Question

3. Consider a system whose Hamiltonian H, admits two eigenstates y, (with eigenvalues F) and v, (with eigenvalues E,). Assume E, E, and they are () orthogonal, (ifi) normalized and (ii) non-degenerate. After the perturbation is on the diagonal matrix elements become zero ie, <4, l H'l Ψ)-(4, I H'ly,)-0, while the off diagonal equals to a constant value ie. (v, l H'l%)-(wil H'ly)-c Using the 2nd order perturbation theory evaluate the energy of the perturbed system.

Answer 1

Given H psi_a = E_a psi_a H psi_b = E_b psi_b The second order correlation to the energy E^(2)_a = sigma_m notequalto a Vertical-bar < psi_m Vertical-bar H' Vertical-bar psi_a > Vertical-bar ^2/E_a - E_m = Vertical-bar < psi_b Vertical-bar H' Vertical-bar psi_a > Vertical-bar ^2/E_s - E_b = Vertical-bar alpha alpha*]^2/E_a - E_b = ( Vertical-bar alpha Vertical-bar ^2)^2/E_a - E_b = -( Vertical-bar alpha Vertical-bar ^2)^2/E_b - E_a parallel, E^(2)_b = sigma_m notequalto b Vertical-bar < psi_m Vertical-bar H' Vertical-bar psi_b > Vertical-bar ^2/E_b - E_a = + ( Vertical-bar alpha Vertical-bar ^2)^2/E_b - E_a therefore E^potential_a = E_a - ( Vertical-bar alpha Vertical-bar ^2)^2/E_b - E_a E^kg/m_b = E_b + ( Vertical-bar alpha Vertical-bar ^2)^2/E_b - E_a.