Advanced Physical Chemistry

G. H. CHENDepartment of Chemistry

University of Hong Kong

Quantum Chemistry

G. H. ChenDepartment of Chemistry

University of Hong Kong

Emphasis Hartree-Fock methodConcepts Hands-on experience

Text Book “Quantum Chemistry”, 4th Ed. Ira N. Levine

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

In 1929, Dirac declared, “The underlying physical laws necessary for the mathematical theory of ...thewhole 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

Quantum Chemistry Methods

• Ab initio molecular orbital methods

• Semiempirical molecular orbital methods

• Density functional method

H E

SchrÖdinger Equation

HamiltonianH = (h2/2m

h2/2me)ii2

+ ZZeri e2/ri

ije2/rij

Wavefunction

Energy

Contents 1. Variation Method2. Hartree-Fock Self-Consistent Field Method

The Variation Method

Consider a system whose Hamiltonian operatorH is time independent and whose lowest-energy eigenvalue is E1. If is any normalized, well-

behaved function that satisfies the boundary conditions of the problem, then

* H dE1

The variation theorem

Proof:Expand in the basis set { k}

= k kk

where {k} are coefficients

Hk = Ekk

then* H dk j k

*j Ej kj

= k |k|2 Ek E 1 k |k|

2 = E1

Since is normalized, *dk |k|

2 = 1

i. : trial function is used to evaluate the upper limit of ground state energy E1

ii. = ground state wave function, * H dE1

iii. optimize paramemters in by minimizing * H d * d

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

* H dx = -(h2/82m) (lx-x2) d2(lx-x2)/dx2 dx = h2/(42m) (x2 - lx) dx = h2l3/(242m)

* dx = x2 (l-x)2 dx = l5/30

E = 5h2/(42l2m) h2/(8ml2) = E1

(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

 Variational Method

Example: one-dimensional harmonic oscillator Potential: V(x) = (1/2) kx2 = (1/2) m2x2 = 22m2x2

Trial wave function for the ground state:

(x) = exp(-cx2)

* H dx = -(h2/82m) exp(-cx2) d2[exp(-cx2)]/dx2

dx + 22m2 x2 exp(-2cx2) dx = (h2/42m) (c/8)1/2 + 2m2 (/8c3)1/2

* dx = exp(-2cx2) dx = (/2)1/2 c-1/2

E = W = (h2/82m)c + (2/2)m2/c

To minimize W,

0 = dW/dc = h2/82m - (2/2)m2c-2

c = 22m/h

W = (1/2) h

   ...

E3 3

E2 2

E1 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 * > E2

the first excited state energy

The trial wave function : d *1 = 0 k=1 ak k

 

d *1 = |a1|2 = 0

 E = d *H/ d * = k=2|ak|

2Ek / k=2|ak|2

> k=2|ak|2E2 / k=2|ak|

2 = E2

e

 

 

+ +

 

1

2

c1

1 + c

2

2

W = H d d= (c1

2 H11 + 2c1 c2 H12 + c22 H22 )

/ (c12 + 2c1 c2 S + c2

2 )  

W (c12 + 2c1 c2 S + c2

2) = c12 H11 + 2c1 c2 H12 + c2

2 H22

Application to H2+

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 < 0.

 Ground State: Eg = W1 = (H11+H12) / (1+S)

= () / 2(1+S)1/2

Excited State: Ee = W2 = (H11-H12) / (1-S)

= () / 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

Trial wave function: k3/2 -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

1s 2pInclusion of other atomic orbitals

Further Improvements H -1/2 exp(-r)He+ 23/2 -1/2 exp(-2r)

Optimization of 1s orbitals

 

 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

Introducing determinant: 

a11 a12

= a11a22-a12a21

a21 a22

  a11 a12 b1 a12

x1 =

a21 a22 b2 a22

a11 a12 a11 b1

x2 =

a21 a22 a21 b2

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

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

  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)

