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

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

4. 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

5. 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.” Beginning of Computational Chemistry Dirac

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

7. H  E  Schr Ö dinger Equation Hamiltonian H =   (  h 2 /2m      h 2 /2m e )  i  i 2 +     Z  Z  e   r   i     e 2 /r i   i  j  e 2 /r ij Wavefunction Energy Contents 1. Variation Method 2. Hartree-Fock Self-Consistent Field Method

8. The Variation Method Consider a system whose Hamiltonian operator H is time independent and whose lowest-energy eigenvalue is E 1. If  is any normalized, well- behaved function that satisfies the boundary conditions of the problem, then  * H  d  E 1 The variation theorem

9. Proof: Expand ii n the basis set {  k }  =  k  k  k where {  k } are coefficients H  k = E k  k then  * H dd  k  j  k *  j E j  kj =  k | k | 2 E k  E 1  k | k | 2 = E 1 Since is normalized,  * dd  k |  k | 2 = 1

10. i.  : trial function is used to evaluate the upper limit of ground state energy E 1 ii.  = ground state wave function,   * H  d  E 1 iii. optimize paramemters in  by minimizing   * H  d    *  d 

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

12.   * H  dx = -(h 2 /8  2 m)  (lx-x 2 ) d 2 (lx-x 2 )/dx 2 dx = h 2 /(4  2 m)  (x 2 - lx) dx = h 2 l 3 /(24  2 m)   *  dx =  x 2 (l-x) 2 dx = l 5 /30 E  = 5h 2 /(4  2 l 2 m)  h 2 /(8ml 2 ) = E 1

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

14. Example: one-dimensional harmonic oscillator Potential: V(x) = (1/2) kx 2 = (1/2) m  2 x 2 = 2  2 m 2 x 2 Trial wave function for the ground state:  (x) = exp(-cx 2 )   * H  dx = -(h 2 /8  2 m)  exp(-cx 2 ) d 2 [exp(-cx 2 )]/dx 2 dx + 2  2 m 2  x 2 exp(-2cx 2 ) dx = (h 2 /4  2 m) (  c/8) 1/2 +  2 m 2 (  /8c 3 ) 1/2   *  dx =  exp(-2cx 2 ) dx = (  /2) 1/2 c -1/2 E  = W = (h 2 /8  2 m)c + (  2 /2)m 2 /c

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

16. . E 3  3 E 2  2 E 1  1 Extension of Variation Method For a wave function  which is orthogonal to the ground state wave function  1, i.e.  d   *  1 = 0 E  =  d   * H  /  d   *  > E 2 the first excited state energy

17. The trial wave function  :  d   *  1 = 0  k=1 a k  k  d   *  1 = |a 1 | 2 = 0 E  =  d   * H  /  d   *  =  k=2 |a k | 2 E k /  k=2 |a k | 2 >  k=2 |a k | 2 E 2 /  k=2 |a k | 2 = E 2

18. e + +  1  2  c 1  1 + c 2  2 W =    H  d     d   = (c 1 2 H 11 + 2c 1 c 2 H 12 + c 2 2 H 22 ) / (c 1 2 + 2c 1 c 2 S + c 2 2 ) W (c 1 2 + 2c 1 c 2 S + c 2 2 ) = c 1 2 H 11 + 2c 1 c 2 H 12 + c 2 2 H 22 Application to H 2 +

19. Partial derivative with respect to c 1 (  W/  c 1 = 0) : W (c 1 + S c 2 ) = c 1 H 11 + c 2 H 12 Partial derivative with respect to c 2 (  W/  c 2 = 0) : W (S c 1 + c 2 ) = c 1 H 12 + c 2 H 22 (H 11 - W) c 1 + (H 12 - S W) c 2 = 0 (H 12 - S W) c 1 + (H 22 - W) c 2 = 0

20. To have nontrivial solution: H 11 - WH 12 - S W H 12 - S WH 22 - W For H 2 +, H 11 = H 22 ; H 12 < 0. Ground State: E g = W 1 = (H 11 +H 12 ) / (1+S)    = (     ) /  2(1+S) 1/2 Excited State: E e = W 2 = (H 11 -H 12 ) / (1-S)    = (     ) /  2(1-S) 1/2 = 0 bonding orbital Anti-bonding orbital

21. Results: D e = 1.76 eV, R e = 1.32 A Exact: D e = 2.79 eV, R e = 1.06 A 1 eV = 23.0605 kcal / mol

22. Trial wave function: k 3/2  -1/2 exp(-kr) E g = W 1 (k,R) at each R, choose k so that  W 1 /  k = 0 Results: D e = 2.36 eV, R e = 1.06 A Resutls: D e = 2.73 eV, R e = 1.06 A 1s1s 2p2p Inclusion of other atomic orbitals Further Improvements H  -1/2 exp(-r) He + 2 3/2  -1/2 exp(-2r) Optimization of 1s orbitals

23. a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 (a 11 a 22 -a 12 a 21 ) x 1 = b 1 a 22 -b 2 a 12 (a 11 a 22 -a 12 a 21 ) x 2 = b 2 a 11 -b 1 a 21 Linear Equations 1. two linear equations for two unknown, x 1 and x 2

24. Introducing determinant: a 11 a 12 = a 11 a 22 -a 12 a 21 a 21 a 22 a 11 a 12 b 1 a 12 x 1 = a 21 a 22 b 2 a 22 a 11 a 12 a 11 b 1 x 2 = a 21 a 22 a 21 b 2

