²c, = 0
(4)
4. Solve this ODE (Hint: assume c, = et). Then you should get the following general
solution,
Cb = e-ißt/2(Aest + Be¬it)
(5)
where A and B are integration constant, and 2 = Va? + (B/2)2 .
5. Based on the initial state, what is c(0)? Show that this value leads to B = -A in
Еq (5).
Transcribed Image Text:6. Show that Eq.(5) can now be written as
C, = 2iAe-ist/2 sin t
(6)
7. Differentiate Eq.(6) and then plug that into Eq.-(2) to find ca. You should get
2A
Ca =
- sin Qt + N cos Nt
(7)
8. Based on the initial state, what is c,(0)? Use this value to find A in Eq.(7).
9. Now put all the pieces of information together to show that the solution of Rabi's
model is
w - wo
Ca(t) = e"(w-ww)t/2
sin Nt
(8a)
cos Nt - i
2Ω
C(t) = e-i(w-wo)t/2
(8b)
sin Nt
where
w - wo
N = 1a2 +
(9)
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