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Advanced Physical Chemistry

Advanced Physical Chemistry. G. H. CHEN Department of Chemistry University of Hong Kong. Quantum Chemistry. G. H. Chen Department of Chemistry University of Hong Kong. Emphasis Hartree-Fock method Concepts Hands-on experience. Text Book “Quantum Chemistry”, 4th Ed.

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Advanced Physical Chemistry

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  1. Advanced Physical Chemistry G. H. CHEN Department of Chemistry University of Hong Kong

  2. Quantum Chemistry G. H. Chen Department of Chemistry University of Hong Kong

  3. Emphasis Hartree-Fock method Concepts Hands-on experience Text Book “Quantum Chemistry”, 4th Ed. Ira N. Levine http://yangtze.hku.hk/lecture/chem3504-3.ppt

  4. Beginning of Computational Chemistry In 1929, Dirac declared, “The underlying physical laws necessary for the mathematical theory of ...the whole of chemistry are thus completely know, and the difficulty is only that the exact application of these laws leads to equations much too complicated to be soluble.” Dirac

  5. Quantum Chemistry Methods • Ab initio molecular orbital methods • Semiempirical molecular orbital methods • Density functional method

  6. SchrÖdinger Equation Hy = Ey Wavefunction Hamiltonian H = (-h2/2ma)2 - (h2/2me)ii2 +  ZaZbe2/rab - i Zae2/ria + ije2/rij Energy Contents 1. Variation Method 2. Hartree-Fock Self-Consistent Field Method

  7. The Variation Method The variation theorem Consider a system whose Hamiltonian operator H is time independent and whose lowest-energy eigenvalue is E1. If f is any normalized, well- behaved function that satisfies the boundary conditions of the problem, then  f* Hf dt >E1

  8. Contributors: Hartree, Fock, Slater, Hund, Mulliken, Lennard-Jones, Heitler, London, Brillouin, Koopmans, Pople, Kohn Application: Chemistry, Condensed Matter Physics, Molecular Biology, Materials Science, Drug Discovery

  9. Proof: Expand f in the basis set { yk} f = kakyk where {ak} are coefficients Hyk = Ekyk then f* Hf dt = kjak* aj Ej dkj = k | ak|2Ek> E 1k | ak|2 = E1 Since is normalized, f*f dt = k | ak|2 = 1

  10. i. f : trial function is used to evaluate the upper limit of ground state energy E1 ii. f= ground state wave function,  f* Hf dt = E1 iii. optimize paramemters in f by minimizing  f* Hf dt / f* f dt

  11. Application to a particle in a box of infinite depth l 0 Requirements for the trial wave function: i. zero at boundary; ii. smoothness  a maximum in the center. Trial wave function: f = x (l - x)

  12. * H  dx = -(h2/82m)  (lx-x2) d2(lx-x2)/dx2 dx = h2/(42m)  (x2 - lx)dx = h2l3/(242m) * dx =  x2 (l-x)2 dx = l5/30 E = 5h2/(42l2m)  h2/(8ml2) = E1

  13. Variational Method (1) Construct a wave function (c1,c2,,cm) (2) Calculate the energy of : E E(c1,c2,,cm) (3) Choose {cj*} (i=1,2,,m) so that Eis minimum

  14. Example: one-dimensional harmonic oscillator Potential: V(x) = (1/2) kx2 = (1/2) m2x2 = 22m2x2 Trial wave function for the ground state: (x) = exp(-cx2) * H  dx = -(h2/82m)  exp(-cx2) d2[exp(-cx2)]/dx2 dx + 22m2  x2 exp(-2cx2) dx = (h2/42m) (c/8)1/2 + 2m2 (/8c3)1/2 * dx =  exp(-2cx2) dx = (/2)1/2 c-1/2 E= W = (h2/82m)c + (2/2)m2/c

  15. To minimize W, 0 = dW/dc = h2/82m - (2/2)m2c-2 c = 22m/h W= (1/2) h

  16. Extension of Variation Method    . . . E3y3 E2y2 E1y1 For a wave function f which is orthogonal to the ground state wave function y1, i.e. dtf*y1 = 0 Ef = dtf*Hf / dtf*f>E2 the first excited state energy

  17. The trial wave function f: dtf*y1 = 0 f = k=1 akyk dtf*y1 = |a1|2 = 0 Ef = dtf*Hf / dtf*f = k=2|ak|2Ek / k=2|ak|2 >k=2|ak|2E2 / k=2|ak|2 = E2

  18. Application to H2+ e + + y1 y2 f = c1y1 + c2y2 W = f*H f dt / f*f dt = (c12H11 + 2c1 c2H12+ c22H22 ) / (c12 + 2c1 c2S + c22 ) W (c12 + 2c1 c2S + c22) = c12H11 + 2c1 c2H12+ c22H22

  19. Partial derivative with respect to c1(W/c1 = 0) : W (c1 + S c2) = c1H11 + c2H12 Partial derivative with respect to c2(W/c2 = 0) : W (S c1 + c2) = c1H12 + c2H22 (H11 - W) c1 + (H12 - S W) c2 = 0 (H12 - S W) c1 + (H22 -W) c2 = 0

