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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.

Ira N. Levine

http//yangtze.hku.hk/lecture/chem3504-3.ppt

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

Quantum Chemistry Methods

- Ab initio molecular orbital methods
- Semiempirical molecular orbital methods
- Density functional method

SchrÖdinger Equation

H y E y

Wavefunction

Hamiltonian H ??(-h2/2ma)??2 - (h2/2me)?i?i2

???? ZaZbe2/rab - ?i ?? Zae2/ria ?i ?j

e2/rij

Energy

Contents 1. Variation Method 2. Hartree-Fock

Self-Consistent Field Method

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 H f dt gt E1

Proof Expand f in the basis set yk f

?k akyk where ak are coefficients Hyk

Ekyk then ? f H f dt ?k ?j ak aj Ej dkj

?k ak2 Ek gt E 1 ?k ak2 E1

Since is normalized, ? ff dt ?k ak2

1

i. f trial function is used to evaluate the

upper limit of ground state energy E1 ii. f

ground state wave function, ? f H f dt

E1 iii. optimize paramemters in f by

minimizing ? f H f dt / ? f f dt

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)

? ? 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

Variational Method

(1) Construct a wave function ?(c1,c2,???,cm) (2)

Calculate the energy of ? E? ?

E?(c1,c2,???,cm) (3) Choose cj (i1,2,???,m)

so that E? is minimum

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) d2exp(-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

To minimize W, 0 dW/dc h2/8?2m -

(?2/2)m?2c-2 c 2?2?m/h W (1/2) h?

Extension of Variation Method

. . . E3 y3 E2 y2 E1 y1

For a wave function f which is orthogonal to the

ground state wave function y1, i.e. ?dt fy1

0 Ef ?dt fHf / ?dt ff gt E2 the first

excited state energy

The trial wave function f ?dt fy1 0 f

?k1 ak yk ?dt fy1 a12 0 Ef ?dt fHf

/ ?dt ff ?k2ak2Ek / ?k2ak2 gt

?k2ak2E2 / ?k2ak2 E2

Application to H2

e

y1

y2 f c1y1 c2y2

W ? fH f dt / ? ff dt (c12 H11 2c1

c2 H12 c22 H22 ) / (c12 2c1 c2 S

c22 ) W (c12 2c1 c2 S c22) c12 H11

2c1 c2 H12 c22 H22

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

To have nontrivial solution H11 - W H12 - S

W H12 - S W H22 - W For H2, H11 H22 H12 lt

0. Ground State Eg W1 (H11H12) / (1S)

f1 (y1y2) / ?2(1S)1/2 Excited

State Ee W2 (H11-H12) / (1-S) f2

(y1-y2) / ?2(1-S)1/2

0

bonding orbital

Anti-bonding orbital

Results De 1.76 eV, Re 1.32 A Exact

De 2.79 eV, Re 1.06 A

1 eV 23.0605 kcal / mol

Further Improvements H p-1/2 exp(-r) He

23/2 p-1/2 exp(-2r)

Optimization of 1s orbitals

Trial wave function k3/2 p-1/2 exp(-kr)

Eg W1(k,R) at each R, choose k so 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

a11x1 a12x2 b1 a21x1 a22x2

b2 (a11a22-a12a21) x1 b1a22-b2a12 (a11a22-a12a

21) x2 b2a11-b1a21

Linear Equations

1. two linear equations for two unknown, x1 and x2

(No Transcript)

n linear equations for n unknown variables

a11x1 a12x2 ... a1nxn b1 a21x1 a22x2

... a2nxn b2 ..................................

.......... an1x1 an2x2 ... annxn bn

a11 a12 ... a1,k-1 b1 a1,k1 ... a1n

a21 a22 ... a2,k-1 b2 a2,k1 ...

a2n det(aij) xk . . ... . .

. ... . an1 an2 ... an,k-1

b2 an,k1 ... ann where,

a11 a12 ... a1n a21 a22 ... a2n det(aij)

. . ... . an1 an2 ... ann

inhomogeneous case bk 0 for at least one k

a11 a12 ... a1,k-1 b1 a1,k1 ... a1n

a21 a22 ... a2,k-1 b2 a2,k1 ... a2n .