25. Our case: b 1 = b 2 = 0, homogeneous 1. trivial solution: x 1 = x 2 = 0 2. nontrivial solution: a 11 a 12 = 0 a 21 a 22 n linear equations for n unknown variables a 11 x 1 + a 12 x 2 +... + a 1n x n = b 1 a 21 x 1 + a 22 x 2 +... + a 2n x n = b 2............................................ a n1 x 1 + a n2 x 2 +... + a nn x n = b n

26. a 11 a 12... a 1,k-1 b 1 a 1,k+1... a 1n a 21 a 22... a 2,k-1 b 2 a 2,k+1... a 2n det(a ij ) x k =............ a n1 a n2... a n,k-1 b 2 a n,k+1... a nn where, a 11 a 12...a 1n a 21 a 22...a 2n det(a ij ) =...... a n1 a n2...a nn

27. a 11 a 12...a 1,k-1 b 1 a 1,k+1...a 1n a 21 a 22...a 2,k-1 b 2 a 2,k+1...a 2n............ a n1 a n2...a n,k-1 b 2 a n,k+1...a nn x k = det(a ij ) inhomogeneous case: b k = 0 for at least one k

28. (a) travial case: x k = 0, k = 1, 2,..., n (b) nontravial case: det(a ij ) = 0 homogeneous case: b k = 0, k = 1, 2,..., n For a n-th order determinant, n det(a ij ) =  a lk C lk l=1 where, C lk is called cofactor

29. Trial wave function  is a variation function which is a combination of n linear independent functions { f 1, f 2,... f n },  c 1 f 1 + c 2 f 2 +... + c n f n n   [( H ik - S ik W ) c k ] = 0 i=1,2,...,n k=1 S ik   d  f i f k H ik   d  f i H f k W   d  H  d 

30. (i) W 1  W 2 ...  W n are n roots of Eq.(1), (ii) E 1  E 2 ...  E n  E n+1 ... are energies of eigenstates; then, W 1  E 1, W 2  E 2,..., W n  E n Linear variational theorem

31. Molecular Orbital (MO):  = c 1  1 + c 2  2 ( H 11 - W ) c 1 + ( H 12 - SW ) c 2 = 0 S 11 =1 ( H 21 - SW ) c 1 + ( H 22 - W ) c 2 = 0 S 22 =1 Generally :  i  a set of atomic orbitals, basis set LCAO-MO  = c 1  1 + c 2  2 +...... + c n  n linear combination of atomic orbitals n  ( H ik - S ik W ) c k = 0 i = 1, 2,......, n k=1 H ik   d   i * H  k S ik   d   i *  k S kk = 1

32. Hamiltonian H =   (  h 2 /2m      h 2 /2m e )  i  i 2 +     Z  Z  e   r   i     e 2 /r i   i  j  e 2 /r ij H  r i ;r   r i ;r   The Born-Oppenheimer Approximation

33.  r i ;r   el  r i ;r   N  r    el ( r  ) =  h 2 /2m e )  i  i 2  i     e 2 /r i   i  j  e 2 /r ij V NN =     Z  Z  e   r  H el ( r  )  el  r i ;r   el ( r  )  el  r i ;r   (3) H N =   (  h 2 /2m      U( r  ) U( r  ) =  el ( r  ) + V NN H N ( r  )  N  r   N  r   The Born-Oppenheimer Approximation:

34. Assignment Calculate the ground state energy and bond length of H 2 using the HyperChem with the 6-31G (Hint: Born-Oppenheimer Approximation)

35. e + + e two electrons cannot be in the same state. Hydrogen Molecule H 2 The Pauli principle

36. Since two wave functions that correspond to the same state can differ at most by a constant factor  = c 2   a  b  c 1  a  b  =c 2  a  b  +c 2 c 1  a  b  c 1 = c 2 c 2 c 1 = 1 Therefore: c 1 = c 2 =  1 According to the Pauli principle, c 1 = c 2 =  1 Wave function:  =  a  b  + c 1  a  b   =  a  b  + c 1  a  b 

37. Wave function  f H 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. E  =2  d      T e +V eN )  + V NN +  d   d      e 2 /r 12 |    =  i=1,2 f ii + J 12 + V NN To minimize E   under the constraint  d   |   use Lagrange’s method: L = E  d         L =  E  d       4  d      T e +V eN )   +4  d   d        e 2 /r 12   d       Energy: E 

39. [ T e +V eN +   d      e 2 /r 12   f + J  f(1) = T e (1)+V eN (1) one electron operator J(1) =   d      e 2 /r 12  two electron Coulomb operator Average Hamiltonian Hartree-Fock equation

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;  is the energy of orbital  LCAO-MO:  c 1  1 + c 2  2 Multiple  1 from the left and then integrate : c 1 F 11 + c 2 F 12 =  (c 1 + S c 2 )

41. Multiple  2 from the left and then integrate : c 1 F 12 + c 2 F 22 =  (S c 1 + c 2 ) where, F ij =  d  i * ( f + J )  j = H ij +  d  i * J  j S =  d  1  2 (F 11 -  ) c 1 + (F 12 - S  ) c 2 = 0 (F 12 - S  ) c 1 + (F 22 -  ) c 2 = 0

42. Secular Equation: F 11 -  F 12 - S   F 12 - S  F 22 -  bonding orbital:  1 = (F 11 +F 12 ) / (1+S)    = (     ) /  2(1+S) 1/2 antibonding orbital:  2 = (F 11 -F 12 ) / (1-S )    = (     ) /  2(1-S) 1/2