 

inhomogeneous case: bk = 0 for at least one k

(a) travial case: xk = 0, k = 1, 2, ... , n

(b) nontravial case: det(aij) = 0 

homogeneous case: bk = 0, k = 1, 2, ... , n

For a n-th order determinant, n

det(aij) = alk Clk

l=1

where, Clk is called cofactor

Trial wave function is a variation function which is a combination of n linear independent functions { f1 , f2 , ... fn},

 c1f1 + c2f2 + ... + cnfn

  n [( Hik - SikW ) ck ] = 0 i=1,2,...,n

k=1

Sik d fi fk

Hik d fi H fk

W dH d

  (i) W1 W2 ... Wn are 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

Molecular Orbital (MO): = c11 + c22

  ( 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 = c11 + c22 + ...... + cnn

linear combination of atomic orbitals

n

( Hik - SikW ) ck = 0 i = 1, 2, ......, nk=1

Hik d i* H k Sik d i

*k Skk = 1

Hamiltonian

H = (h2/2mh2/2me)ii

2 + ZZeri e2/ri

ije2/rij   

H ri;rri;r

The Born-Oppenheimer Approximation

ri;relri;rNr

el(r )= h2/2me)ii2

ie2/ri

ije2/rij VNN = ZZer

Hel(r) elri;rel(r)elri;r

(3) HN = (h2/2m U(r)

U(r) = el(r) + VNN

HN(r) NrNr

The Born-Oppenheimer Approximation:

Assignment

Calculate the ground state energy and bond length of H2

using the HyperChem with the 6-31G(Hint: Born-Oppenheimer Approximation)

e  + + 

e

two electrons cannot be in the same state.

Hydrogen Molecule H2

The Pauli principle

Since two wave functions that correspond to the same state can differ at most by a constant factor = c2 abc1ab=c2ab+c2c1ab

c1 = c2 c2c1 = 1Therefore: c1 = c2 = 1According to the Pauli principle, c1 = c2 =1

Wave function:= ab+ c1ab= ab+ c1ab

 Wave function f H2 : ! [

!

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

E=2 dTe+VeN) + VNN

+ dde2/r12 | = i=1,2 fii + J12 + VNN

To minimize Eunder the constraint d|use Lagrange’s method:  L = E dL = E d

4 dTe+VeN) +4 dde2/r12

d 

Energy: E

[ Te+VeN +de2/r12

 

f + Jf(1) = Te(1)+VeN(1) one electron operator

J(1) =de2/r12 two electron

Coulomb operator

Average Hamiltonian

Hartree-Fock equation

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: c11 + c22

 Multiple 1 from the left and then integrate :

c1F11 + c2F12 = (c1 + S c2)

Multiple 2 from the left and then integrate : 

c1F12 + c2F22 = (S c1 + c2) where,

Fij = di* ( f + J ) j = Hij + di

* J j

S = d1 2

(F11 - ) c1 + (F12 - S ) c2 = 0

(F12 - S ) c1 + (F22 - ) c2 = 0

Secular Equation:  F11 - F12 - S F12 - S F22 -  

bonding orbital: 1 = (F11+F12) / (1+S)

= () / 2(1+S)1/2

 antibonding orbital: 2 = (F11-F12) / (1-S )

= () / 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

 ----------------             1

  ---------------- 2

12 12 = 1/2 [122 1

ddH

dd(T1+V1N+T2+V2N+V12

+VNN)

1 T1+V1N|12 T2+V2N|2 + 12 V12 1212 V12 12 +

VNN

= i i T1+V1N |i+ 12 V12 1212 V12 12 + VNN

= i=1,2 fii + J12 K12 + VNN

Particle One: f(1) + J2(1) K2(1)Particle Two: f(2) + J1(2) K1(2)

  f(j) h2/2me)j

2 Zrj

Jj(1) drj*

e2/r12j

Kj(1) j drj*

e2/r12

Average Hamiltonian

f(1)+ J2(1) K2(1)1(1)11(1)f(2)+ J1(2) K1(2)2(2)22(2)

F(1) f(1)+ J2(1) K2(1) Fock operator for 1F(2) f(2)+ J1(2) K1(2) Fock operator for 2

Hartree-Fock Equation:

Fock Operator:

1. Many-Body Wave Function is approximated by Slater Determinant

2. Hartree-Fock EquationF 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

3. Roothaan Method (introduction of Basis functions)i = k cki k LCAO-MO

  {k } is a set of atomic orbitals (or basis functions)

4. Hartree-Fock-Roothaan equation j ( Fij - i Sij ) cji = 0

  Fij iF j Sij ij

5. Solve the Hartree-Fock-Roothaan equation self-consistently

Assignment one8.40, 10.5, 10.6, 10.7, 10.8,

11.37, 13.37

1. At the Hartree-Fock Level there are two possible Coulomb integrals contributing the energy betweentwo 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.

Summary

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)

 

Nelectrons spin up and Nelectrons spin down. 