. ... . . . ... .

an1 an2 ... an,k-1 b2 an,k1 ... ann xk

det(aij)

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 l1 where, Clk is

called cofactor

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 i1,2,...,n k1 Sik

? ?dt fi fk Hik ? ? dt fi H fk W ? ? dt f H f

/ ? dt f f

(i) W1 ? W2 ? ... ? Wn are n roots of

Eq.(1), (ii) E1 ? E2 ? ... ? En ? En1 ? ...

are energies of eigenstates then, W1 ?

E1, W2 ? E2, ..., Wn ? En

Linear variational theorem

Molecular Orbital (MO) ? c1?1 c2?2

( H11 - W ) c1 ( H12 - SW ) c2 0

S111 ( H21 - SW ) c1 ( H22 - W

) c2 0

S221 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 k1 Hik ? ? dt ?i H ?k Sik ?

?dt ?i?k Skk 1

The Born-Oppenheimer Approximation

Hamiltonian H ??(-h2/2ma)??2 -

(h2/2me)?i?i2 ???? ZaZbe2/rab -

?i ?? Zae2/ria ?i ?j e2/rij

H y(rira) E y(rira)

The Born-Oppenheimer Approximation

- (1) y(rira) yel(rira) yN(ra)
- (2) Hel(ra ) - (h2/2me)?i?i2 - ?i?? Zae2/ria
- ?i?j e2/rij
- VNN ???b ZaZbe2/rab
- Hel(ra) yel(rira) Eel(ra) yel(rira)
- (3) HN ??(-h2/2ma)??2 U(ra)
- U(ra) Eel(ra) VNN
- HN(ra) yN(ra) E yN(ra)

Assignment Calculate the ground state energy and

bond length of H2 using the HyperChem with the

6-31G (Hint Born-Oppenheimer Approximation)

Hydrogen Molecule H2

e

e two electrons cannot be

in the same state.

The Pauli principle

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) c2 f(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

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

Energy E?

- E?2? dt1 f(1) (TeVeN) f(1) VNN
- ? dt1 dt2 f2(1) e2/r12 f2(2)
- ?i1,2 fii J12 VNN
- To minimize E? under the constraint ? dt f2

1, - use Lagranges method
- L E? - 2 e ? dt1 f2(1) - 1
- dL dE? - 4 e ? dt1 f(1)df(1)
- 4? dt1 df(1)(TeVeN)f(1)
- 4? dt1 dt2 f(1)f(2) e2/r12 f(2)df(1)
- - 4 e ? dt1 f(1)df(1)
- 0

TeVeN ? 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

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 e is the energy of orbital f.

LCAO-MO f c1y1 c2y2 Multiple y1 from the

left and then integrate c1F11 c2F12 e (c1

S c2)

Multiple y2 from the left and then integrate

c1F12 c2F22 e (S c1 c2) where,

Fij ? dt yi ( f J ) yj Hij ? dt yi J

yj S ? dt y1 y2 (F11 - e) c1 (F12

- S e) c2 0 (F12 - S e) c1 (F22 - e) c2 0

bonding orbital e1 (F11F12) / (1S)

f1 (y1y2) / ?2(1S)1/2 antibonding

orbital e2 (F11-F12) / (1-S ) f2

(y1-y2) / ?2(1-S)1/2

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.

Bond Order one-half the difference between the

number of bonding and antibonding electrons

--------?-------- f1

--------?-------- f2

Ey ? dt1dt2 y H y ? dt1dt2 y

(T1V1NT2V2NV12VNN) y ltf1(1)

T1V1Nf1(1)gt ltf2(2) T2V2Nf2(2)gt

ltf1(1) f2(2) V12 f1(1) f2(2)gt

- ltf1(2) f2(1) V12 f1(1) f2(2)gt VNN

?i ltfi(1) T1V1N fi(1)gt ltf1(1)

f2(2) V12 f1(1) f2(2)gt - ltf1(2)

f2(1) V12 f1(1) f2(2)gt VNN

?i1,2 fii J12 - K12 VNN

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)

