Ch121a Atomic Level Simulations of Materials and
Molecules
Room BI 115
Hours: Monday, Wednesday, Friday 2-3pm
Lecture 1, March 31, 2014
QM-1: HF
Presented by Dr. Julius Su/SKIES
William A. Goddard III, wag@wag.caltech.edu
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
TA’s Caitlin Scott and Andrea Kirkpatrick
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
1
CH121a Atomic Level Simulations of Materials and
Molecules
Instructor: William A. Goddard III
Prerequisites: some knowledge of quantum mechanics, classical
mechanics, thermodynamics, chemistry, Unix. At least at the Ch2a
level
Ch121a is meant to be a practical hands-on introduction to
expose students to the tools of modern computational
chemistry and computational materials science relevant to
atomistic descriptions of the structures and properties of
chemical, biological, and materials systems.
This course is aimed at experimentalists (and theorists) in
chemistry, materials science, chemical engineering, applied
physics, biochemistry, physics, geophysics, and mechanical
engineering with an interest in characterizing and designing
molecules, drugs, and materials.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
2
Motivation: Design Materials, Catalysts, Pharma from 1st
Principles so can do design prior to experiment
To connect 1st Principles (QM) to Macro work use an overlapping hierarchy of
methods (paradigms) (fine scale to coarse) so that parameters of coarse level
are determined by fine scale calculations.
Thus all simulations are first-principles based
time
ELECTRONS ATOMS
GRAINS
GRIDS
hours
Continuum
(FEM)
millisec
Micromechanical modeling
Protein clusters
MESO
nanosec
MD
picosec
Deformation and Failure
Protein Structure and Function
QM
femtosec
simulations real devices
full cell (systems biology)
distance
Å
nm
micron
mm
Big breakthrough making FC simulations
yards practical:
Accurate calculations for bulk phases
reactive force fields based on QM
and molecules (EOS, bond dissociation)
Describes: chemistry,charge transfer, etc. For
3
Chemical
Reactions (P-450 ©
oxidation)
metals,
oxides,III,organics.
Lecture
1Ch121a-Goddard-L01
copyright 2012 William
A. Goddard
all rights reserved\
Lectures
The lectures cover the basics of the fundamental methods:
quantum mechanics,
force fields,
molecular dynamics,
Monte Carlo,
statistical mechanics, etc.
required to understand the theoretical basis for the simulations
the homework applies these principles to practical problems
making use of modern generally available software.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
4
Homework and Research Project
First 5 weeks: The homework each week uses generally available
computer software implementing the basic methods on
applications aimed at exposing the students to understanding how
to use atomistic simulations to solve problems.
Each calculation requires making decisions on the specific
approaches and parameters relevant and how to analyze the
results.
Midterm: each student submits proposal for a project using the
methods of Ch121a to solve a research problem that can be
completed in the final 5 weeks.