43. Molecular Orbital Configurations of Homo nuclear Diatomic Molecules H 2, Li 2, O, He 2, etc Moecule Bond order De/eV H 2 +  2.79 H 2 1 4.75 He 2 +  1.08 He 2 0 0.0009 Li 2 1 1.07 Be 2 0 0.10 C 2 2 6.3 N 2 +  8.85 N 2 3 9.91 O 2 + 2  6.78 O 2 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. --------  --------  1 --------  --------  2   1  2    1  2  = 1/  2 [  1  2  2   1 

46.     d   d     H   d   d     (T 1 +V 1N +T 2 +V 2N +V 12 +V NN )   1  T 1 +V 1N |  1   2  T 2 +V 2N |  2  +  1  2  V 12  1  2   1  2  V 12  1  2  + V NN =  i  i  T 1 +V 1N |  i   +  1  2  V 12  1  2   1  2  V 12  1  2  + V NN =  i=1,2 f ii + J 12  K 12 + V NN

47. Particle One: f(1) + J 2 (1)  K 2 (1) Particle Two: f(2) + J 1 (2)  K 1 (2) f(j)   h 2 /2m e )  j 2    Z   r j  J j (1)  dr   j *  e 2 /r 12  j  K j (1)  j  dr   j *  e 2 /r 12  Average Hamiltonian

48.  f(1)+ J 2 (1)  K 2 (1)  1 (1)  1  1 (1)  f(2)+ J 1 (2)  K 1 (2)  2 (2)  2  2 (2) F(1)  f(1)+ J 2 (1)  K 2 (1) Fock operator for 1 F(2)  f(2)+ J 1 (2)  K 1 (2) Fock operator for 2 Hartree-Fock Equation: Fock Operator:

49. 1. Many-Body Wave Function is approximated by Slater Determinant 2. Hartree-Fock Equation F  i =  i  i F Fock operator  i the i-th Hartree-Fock orbital  i the energy of the i-th Hartree-Fock orbital Hartree-Fock Method

50. 3. Roothaan Method (introduction of Basis functions)  i =  k c ki  k LCAO-MO {  k } is a set of atomic orbitals (or basis functions) 4. Hartree-Fock-Roothaan equation  j ( F ij -  i S ij ) c ji = 0 F ij  i  F  j  S ij  i  j  5. Solve the Hartree-Fock-Roothaan equation self-consistently

51. Assignment one 8.40, 10.5, 10.6, 10.7, 10.8, 11.37, 13.37

52. 1.At the Hartree-Fock Level there are two possible Coulomb integrals contributing the energy between two electrons i and j: Coulomb integrals J ij and exchange integral K ij ; 2.For two electrons with different spins, there is only Coulomb integral J ij ; 3. For two electrons with the same spins, both Coulomb and exchange integrals exist. Summary

53. 4.Total Hartree-Fock energy consists of the contributions from one-electron integrals f ii and two-electron Coulomb integrals J ij and exchange integrals K ij ; 5.At the Hartree-Fock Level there are two possible Coulomb potentials (or operators) between two electrons i and j: Coulomb operator and exchange operator; J j (i) is the Coulomb potential (operator) that i feels from j, and K j (i) is the exchange potential (operator) that that i feels from j.

54. 6. Fock operator (or, average Hamiltonian) consists of one-electron operators f(i) and Coulomb operators J j (i) and exchange operators K j (i)    

55. N   electrons spin up and N   electrons spin down. Fock matrix for an electron 1  with spin up: F  (1  ) = f  (1  ) +  j [ J j  (1  )  K j  (1  ) ] +  j J j  (1  ) j=1 ,N  j=1 ,N  Fock matrix for an electron 1  with spin down: F  (1  ) = f  (1  ) +  j [ J j  (1  )  K j  (1  ) ] +  j J j  (1  ) j=1 ,N   j=1 ,N 

56. f(1)   h 2 /2m e )  1 2   N Z N  r 1N J j  (1)  dr   j   e 2 /r 12  j   K j  (1)  j   dr   j  *  e 2 /r 12  Energy =  j  f jj  +  j  f jj  +(1/2)  i   j  ( J ij   K ij  ) + (1/2)  i   j  ( J ij   K ij  ) +  i   j  J ij   + V NN i=1,N   j=1,N 

57. f jj   f jj   j   f  j   J ij   J ij    j (2)  J i     j (2)  K ij   K ij    j (2)  K i     j (2)  J ij   J ij    j (2)  J i     j (2)  F(1) = f (1) +  j=1,n/2 [ 2J j (1)  K j (1) ] Energy = 2  j=1,n/2 f jj +  i=1,n/2  j=1,n/2 ( 2J ij  K ij ) + V NN Close subshell case: ( N  = N  = n/2 )

58.   a  b  c  d  n  f(1)  e  f  g  h  n   a  f(1)  e  b  c  d  n  f  g  h  n  =  a  f(1)  e  if b=f, c=g,..., d=h; 0, otherwise  a  b  c  d  n  V 12 |  e  f  g  h  n   a  b  V 12  e  f  c  d  n  g  h  n  =  a  b  V 12  e  f  if c=g,..., d=h; 0, otherwise The Condon-Slater Rules

