Question

Problem #4 - 20 PTS → Evaluate the first order correction to the energy of the nth state of one-dimensional harmonic oscillator having the potential energy V = mox? + bx*, where bx* « mox? using raising and lowering operators At and A. Remember x = ( '(A+ A+). 11/2

Answer 1

Given perturbation, H'=bx4 where, * below 2 A++*+) 2 mw ::x= [(A+++ (a*%+2 AA] (A- A+ = ( ) zinte [ (A+)**(**)*+2. ] - [14]++at} + ] stapa? y = Get [la 94 600] +13 [use a thug)ab*+c++2(at +be+ac] :: van 'n va a far nga +(4+72+ (4*)]] the two cars3"t (bytu 1 221 1) +2(4=)? +20A+)2] 2-то where B = (A=3"+(4+]+ 2[(A= 3°4 (A+)"}

Now, the first order correction in the nth energy level is E") = (n Hlln) z (nIbx4 in) = (npb 62 (2 + a)n) using, Atiny = cIn+1) A ln>=6-10-17 where, ct, c are constants & orthonomality <nlmy=0 for min 1. 1 for man when B is operated on ket my it definitely changes due to all operators say an> ln'y nan" - Knin') = 0; Only / term will contribute to the correction Ona m EFZ b (bit) I z inw! . = 36 k2 8 mu?