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

Hartree-Fock Method

1. Many-Body Wave Function is approximated by

Slater Determinant 2. Hartree-Fock Equation F

fi ei fi F Fock operator fi the i-th

Hartree-Fock orbital ei the energy of the i-th

Hartree-Fock orbital

3. Roothaan Method (introduction of Basis

functions) fi ?k cki yk LCAO-MO

yk is a set of atomic orbitals (or basis

functions) 4. Hartree-Fock-Roothaan equation

?j ( Fij - ei Sij ) cji 0 Fij ? lt ?i F

?j gt Sij ? lt ?i ?j gt 5. Solve the

Hartree-Fock-Roothaan equation

self-consistently

Assignment one 8.40, 10.5, 10.6, 10.7, 10.8,

11.37, 13.37

Summary

1. At the Hartree-Fock Level there are two

possible Coulomb integrals contributing the

energy between two electrons i and j Coulomb

integrals Jij and exchange integral

Kij 2. For two electrons with different spins,

there is only Coulomb integral Jij 3. For two

electrons with the same spins, both Coulomb and

exchange integrals exist.

4. Total Hartree-Fock energy consists of the

contributions from one-electron integrals fii

and two-electron Coulomb integrals Jij and

exchange integrals Kij 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 Jj(i) is the Coulomb

potential (operator) that i feels from j, and

Kj(i) is the exchange potential (operator) that

that i feels from j.

6. Fock operator (or, average Hamiltonian)

consists of one-electron operators f(i) and

Coulomb operators Jj(i) and exchange

operators Kj(i)

?

? ?

? ?

?

? ?

? ? ?

? ?

Fock matrix for an electron 1? with spin

down Fb(1?) f b(1?) ?j Jjb(1?) - Kjb(1?)

?j Jja(1?) j1?,Nb

j1?,Na

f(1) ? -(h2/2me)?12 - ?N ZN/r1N Jja(1) ? ? dr2

fja(2) e2/r12 fja(2) Kja(1) q(1) ? fja(1) ? dr2

fja(2) e2/r12 q(2)

i1,Na j1,Nb

fjj ? fjja ? ltfja f fjagt Jij ? Jijaa ?

ltfaj(2) Jia(1) faj(2)gt Kij ? Kijaa ? ltfaj(2)

Kia(1) faj(2)gt Jij ? Jijab ? ltfbj(2) Jia(1)

fbj(2)gt

F(1) f (1) ?j1,n/2 2Jj(1) - Kj(1)

Energy 2 ?j1,n/2 fjj ?i1,n/2

?j1,n/2 ( 2Jij - Kij ) VNN

Close subshell case ( Na Nb n/2 )

The Condon-Slater Rules

ltfa(1)fb(2)fc(3)...fd(n) f(1)

fe(1)ff(2)fg(3)...fh(n)gt ltfa(1) f(1) fe(1)gt

lt fb(2)fc(3)...fd(n) ff(2)fg(3)...fh(n)gt

ltfa(1) f(1) fe(1)gt if bf, cg, ...,

dh 0, otherwise ltfa(1)fb(2)fc(3)...f

d(n) V12 fe(1)ff(2)fg(3)...fh(n)gt ltfa(1)

fb(2) V12 fe(1) ff(2)gt lt fc(3)...fd(n)

fg(3)...fh(n)gt ltfa(1) fb(2) V12 fe(1)

ff(2)gt if cg, ..., dh 0, otherwise

------- the lowest

unoccupied molecular orbital ? -------

the highest occupied molecular orbital ?

-------

-------

LUMO

HOMO

Koopmans Theorem

The energy required to remove an electron from

a closed-shell atom or molecules is well

approximated by minus the orbital energy e of the

AO or MO from which the electron is removed.

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

Basis Set ?i ?p cip ?p

Gaussian type functions gijk N xi yj zk

exp(-ar2) (primitive Gaussian function) ?p ?u

dup gu (contracted Gaussian-type function,

CGTF) u ijk p nlm

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

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 6-31G, 6-31G,

6-31G and 6-31G basis sets. Double-zeta

(DZ) basis set two STO for each AO

6-31G for a carbon atom (10s4p) ? 3s2p

1s 2s 2pi (ix,y,z) 6GTFs

3GTFs 1GTF 3GTFs 1GTF

1CGTF 1CGTF 1CGTF 1CGTF 1CGTF

(s) (s) (s) (p)

(p)

Minimal basis set One STO for

each inner-shell and

valence-shell AO of each atom

example C2H2 (2S1P/1S)

C 1S, 2S, 2Px,2Py,2Pz

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 C2H2

(4S2P/2S) C two

1S, two 2S,

two 2Px, two 2Py,two 2Pz

H two 1S (STOs)

total 24 STOs as Basis set

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(DZP, or

DZP) Additional STO w/l quantum

number larger

than the lmax of the valence - shell

? ( 2Px, 2Py ,2Pz ) to H

? Five 3d Aos to Li - Ne , Na -Ar

? C2H5 O Si H3

(6s4p1d/4s2p1d/2s1p)

Si C,O H

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.