59. ------- the lowest unoccupied molecular orbital  ------- the highest occupied molecular orbital  ------- ------- The energy required to remove an electron from a closed-shell atom or molecules is well approximated by minus the orbital energy  of the AO or MO from which the electron is removed. HOMO LUMO Koopman’s Theorem

60. # HF/6-31G(d) Route section water energy Title 0 1 Molecule Specification O -0.464 0.177 0.0 (in Cartesian coordinates H -0.464 1.137 0.0 H 0.441 -0.143 0.0

61. Slater-type orbitals (STO)  nlm = N r n-1 exp(  r/a 0 ) Y lm ( ,  )  the orbital  exponent *  is used instead of  in the textbook Gaussian type functions g ijk = N x i y j z k exp(-  r 2 ) (primitive Gaussian function)  p =  u d up g u (contracted Gaussian-type function, CGTF) u = {ijk}p = {nlm} Basis Set  i =  p c ip  p

62. Basis set of GTFs STO-3G, 3-21G, 4-31G, 6-31G, 6-31G*, 6-31G** -------------------------------------------------------------------------------------  complexity & accuracy Minimal basis set: one STO for each atomic orbital (AO) STO-3G: 3 GTFs for each atomic orbital 3-21G: 3 GTFs for each inner shell AO 2 CGTFs (w/ 2 & 1 GTFs) for each valence AO 6-31G: 6 GTFs for each inner shell AO 2 CGTFs (w/ 3 & 1 GTFs) for each valence AO 6-31G*: adds a set of d orbitals to atoms in 2nd & 3rd rows 6-31G**: adds a set of d orbitals to atoms in 2nd & 3rd rows and a set of p functions to hydrogen Polarization Function

63. Diffuse Basis Sets: For excited states and in anions where electronic density is more spread out, additional basis functions are needed. Diffuse functions to 6-31G basis set as follows: 6-31G* - adds a set of diffuse s & p orbitals to atoms in 1st & 2nd rows (Li - Cl). 6-31G** - adds a set of diffuse s and p orbitals to atoms in 1st & 2nd rows (Li- Cl) and a set of diffuse s functions to H Diffuse functions + polarisation functions: 6-31+G*, 6-31++G*, 6-31+G** and 6-31++G** basis sets. Double-zeta (DZ) basis set: two STO for each AO

64. 6-31G for a carbon atom:(10s4p)  [3s2p] 1s2s2p i (i=x,y,z) 6GTFs 3GTFs 1GTF3GTFs 1GTF 1CGTF 1CGTF 1CGTF 1CGTF 1CGTF (s)(s) (s) (p) (p)

65. Minimal basis set: One STO for each inner-shell and valence-shell AO of each atom example: C 2 H 2 (2S1P/1S) C: 1S, 2S, 2P x,2P y,2P z H: 1S total 12 STOs as Basis set Double-Zeta (DZ) basis set: two STOs for each and valence-shell AO of each atom example: C 2 H 2 (4S2P/2S) C: two 1S, two 2S, two 2P x, two 2P y,two 2P z H: two 1S (STOs) total 24 STOs as Basis set

66. Split -Valence (SV) basis set Two STOs for each inner-shell and valence-shell AO One STO for each inner-shell AO Double-zeta plus polarization set(DZ+P, or DZP) Additional STO w/l quantum number larger than the l max of the valence - shell  ( 2P x, 2P y,2P z ) to H  Five 3d Aos to Li - Ne, Na -Ar  C 2 H 5 O Si H 3 : (6s4p1d/4s2p1d/2s1p) Si C,O H

67. Assignment two: Calculate the structure, ground state energy, molecular orbital energies, and vibrational modes and frequencies of a water molecule using Hartree-Fock method with 3-21G basis set.

68. 1. L-Click on (click on left button of Mouse) “Startup”, and select and L-Click on “Program/Hyperchem”. 2. Select “Build’’ and turn on “Explicit Hydrogens”. 3. Select “Display” and make sure that “Show Hydrogens” is on; L-Click on “Rendering” and double L-Click “Spheres”. 4. Double L-Click on “Draw” tool box and double L-Click on “O”. 5. Move the cursor to the workspace, and L-Click & release. 6. L-Click on “Magnify/Shrink” tool box, move the cursor to the workspace; L-press and move the cursor inward to reduce the size of oxygen atom. 7. Double L-Click on “Draw” tool box, and double L-Click on “H”; Move the cursor close to oxygen atom and L-Click & release. A hydrogen atom appears. Draw second hydrogen atom using the same procedure. Ab Initio Molecular Orbital Calculation: H 2 O (using HyperChem)

69. 8. L-Click on “Setup” & select “Ab Initio”; double L-Click on 3-21G; then L-Click on “Option”, select “UHF”, and set “Charge” to 0 and “Multiplicity” to 1. 9. L-Click “Compute”, and select “Geometry Optimization”, and L-Click on “OK”; repeat the step till “Conv=YES” appears in the bottom bar. Record the energy. 10.L-Click “Compute” and L-Click “Orbitals”; select a energy level, record the energy of each molecular orbitals (MO), and L-Click “OK” to observe the contour plots of the orbitals. 11.L-Click “Compute” and select “Vibrations”. 12.Make sure that “Rendering/Sphere” is on; L-Click “Compute” and select “Vibrational Spectrum”. Note that frequencies of different vibrational modes. 13.Turn on “Animate vibrations”, select one of the three modes, and L-Click “OK”. Water molecule begins to vibrate. To suspend the animation, L-Click on “Cancel”.