The homework for the last 5 weeks is to turn in a one page report
on progress with the project
The final is a research report describing the calculations and
conclusions
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
5
Methods to be covered in the lectures include:
Quantum Mechanics: Hartree Fock and Density Function
methods
Force Fields standard FF commonly used for simulations of
organic, biological, inorganic, metallic systems, reactions;
ReaxFF reactive force field: for describing chemical reactions,
shock decomposition, synthesis of films and nanotubes, catalysis
Molecular Dynamics: structure optimization, vibrations, phonons,
elastic moduli, Verlet, microcanonical, Nose, Gibbs
Monte Carlo and Statistical thermodynamics Growth
amorphous structures, Kubo relations, correlation functions, RIS,
CCBB, FH methods growth chains, Gauss coil, theta temp
Coarse grain approaches
eFF for electron dynamics
Tight Binding for electronic properties
solvation, diffusion,
mesoscale force fields
6
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Applications will include prototype examples
involving such materials as:
Organic molecules (structures, reactions);
Semiconductors (IV, III-V, surface reconstruction)
Ceramics (BaTiO3, LaSrCuOx)
Metal alloys (crystalline, amorphous, plasticity)
Polymers (amorphous, crystalline, RIS theory, block);
Protein structure, ligand docking
DNA-structure, ligand docking
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
7
The stratospheric review of QM
You should have already been exposed to all the
material on the next xx slides
This is just a review to remind you of the key points
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
8
Starting point for First Principles QM
Classical Mechanics
Energy = Kinetic energy + Potential energy
Kinetic energy =
+
A
1
1 2
Z Z
Z
1
p
2A +
p
i A B A
2M A
i 2
A B R AB
i
A R Ai
i j rij
electrons
atoms
1
1 2
Z AZ B
ZA
1
2
A 2M energy
i 2 =i
A
Potential
R
R
A
B
i
A
i j rij
A
AB
Ai
Nucleus-Nucleus
repulsion
Nucleus-Electron
attraction
Electron-Electron
repulsion
Can optimize electron coordinates and momenta separately,
thus lowest energy: all p=0 KE =0
All electrons on nuclei: PE = - infinity
Makes for dull world
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
9
Starting point for First Principles
Ab Initio, quantum mechanics
The wavefunction Ψ(r1,r2,…,rN) contains all
information of system determine KE and PE
Energy = < Ψ|KE operator|Ψ> + < Ψ|PE operator|Ψ>
Kinetic energy op =
A
1
1
Z Z
Z
1
2A i2 A B A
2M A
i 2
A B RAB
i
A RAi
i j rij
atoms
electrons
1
1 2
Z AZ B
ZA
1
2
Potential
A 2M energy
i 2= i
A
R
R
A B
i
A
i j rij
A
AB
Ai
Optimize Ψ, get HelΨ=EΨ
11 22 1Z A2Z B Z A Z B Z A Z A 1
1
i
A H
i
2A =A
2i Mel
Ai B2 R AB A B iRABA R Ai
jAirij i j rij
i
A iR
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
10
Schrodinger Equation
HelΨ=EΨ
1 1 22
1 Z A2Z B Z A Z B Z A
ZA 1
1
H =
Ai i
2 RAB A B Ri AB A RiAi A iR Ai
A 2M
i B
iel 2
A
j rij i j rij
A
2
A
Solving SE gives exact properties of molecules, solids,
enzymes, etc
History
H atom, Schrodinger 1925-26
H2 Simple (Valence bond) 1927, accurate 1937
C2H6 simple 1963, accurate 1980’s
2008: can get accurate wavefunctions for ~100-200
atoms
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
11
Topic 1: Practical Quantum Chemistry
Solve Schrödinger Equation HelΨ=EΨ
1 1 22
1 Z A2Z B Z A Z B Z A
ZA 1
1
H
=
el Ai i
2M
i 2
A
2
i B
A
RAB A B Ri AB A RiAi
R r
A i Ai
j ij i j
rij
For benzene we have 12 nuclei and
hence 3*12=36 degrees of freedom
(dof) and 42 electrons or 3*42 dof
For each set of nuclear dof, we solve HelΨ=EΨ to
calculate Ψ, the probability amplitudes for finding the 42
electrons at various locations
1st term: kinetic energy operator HOW MANY
2nd term: attraction of electrons to nuclei: HOW MANY
3rd term: electron-electron repulsion HOW MANY
th term: what is missing?
12
4Lecture
1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
The Schrödinger Equation: Kinetic Energy
Solve Schrödinger Equation HelΨ=EΨ
1 1 22
1 Z A2Z B Z A Z B Z A
ZA 1
1
H
=
el Ai i
2M
i 2
A
2
i B
A
RAB A B Ri AB A RiAi
R r
A i Ai
j ij i j
rij
For benzene we have 12 nuclei and
hence 3*12=36 degrees of freedom
(dof) and 42 electrons or 3*42 dof
For each set of nuclear dof, we solve HelΨ=EΨ to
calculate Ψ, the probability amplitudes for finding the 42
electrons at various locations
1st term: kinetic energy operator: 42 terms
2nd term: attraction of electrons to nuclei: HOW MANY
3rd term: electron-electron repulsion HOW MANY
th term: what is missing?
