Question

3. A one dimensional system of two distinguishable particles has the Hamiltonian $H(t) = H_0 + H'(t)$ where $H_0 = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{1}{2}m_1\omega_1^2x_1^2 + \frac{1}{2}m_2\omega_2^2x_2^2$, $H'(t) = \lambda(at + b)x_1x_2$. Initially at time $t = 0$ the system is in the ground state of $H_0$. Treating $H'(t)$ as a perturbation, determine at time $t$, all first order transition probabilities to the excited states of $H_0$.

          3. A one dimensional system of two distinguishable particles has the Hamiltonian $H(t) = H_0 + H'(t)$ where

$H_0 = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{1}{2}m_1\omega_1^2x_1^2 + \frac{1}{2}m_2\omega_2^2x_2^2$,
$H'(t) = \lambda(at + b)x_1x_2$.

Initially at time $t = 0$ the system is in the ground state of $H_0$. Treating
$H'(t)$ as a perturbation, determine at time $t$, all first order transition
probabilities to the excited states of $H_0$.

3. A one dimensional system of two distinguishable particles has the Hamiltonian H(t) = H0 + H'(t) where

H0 = (p1^2)/(2m1) + (p2^2)/(2m2) + (1)/(2)m1ω1^2x1^2 + (1)/(2)m2ω2^2x2^2,
H'(t) = λ(at + b)x1x2.

Initially at time t = 0 the system is in the ground state of H0. Treating
H'(t) as a perturbation, determine at time t, all first order transition
probabilities to the excited states of H0.

Added by Gary L.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Frank Deng

06/06/2022

Step 1

Step 1: We are given the Hamiltonian of the system, H(t) = Ho + H'(t), where Ho represents the unperturbed Hamiltonian and H'(t) represents the perturbation. Show more…

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3. A one dimensional system of two distinguishable particles has the Hamil tonian H(t)=Ho+H't)where Pi p2 1 Ho 2m1 2m2 2 1 22 -=m2W2 2 Ht=Xat+bxx2 Initially at time t =0 the system is in the ground state of Ho. Treating H'(t) as a perturbation, determine at time t, all first order transition probabilities to the excited states of Ho.