70. The Hartree-Fock treatment of H 2 + e-e- + e-e-

71. f 1 =  1 (1)  2 (2) f 2 =  1 (2)  2 (1)  = c 1 f 1 + c 2 f 2 H 11 - WH 12 - S W H 21 - S WH 22 - W H 11 = H 22 = H 12 = H 21 = S = [ = S 2 ] The Heitler-London ground-state wave function {[  1 (1)  2 (2) +  1 (2)  2 (1)]/  2(1+S) 1/2 } [  (1)  (2)  (2)  (1)]/  2 = 0 The Valence-Bond Treatment of H 2

72. Comparison of the HF and VB Treatments HF LCAO-MO wave function for H 2 [  1 (1) +  2 (1)] [  1 (2) +  2 (2)] =  1 (1)  1 (2) +  1 (1)  2 (2) +  2 (1)  1 (2) +  2 (1)  2 (2) H  H  H H H H H  H  VB wave function for H 2  1 (1)  2 (2) +  2 (1)  1 (2) H H H H

73. At large distance, the system becomes H............ H MO: 50% H............ H 50% H +............ H  VB: 100% H............ H The VB is computationally expensive and requires chemical intuition in implementation. The Generalized valence-bond (GVB) method is a variational method, and thus computationally feasible. (William A. Goddard III)

74. The Heitler-London ground-state wave function

75. Electron Correlation Human Repulsive Correlation

76. Electron Correlation: avoiding each other Two reasons of the instantaneous correlation: (1) Pauli Exclusion Principle (HF includes the effect) (2) Coulomb repulsion (not included in the HF) Beyond the Hartree-Fock Configuration Interaction (CI)* Perturbation theory* Coupled Cluster Method Density functional theory

77. -e -e r 12 r 2 r 1 +2e H = - (h 2 /2m e )  1 2 - 2e 2 /r 1 - (h 2 /2m e )  2 2 - 2e 2 /r 2 + e 2 /r 12 H 1 0 H 2 0 H’

78. H 0 = H 1 0 + H 2 0  (0) (1,2) = F 1 (1) F 2 (2) H 1 0 F 1 (1) = E 1 F 1 (1) H 2 0 F 2 (1) = E 2 F 2 (1) E 1 = -2e 2 /n 1 2 a 0 n 1 = 1, 2, 3,... E 2 = -2e 2 /n 2 2 a 0 n 2 = 1, 2, 3,...  (0) (1,2) = (1/    2  a 0 ) 3/2 exp(-2r 1 /a 0 )  (1/    2  a 0 ) 3/2 exp(-2r 1 /a 0 ) E (0) =  4e 2 /a 0 E (1) = = 5e 2 /4a 0 E  E (0) + E (1) = -108.8 + 34.0 = -74.8 (eV) [compared with exp. -79.0 eV] Ground state wave function

79. H = H 0 + H’ H 0  n (0) = E n (0)  n (0)  n (0) is an eigenstate for unperturbed system H’ is small compared with H 0 Nondegenerate Perturbation Theory (for Non-Degenerate Energy Levels)

80. H(  = H 0 + H’ H  n  = E n  n   n  n   n     n   k  n (k)   n  n   n     n   k  n (k)   the original Hamiltonian Introducing a parameter  n  n   n   n   n (k)   n  n   n   n   n (k)  Where, = 0, j=1,2,...,k,...

81. H   n  = E n   n   solving for E n   n  H   n   H’  n  = E n   n   n   n   solving for E n   n  H   n   H’  n  = E n   n   n   n   n   n   solving for E n   n 

82. Multiplied  m (0) from the left and integrate, + = E n   E n   mn [E m   E n   + = E n   mn For m = n, For m  n, = / [E n   E m   If we expand  n (1) =  c nm  m (0), c nm = / [E n   E m   for m  n; c nn = 0.  n (1) =  m / [E n   E m   m (0) Eq.(2) The first order: E n   Eq.(1)

83. The second order: + = <  m (0)  n (2) >E n   E n   E n   mn Set m = n, we have E n  =  m  n |  m (0)  H'  n (0) >| 2 / [E n   E m   q.(3)

84. a. Eq.(2) shows that the effect of the perturbation on the wave function  n (0) is to mix in contributions from the other zero-th order states  m (0) m  n. Because of the factor 1/(E n (0) -E m (0) ), the most important contributions to the  n (1) come from the states nearest in energy to state n. b. To evaluate the first-order correction in energy, we need only to evaluate a single integral H’ nn ; to evaluate the second-order energy correction, we must evalute the matrix elements H’ between the n-th and all other states m. c. The summation in Eq.(2), (3) is over all the states, not the energy levels. Discussion: (Text Book: page 522-527)

85. Moller-Plesset (MP) Perturbation Theory The MP unperturbed Hamiltonian H 0 H 0 =  m F(m) where F(m) is the Fock operator for electron m. And thus, the perturbation H ’ H ’ = H - H 0 Therefore, the unperturbed wave function is simply the Hartree-Fock wave function . Ab initio methods: MP2, MP4

86. Example One: Consider the one-particle, one-dimensional system with potential-energy function V = b for L/4 < x < 3L/4, V = 0 for 0 < x  L/4 & 3L/4  x < L and V =  elsewhere. Assume that the magnitude of b is small, and can be treated as a perturbation. Find the first-order energy correction for the ground and first excited states. The unperturbed wave functions of the ground and first excited states are  1 = (2/L) 1/2 sin(  x/L) and  2 = (2/L) 1/2 sin(2  x/L), respectively.