Fock matrix for an electron 1 with spin up:

 F(1) = f (1) + j [ Jj(1) Kj

(1) ] + j Jj(1)

j=1,N j=1,N

Fock matrix for an electron 1 with spin down: F(1) = f (1) + j [ Jj

(1) Kj(1) ] + j

Jj(1)

j=1,Nj=1,N 

f(1) h2/2me)12 N ZNr1N

Jj(1) drj

e2/r12j

Kj(1) j

drj

*e2/r12

Energy = j fjj

+j fjj

+(1/2) i j

( Jij Kij

)

+ (1/2) i j

( Jij Kij

) + i j

Jij

+ VNN

i=1,Nj=1,N

fjjfjj

jf j

JijJij

j(2)Ji

j(2)Kij

Kij

j(2)Ki

j(2)

JijJij

j(2)Ji

j(2) F(1) = f (1) + j=1,n/2 [ 2Jj(1) Kj(1) ] Energy = 2 j=1,n/2 fjj + i=1,n/2 j=1,n/2 ( 2Jij Kij ) +VNN

Close subshell case: ( N= N= n/2 )

abcdnf(1) efghnaf(1)

ebcdnfghn= af(1) eif b=f, c=g, ..., d=h; 0, otherwise abcdnV12 |efghnabV12

efcdnghn= abV12 efif c=g, ..., d=h; 0, otherwise

The Condon-Slater Rules

-------the lowest unoccupied molecular orbital -------

the highest occupied molecular orbital ------- -------

The energy required to remove an electron from aclosed-shell atom or molecules is well approximatedby minus the orbital energy of the AO or MO fromwhich the electron is removed.

HOMO

LUMO

Koopman’s Theorem

# HF/6-31G(d) Route section water energy Title

0 1 Molecule Specification O -0.464 0.177 0.0 (in Cartesian coordinatesH -0.464 1.137 0.0H 0.441 -0.143 0.0

Slater-type orbitals (STO)  nlm = N rn-1exp(r/a0) Ylm(,)

 the orbitalexponent* is used instead of in the textbook

Gaussian type functionsgijk = N xi yj zk exp(-r2)

(primitive Gaussian function)p = u dup gu

(contracted Gaussian-type function, CGTF)u = {ijk} p = {nlm}

Basis Set i = p cip p

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 orbital3-21G: 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 rows6-31G**: 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 densityis 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

6-31G for a carbon atom: (10s4p) [3s2p]

1s 2s 2pi (i=x,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(DZ+P, 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, groundstate energy, molecular orbital energies, and vibrational modes and frequencies of a water molecule using Hartree-Fock method with 3-21G basis set.

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: H2O

(using HyperChem)

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”.

The Hartree-Fock treatment of H2

+

e-

+

e-

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 = <1(1) 2(2)|H|1(1) 2(2)>

H12 = H21 = <1(1) 2(2)|H|1(2) 2(1)>

S = <1(1) 2(2)|1(2) 2(1)> [ = S2 ]

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 H2

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 ............ HMO: 50% H ............ H 50% H+............ H

VB: 100% H ............ H

The VB is computationally expensive and requireschemical intuition in implementation.

The Generalized valence-bond (GVB) method is avariational method, and thus computationally feasible.(William A. Goddard III)

)1()2(

)2()1(

2

1

f

f2211 fcfc

022

12

21

11

WH

SWH

SWH

WH

22121

21212112

21212211

)1()2()2()1(

)1()2()2()1(

)2()1()2()1(

SS

HHH

HHH

The Heitler-London ground-state wave function

2/)1()2()2()1()1(2/)1()2()2()1( 2121 S

R

R

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-FockConfiguration Interaction (CI)*Perturbation theory*Coupled Cluster MethodDensity functional theory

-e -e r12

r2 r1

+2e

H = - (h2/2me)12 - 2e2/r1 - (h

2/2me)22 - 2e2/r2 + e2/r12

H10 H2

0 H’

H0 = H10 + H2

0

(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, ...

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

3/2exp(-2r1/a0)

E(0) = 4e2/a0 

E(1) = <(0)(1,2)| H’ |(0)(1,2)> = 5e2/4a0

E E(0) + E(1) = -108.8 + 34.0 = -74.8 (eV) [compared with exp. -79.0 eV]

Ground state wave function

H = H0 + H’H0n

(0) = En(0)n

(0)

n(0) is an eigenstate for unperturbed system

H’ is small compared with H0

Nondegenerate Perturbation Theory (for Non-Degenerate Energy Levels)

H( = H0 + H’Hn = Ennnn

nn

kn(k)

nnn

nkn

(k)

the original Hamiltonian

Introducing a parameter

nnn

nn

(k)nn

nn

n(k)

Where, < nn

(j) > = 0, j=1,2,...,k,...

