7. Differentiate Eq.(6) and then plug that into Eq.(2) to find ca. You should get 24 94/2 ( sin t +N cos Nt (7) 8. Based on the initial state, what is c,a(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)}/2 sin t (8a) ca(t) sin t (8b) where wo N = a2 + (9)
7. Differentiate Eq.(6) and then plug that into Eq.(2) to find ca. You should get 24 94/2 ( sin t +N cos Nt (7) 8. Based on the initial state, what is c,a(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)}/2 sin t (8a) ca(t) sin t (8b) where wo N = a2 + (9)
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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Please do 7, 8, and 9
![### Equation Derivations for Rabi's Model
#### Step 6
Show that Eq. (5) can now be written as:
\[ c_b = 2iA e^{-i \delta t/2} \sin \Omega t \tag{6} \]
#### Step 7
Differentiate Eq. (6) and then plug that into Eq. (2) to find \( c_a \). You should get:
\[ c_a = -\frac{2A}{\alpha} e^{i \delta t/2} \left( -\frac{i \beta}{2} \sin \Omega t + \Omega \cos \Omega t \right) \tag{7} \]
#### Step 8
Based on the initial state, what is \( c_a(0) \)? Use this value to find \( A \) in Eq. (7).
#### Step 9
Now put all the pieces of information together to show that the solution of Rabi’s model is:
\[
c_a(t) = e^{i(\omega - \omega_0) t/2} \left( \cos \Omega t - i \frac{\left( \omega - \omega_0 \right)}{2 \Omega} \sin \Omega t \right) \tag{8a}
\]
\[
c_b(t) = e^{-i(\omega - \omega_0) t/2} \left( \frac{\alpha}{i \Omega} \right) \sin \Omega t \tag{8b}
\]
Where:
\[ \Omega = \sqrt{\alpha^2 + \left( \frac{\omega - \omega_0}{2} \right)^2} \tag{9} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa4426dc-92c3-4ac7-bd5a-9ba1dd10b9af%2F42a0858f-a940-41e5-9963-c4cb64e595d4%2F3z4w3jb_processed.png&w=3840&q=75)
Transcribed Image Text:### Equation Derivations for Rabi's Model
#### Step 6
Show that Eq. (5) can now be written as:
\[ c_b = 2iA e^{-i \delta t/2} \sin \Omega t \tag{6} \]
#### Step 7
Differentiate Eq. (6) and then plug that into Eq. (2) to find \( c_a \). You should get:
\[ c_a = -\frac{2A}{\alpha} e^{i \delta t/2} \left( -\frac{i \beta}{2} \sin \Omega t + \Omega \cos \Omega t \right) \tag{7} \]
#### Step 8
Based on the initial state, what is \( c_a(0) \)? Use this value to find \( A \) in Eq. (7).
#### Step 9
Now put all the pieces of information together to show that the solution of Rabi’s model is:
\[
c_a(t) = e^{i(\omega - \omega_0) t/2} \left( \cos \Omega t - i \frac{\left( \omega - \omega_0 \right)}{2 \Omega} \sin \Omega t \right) \tag{8a}
\]
\[
c_b(t) = e^{-i(\omega - \omega_0) t/2} \left( \frac{\alpha}{i \Omega} \right) \sin \Omega t \tag{8b}
\]
Where:
\[ \Omega = \sqrt{\alpha^2 + \left( \frac{\omega - \omega_0}{2} \right)^2} \tag{9} \]
![**Exercise # 8**
In class, we introduced Rabi’s model, a two-level system with a time-dependent perturbation:
\[ H'_{aa} = H'_{bb} = 0 \quad \text{and} \quad H'_{ab} = (H'_{ba})^* = ahe^{i\omega t} \]
where \( \alpha \) and \( \omega \) are real parameters. This leads to the dynamical equation:
\[ i \hbar \frac{d}{dt} \begin{bmatrix} c_a \\ c_b \end{bmatrix} = \begin{bmatrix} 0 & ah e^{i(\omega - \omega_0)t} \\ ah e^{-i(\omega - \omega_0)t} & 0 \end{bmatrix} \begin{bmatrix} c_a \\ c_b \end{bmatrix} \tag{1} \]
where \( \omega_0 \equiv (E_b - E_a)/\hbar \) is a positive constant provided \( E_a < E_b \). Then the state vector is written in terms of the solutions, \( c_a(t) \) and \( c_b(t) \),
\[ |\Psi(t)\rangle = c_a(t)e^{-iE_a t/\hbar}|E_a\rangle + c_b(t) e^{-iE_b t/\hbar}|E_b\rangle. \]
In what follows, you solve Eq. (1) for the initial state
\[ |\Psi(0)\rangle = |E_a\rangle. \]
1. Let \( \beta \equiv \omega - \omega_0 \) and write down Eq.(1) in components. You should get
\[ \dot{c}_a = -i\alpha e^{i\beta t} c_b \tag{2a} \]
\[ \dot{c}_b = -i\alpha e^{-i\beta t} c_a \tag{2b} \]
2. Take the time derivative of the above. You get two coupled second-order equations,
\[ \ddot{c}_a = \alpha e^{i\beta t} (\beta c_b - i\dot{c}_b) \tag{3a} \]
\[ \ddot{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa4426dc-92c3-4ac7-bd5a-9ba1dd10b9af%2F42a0858f-a940-41e5-9963-c4cb64e595d4%2Fplf3ho_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:
\[ H'_{aa} = H'_{bb} = 0 \quad \text{and} \quad H'_{ab} = (H'_{ba})^* = ahe^{i\omega t} \]
where \( \alpha \) and \( \omega \) are real parameters. This leads to the dynamical equation:
\[ i \hbar \frac{d}{dt} \begin{bmatrix} c_a \\ c_b \end{bmatrix} = \begin{bmatrix} 0 & ah e^{i(\omega - \omega_0)t} \\ ah e^{-i(\omega - \omega_0)t} & 0 \end{bmatrix} \begin{bmatrix} c_a \\ c_b \end{bmatrix} \tag{1} \]
where \( \omega_0 \equiv (E_b - E_a)/\hbar \) is a positive constant provided \( E_a < E_b \). Then the state vector is written in terms of the solutions, \( c_a(t) \) and \( c_b(t) \),
\[ |\Psi(t)\rangle = c_a(t)e^{-iE_a t/\hbar}|E_a\rangle + c_b(t) e^{-iE_b t/\hbar}|E_b\rangle. \]
In what follows, you solve Eq. (1) for the initial state
\[ |\Psi(0)\rangle = |E_a\rangle. \]
1. Let \( \beta \equiv \omega - \omega_0 \) and write down Eq.(1) in components. You should get
\[ \dot{c}_a = -i\alpha e^{i\beta t} c_b \tag{2a} \]
\[ \dot{c}_b = -i\alpha e^{-i\beta t} c_a \tag{2b} \]
2. Take the time derivative of the above. You get two coupled second-order equations,
\[ \ddot{c}_a = \alpha e^{i\beta t} (\beta c_b - i\dot{c}_b) \tag{3a} \]
\[ \ddot{
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