13
4Lecture
1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
The Schrödinger Equation: Nuclear-Electron
Solve Schrödinger Equation HelΨ=EΨ
1 1 22
1 Z A2Z B Z A Z B Z A
ZA 1
1
H
=
el Ai i
2M
i 2
A
2
i B
A
RAB A B Ri AB A RiAi
R r
A i Ai
j ij i j
rij
For benzene we have 12 nuclei and
hence 3*12=36 degrees of freedom
(dof) and 42 electrons or 3*42 dof
For each set of nuclear dof, we solve HelΨ=EΨ to
calculate Ψ, the probability amplitudes for finding the 42
electrons at various locations
1st term: kinetic energy operator: 42 terms
2nd term: attraction of electrons to nuclei: 42*12= 504
3rd term: electron-electron repulsion HOW MANY
th term: what is missing?
14
4Lecture
1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
The Schrödinger Equation: Electron-Electron
Solve Schrödinger Equation HelΨ=EΨ
1 1 22
1 Z A2Z B Z A Z B Z A
ZA 1
1
H
=
el Ai i
2M
i 2
A
2
i B
A
RAB A B Ri AB A RiAi
R r
A i Ai
j ij i j
rij
For benzene we have 12 nuclei and
hence 3*12=36 degrees of freedom
(dof) and 42 electrons or 3*42 dof
For each set of nuclear dof, we solve HelΨ=EΨ to
calculate Ψ, the probability amplitudes for finding the 42
electrons at various locations
1st term: kinetic energy operator: 42 terms
2nd term: attraction of electrons to nuclei: 42*12= 504
3rd term: electron-electron repulsion 42*41/2= 861 terms
th term: what is missing?
15
4Lecture
1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
The Schrödinger Equation: Nuclear-Nuclear
Solve Schrödinger Equation HelΨ=EΨ
1 1 22
1 Z A2Z B Z A Z B Z A
ZA 1
1
H
=
2
i
el Ai
2
R
R
r
M
2
R
R
i B
iAi A i Ai
i A
A
j ij i j rij
AB A B i AB A
Missing is the nuclear-nuclear repulsion
Enn = SA<B ZA*ZB/RAB
This does not depend on electron coordinates, but it
does affect the total energy
Eel = <Ψ|Hel|Ψ>/<Ψ|Ψ>=
Etotal = Eel + Enn
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
16
Closed shell Hartree Fock (HF)
For benzene with 42 electrons, the ground state HF wavefunction
has 21 doubly occupied orbitals, φ1,.. φi,.. φ21
And we want to determine the optimum shape and energy for
these orbitals
First consider the componets of the total energy
Σ i=1,21< φi|h|φi> from the 21 up spin orbitals
Σ i=1,21< φi|h|φi> from the 21 down spin orbitals
Σ I<j=1,21 [Jij – Kij] interactions between the 21 up spin orbitals
Σ I<j=1,21 [Jij – Kij] interactions between the 21 down spin orbitals
Σ I≠j=1,21 [Jij] interactions of the 21 up spin orbitals with the 21
down spin orbitals
Enn = Σ A<B=1,12 ZAZB/RAB nuclear-nuclear repulsion
Combining these terms leads to
E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I≠j=1,21 2[2Jij-Kij] + Σ I=1,21 [2Jii]
But Jii = Kii so we can rewrite this as
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
17
The HF orbitals of H2O
TAs put energies of 5
occupied orbitals plus
lowest 2 unoccupied
orbitals, use correct
symmetry notation
Show orbitals
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
18
The HF orbitals of ethylene
TAs put energies of 8
occupied orbitals plus
lowest 2 unoccupied
orbitals, use correct
symmetry notation
Show orbitals
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
19
Results for Benzene
The energy of the C1s orbital is ~ - Zeff2/2
where Zeff = 6 – 0.3125 = 5.6875
Thus e1s ~ -16.1738 h0 = - 440.12 eV.