Hn = En

n

solving for Enn

HnH’n

= Enn

nn

solving for Enn

HnH’n

= Enn

nn

nn

solving for Enn

 Multiplied m

(0) from the left and integrate,<m

(0) Hn(1) > + <m

(0) H'n(0) > = <m

(0)n(1) >En

Enmn

<m(0)n

(1) > [EmEn

+ <m(0) H'n

(0) > = Enmn

For m = n,

For m n, <m(0)n

(1) > = <m(0) H'n

(0) > /

[EnEm

If we expand n(1) = cnmm

(0),

cnm = <m(0) H'n

(0) > / [EnEm

for m n;

cnn = 0.

n(1) = m <m

(0) H'n(0) > / [En

Emm

(0) Eq.(2)

The first order:

En<n

(0) H'n(0) > Eq.(1)

The second order:

<m(0)Hn

(2) > + <m(0)H'n

(1) > = <m(0)n

(2)

>En<m

(0)n(1) >En

Enmn

 Set m = n, we have

En= m n |m

(0) H'n(0) >|2 / [En

Emq.(3)

a. Eq.(2) shows that the effect of the perturbationon the wave function n

(0) is to mix in

contributions from the other zero-th order states m

(0) mn. Because of the factor 1/(En(0)-Em

(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)

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 systemwith potential-energy function V = b for L/4 < x < 3L/4,V = 0 for 0 < x L/4 & 3L/4 x < Land 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 groundand 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(2x/L),

respectively.

Example Two:As the first step of the Moller-Plesset perturbation theory, Hartree-Fock method gives the zeroth-orderenergy. 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 <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%

Statistical Mechanics

Content:Ensembles and Their DistributionsQuantum StatisticsCanonical Partition FunctionNon-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 thecurrent state of the system. In other words, a statefunction is some function of the state of the system.

State Functions: E, N, T, V, P, ......

When a system reaches its equilibrium, its statefunctions 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.

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.

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 isolatedsystem 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 systemsamong 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 weightof 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 +

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.

Utilizing the Stirling’s approximation, ln x! = x ln x - x

lnW/ai = - ln ai/A = - ,

Distribution of a Canonical Ensemble

State 1 2 3 … k …Energy E1 E2 E3 … Ek …

Occupation 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 iwith the energy Ei ,

pi = ai/A = exp( -Ei)

The above formula is the canonical distributionof 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 GrandCanonical 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 agrand 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 samestate at the same time.

Question: what is the average number particles or occupation of a quantum state?

Fermi-Dirac Statistics

System: a fermion’s state with an energy ( - / )

------------------ --------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}

exp1

exp

exp0exp

exp

exp0exp

exp10exp0

εn

Therefore, the average occupation number n() of a fermion state whose energy is ,

n() = 1 / {exp[(-)] + 1}

is the chemical potential. When = , n = 1/2For instance, distribution of electrons

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 theboson’s state,

--exp-1

--exp

...3exp2expexp0exp

...3exp32exp2exp10exp0

n

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 negativeoccupation 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() << 1. In another word, exp[(-)] >> 1.Neglecting +1 or -1 in the denominators, both Fermi-Dirac and Bose-Einstein Statistics become

n() = exp[-(-)]

The Boltzmann distribution!

Canonical Partition Functionthe 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[-i(1) - i(2) - i(3) - i(4) -…

- i(N)]

= {i exp[-i(1)]} {i exp[-i(2)]} …

{i exp[-i(N)]}

= qN

Distinguishable and Indistinguishable Molecules

for distinguishable molecules:

for indistinguishable molecules:

Q = qN

Q = qN/N!

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

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 Pressurep = -(A/V)T

p = kT( lnQ/V)T

Non-Ideal Gas

Now let’s derive the equation of state for realgases.

Consider a real gas with N monatomic moleculesin a volume V. Assuming the temperature is T,and the mass of each molecule is m. So the canonical partition function Q can be expressedas

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 …

drNwhere, H = (1/2m) i pi2 + i>j V(ri,rj)

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

ZN = … exp(-i>j V(ri,rj) / kT) dr1 …

drN [ note: for ideal gas, ZN = VN , and

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

ZN = VN

Q = (1/N!) (2m 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(-i>j V(ri,rj) / kT) - 1 ] } dr1 … drN