  20. To have nontrivial solution: H11 - W H12 - S W H12 - S W H22 -W For H2+,H11 = H22; H12 < 0. Ground State: Eg = W1 = (H11+H12) / (1+S) f1= (y1+y2) / 2(1+S)1/2 Excited State: Ee = W2 = (H11-H12) / (1-S) f2= (y1-y2) / 2(1-S)1/2 = 0 bonding orbital Anti-bonding orbital

  21. Results: De = 1.76 eV, Re = 1.32 A Exact: De = 2.79 eV, Re = 1.06 A 1 eV = 23.0605 kcal / mol

  22. 2p 1s Further Improvements H p-1/2exp(-r) He+ 23/2p-1/2exp(-2r) Optimization of 1s orbitals Trial wave function: k3/2p-1/2exp(-kr)  Eg = W1(k,R) at each R, choose kso thatW1/k = 0 Results: De = 2.36 eV, Re = 1.06 A Resutls: De = 2.73 eV, Re = 1.06 A Inclusion of other atomic orbitals

  23. a11x1 + a12x2 = b1 a21x1 + a22x2 = b2 (a11a22-a12a21) x1 = b1a22-b2a12 (a11a22-a12a21) x2 = b2a11-b1a21 Linear Equations 1. two linear equations for two unknown, x1 and x2

  24. Introducing determinant: a11 a12 = a11a22-a12a21 a21 a22 a11 a12b1 a12 x1 = a21 a22 b2 a22 a11 a12a11 b1 x2 = a21 a22a21 b2

  25. Our case: b1 = b2 = 0, homogeneous   1. trivial solution: x1 = x2 = 0   2. nontrivial solution:   a11 a12 = 0 a21 a22 n linear equations for n unknown variables a11x1 + a12x2 + ... + a1nxn= b1 a21x1 + a22x2 + ... + a2nxn= b2 ............................................ an1x1 + an2x2 + ... + annxn= bn

  26. a11 a12 ... a1,k-1 b1 a1,k+1 ... a1n a21 a22 ... a2,k-1 b2 a2,k+1 ... a2n det(aij) xk= . . ... . . . ... . an1 an2 ... an,k-1 b2 an,k+1 ... ann where, a11 a12 ... a1n a21 a22 ... a2n det(aij) = . . ... . an1 an2 ... ann

  27. inhomogeneous case: bk = 0 for at least one k a11 a12 ... a1,k-1 b1 a1,k+1 ... a1n a21 a22 ... a2,k-1 b2 a2,k+1 ... a2n . . ... . . . ... . an1 an2 ... an,k-1 b2 an,k+1 ... ann xk = det(aij)

  28. homogeneous case: bk = 0, k = 1, 2, ... , n (a) travial case: xk = 0, k = 1, 2, ... , n (b) nontravial case: det(aij) = 0 For a n-th order determinant, n det(aij) =  alk Clk l=1 where, Clk is called cofactor

  29. Trial wave function f is a variation function which is a combination of n linear independent functions { f1 , f2 , ... fn}, f = c1f1 + c2f2 + ... + cnfn n  [( Hik - SikW ) ck ] = 0 i=1,2,...,n k=1 Sikdtfi fk Hikdtfi H fk W  dt f Hf /  dt f f

  30. (i) W1W2 ... Wnare n roots of Eq.(1), (ii) E1E2 ... En En+1  ... are energies of eigenstates; then, W1E1, W2E2, ..., WnEn Linear variational theorem

  31. Molecular Orbital (MO):  = c11 + c22   ( H11 - W ) c1 + ( H12 - SW ) c2 = 0 S11=1 ( H21 - SW ) c1 + ( H22 - W ) c2 = 0 S22=1 Generally : i a set of atomic orbitals, basis set LCAO-MO  = c11 + c22 + ...... + cnn linear combination of atomic orbitals n  ( Hik - SikW ) ck = 0 i = 1, 2, ......, n k=1 Hikdti* H k Sikdti*k Skk = 1

  32. The Born-Oppenheimer Approximation Hamiltonian H =(-h2/2ma)2 - (h2/2me)ii2 +  ZaZbe2/rab - i Zae2/ria + ije2/rij H y(ri;ra) = E y(ri;ra)

  33. The Born-Oppenheimer Approximation: • (1) y(ri;ra) = yel(ri;ra) yN(ra) • (2) Hel(ra )= - (h2/2me)ii2- i Zae2/ria • + ije2/rij • VNN = b ZaZbe2/rab • Hel(ra)yel(ri;ra) = Eel(ra)yel(ri;ra) • (3) HN =(-h2/2ma)2 +U(ra) • U(ra) = Eel(ra) + VNN • HN(ra)yN(ra) = E yN(ra)

  34. Assignment Calculate the ground state energy and bond length of H2 using the HyperChem with the 6-31G (Hint: Born-Oppenheimer Approximation)

