Question

9.5 A particle of mass m is in the ground state in the harmonic oscillator potential A small perturbation Bx6 is added to this potential (a) What are the units of ?? (b) How small must B be in order for perturbation theory to be valid? (c) Calculate the first-order change in the energy of the particle.

Answer 1

Given that (a) implies The harmonic oscillators Potential for ground state v (x) = 1/2 kx^2 Now Hamiltonian for ground state implies H = -h^2 /2m partial differential^2/partial differential x^2 + v (x) + H'_p H'_p = beta x^6 v (x) = 1/2 kx^2 H = -h^2 /2m partial differential^2/partial differential x^2 + 1/2 kx^2 + beta x^6 implies beta x^6 has the units of energy is kg - m^2 s^-2 x has units of length (m) beta (m)^6 = kg m^2 s^-2 beta = kg - m^-4 s^-2 units of beta is = kg - m^-4 s^-2 (b) The Theory beta should belong to 0 < beta < L because. Beta greaterthan 1 Taylor expansion of potential about x = 0 will diverge x rightarrow plusminus infinity so beta should be 0 < beta < L such that expansion is conversing (c) The first-order change in the energy particle in ground state < 0/beta x^6 0 > implies beta < 0 mod (x^6) 0 >