Question

1. Consider a charged particle bound in the harmonic oscillator potential V(x) = mw x2. A weak electric field is applied to the system such that the potential energy, U(X), now has an extra term: V(x) = -qEx. We write the full Hamiltonian as H = Ho +V(x) where Ho = Px +mw x2 V(x) = –qEx. (a) Write down the unperturbed energies, EO. (b) Find the first-order correction to E . (c) Calculate the second-order correction to E . (d) Show that the perturbed system can be solved exactly by completing the square in the Hamiltonian. (e) Compare the exact energies to the perturbation results.

Answer 1

@ Let us define two operator at and a att Varmu (mwz-ip) as vos mw (mwx tip) : ata= akimw (mwx-cp)(mwx+ cp) . Low [mwk'+p°t i mw(* P-px)] w [p'z fmwx)"] +%*[ .P] - Hw-V (As [XP]z in and H="m [P+(mwx)"] —O His Hamiltonian we also define the number operator 1 N= ata We denote energy eigenket of N by its eigenvalue n, so Nin) = nin) From equation we get H = fata +93) w = (N+18) *W [as N='ata] or Hin) = (N+y)*w*) = (1+y):5Win) :. Eri S = (+13) *W [EXC) is the unperturbed energy : 603-YW ground state energy E, a) = 34 mW [ First - Excited state enumiry)

© We know that first order.correction to Erio) is = En = {1V(x))n) where V(x) is the perturbed term In question y(x) = -98x :- Ena -(19€x1n) :-98<mixing 5-98<ni v Wow (a+at)in] [As x= V kw (atat)] = -9€ (nlarat/n) 5 mw TAS <niatin) = Voti <nin+1) = 0] (0]@17) - n (n]n-l} = 0

© We know that second order correction to Friis E(?)- (QE)' kxmlin) man (nom) ka m is the mth state » nth state n 1 = (98) W , 5 [Voti Sm, nhi trn Sm, n-17% to 2mw Z. man (nom) - (98) h. [ln +1) 8minti+n6m,na!) min nom = les operation = (95)". [-(n+1)+n] try] =-(9€) zmwl o Full Hamiltonian is He HotV(2) or H = Pxom +% mwe and abstract - Pozamye mw-(x'-246%) = Pilzm+ 12 mw4yx- 2016. a. carm.) - hitta) [Add ( ) = Päm+ y mow(X-WW) - emily ymwe [ completing the - Påfm+ v mestay [sad ya x-qe Harmitonun] = Ho - oer — square termin - he is the perturbed term in 2.mw Hamiltonian

e) so the energy due to perturbation is Ens En C) + Ey(") + Ent?) = {Y3)#w+0+ ( a ) =(n+1) 5w - (98) žmon from equation 4 we get H = Ho - (99% mo :: E = {n+%) *W-(92) mwr it is the exact energy both are same.