  35. Hydrogen Molecule H2 e + + e two electrons cannot be in the same state. The Pauli principle

  36. Wave function: f(1,2) = ja(1)jb(2) + c1 ja(2)jb(1) f(2,1) = ja(2)jb(1) + c1 ja(1)jb(2) Since two wave functions that correspond to the same state can differ at most by a constant factor f(1,2) = c2f(2,1) ja(1)jb(2) + c1ja(2)jb(1) =c2ja(2)jb(1) +c2c1ja(1)jb(2) c1 = c2 c2c1 = 1 Therefore:c1 = c2 = 1 According to the Pauli principle, c1 = c2 =- 1

  37. Wave function f of H2 : y(1,2) = 1/2! [f(1)a(1)f(2)b(2) - f(2)a(2)f(1)b(1)] f(1)a(1) f(2)a(2) = 1/2! f(1)b(1) f(2)b(2) The Pauli principle (different version) the wave function of a system of electrons must be antisymmetric with respect to interchanging of any two electrons. Slater Determinant

  38. Energy: E • E=2dt1 f*(1) (Te+VeN) f(1) + VNN • + dt1 dt2 |f2(1)| e2/r12 |f2(2)| • = i=1,2 fii + J12 + VNN • To minimize Eunder the constraintdt |f2|= 1, • useLagrange’s method: • L = E - 2 e [dt1 |f2(1)|- 1] • dL = dE - 4 e dt1 f*(1)df(1) = 4dt1 df*(1)(Te+VeN)f(1) • +4dt1 dt2 f*(1)f*(2) e2/r12 f(2)df(1) • - 4 e dt1 f*(1)df(1) • = 0

  39. [ Te+VeN +dt2 f*(2) e2/r12 f(2) ] f(1) = e f(1) Average Hamiltonian Hartree-Fock equation ( f + J ) f = e f f(1) = Te(1)+VeN(1) one electron operator J(1) =dt2 f*(2) e2/r12 f(2) two electron Coulomb operator

  40. f(1) is the Hamiltonian of electron 1 in the absence of electron 2; J(1) is the mean Coulomb repulsion exerted on electron 1 by 2; eis the energy of orbital f. LCAO-MO: f = c1y1 + c2y2 Multiple y1 from the left and then integrate : c1F11 + c2F12 = e (c1 + S c2)

  41. Multiple y2from the left and then integrate : c1F12 + c2F22 = e (S c1 + c2) where, Fij = dt yi*( f + J) yj = Hij + dt yi*Jyj S = dt y1y2 (F11 - e) c1 + (F12 - S e) c2 = 0 (F12 - S e) c1 + (F22 -e) c2 = 0

  42. Secular Equation: F11 - eF12 - S e = 0 F12 - SeF22 -e bonding orbital: e1 = (F11+F12) / (1+S) f1= (y1+y2) / 2(1+S)1/2 antibonding orbital: e2 = (F11-F12) / (1-S ) f2= (y1-y2) / 2(1-S)1/2

  43. Molecular Orbital Configurations of Homo nuclear Diatomic Molecules H2, Li2, O, He2, etc Moecule Bond order De/eV H2+ 2.79 H2 1 4.75 He2+  1.08 He2 0 0.0009 Li2 1 1.07 Be2 0 0.10 C2 2 6.3 N2+ 8.85 N2 3 9.91 O2+ 2 6.78 O2 2 5.21 The more the Bond Order is, the stronger the chemical bond is.

  44. Bond Order: one-half the difference between the number of bonding and antibonding electrons

  45. f1(1)a(1) f2(1)a(1) y(1,2) = 1 /2 f1(2)a(2) f2(2)a(2) = 1/2 [f1(1) f2(2) - f2(1)f1(2)] a(1) a(2) ---------------- f1 ---------------- f2

  46. Ey = dt1dt2 y* H y = dt1dt2 y* (T1+V1N+T2+V2N+V12+VNN) y = <f1(1)| T1+V1N|f1(1)> + <f2(2)| T2+V2N|f2(2)> + <f1(1) f2(2)| V12 | f1(1) f2(2)> - <f1(2) f2(1)| V12 | f1(1) f2(2)> + VNN = i<fi(1)| T1+V1N |fi(1)> + <f1(1) f2(2)| V12 | f1(1) f2(2)> - <f1(2) f2(1)| V12 | f1(1) f2(2)> +VNN = i=1,2 fii + J12 -K12 + VNN

  47. Average Hamiltonian Particle One: f(1) + J2(1)-K2(1) Particle Two: f(2) + J1(2)-K1(2) f(j) -(h2/2me)j2 - Za/rja Jj(1) q(1)  q(1)  dr2 fj*(2) e2/r12 fj(2) Kj(1) q(1)  fj(1) dr2 fj*(2) e2/r12 q(2)

  48. Hartree-Fock Equation: [ f(1)+ J2(1) -K2(1)] f1(1) = e1 f1(1) [ f(2)+ J1(2) -K1(2)] f2(2) = e2 f2(2) Fock Operator: F(1) f(1)+ J2(1) -K2(1) Fock operator for 1 F(2) f(2)+ J1(2) -K1(2) Fock operator for 2

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