Advanced Physical Chemistry

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Advanced Physical Chemistry
G. H. CHEN Department of Chemistry University of
Hong Kong
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Quantum Chemistry
G. H. Chen Department of Chemistry University of
Hong Kong
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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
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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
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Quantum Chemistry Methods

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

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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
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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
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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
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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
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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)
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? ? 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
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 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
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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
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To minimize W, 0 dW/dc h2/8?2m -
(?2/2)m?2c-2 c 2?2?m/h W (1/2) h?
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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
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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
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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
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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
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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
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Results De 1.76 eV, Re 1.32 A   Exact
De 2.79 eV, Re 1.06 A  
1 eV 23.0605 kcal / mol
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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
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    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
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(No Transcript)
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n linear equations for n unknown variables
a11x1 a12x2 ... a1nxn b1 a21x1 a22x2
... a2nxn b2 ..................................
.......... an1x1 an2x2 ... annxn bn
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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
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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)  
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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
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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
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  (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
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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
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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)
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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)

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

  e two electrons cannot be
in the same state.
The Pauli principle
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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
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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
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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 

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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
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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)
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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
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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
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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.
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Bond Order one-half the difference between the
number of bonding and antibonding electrons
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--------?-------- f1  
--------?-------- f2
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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
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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)
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Hartree-Fock Equation
f(1) J2(1) - K2(1) f1(1) e1 f1(1) f(2)
J1(2) - K1(2) f2(2) e2 f2(2)
Fock Operator
F(1) ? f(1) J2(1) - K2(1) Fock operator for
1 F(2) ? f(2) J1(2) - K1(2) Fock operator for 2
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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
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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
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Assignment one 8.40, 10.5, 10.6, 10.7, 10.8,
11.37, 13.37
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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.
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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.
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6. Fock operator (or, average Hamiltonian)
consists of one-electron operators f(i) and
Coulomb operators Jj(i) and exchange
operators Kj(i)
?
? ?
? ?
?  
? ?
? ? ?
? ?
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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 
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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
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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 )
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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
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------- 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.
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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
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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
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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
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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
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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)
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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
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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
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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.
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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.
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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.
69
The Hartree-Fock treatment of H2
70
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
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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
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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)
73
The Heitler-London ground-state wave function
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Electron Correlation
Human Repulsive Correlation
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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
76
H - (h2/2me)?12 - 2e2/r1 - (h2/2me)?22 - 2e2/r2
e2/r12 H10
H20
H
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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
78
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
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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,...
80

• 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)
81
  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)
82
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)
83
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.
84
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
85
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.
86
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.
87
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 ? ?
88
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
89
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.
90
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.
91
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.
92
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93
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94
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96
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.
97
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
98
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, .......
99
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
100
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 ?
101
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.
102
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!
103
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)
104
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).
105
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
106
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!
107
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)
108
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.
109
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?
110
Fermi-Dirac Statistics
------------------ --------l---------
occupation n 0 n
1 energy 0 ?
probability exp(0) exp-?(?-?)
111
There are only two states because of the Pauli
exclusion principle.
Thus, the average occupation of the quantum
state ?,
1 / exp?(?-?) 1
112
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
113
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,
114
1 / exp?(?-?) - 1

Therefore, the average occupation number n(?) of
a boson state whose energy is ?,
n(?) 1 / exp?(?-?) - 1
115
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.
116
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!
117
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.
118
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 ?.
119
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
120
Distinguishable and Indistinguishable Molecules
for distinguishable molecules for
indistinguishable molecules
Q qN
Q qN/N!
121
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
122
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
123
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.
124
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
125
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
126
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.
127
Thus,
128
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
129
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 )
130
C. LENNARD-JONES POTENTIAL
U(r) 4? (?/r)12 - (?/r)6
With x ?/r, T kT / ?
131
Maxwells Demon (1867)
132
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
133
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
134
Measurement via light signals (L. Brillouin, 1951)
h n gtgt k T
A Temporary Resolution !!!???
135
Mechanical Detection of the Molecule
Counter-clockwise rotation always !!! A Perpetual
Machine of second kind ???
136
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
137
Feynmans Ratchet and Pawl System (1961)
T1T2, no net rotation
138
A honeybee stinger
139
A Simplest Maxwells demon
140
Average over 200 trajectoriesNo temperature
difference!!!
T
t
141
A cooler demon
T1 gt T2
door
TL gt TR !!!
142
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
143
Feynmans Ratchet and Pawl System (1961)
T1T2, no net rotation T1 gt T2, counter-clockwise
rotation T1 gt T2, clockwise rotation
Mechanical Rectifier
144
A two-chamber design an analogy to Feynmans
Ratchet and Pawl
145
Potential of the pawl
146
Feynmans ratchet-pawl system
147
Feynmans Ratchet and Pawl
148
Micro-reversibility
149
Determination of temperature at equilibrium
150
Simulation results
The ratchet moves when the leg is cooled down.
151
Angular velocity versus TL - TB
152
The Ratchet and Pawl as an engine
TB80 TL20 (TB- TL) / TL 75
153
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
154
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
155
Oscillation
Hibernation
Revival
156
Non-ergodic Zone
Entropy reduction?
D.J. Evans et al. Prob(dS)/Prob(-dS) Exp(dS)
gt
157
Revival
Hibernation
Entropy Q Partition Function S k lnW - Nk
?i pi ln pi k lnQ - (?lnQ/??)V / T