Question

(a) As suggested in class one can model two adjacent inert gas atoms as linear oscillating electric dipoles. Fill in the mathematics between the following two equations: $H_1 = \frac{e^2}{4\pi\epsilon_0} \left[ \frac{1}{R} + \frac{1}{R + x_1 - x_2} - \frac{1}{R + x_1} - \frac{1}{R - x_2} \right]$ and $H_1 \approx -\frac{e^2}{4\pi\epsilon_0} \frac{2x_1x_2}{R^3}$ (b) As suggested in the lecture, a normal mode transformation can change the coupled oscillating dipole problem into a sum of two independent oscillators with slightly modified energies and spring constants. Do the math which uses the normal mode approximation to go from the coupled oscillator hamiltonian $H_0 + H_1 = \frac{p_1^2}{2m} + \frac{Cx_1^2}{2} + \frac{p_2^2}{2m} + \frac{Cx_2^2}{2} - \frac{2e^2x_1x_2}{4\pi\epsilon_0R^3}$ to two uncoupled oscillators $H_0 + H_1 = \frac{P_a^2}{2m} + \frac{C_ax_a^2}{2} + \frac{P_s^2}{2m} + \frac{C_sx_s^2}{2}$ (c) In lecture we argued that the coupling lowers the ground state energy of the two oscillator system as follows: $\hbar\omega_s + \hbar\omega_a = \hbar\omega_0 - \frac{\hbar\omega_0}{8} \left[\frac{2e^2}{\pi\epsilon_0CR^3}\right]^2 = \hbar\omega_0 - \frac{A}{R^6}$ Prove this statement.

          (a) As suggested in class one can model two adjacent inert gas atoms as linear oscillating electric dipoles. Fill in the mathematics between the following two equations:
$H_1 = \frac{e^2}{4\pi\epsilon_0} \left[ \frac{1}{R} + \frac{1}{R + x_1 - x_2} - \frac{1}{R + x_1} - \frac{1}{R - x_2} \right]$
and
$H_1 \approx -\frac{e^2}{4\pi\epsilon_0} \frac{2x_1x_2}{R^3}$
(b) As suggested in the lecture, a normal mode transformation can change the coupled oscillating dipole problem into a sum of two independent oscillators with slightly modified energies and spring constants. Do the math which uses the normal mode approximation to go from the coupled oscillator hamiltonian
$H_0 + H_1 = \frac{p_1^2}{2m} + \frac{Cx_1^2}{2} + \frac{p_2^2}{2m} + \frac{Cx_2^2}{2} - \frac{2e^2x_1x_2}{4\pi\epsilon_0R^3}$
to two uncoupled oscillators
$H_0 + H_1 = \frac{P_a^2}{2m} + \frac{C_ax_a^2}{2} + \frac{P_s^2}{2m} + \frac{C_sx_s^2}{2}$
(c) In lecture we argued that the coupling lowers the ground state energy of the two oscillator system as follows:
$\hbar\omega_s + \hbar\omega_a = \hbar\omega_0 - \frac{\hbar\omega_0}{8} \left[\frac{2e^2}{\pi\epsilon_0CR^3}\right]^2 = \hbar\omega_0 - \frac{A}{R^6}$
Prove this statement.

(a) As suggested in class one can model two adjacent inert gas atoms as linear oscillating electric dipoles. Fill in the mathematics between the following two equations:
H1 = (e^2)/(4πϵ0)[ (1)/(R) + (1)/(R + x1 - x2) - (1)/(R + x1) - (1)/(R - x2)]
and
H1 ≈ -(e^2)/(4πϵ0)(2x1x2)/(R^3)
(b) As suggested in the lecture, a normal mode transformation can change the coupled oscillating dipole problem into a sum of two independent oscillators with slightly modified energies and spring constants. Do the math which uses the normal mode approximation to go from the coupled oscillator hamiltonian
H0 + H1 = (p1^2)/(2m) + (Cx1^2)/(2) + (p2^2)/(2m) + (Cx2^2)/(2) - (2e^2x1x2)/(4πϵ0R^3)
to two uncoupled oscillators
H0 + H1 = (Pa^2)/(2m) + (Caxa^2)/(2) + (Ps^2)/(2m) + (Csxs^2)/(2)
(c) In lecture we argued that the coupling lowers the ground state energy of the two oscillator system as follows:
ħ + ħ = ħω0 - (ħω0)/(8)[(2e^2)/(πϵ0CR^3)]^2 = ħω0 - (A)/(R^6)
Prove this statement.

Added by Josefina K.

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