87. Example Two: As the first step of the Moller-Plesset perturbation theory, Hartree-Fock method gives the zeroth-order energy. Is the above statement correct? Example Three: Show that, for any perturbation H’, E 1 (0) + E 1 (1)  E 1 where E 1 (0) and E 1 (1) are the zero-th order energy and the first order energy correction, and E 1 is the ground state energy of the full Hamiltonian H 0 + H’. Example Four: Calculate the bond orders of Li 2 and Li 2 +.

88. Ground State Excited State CPU Time Correlation Geometry Size Consistent (CH 3 NH 2,6-31G*) HFSCF   1 0 OK  DFT   ~1   CIS   <10 OK  CISD   17 80-90%   (20 electrons) CISDTQ   very large 98-99%   MP2   1.5 85-95%   (DZ+P) MP4   5.8 >90%   CCD   large >90%   CCSDT   very large ~100%  

89. Statistical Mechanics Content: Ensembles and Their Distributions Quantum Statistics Canonical Partition Function Non-Ideal Gas References: 1. Grasser & Richards, “An Introduction to Statistical Thermodynamcis” 2. Atkins, “Physical Chemistry”

90. Ensembles and Their Distributions State Functions The value of a state function depends only on the current state of the system. In other words, a state function is some function of the state of the system.

91. State Functions: E, N, T, V, P,...... When a system reaches its equilibrium, its state functions E, N, T, V, P and others no longer vary.

92. Ensemble An ensemble is a collection of systems. A Thought Experiment to construct an ensemble To set up an ensemble, we take a closed system of specific volume, composition, and temperature, and then, replicate it A times. We have A such systems. The collection of these systems is an ensemble. The systems in an ensemble may or may not exchange energy, molecules or atoms.

96. Microcanonical Ensemble: N, V, E are common; Canonical Ensemble: N, V, T are common; Grand Canonical Ensemble: , V, T are common. Microcanonical System: N, E are fixed; Canonical System: N is fixed, but E varies; Grand Canonical System: N, E vary.

97. Example: What kind of system is each of the following systems: (1) an isolated molecular system; (2) an equilibrium system enclosed by a heat conducting wall; (3) a pond; (4) a system surrounded by a rigid and insulating material.

98. Principle of Equal A Priori Probabilities Probabilities of all accessible states of an isolated system are equal. For instance, four molecules in a three-level system: the following two conformations have the same probability. ---------l-l-------- 2  ---------l--------- 2  ---------l----------  ---------1-1-1----  ---------l---------- 0------------------- 0

99. Configurations and Weights Imagine that an ensemble contains total A systems among which a 1 systems with energy E 1 and N 1 molecules, a 2 systems with energy E 2 and N 2 molecules, a 3 systems with energy E 3 and N 3 molecules, with energy  1, and so on. The specific distribution of systems in the ensemble is called configuration of the system, denoted as { a 1, a 2, a 3,......}.

100. A configuration { a 1, a 2, a 3,......} can be achieved in W different ways, where W is called the weight of the configuration. And W can be evaluated as follows, W = A! / (a 1 ! a 2 ! a 3 !...) Distribution of a Microcanonical Ensemble State123…k… EnergyEEE…E… Occupationa 1 a 2 a 3 …a k …

101. Constraint  i a i = A W = A! / a 1 !a 2 ! a 3 !… To maximize lnW under the constraint, we construct a Lagrangian L = lnW +   i a i Thus, 0 =  L/  a i =  lnW/  a i + 

102. the probability of a system being found in state i, p i = a i /A = exp(  ) = constant or, in another word, the probabilities of all states with the same energy are equal. Utilizing the Stirling’s approximation, ln x! = x ln x - x  lnW/  a i = - ln a i /A = - ,

103. Distribution of a Canonical Ensemble State123…k… EnergyE 1 E 2 E 3 …E k … Occupationa 1 a 2 a 3 …a k … Constraints:  i a i = A  i a i E i =  where,  is the total energy in the ensemble. W = A! / a 1 !a 2 ! a 3 !…

104. To maximize lnW under the above constraints, construct a Lagrangian L = lnW +   i a i -   i a i E i 0 =  L/  a i =  lnW/  a i +  -  E i ln a i /A =  -  E i the probability of a system being found in state i with the energy E i, p i = a i /A = exp(  -  E i )

105. The above formula is the canonical distribution of a system. Different from the Boltzmann distribution of independent molecules, the canonical distribution applies to an entire system as well as individual molecule. The molecules in this system can be independent of each other, or interact among themselves. Thus, the canonical distribution is more general than the Boltzmann distribution. (note, in the literature the canonical distribution and the Boltzmann distribution are sometimes interchangeable).

106. Distribution of a Grand Canonical Ensemble State123…k… EnergyE 1 E 2 E 3 …E k … Mol. No.N 1 N 2 N 3 …N k … Occupationa 1 a 2 a 3 …a k …

107. Constraints:  i a i = A  i a i E i =   i a i N i = N where,  and N are the total energy and total number of molecules in the ensemble, respectively. W = A! / a 1 !a 2 ! a 3 !…

108. To maximize lnW under the above constraints, construct a Lagrangian L = lnW +   i a i -   i a i E i -   i a i N i 0 =  L/  a i =  lnW/  a i +  -  E i -  N i ln a i /A =  -  E i -  N i the probability of a system being found in state i with the energy E i and the number of particles N i, p i = a i /A = exp(  -  E i -  N i )

109. The above formula describes the distribution of a grand canonical system, and is called the grand canonical distribution. When N i is fixed, the above distribution becomes the canonical distribution. Thus, the grand canonical distribution is most general.