Ab Initio Molecular Orbital Calculation H2O

(using HyperChem)

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.

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

ConvYES 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.

The Hartree-Fock treatment of H2

The Valence-Bond Treatment of H2

f1 ?1(1) ?2(2) f2 ?1(2) ?2(1) ? c1 f1

c2 f2 H11 - W H12 - S W H21 - S W H22 -

W H11 H22 lt?1(1) ?2(2)H?1(1) ?2(2)gt H12

H21 lt?1(1) ?2(2)H?1(2) ?2(1)gt S lt?1(1)

?2(2)?1(2) ?2(1)gt S2 The Heitler-London

ground-state wave function ?1(1) ?2(2) ?1(2)

?2(1)/?2(1S)1/2 a(1)b(2)-a(2)b(1)/?2

0

Comparison of the HF and VB Treatments

HF LCAO-MO wave function for H2 ?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

H2 ?1(1) ?2(2) ?2(1) ?1(2) H

H H H

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)

The Heitler-London ground-state wave function

Electron Correlation

Human Repulsive Correlation

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

H - (h2/2me)?12 - 2e2/r1 - (h2/2me)?22 - 2e2/r2

e2/r12 H10

H20

H

H0 H10 H20 y(0)(1,2) F1(1) F2(2) H10 F1(1)

E1 F1(1) H20 F2(1) E2 F2(1) E1 -2e2/n12a0

n1 1, 2, 3, ... E2 -2e2/n22a0 n2 1, 2, 3,

...

Ground state wave function

y(0)(1,2) (1/p1/2)(2/a0)3/2exp(-2r1/a0)

(1/p1/2)(2/a0)3/2exp(-2r1/a0) E(0) -

4e2/a0 E(1) lty(0)(1,2) H y(0)(1,2)gt

5e2/4a0 E ? E(0) E(1) -108.8 34.0

-74.8 (eV) compared with exp. -79.0 eV

Nondegenerate Perturbation Theory (for

Non-Degenerate Energy Levels)

H H0 H H0yn(0) En(0) yn(0) yn(0) is an

eigenstate for unperturbed system H is small

compared with H0

Introducing a parameter l

H(l) H0 lH H(l) yn(l) En(l) yn(l) yn(l)

yn(0) l yn(1) l2 yn(2) ... lk yn(k)

... En(l) En(0) l En(1) l2 En(2) ... lk

En(k) ...

l 1, the original Hamiltonian

yn yn(0) yn(1) yn(2) ... yn(k) ... En

En(0) En(1) En(2) ... En(k) ...

Where, lt yn(0) yn(j) gt 0, j1,2,...,k,...

- H0yn(0) En(0) yn(0)
- solving for En(0), yn(0)

- H0yn(1) H yn(0) En(0) yn(1) En(1)yn(0)
- solving for En(1), yn(1)

H0yn(2) H yn(1) En(0) yn(2) En(1)yn(1)

En(2)yn(0) ? solving for En(2),yn(2)

Multiplied ym(0) from the left and

integrate, ltym(0) H0 yn(1) gt lt ym(0) H'

yn(0) gt lt ym(0)yn(1) gtEn(0) En(1) ?mn

ltym(0)yn(1) gt Em(0)- En(0) lt ym(0) H'

yn(0) gt En(1) ?mn

The first order

For m n,

En(1) lt yn(0) H' yn(0) gt Eq.(1)

For m ? n, ltym(0)yn(1) gt lt ym(0) H' yn(0)

gt / En(0)- Em(0) If we expand yn(1) ? cnm

ym(0), cnm lt ym(0) H' yn(0) gt / En(0)-

Em(0) for m ? n cnn 0.

yn(1) ?m lt ym(0) H' yn(0) gt / En(0)-

Em(0) ym(0) Eq.(2)

The second order

ltym(0)H0yn(2) gt lt ym(0)H'yn(1) gt lt

ym(0)yn(2) gtEn(0) lt ym(0)yn(1) gtEn(1)

En(1) ?mn Set m n, we have

En(2) ?m ? n ltym(0) H' yn(0) gt2 /

En(0)- Em(0) Eq.(3)

Discussion (Text Book page 522-527)

a. Eq.(2) shows that the effect of the

perturbation on the wave function yn(0) is to

mix in contributions from the other zero-th

order states ym(0) m?n. Because of the factor

1/(En(0)-Em(0)), the most important

contributions to the yn(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 Hnn 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.