This leads to 6 orbitals all with very similar energies.
This lowest has the + combination of all 6 1s orbitals,
while the highest alternates with 3 nodal planes.
There are 6 CH bonds and 6 CC bonds that are
symmetric with respect to the benzene plane, leading to
12 sigma MOs
The highest MOs involve the p electrons. Here there are
6 electrons and 6 pp atomic orbitals leading to 3 doubly
occupied and 3 empty orbitals with the pattern
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
20
The HF orbitals of benzene
TAs put energies of
21 occupied orbitals
plus lowest 4
unoccupied orbitals,
use correct symmetry
notation
Show orbitals
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
21
Pi orbitals of benzene
Top view
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
22
The HF orbitals of
N2
With 14 electrons we
get M=7 doubly
occupied HF orbitals
We can visualize this
as a triple NN bond
plus valence lone
pairs
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
23
The energy diagram for N2
TAs put energies of 7
occupied orbitals plus
lowest 2 unoccupied
orbitals, use correct
symmetry notation
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
24
The energy expression for closed shell HF
E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I<j=1,21 2[2Jij-Kij] + Σ I=1,21 [Jii]
This says for any two different orbitals we get 4 coulomb
interactions and 2 exchange interactions, but the two electrons in
the same orbital only lead to a single Coulomb term
Since Jii = Kii (self coulomb = self exchange) we can write
E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I≠j=1,21 [2Jij-Kij] + Σ I=1,21 [2Jii-Kii]
and hence
E = Σ i=1,21 2< φi|h|φi> + Enn + Σ I,j=1,21 [2Jij-Kij]
which is the final expression for Closed Shell HF
Now we need to apply the variational principle to find the
equations determining the optimum orbitals, the HF orbitals
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
25
Consider the case of 4 electrons in 2 orbitals
E = 2<φ1|h|φ1> + 2< φ2|h|φ2> + Enn
+ [2J11-K11] +2[2J12-K12] + [2J22-K22]
Here we can write Jij = (ii|jj) where the first two indices go with
electron 1 and the other two with electron 2
Also we write Jij = (ii|jj) = <i|Jj|i>, where Jj is the coulomb potetial
seen by electron 1 due to the electron in orbital j.
Thus if we change φ1 to φ1 + dφ1 the change in the energy is
dE = 4<dφ1|h|φ1> + 4 <dφ1|2J1-K1|φ1> + 4 <dφ1|2J2-K2|φ1>
= 4 <dφ1|HHF|φ1>
Where HHF = h + Σj=1,2 [2Jj-Kj] is called the HF Hamiltonian
In the above expression we assume that φ1 was normalized,
<φ1|h|φ1> = 1.
Imposing this constraint (a Lagrange multiplier) leads to
<dφ1|HHF – l1|φ1> = 0 and <dφ2|HHF – l2|φ2> = 0
Thus the optimum orbitals satisfy HHFφk = lk φk the HF equations
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
26
The general case of 2M electrons
For the general case the HF closed shell equations are
HHFφk = lk φk where we solve for k=1,M occupied orbitals
HHF = h + Σj=1,M [2Jj-Kj]
This is the same as the Hamiltonian for a one electron system
moving in the average electrostatic and exchange potential, 2Jj-Kj
due to the other N-1 = 2M-1 electrons
Problem: sum over 2Jj leads to 2M Coulomb terms, not 2M-1
This is because we added the self Coulomb and exchange terms
But (2Jk-Kk) φk = (Jk) φk so that these self terms cancel.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
27
Analyze HF equations
The optimum orbitals for the 4 electron closed shell wavefunction
Ψ(1,2,3,4) = A[(aa)(ab)(ba)(bb)]
Are eigenstate of the HF equations
HHFφk = lk φk for k=1,2
where HHF = h + Σj=1,2 [2Jj-Kj]
This looks like a one-electron Hamiltonian but it involves the
average Coulomb potential of 2 electrons in φa plus 2 electrons in
φb plus exchange interactions with one electron in φa plus one
electron in φb
It seems wrong that there should be 4 coulomb interactions
whereas each electron sees only 3 other electrons and that there
are two exchange interactions whereas each electron sees only
one other with the same spin.