= VN + … [ exp(-i>j 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,

12

12

2121

2

exp12

1

,exp1

2

1

drkT

rV

drdrkT

rrVVBTB

A. HARD-SPHERE POTENTIAL

r12 < U(r12) =

0 r12 >

B2(T) = (1/2) 0 4r2 dr

  = 23/3

= (1/2) 0 4r2

dr= 23/3

B. SQUARE-WELL POTENTIAL

r12 < U(r12) = - < r12 <

0 r12 >

B2(T) = (1/2) 0 4r2 dr

  = (23/3) [1 - (3 -1) ( e - 1 )] = (23/3) [1 - (3 -1) ( e - 1 )]= (1/2) 0

4r2

dr

C. LENNARD-JONES POTENTIAL

U(r) = 4 [ (/r)12 - (/r)6 ]

With x = /r, T* = kT /

= 0 { 1 - exp[(-4/T*) ( x12 - x6 )] } x2 dx

3

2

2TB

B2(T) = ( ) 0 { 1 - exp{( ) [ ]} } 4r2

dr

612

rr

2

1

kT

4

Maxwell’s Demon (1867)

Thermal Fluctuation (Smolochowski, 1912)

In his talk “Experimentally Verifiable Molecular Phenomena thatContradict Ordinary Thermodynamics”,… Smoluchowski showedThat one could observe violations of almost all the usual statementsOf 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. Nodevice could be ever made that would use the existing fluctuationsto convert heat completely into work on a macroscopic scale … subject to the same chance fluctuations….

-----H.S. Leff & A.F. Rex, “Maxwell’s Demon”

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

Measurement via light signals (L. Brillouin, 1951)

h k T

A Temporary Resolution !!!???

Mechanical Detection of the Molecule

Counter-clockwise rotation always !!!A Perpetual Machine of second kind ???

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)

Feynman’s Ratchet and Pawl System (1961)

T1=T2, no net rotation

Feynman’s Lecture Notes

A honeybee stinger

potential

coordinate

A Simplest Maxwell’s demon

door

Average over 200 trajectoriesNo temperature difference!!!

T

t

A cooler demon

T1 > T2

door

TL > TR !!!

Our simple demonNo. of particles: 60The door’s moment of inertia: 0.2Force constant of the string: 10

Maxwell’s demonNo. of particles: 60Threshold energy: 20

TL

TR

Number of particles in left side

rate

Feynman’s Ratchet and Pawl System (1961)

T1=T2, no net rotationT1 > T2, counter-clockwise rotationT1 > T2, clockwise rotation

Mechanical Rectifier

Feynman’s Lecture Notes

A two-chamber design: an analogy to Feynman’s Ratchet and Pawl

Feynman’s Lecture Notes Our two-chamber design

string

Potential of the pawl

string

radian

potential

Feynman’s ratchet-pawl system

Feynman’s Ratchet and Pawl

TL = TB

Micro-reversibility

),(),( qpPqpP

Pawl Pawl

a transition state

Determination of temperature at equilibrium

ratchetgas TT

Simulation results

The ratchet moves when the leg is cooled down.

Angular velocity versus TL - TB

kB

rad

ian

The Ratchet and Pawl as an engine

0 2 4 6 8 10

0.0000

0.0005

0.0010

0.0015

0.0020

0.0025

effi

cie

ncy

Applied torque

Calculated points

TB=80 TL=20

(TB- TL) / TL =75%

Density of distribution in the phase space

q1…qf,p1…pf) q1qfp1pf

Liouville’s Theorem: d/dt = 0

Coarse-grained density over q1qfp1pf at q1…qf,p1…pf) :

P = … q1…qf,p1…pf) dq1dqf dp1 ... dpf / q1qfp1pf

Boltzmann’s H: H = … P log P dq1dqf dp1 ... dpf

whereq1…qf,p1…pf) is fine-grained density at q1…qf,p1…pf)

d(… log dq1dqf dp1...dpf)/dt = 0

Q = log - log P - + P 0

At t1, 1P1

H1= … 1 log 1 dq1dqf dp1 ... dpf

At t2, 2P2

H2= … P2 log P2 dq1dqf dp1 ... dpf

H1 - H2 0

Boltzmann’s H-Theorem

0 500 1000 1500 2000

-30

-20

-10

0

10

20

30

40

Re

lativ

e d

ista

nce

(A

ng

stro

m)

Time (ps)

(5,0)@(14,0)55A @ 70A, 500K

Oscillation Hibernation Revival

0 500 1000 1500 2000

-30

-20

-10

0

10

20

30

40R

ela

tive

dis

tan

ce (

An

gst

rom

)

Time (ps)

(5,0)@(14,0)55A @ 70A, 500K

Hibernation

Revival

Entropy [Q: Partition Function]S = k lnW = - Nk i pi ln pi = k lnQ - (lnQ/)V / T