Moller-Plesset (MP) Perturbation Theory The MP

unperturbed Hamiltonian H0 H0 ?m

F(m) where F(m) is the Fock operator for

electron m. And thus, the perturbation H

H H - H0 Therefore, the unperturbed

wave function is simply the Hartree-Fock wave

function ?. Ab initio methods MP2, MP4

Example One Consider the one-particle,

one-dimensional system with potential-energy

function V b for L/4 lt x lt 3L/4, V 0 for

0 lt x ? L/4 3L/4 ? x lt 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.

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, E1(0) E1(1) ? E1 where E1(0) and E1(1)

are the zero-th order energy and the first order

energy correction, and E1 is the ground state

energy of the full Hamiltonian H0 H.

Example Four Calculate the bond orders of Li2

and Li2.

Ground State Excited State CPU Time

Correlation Geometry Size Consistent

(CH3NH2,6-31G) HFSCF ?

? 1 0

OK ? DFT

? ?

1 ?

? CIS ?

? lt10

OK ? CISD

? ?

17 80-90 ?

?

(20

electrons) CISDTQ ? ?

very large 98-99 ?

? MP2 ?

? 1.5

85-95 ? ?

(DZP) MP4

? ?

5.8 gt90 ?

? CCD ? ?

large gt90

? ? CCSDT ?

? very large

100 ? ?

Statistical Mechanics

Content Ensembles and Their Distributions Quant

um Statistics Canonical Partition

Function Non-Ideal Gas

References 1. Grasser Richards, An

Introduction to Statistical

Thermodynamcis 2. Atkins, Physical Chemistry

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.

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.

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.

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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.

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

Configurations and Weights

Imagine that an ensemble contains total A

systems among which a1 systems with energy E1 and

N1 molecules, a2 systems with energy E2 and N2

molecules, a3 systems with energy E3 and N3

molecules, with energy ?1, and so on. The

specific distribution of systems in the ensemble

is called configuration of the system, denoted

as a1, a2, a3, .......

A configuration a1, a2, a3, ...... 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! / (a1! a2! a3! ...)

Distribution of a Microcanonical Ensemble

State 1 2 3 k Energy E E E E Occupation

a1 a2 a3 ak

Constraint ?i ai A

W A! / a1! a2! a3!

To maximize lnW under the constraint, we

construct a Lagrangian

L lnW ? ?i ai

Thus,

0 ?L/?ai ?lnW/?ai ?

Utilizing the Stirlings approximation, ln x!

x ln x - x

?lnW/?ai - ln ai/A - ?,

the probability of a system being found in state

i,

pi ai/A exp(?)

constant

or, in another word, the probabilities of all

states with the same energy are equal.

Distribution of a Canonical Ensemble

State 1 2 3 k Energy E1 E2 E3 Ek Occupa

tion a1 a2 a3 ak

Constraints ?i ai A ?i ai Ei ?

where, ? is the total energy in the ensemble. W

A! / a1! a2! a3!

To maximize lnW under the above

constraints, construct a Lagrangian

L lnW ? ?i ai - ? ?i ai Ei 0 ?L/?ai

?lnW/?ai ? - ?Ei

ln ai/A ? - ?Ei

the probability of a system being found in state

i with the energy Ei ,

pi ai/A exp(? -?Ei)

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).

Distribution of a Grand Canonical Ensemble

State 1 2 3 k Energy E1 E2 E3 Ek Mol.

No. N1 N2 N3 Nk Occupation a1 a2 a3 ak

Constraints ?i ai A ?i ai Ei ? ?i ai

Ni N

where, ? and N are the total energy and total

number of molecules in the ensemble,

respectively.

W A! / a1!a2! a3!