This arises because we added and subtracted a self term in the
total energy
Since (Jk-Kk)φk = 0 there spurious terms cancel.
28
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Main practical applications of QM
Determine the Optimum geometric structure and
energies of molecules and solids
Determine geometric structure and energies of
reaction intermediates and transition states for
various reaction steps
Determine properties of the optimized
geometries: bond lengths, energies,
frequencies, electronic spectra, charges
Determine reaction mechanism: detailed
sequence of steps from reactants to products
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
29
The Matrix HF equations
The HF equations are actually quite complicated because Kj
is an integral operator, Kj φk(1) = φj(1) ʃ d3r2 [φj(2) φk(2)/r12]
The practical solution involves expanding the orbitals in terms
of a basis set consisting of atomic-like orbitals,
φk(1) = Σm Cm Xm, where the basis functions, {Xm, m=1, MBF}
are chosen as atomic like functions on the various centers
As a result the HF equations HHFφk = lk φk
Reduce to a set of Matrix equations
ΣjmHjmCmk = ΣjmSjmCmkEk
This is still complicated since the Hjm operator includes
Exchange terms
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
30
Minimal Basis set – STO-3G
For benzene the smallest possible basis set is to use a 1s-like
single exponential function, exp(-zr) called a Slater function,
centered on each the 6 H atoms and
C1s, C2s, C2pz, C2py, C2pz functions on each of the 6 C atoms
This leads to 42 basis functions to describe the 21 occupied MOs
and is refered to as a minimal basis set.
In practice the use of exponetial functions, such as exp(-zr),
leads to huge computational costs for multicenter molecules and
we replace these by an expansion in terms of Gaussian basis
functions, such as exp(-ar2).
The most popular MBS is the STO-3G set of Pople in which 3
gaussian functions are combined to describe each Slater function
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
31
Double zeta + polarization Basis sets – 6-31G**
To allow the atomic orbitals to contract as atoms are brought
together to form bonds, we introduce 2 basis functions of the
same character as each of the atomic orbitals:
Thus 2 each of 1s, 2s, 2px, 2py, and 2pz for C
This is referred to as double zeta. If properly chosen this leads to
a good description of the contraction as bonds form.
Often only a single function is used for the C1s, called split
valence
In addition it is necessary to provide one level higher angular
momentum atomic orbitals to describe the polarization involved in
bonding
Thus add a set of 2p basis functions to each H and a set of 3d
functions to each C.
The most popular such basis is referred to as 6-31G**
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
32
6-31G** and 6-311G**
6-31G** means that the 1s is described with 6 Gaussians,
the two valence basis functions use 3 gaussians for the
inner one and 1 Gaussian for the outer function
The first * use of a single d set on each heavy atom
(C,O etc)
The second * use of a single set of p functions on each
H
The 6-311G** is similar but allows 3 valence-like functions
on each atom.
There are addition basis sets including diffuse functions (+)
and additional polarization function (2d, f) (3d,2f,g), but
these will not be relvent to EES810
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
33
Effective Core Potentials (ECP, psuedopotentials)
For very heavy atoms, say starting with Sc, it is computationally
convenient and accurate to replace the inner core electrons
with effective core potentials
For example one might describe:
• Si with just the 4 valence orbitals, replacing the Ne core with
an ECP or
• Ge with just 4 electrons, replacing the Ni core
• Alternatively, Ge might be described with 14 electrons with the
ECP replacing the Ar core. This leads to increased accuracy
because the
• For transition metal atoms, Fe might be described with 8
electrons replacing the Ar core with the ECP.