110. Quantum Statistics Quantum Particle: Fermion(S = 1/2, 3/2, 5/2,...) e.g. electron, proton, neutron, 3 He nuclei Boson(S = 0, 1, 2,...) e.g.deuteron, photon, phonon, 4 He nuclei Pauli Exclusion Principle: Two identical fermions can not occupy the same state at the same time. Question: what is the average number particles or occupation of a quantum state?

111. Fermi-Dirac Statistics System: a fermion’s state with an energy  (   -  /  ) ------------------ --------l--------- occupation n = 0 n = 1 energy 0  probability exp(0) exp[-  (  -  )]

112. There are only two states because of the Pauli exclusion principle. Thus, the average occupation of the quantum state , 1 / {exp[  (  -  )] + 1}

113. Therefore, the average occupation number n(  ) of a fermion state whose energy is , n(  ) = 1 / {exp[  (  -  )] + 1}  is the chemical potential. When  = , n = 1/2 For instance, distribution of electrons

114. Bose-Einstein Statistics System: a boson’s state with an energy  Occupation of the system may be 0, 1, 2, 3, …, and correspondingly, the energy may be 0, , 2 , 3 , …. Therefore, the average occupation of the boson’s state,

115. 1 / {exp[  (  -  )] - 1} = Therefore, the average occupation number n(  ) of a boson state whose energy is , n(  ) = 1 / {exp[  (  -  )] - 1}

116. the chemical potential  must less than or equal to the ground state energy of a boson, i.e.    0, where  0 is the ground state energy of a boson. This is because that otherwise there is a negative occupation which is not physical. When  =  0, n(  )  , i.e., the occupation number is a macroscopic number. This phenomena is called Bose-Einstein Condensation! 4 He superfluid: when T  T c = 2.17K, 4 He fluid flows with no viscosity.

117. Classical or Chemical Statistics When the temperature T is high enough or the density is very dilute, n(  ) becomes very small, i.e. n(  ) > 1. Neglecting +1 or -1 in the denominators, both Fermi-Dirac and Bose-Einstein Statistics become n(  ) = exp[-  (  -  )] The Boltzmann distribution!

118. Canonical Partition Function the canonical distribution p i = exp(-  -  E i ) Sum over all the states,  i p i = 1. Thus, p i = exp(-  E i ) / Q where, Q   i exp(-  E i ) is called the canonical partition function.

119. An interpretation of the partition function: If we set the ground state energy E 0 to zero, As T  0, Q  the number of ground state, usually 1; As T , Q  the total number of states, usually .

120. Independent Molecules Total energy of a state i of the system, E i =  i (1) +  i (2) +  i (3) +  i (4) +…+  i (N) Q =  i exp[-  i (1) -  i (2) -  i (3) -  i (4) -… -  i (N)] = {  i exp[-  i (1)]} {  i exp[-  i (2)]} … {  i exp[-  i (N)]} = q N

121. Distinguishable and Indistinguishable Molecules for distinguishable molecules: for indistinguishable molecules: Q = q N Q = q N /N!

122. The Relation between entropy S and partition function Q S = [U-U(0)] / T + k lnQ The Helmholtz energy A - A(0) = -kT ln Q Fundamental Thermodynamic Relationships Relation between energy and partition function U = U(0) - (  lnQ/  ) V

123. The Enthalpy H - H(0) = -(  lnQ/  ) V + kTV(  lnQ/  V) T The Gibbs energy G - G(0) = - kT ln Q + kTV(  lnQ/  V) T The Pressure p = -(  A/  V) T p = kT(  lnQ/  V) T

124. Non-Ideal Gas Now let’s derive the equation of state for real gases. Consider a real gas with N monatomic molecules in a volume V. Assuming the temperature is T, and the mass of each molecule is m. So the canonical partition function Q can be expressed as Q =  i exp(-E i / kT) where the sum is over all possible state i, and E i is the energy of state i.

125. In the classical limit, Q may be expressed as Q =(1/N!h 3N )  …  exp(-H / kT) dp 1 … dp N dr 1 … dr N where, H = (1/2m)  i p i 2 +  i>j V(r i,r j ) Q = (1/N!) (2  mkT / h 2 ) 3N/2 Z N Z N =  …  exp(-  i>j V(r i,r j ) / kT) dr 1 … dr N [ note: for ideal gas, Z N = V N, and Q = (1/N!) (2  m kT / h 2 ) 3N/2 V N ] Z N = V N Q = (1/N!) (2  m kT / h 2 ) 3N/2 V N

126. The equation of state may be obtained via p = kT(  lnQ/  V) T We have thus, p / kT = (  lnQ/  V) T = (  lnZ N /  V) T (  lnZ N /  V) T Z N =  …  { 1 + [ exp(-  i>j V(r i,r j ) / kT) - 1 ] } dr 1 … dr N = V N +  …  [ exp(-  i>j V(r i,r j ) / kT) - 1 ] dr 1 … dr N  V N + (1/2) V N-2 N(N-1)  [ exp(- V(r 1,r 2 ) / kT) - 1 ] dr 1 dr 2  V N { 1 - (1/2V 2 ) N 2  [ 1 - exp(- V(r 1,r 2 ) / kT) ] dr 1 dr 2 } = V N { 1 - B N 2 / V } where, B = (1/2V)  [ 1 - exp(- V(r 1,r 2 ) / kT) ] dr 1 dr 2 B = (1/2V)  [ 1 - exp(- V(r 1,r 2 ) / kT) ] dr 1 dr 2

127. Therefore, the equation of state for our gas: p / kT = N / V + (N / V) 2 B = n + B n 2 Comparison to the Virial Equation of State The equation of state for a real gas P / kT = n + B 2 (T) n 2 + B 3 (T) n 3 + … This is the virial equation of state, and the quantities B 2 (T), B 3 (T), … are called the second, third, … virial coefficients.

