Question

1. The analysis of the two state system in time-dependent perturbation theory can be simplified by taking the perturbation Hamiltonian to be $\frac{V}{2}e^{-i\omega t}$ $H' = $ (1) and the matrix elements between the two states $|a\rangle$ and $|b\rangle$ to be $H_{ba}' = \frac{V_{ba}}{2}e^{-i\omega t}$, $H_{ab}' = \frac{V_{ab}}{2}e^{i\omega t}$. (2) (The perturbation Hamiltonian is complex in this \"rotating wave ap- proximation\" but the matrix elements are taken so that the perturba- tion matrix is Hermitian.) a. If, as in the lecture, we write $\psi = c_a(t)\psi_a e^{-iE_at/\hbar} + c_b(t)\psi_b e^{-iE_bt/\hbar}$ (3) the equations for the coefficients $c_a$ and $c_b$ are $\dot{c}_a = -\frac{i}{\hbar}H_{ba}'e^{-i\omega_0 t}c_b$, $\dot{c}_b = -\frac{i}{\hbar}H_{ab}'e^{i\omega_0 t}c_a$ (4) Solve the equations for the time-dependent coefficients with the initial conditions $c_a(0) = 1$, $c_b(0) = 0$. Express your results for $c_a(t)$ and $c_b(t)$ in terms of the Rabi flopping frequency $\omega_r = \frac{1}{2}\sqrt{(\omega - \omega_0)^2 + (|V_{ab}|/\hbar)^2}$ (5) b. Determine the transition probability, $P_{a\to b}(t)$, and show that it never exceeds 1. Confirm that $|c_a(t)|^2 + |c_b(t)|^2 = 1$. c. Check that $P_{a\to b}(t)$ reduces to the perturbation theory result, $P_{a\to b}(t) = |c_b(t)|^2 \approx \frac{|V_{ab}|^2}{\hbar^2}\frac{\sin^2[(\omega_0 - \omega)t/2]}{(\omega_0 - \omega)^2}$ (6) when the perturbation is \"small\", and state precisely what small means in this context, as a constraint on V. d. At what time does the system first return to its initial state?

          1. The analysis of the two state system in time-dependent perturbation
theory can be simplified by taking the perturbation Hamiltonian to be
$\frac{V}{2}e^{-i\omega t}$
$H' = $
(1)
and the matrix elements between the two states $|a\rangle$ and $|b\rangle$ to be
$H_{ba}' = \frac{V_{ba}}{2}e^{-i\omega t}$, $H_{ab}' = \frac{V_{ab}}{2}e^{i\omega t}$.
(2)
(The perturbation Hamiltonian is complex in this \"rotating wave ap-
proximation\" but the matrix elements are taken so that the perturba-
tion matrix is Hermitian.)
a. If, as in the lecture, we write
$\psi = c_a(t)\psi_a e^{-iE_at/\hbar} + c_b(t)\psi_b e^{-iE_bt/\hbar}$ (3)
the equations for the coefficients $c_a$ and $c_b$ are
$\dot{c}_a = -\frac{i}{\hbar}H_{ba}'e^{-i\omega_0 t}c_b$, $\dot{c}_b = -\frac{i}{\hbar}H_{ab}'e^{i\omega_0 t}c_a$
(4)
Solve the equations for the time-dependent coefficients with the
initial conditions $c_a(0) = 1$, $c_b(0) = 0$. Express your results for
$c_a(t)$ and $c_b(t)$ in terms of the Rabi flopping frequency
$\omega_r = \frac{1}{2}\sqrt{(\omega - \omega_0)^2 + (|V_{ab}|/\hbar)^2}$ (5)
b. Determine the transition probability, $P_{a\to b}(t)$, and show that it
never exceeds 1. Confirm that $|c_a(t)|^2 + |c_b(t)|^2 = 1$.
c. Check that $P_{a\to b}(t)$ reduces to the perturbation theory result,
$P_{a\to b}(t) = |c_b(t)|^2 \approx \frac{|V_{ab}|^2}{\hbar^2}\frac{\sin^2[(\omega_0 - \omega)t/2]}{(\omega_0 - \omega)^2}$ (6)
when the perturbation is \"small\", and state precisely what small
means in this context, as a constraint on V.
d. At what time does the system first return to its initial state?

1. The analysis of the two state system in time-dependent perturbation
theory can be simplified by taking the perturbation Hamiltonian to be
(V)/(2)e^-iω t
H' =
(1)
and the matrix elements between the two states |a⟩ and |b⟩ to be
Hba' = (Vba)/(2)e^-iω t, Hab' = (Vab)/(2)e^iω t.
(2)
(The perturbation Hamiltonian is complex in this r̈otating wave ap-
proximationb̈ut the matrix elements are taken so that the perturba-
tion matrix is Hermitian.)
a. If, as in the lecture, we write
ψ = ca(t) e^-iEat/ħ + cb(t) e^-iEbt/ħ (3)
the equations for the coefficients ca and cb are
ċa = -(i)/(ħ)Hba'e^-iω0 tcb, ċb = -(i)/(ħ)Hab'e^iω0 tca
(4)
Solve the equations for the time-dependent coefficients with the
initial conditions ca(0) = 1, cb(0) = 0. Express your results for
ca(t) and cb(t) in terms of the Rabi flopping frequency
= (1)/(2)√((ω - ω0)^2 + (|Vab|/ħ)^2) (5)
b. Determine the transition probability, Pa→ b(t), and show that it
never exceeds 1. Confirm that |ca(t)|^2 + |cb(t)|^2 = 1.
c. Check that Pa→ b(t) reduces to the perturbation theory result,
Pa→ b(t) = |cb(t)|^2 ≈(|Vab|^2)/(ħ^2)(sin^2[(ω0 - ω)t/2])/((ω0 - ω)^2) (6)
when the perturbation is s̈mall,̈ and state precisely what small
means in this context, as a constraint on V.
d. At what time does the system first return to its initial state?

Added by Veronica M.

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