Exercise # 8 In class we introduced Rabi's model, a two level system with a time-dependent perturbation H = H, = 0 and H = (Hm)" = aħe"st where a and w are real parameters. This leads to the dynamical equation (1) where wn = (Es – E.)/h is a positive constant provided E, < Eg. Then the state vector is written in terms of the solutions, ca(t) and ca(t), as |V(t)) = ca(t) e¬iEst/E.) + cs(t) e¬iEst/h\E). In what follows, you solve Eq. (1) for the initial state |¥(0)) = |E.). 1. Let 3 = w – wo and write down Eq.(1) in components. You should get a = -iaest 4 = -iae-ist c, (2a) (2b) 2. Take the time derivative of the above. You get two coupled second-order equations, č = ae* (3c, – iċ) ö, = -ae-i*(Bc, + iċ.) (3a) %3D (3b) 3. Eliminate ca and ca from Eq.(3b) by using Eq.(2). You should get the following: ä + ißi, + a²q = 0 (4)
Exercise # 8 In class we introduced Rabi's model, a two level system with a time-dependent perturbation H = H, = 0 and H = (Hm)" = aħe"st where a and w are real parameters. This leads to the dynamical equation (1) where wn = (Es – E.)/h is a positive constant provided E, < Eg. Then the state vector is written in terms of the solutions, ca(t) and ca(t), as |V(t)) = ca(t) e¬iEst/E.) + cs(t) e¬iEst/h\E). In what follows, you solve Eq. (1) for the initial state |¥(0)) = |E.). 1. Let 3 = w – wo and write down Eq.(1) in components. You should get a = -iaest 4 = -iae-ist c, (2a) (2b) 2. Take the time derivative of the above. You get two coupled second-order equations, č = ae* (3c, – iċ) ö, = -ae-i*(Bc, + iċ.) (3a) %3D (3b) 3. Eliminate ca and ca from Eq.(3b) by using Eq.(2). You should get the following: ä + ißi, + a²q = 0 (4)
Related questions
Question
Plese do 1, 2, and 3
![Exercise #8
In class we introduced Rabi’s model, a two-level system with a time-dependent perturbation
\[ H'_{aa} = H'_{bb} = 0 \quad \text{and} \quad H'_{ab} = (H'_{ba})^* = \alpha h e^{i \omega t} \]
where \(\alpha\) and \(\omega\) are real parameters. This leads to the dynamical equation
\[ i \hbar \frac{d}{dt} \begin{bmatrix} c_a \\ c_b \end{bmatrix} = \begin{bmatrix} 0 & \alpha h e^{i(\omega - \omega_0)t} \\ \alpha h e^{i(\omega_0 - \omega)t} & 0 \end{bmatrix} \begin{bmatrix} c_a \\ c_b \end{bmatrix} \tag{1} \]
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 \(c_b(t)\), as
\[ |\Psi(t)\rangle = c_a(t) e^{-iE_a t/\hbar}|E_a\rangle + c_b(t) e^{-iE_b t/\hbar}|E_b\rangle. \]
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
\[ \dot{c}_a = -i \alpha e^{i \beta t} c_b \quad \tag{2a} \]
\[ \dot{c}_b = -i \alpha e^{-i \beta t} c_a \quad \tag{2b} \]
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) \quad \tag{3a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa4426dc-92c3-4ac7-bd5a-9ba1dd10b9af%2F5569b589-4e15-4f66-9b19-507574d0092a%2Fp04biv7_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
\[ H'_{aa} = H'_{bb} = 0 \quad \text{and} \quad H'_{ab} = (H'_{ba})^* = \alpha h e^{i \omega t} \]
where \(\alpha\) and \(\omega\) are real parameters. This leads to the dynamical equation
\[ i \hbar \frac{d}{dt} \begin{bmatrix} c_a \\ c_b \end{bmatrix} = \begin{bmatrix} 0 & \alpha h e^{i(\omega - \omega_0)t} \\ \alpha h e^{i(\omega_0 - \omega)t} & 0 \end{bmatrix} \begin{bmatrix} c_a \\ c_b \end{bmatrix} \tag{1} \]
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 \(c_b(t)\), as
\[ |\Psi(t)\rangle = c_a(t) e^{-iE_a t/\hbar}|E_a\rangle + c_b(t) e^{-iE_b t/\hbar}|E_b\rangle. \]
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
\[ \dot{c}_a = -i \alpha e^{i \beta t} c_b \quad \tag{2a} \]
\[ \dot{c}_b = -i \alpha e^{-i \beta t} c_a \quad \tag{2b} \]
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) \quad \tag{3a
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