• But much more accurate is to use the small Ne core, explicitly
treating the (3s)2(3p)6 along with the 3d and 4s electrons
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
34
Software packages
Jaguar: Good for organometallics
QChem: very fast for organics
Gaussian: many analysis tools
GAMESS
HyperChem
ADF
Spartan/Titan
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
35
HF wavefunctions
Good distances, geometries, vibrational levels
But
breaking bonds is described extremely poorly
energies of virtual orbitals not good description of
excitation energies
cost scales as 4th power of the size of the
system.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
36
Electron correlation
In fact when the electrons are close (rij small), the electrons
correlate their motions to avoid a large electrostatic repulsion,
1/rij
Thus the error in the HF equation is called electron correlation
For He atom
E = - 2.8477 h0 assuming a hydrogenic orbital exp(-zr)
E = -2.86xx h0 exact HF (TA look up the energy)
E = -2.9037 h0 exact
Thus the elecgtron correlation energy for He atom is 0.04xx h0
= 1.x eV = 24.x kcal/mol.
Thus HF accounts for 98.6% of the total energy
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
37
Configuration interaction
Consider a set of N-electron wavefunctions: {i;
i=1,2, ..M} where < i|j> = dij {=1 if i=j and 0 if i ≠j)
Write approx = S (i=1 to M) Ci i
Then E = < approx|H|approx>/< approx|approx>
E= < Si Ci i |H| Si Cj j >/ < Si Ci i | Si Cj j >
How choose optimum Ci?
Require dE=0 for all dCi get
Sj <i |H| Cj j > - Ei< i | Cj j > = 0 ,which we
write as
HCi = SCiEi in matrix notation, ie ΣjkHjkCki = ΣjkSjkCkiEi
where Hjk = <j|H|k > and Sjk = < j|k > and Ci is a
column vector for the ith eigenstate
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
38
Configuration interaction upper bound theorm
Consider the M solutions of the CI equations
HCi = SCiEi ordered as i=1 lowest to i=M highest
Then the exact ground state energy of the system
Satisfies Eexact ≤ E1
Also the exact first excited state of the system
satisfies
E1st excited ≤ E2
etc
This is called the Hylleraas-Unheim-McDonald
Theorem
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
39
Stop April 1, 2013
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
40
Alternative to Hartree-Fork,
Density Functional Theory
Walter Kohn’s dream:
replace the 3N electronic degrees of freedom needed to define
the N-electron wavefunction Ψ(1,2,…N) with
just the 3 degrees of freedom for the electron density (x,y,z).
It is not obvious that this would be possible but
min
P. Hohenberg and W.
Kohn
Phys.
Rev.
FHK
[
]B
76, 6062
V(1964).
= V - rep
Showed that there exists some functional of the density
that gives the exact energy of the system
Kohn did not specify the nature or form of this functional,
but research over the last 46 years has provided
increasingly accurate approximations to it.
Lecture 1Ch121a-Goddard-L01
Walter Kohn (1923-)
Nobel
Chemistry 1998 41
© copyright 2012 William A. Goddard
III, allPrize
rights reserved\
The Hohenberg-Kohn theorem
The Hohenberg-Kohn theorem states that if N interacting
electrons move in an external potential, Vext(1..N), the
ground-state electron density (xyz)=(r) minimizes the
functional
E[] = F[] + ʃ (r) Vext(r) d3r
where F[] is a universal functional of and the minimum
value of the functional, E, is E0, the exact ground-state
electronic energy.
Here we take Vext(1..N) = Si=1,..N SA=1..Z [-ZA/rAi], which is the
electron-nuclear attraction part of our Hamiltonian. HK do
NOT tell us what the form of this universal functional, only of
its existence
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
42
Proof of the Hohenberg-Kohn theorem
Mel Levy provided a particularly simple proof of Hohenberg-Kohn
theorem {M. Levy, Proc. Nat. Acad. Sci. 76, 6062 (1979)}.
