Quantum Physics 2011/12

so√ mω∫γπ¯hx 4 exp (−mωx 2 /¯h) dx = 3γ( ¯h) 22mωBut to do it yourself, its best to use raising and lowering operators.The unperturbed energy eigenstates satisfyĤ 0 |n〉 = E n |n〉 with E n = ( n + 1 2)¯hω, n = 0, 1, 2, . . .We can write ˆx in terms of â and â † :√¯h (âˆx =)+ â†2mωThus∆E n = γ( ) 2¯h〈n| ( â + â †) 4|n〉2mωExpanding the bracket ( â + â †) 4looks pretty awful until you realise that, from theraising and lowering properties and the orthonormality of the energy eigenstates, onlyterms which contain an equal number of raising and lowering operators will give anon-zero contribution to the diagonal matrix element 〈n|∆ ˆV |n〉. Thus∆E n = γ( ) 2¯h〈n| ( â 2 â †2 + ââ † ââ † + ââ † â † â + â † âââ † + â † ââ † â + â †2 â 2) |n〉2mωFor the ground state, n = 0, this simplifies even further because â|0〉 = 0, so the third,fifth and sixth terms all give zero. Evaluating the remaining three terms,Thusâ 2 â †2 |0〉 = â 2 â † |1〉 = √ 2 â 2 |2〉 = √ 2 √ 2 â|1〉 = 2|0〉ââ † ââ † |0〉 = ââ † â|1〉 = ââ † |0〉 = â|1〉 = |0〉â † âââ † |0〉 = â † ââ|1〉 = â † â|0〉 = 0∆E 0 = 3γ( ) 2 ( ) 2¯h¯h〈0|0〉 = 3γ2mω2mωFor the general case, we have to work a little harder;â 2 â †2 |n〉 = √ n + 1 â 2 â † |n + 1〉 =√(n + 1)(n + 2) â 2 |n + 2〉= √ n + 1 (n + 2)â|n + 1〉 = (n + 1)(n + 2)|n〉ââ † ââ † |n〉 = √ n + 1 ââ † â|n + 1〉 = (n + 1) ââ † |n〉 = (n + 1) √ n + 1 â|n + 1〉= (n + 1) 2 |n〉â † âââ † |n〉 = √ n + 1 â † ââ|n + 1〉 = (n + 1)â † â|n〉 = (n + 1) √ n â † |n〉= n(n + 1)|n〉