128. Thus,

129. A. HARD-SPHERE POTENTIAL  r 12 <  U(r 12 ) = 0r 12 >  B 2 (T) = (1/2)  0  4  r 2 dr = 2  3 /3 = (1/2)  0  4  r 2 dr = 2  3 /3

130. B. SQUARE-WELL POTENTIAL  r 12 <  U(r 12 ) = -  < r 12 <  0r 12 >  B 2 (T) = (1/2)  0  4  r 2 dr = (2  3 /3) [1 - ( 3 -1) ( e  - 1 )] = (1/2)  0  4  r 2 dr

131. C. LENNARD-JONES POTENTIAL U(r) = 4  [ (  /r) 12 - (  /r) 6 ] Withx =  /r, T * = kT /  =  0  { 1 - exp[(-4/T * ) ( x 12 - x 6 )] } x 2 dx B 2 (T) = ( )  0  { 1 - exp{( ) [ ]} } 4  r 2 dr

132. Maxwell ’ s Demon (1867)

133. Thermal Fluctuation (Smolochowski, 1912) In his talk “Experimentally Verifiable Molecular Phenomena that Contradict Ordinary Thermodynamics”,… Smoluchowski showed That one could observe violations of almost all the usual statements Of the second law by dealing with sufficiently small systems. … the increase of entropy… The one statement that could be upheld… was the impossibility of perpetual motion of the second kind. No device could be ever made that would use the existing fluctuations to convert heat completely into work on a macroscopic scale … subject to the same chance fluctuations…. -----H.S. Leff & A.F. Rex, “Maxwell’s Demon”

134. Szilard ’ s one-molecule gas model (1929) To save the second law, a measure of where-about of the molecule produces at least entropy > k ln2

135. Measurement via light signals (L. Brillouin, 1951) h  k T A Temporary Resolution !!!???

136. Mechanical Detection of the Molecule Counter-clockwise rotation always !!! A Perpetual Machine of second kind ???

137. Bennett ’ s solution (1982) Demon’s memory To complete thermodynamic circle, Demon has to erase its memory !!! Memory eraser needs minimal Entropy production of k ln2 (R. Landauer, 1961)

138. Feynman ’ s Ratchet and Pawl System (1961) T 1 =T 2, no net rotation Feynman’s Lecture Notes

139. A honeybee stinger potential coordinate

140. A Simplest Maxwell’s demon door

141. Average over 200 trajectories No temperature difference!!! T t

142. A cooler demon T 1 > T 2 door T L > T R !!!

143. Our simple demon No. of particles: 60 The door’s moment of inertia: 0.2 Force constant of the string: 10 Maxwell’s demon No. of particles: 60 Threshold energy: 20 TLTL TRTR Number of particles in left side

144.  rate Feynman ’ s Ratchet and Pawl System (1961) T 1 =T 2, no net rotation T 1 > T 2, counter-clockwise rotation T 1 > T 2, clockwise rotation Mechanical Rectifier Feynman’s Lecture Notes

145. A two-chamber design: an analogy to Feynman ’ s Ratchet and Pawl Feynman’s Lecture Notes Our two-chamber design string

146. Potential of the pawl string radian potential

147. Feynman’s ratchet-pawl system

148. Feynman ’ s Ratchet and Pawl T L = T B

149. Micro-reversibility Pawl a transition state

150. Determination of temperature at equilibrium

151. Simulation results The ratchet moves when the leg is cooled down.

152. Angular velocity versus T L - T B kBkB radian

153. The Ratchet and Pawl as an engine T B =80T L =20 (T B - T L ) / T L =75%

154. Density of distribution in the phase space  q 1 …q f,p 1 …p f )  q 1  q f  p 1  p f Liouville’s Theorem: d  /dt = 0 Coarse-grained density over  q 1  q f  p 1  p f at  q 1 …q f,p 1 …p f ) : P =  …   q 1 …q f,p 1 …p f ) dq 1  dq f dp 1... dp f /  q 1  q f  p 1  p f Boltzmann’s H: H =  …  P log P dq 1  dq f dp 1... dp f where  q 1 …q f,p 1 …p f ) is fine-grained density at  q 1 …q f,p 1 …p f )

155. d(  …   log  dq 1  dq f dp 1...dp f )/dt = 0 Q =  log  -  log P -  + P  0 At t 1,  1  P 1 H 1 =  …   1 log  1 dq 1  dq f dp 1... dp f At t 2,  2  P 2 H 2 =  …  P 2 log P 2 dq 1  dq f dp 1... dp f H 1 - H 2  0 Boltzmann’s H-Theorem

156. OscillationHibernationRevival

158. Hibernation Revival Entropy [Q: Partition Function] S = k lnW = - Nk  i p i ln p i = k lnQ - (  lnQ/  ) V / T