To maximize lnW under the above constraints,

construct a Lagrangian

L lnW ? ?i ai - ? ?i ai Ei - ? ?i ai Ni 0

?L/?ai ?lnW/?ai ? - ?Ei - ? Ni ln ai/A ? -

?Ei - ? Ni

the probability of a system being found in state

i with the energy Ei and the number of

particles Ni,

pi ai/A exp(? -?Ei -? Ni)

The above formula describes the distribution of

a grand canonical system, and is called the grand

canonical distribution. When Ni is fixed, the

above distribution becomes the canonical

distribution. Thus, the grand canonical

distribution is most general.

Quantum Statistics

Quantum Particle Fermion (S 1/2, 3/2, 5/2,

...) e.g. electron, proton, neutron, 3He nuclei

Boson (S 0, 1, 2, ...) e.g. deuteron,

photon, phonon, 4He 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?

Fermi-Dirac Statistics

------------------ --------l---------

occupation n 0 n

1 energy 0 ?

probability exp(0) exp-?(?-?)

There are only two states because of the Pauli

exclusion principle.

Thus, the average occupation of the quantum

state ?,

1 / exp?(?-?) 1

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

Bose-Einstein Statistics

System a bosons 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 bosons state,

1 / exp?(?-?) - 1

Therefore, the average occupation number n(?) of

a boson state whose energy is ?,

n(?) 1 / exp?(?-?) - 1

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!

4He superfluid when T ? Tc 2.17K, 4He fluid

flows with no viscosity.

Classical or Chemical Statistics

When the temperature T is high enough or the

density is very dilute, n(?) becomes very

small, i.e. n(?) ltlt 1. In another word,

exp?(?-?) gtgt 1. Neglecting 1 or -1 in the

denominators, both Fermi-Dirac and Bose-Einstein

Statistics become

n(?) exp-?(?-?)

The Boltzmann distribution!

Canonical Partition Function

the canonical distribution

pi exp(-?-?Ei)

Sum over all the states, ?i pi 1. Thus,

pi exp(-?Ei) / Q

where, Q ? ?i exp(-?Ei) is called the

canonical partition function.

An interpretation of the partition function

If we set the ground state energy E0 to zero,

As T ? 0, Q ? the number of ground state,

usually 1 As T ??, Q ? the total number of

states, usually ?.

Independent Molecules

Total energy of a state i of the system, Ei

?i(1) ?i(2) ?i(3) ?i(4) ?i(N) Q

?i exp-b?i(1) - b?i(2) - b?i(3) - b?i(4) -

- b?i(N) ?i exp-b?i(1) ?i

exp-b?i(2) ?i exp-b?i(N) qN

Distinguishable and Indistinguishable Molecules

for distinguishable molecules for

indistinguishable molecules

Q qN

Q qN/N!

Fundamental Thermodynamic Relationships

Relation between energy and partition function

U U(0) - (?lnQ/??)V

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

The Pressure

p -(?A/?V)T

p kT(? lnQ/?V)T

The Enthalpy

H - H(0) -(? lnQ/??)V kTV(? lnQ/?V)T

The Gibbs energy

G - G(0) - kT ln Q kTV(? lnQ/?V)T

Non-Ideal Gas

Now lets 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(-Ei / kT)

where the sum is over all possible state i, and

Ei is the energy of state i.

In the classical limit, Q may be expressed as

Q (1/N!h3N) ?? exp(-H / kT) dp1 dpN dr1 drN

where, H (1/2m) ?i pi2 ?igtj V(ri,rj)

Q (1/N!) (2?mkT / h2)3N/2 ZN

ZN ?? exp(-?igtj V(ri,rj) / kT) dr1 drN

ZN VN

note for ideal gas, ZN VN , and Q (1/N!)

(2?m kT / h2)3N/2 VN

Q (1/N!) (2?m kT / h2)3N/2 VN

The equation of state may be obtained via

p kT(? lnQ/?V)T

We have thus,

p / kT (? lnQ/?V)T (? lnZN / ?V)T

(? lnZN / ?V)T

ZN ?? 1 exp(-?igtj V(ri,rj) / kT) - 1

dr1 drN VN ?? exp(-?igtj V(ri,rj) /

kT) - 1 dr1 drN ? VN (1/2) VN-2 N(N-1) ??

exp(- V(r1,r2) / kT) - 1 dr1 dr2 ? VN 1 -

(1/2V2) N2 ?? 1 - exp(- V(r1,r2) / kT) dr1

dr2 VN 1 - B N2 / V where, B (1/2V) ??

