Question

We study the vibrations in a diatomic molecule with the reduced mass m. Let x = R − R_e, which is the bonding distance deviation from equilibrium distance. Hamiltonian operator consist of two parts: H = H⁽⁰⁾ + H⁽¹⁾, where H⁽⁰⁾ is the Hamiltonian operator to a harmonic oscillator with force constant k, and H⁽¹⁾ = λx³ (λ is a constant < 0).

* Calculate the first order correction to the energy state v.

Answer 1

I apologise but i could not understand the form of hamiltonian. It could be either of 3 times lambda or lambda times x cubed. I solved for both. Hope this helps.

A the given problem the wave function of the allator has not been provided. so, we & hall esume the most accepted form of the are function of ground state of Harmonie asun - da/ where a cono unt. Now; when a system, a perturbation the lot order Hill is correction operative on to energy is given by the En ot go 41 where 4 = normalised Thus, in the perturbation is given in problem, a 3 Hamiltonian of ara [ 9 could not 1. differentiale assuming Hl. 10 3.4o da das Now. tia): 3. e-da is an odd-function +(m) = laje alaga Hence ; pt ansvennda 2 t0) = 0 Thus; if f(1) = berturbation x², there shall be no

Assuming to Å (1) = 20 3n; we calculate &" - Stt 32.4 da sistema I wen function 539. (entry as the nada por = 69184) a jedan dan = 61.662 463 Thus; 1st order in energy = 32. perturbation
i calculated using v=0. But from the nature of the problem, even if you solve using a general form it will be same. Below are the results:

O bey tuy bokon for wth state general formula Yuln): of Hill 2 A All the am? Yuda 2 M = Aripta x3, e-?Manda 0 [ Reasoning already provided gf H11) = 32 El a 1 ay 32. Y v de -ZM 32A2 1 e 2 Mn da form 12. =) General