Question

calculation of geometries. Question 5 (8 points (2,3,3)) For a particular problem, the wavefunction is written as a superposition of two orthonormal functions (xi): Y = CX,+C2X2. The coefficients are determined with variation theory. The following integrals are given in eV): H, =Sxix, dt =-10 H2 = Sxixdr = -0.5 H,2 = xixdt =-1.0 a) Give the secular equations. b) Find the energies. c) Find the wavefunction for the state with the lowest energy (if you did not find the energies in the previous question, use E = 0.28 eV).

Answer 1

pagene: 5) Solution: . Given warelength is written as a Superposition of these orthonormal functions (%) : 4 = C,X,+cq x • Now Consider the given integrals H, =Sx by dq = -1,0 Had = f g dq = -05 H2 = Solt, dq = -10 het S12 = Sai = 0 (overlap mat ix Elements) S = Saa = 1 (a) Secular determinent : Hi - Esin #12 - Esial Hai - Esai H₂2 - E5221 Now Apply the values in Secular Equabion 1-10-E - 1.0 1 -10 -0.5-6 (b) het Energies are: (-1.0-6) (-0.5-6) - [1.0) (-1,0) = 0 -) E + 15€ -0.5 = 0 = -1.57 $2.2572 =-1,511 4.25 =) 2 -0.75 +1.03

page no: E = - 1.78 ev Eg = 0.28en (0) For E, = -178 ev :- (-1,0 +178 -1.0 to ) ( :-( :) - 1.0 -0.5+1.8 Let us consider 0.78 x, My = -2, + 1.287,30 2 J whereas, [W, = 1,2882] :: 14>= (1.28) By using normalised care fonction 143 - vol2 (126) a votace (10) ~ 10.79 1 (0.62) #tena pscored X