1 - exp(- V(r1,r2) / kT) dr1 dr2

B (1/2V) ?? 1 - exp(- V(r1,r2) / kT) dr1 dr2

Therefore, the equation of state for our gas

p / kT N / V (N / V)2 B

n B n2

Comparison to the Virial Equation of State

The equation of state for a real gas

P / kT n B2(T) n2 B3(T) n3

This is the virial equation of state, and the

quantities B2(T), B3(T), are called the

second, third, virial coefficients.

Thus,

A. HARD-SPHERE POTENTIAL

? r12 lt ? U(r12) 0 r12 gt ?

B2(T) (1/2) ?0? 4?r2 dr 2??3/3

(1/2) ?0? 4?r2 dr

2??3/3

B. SQUARE-WELL POTENTIAL

? r12 lt ? U(r12) -? ? lt r12 lt

?? 0 r12 gt ??

B2(T) (1/2) ?0? 4?r2 dr (2??3/3) 1 -

(?3 -1) ( e?? - 1 )

(1/2) ?0? 4?r2 dr

(2??3/3) 1 - (?3 -1) ( e?? - 1 )

C. LENNARD-JONES POTENTIAL

U(r) 4? (?/r)12 - (?/r)6

With x ?/r, T kT / ?

Maxwells Demon (1867)

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,

Maxwells Demon

Szilards one-molecule gas model (1929)

To save the second law, a measure of where-about

of the molecule produces at least entropy gt k

ln2

Measurement via light signals (L. Brillouin, 1951)

h n gtgt k T

A Temporary Resolution !!!???

Mechanical Detection of the Molecule

Counter-clockwise rotation always !!! A Perpetual

Machine of second kind ???

Bennetts solution (1982)

To complete thermodynamic circle, Demon has to

erase its memory !!! Memory eraser needs minimal

Entropy production of k ln2 (R. Landauer, 1961)

Demons memory

Feynmans Ratchet and Pawl System (1961)

T1T2, no net rotation

A honeybee stinger

A Simplest Maxwells demon

Average over 200 trajectoriesNo temperature

difference!!!

T

t

A cooler demon

T1 gt T2

door

TL gt TR !!!

TL

TR

Number of particles in left side

Our simple demon No. of particles 60 The doors

moment of inertia 0.2 Force constant of the

string 10

Maxwells demon No. of particles 60 Threshold

energy 20

Feynmans Ratchet and Pawl System (1961)

T1T2, no net rotation T1 gt T2, counter-clockwise

rotation T1 gt T2, clockwise rotation

Mechanical Rectifier

A two-chamber design an analogy to Feynmans

Ratchet and Pawl

Potential of the pawl

Feynmans ratchet-pawl system

Feynmans Ratchet and Pawl

Micro-reversibility

Determination of temperature at equilibrium

Simulation results

The ratchet moves when the leg is cooled down.

Angular velocity versus TL - TB

The Ratchet and Pawl as an engine

TB80 TL20 (TB- TL) / TL 75

Density of distribution in the phase space

dN r(q1qf,p1pf) dq1 ...dqf dp1 ... dpf

where r(q1qf,p1pf) is fine-grained density at

(q1qf,p1pf)

Liouvilles Theorem dr/dt 0

Coarse-grained density over dq1 ...dqf dp1 ...

dpf at (q1qf,p1pf) P ?? r(q1qf,p1pf)

dq1 ...dqf dp1 ... dpf / dq1 ...dqf dp1 ... dpf

Boltzmanns H H ?? P log P dq1 ...dqf

dp1 ... dpf

Boltzmanns H-Theorem

d(?? r log r dq1...dqf dp1...dpf)/dt 0 Q r

log r - r log P - r P ? 0 At t1, r1 P1 H1

?? r1 log r1 dq1 ...dqf dp1 ... dpf At t2, r2

? P2 H2 ?? P2 log P2 dq1 ...dqf dp1 ...

dpf H1 - H2 ? 0

Oscillation

Hibernation

Revival

Non-ergodic Zone

Entropy reduction?

D.J. Evans et al. Prob(dS)/Prob(-dS) Exp(dS)

gt

Revival

Hibernation

Entropy Q Partition Function S k lnW - Nk

?i pi ln pi k lnQ - (?lnQ/??)V / T