MASSACHUSETTS INSTITUTE OF TECHNOLOGY5.73 Quantum Mechanics IFall, 2002Professor Robert W. FieldProblem Set #6DUE: At the start of Lecture on Friday, October 25.Reading: CTDL pp. 290-307, 1148-1155. [optional, 1169-1199]Problems:1. You are going to derive the “x–k” relationships given on pages 17-4 and 17-5.You have worked out the relationships between m, k, a, and b inHp x x x=+++2234212/ mkaband Y00, ωe, ωexe inEhcY n xnneee/(/)(/)=+ + − +00212 12ωω,for a single-oscillator (diatomic) molecule. Now you are going to consider 3N–6anharmonically coupled, anharmonic oscillators in an N-atom polyatomicmolecule. The only thing that is different is that there are many more terms inH(1) and the En()2 terms involve short summations over several combinations ofoscillators. In all of your derivations ignore the hmω12/ factor that makes qdimensionless.Chemistry 5.73 Page 2Problem Set #6A. xii appears in the energy level expression asExnnn n ii iN12 3 6122…−=+()L /.The first term in the equation for xii on page 17-4 comes from one of thetwo strictly diagonal matrix elements of H(1). These are the ∆ni = 0 matrixelements of qi4. Derive this term.B. The second term in xii comes from matrix elements of terms like qqis2.There are several classes of such matrix elements: (∆ni, ∆ns) = (1,0),(–1,0), (1,2), (1,–2), (–1,2), and (–1,–2). The first two have only ± ωi inthe denominator, while the other four have energy denominators of theform ±ωi ± 2ωs. Sum these terms and derive the second term in the xiiequation.C. The first term in xij on page 17-5 comes from another strictly diagonalmatrix element of H(1)Exnnnn n ij i jN12 3 612 12…−=+()+()L //which comes from diagonal (∆ni = 0, ∆nj = 0) matrix elements of qqij22.Derive this contribution to xij.D. The second term in xij on page 17-5 comes from ∆ni = 0, ∆nj = 0 matrixelements of terms like qqit2 and qqjt2. The selection rules for qt is ∆nt = ±1and the energy denominator will be ±ωt. Derive this term.E. [OPTIONAL] The final term in xij comes from matrix elements of termslike qiqjqt. There are eight such terms: (∆ni, ∆nj, ∆nt) = (1,1,1), (–1,1,1),… (–1,–1,–1) with corresponding energy denominators. Derive this term.2. In addition to the x-k relationships by which the vibrational anharmonicityconstants, xij, are related to the cubic and quartic anharmonicity constants of thepotential surface, perturbation theory can be used to derive the relationships of therotational anharmonicity constants, αiA, B, or C[] to the coefficients of the qi3 cubicanharmonicity term in the potential, e.g.BBvvv v eiNiiN12 3613612,,/.…=−−=− +()∑αFor a polyatomic molecule, you need to know the partial derivatives of thereciprocal moments of inertia with respect to each of the normal coordinatedisplacements, and that information comes from a normal coordinate analysis (FChemistry 5.73 Page 3Problem Set #6and G matrices) that is beyond the scope of this class. Here, you will solve thesimpler problem of Bv = Be - αe(v + 1/2) for a diatomic molecule. The rotational“constant” operator is proportional to R–2,xRRRRxRxRBBxRxReeeeveee=−=−++…=−++…−−222212 312 3 .So, by writing H as H(0) + H(1)HaaaaHxx() ≤ ≤/()()/()012221312121821hcckBJJBhcRhc a hc B R J Jeeeee=+()πµ()++=πµ=()−()+and the second-order corrections to EvJ will contain three termsEhcahcvJ v JE E hcBRJJvvE E hcaBhcRJJvv v vEEvJvvJvJeevvJvJeevvJvJ()() () () ()() (22320022222003004121=′−()++()′−()−+()′′−′′′′′′∑∑∑xxxx))()hcwhere the first term is a contribution to ωexe, the second term gives the centrifugaldistortion DBeee≈()432ω, and the third term is the one that will contain the desired(v+1/2)J(J+1) dependence of the αe term. Note that there is also a first ordercorrection to the energy E hcBRJ J vJ vJvJee()12231=+()x. This gives the harmoniccontribution to αe, which is usually smaller and of opposite sign to the cubic term(when a < 0). Derive the two contributions to αe and express them in terms of Be,ωe, µ, and fundamental constants (h, c, etc.).3. CTDL, page 205,