Define the functional O as O[(r)] = min <Ψ|O|Ψ>
|Ψ>(r)
where we consider all wavefunctions Ψ that lead to the same
density, (r), and select the one leading to the lowest expectation
value for <Ψ|O|Ψ>.
F[] is defined as F[(r)] = min <Ψ|F|Ψ>
|Ψ>(r)
where F = Si [- ½ i2] + ½ Si≠k [1/rik].
Thus the usual Hamiltonian is H = F + Vext
Now consider a trial function Ψapp that leads to the density (r)
and which minimizes <Ψ|F|Ψ>
Then E[] = F[] + ʃ (r) Vext(r) d3r = <Ψ|F +Vext|Ψ> = <Ψ|H|Ψ>
Thus E[] ≥ E0 the exact ground state energy.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
43
The Kohn-Sham equations
Walter Kohn and Lou J. Sham. Phys. Rev. 140, A1133 (1965).
Provided a practical methodology to calculate DFT wavefunctions
They partitioned the functional E[] into parts
E[] = KE0 + ½ ʃʃd3r1 d3r2 [(1) (2)/r12 + ʃd3r (r) Vext() + Exc[(r)]
Where
KE0 = Si <φi| [- ½ i2 | φi> is the KE of a non-interacting electron
gas having density (r). This is NOT the KE of the real system.
The 2nd term is the total electrostatic energy for the density (r).
Note that this includes the self interaction of an electron with itself.
The 3rd term is the total electron-nuclear attraction term
The 4th term contains all the unknown aspects of the Density
Functional
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
44
Solving the Kohn-Sham equations
Requiring that ʃ d3r (r) = N the total number of electrons and
applying the variational principle leads to
[d/d(r)] [E[] – m ʃ d3r (r) ] = 0
where the Lagrange multiplier m = dE[]/d = the chemical
potential
Here the notation [d/d(r)] means a functional derivative inside
the integral.
To calculate the ground state wavefunction we solve
HKS φi = [- ½ i2 + Veff(r)] φi = ei φi
self consistently with (r) = S i=1,N <φi|φi>
where Veff (r) = Vext (r) + J(r) + Vxc(r) and Vxc(r) = dEXC[]/d
KS
Lecture 1Ch121a-Goddard-L01
Thus H
© copyright 2012 William
A. Goddard III, all rights reserved\
looks quite analogous
to HHF
45
The Local Density Approximation (LDA)
We approximate Exc[(r)] as
ExcLDA[(r)] = ʃ d3r eXC((r)) (r)
where eXC((r)) is derived from Quantum Monte Carlo
calculations for the uniform electron gas {DM Ceperley and BJ
Alder, Phys.Rev.Lett. 45, 566 (1980)}
It is argued that LDA is accurate for simple metals and simple
semiconductors, where it generally gives good lattice
parameters
It is clearly very poor for molecular complexes (dominated by
London attraction), and hydrogen bonding
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
46
Generalized gradient approximations
The errors in LDA derive from the assumption that the density
varies very slowly with distance.
This is clearly very bad near the nuclei and the error will depend
on the interatomic distances
As the basis of improving over LDA a powerful approach has been
to consider the scaled Hamiltonian
E xc = E x E c
Lecture 1Ch121a-Goddard-L01
E x = ε x ρ(r), ρ(r) ,...]ρ(r) dr
© copyright 2012 William A. Goddard III, all rights reserved\
47
LDA exchange
Here we say that in LDA each electron interacts with all N
electrons but should be N-1. The exchange term cancels this
extra term. If density is uniform then error is proportional to 1/N.
since electron density is = N/V
ε
LDA
x
(ρ) = A xρ(r )
Lecture 1Ch121a-Goddard-L01
1
3
1
3
3 3
A x = - .
4π
© copyright 2012 William A. Goddard III, all rights reserved\
48
Generalized gradient approximations
E xc = E x E c
3.5
3.0
ε
(ρ, ρ ) = ε
LDA
x
F(s)
Becke 88
2.5
F(S)
E x = ε x ρ(r), ρ(r) ,...]ρ(r) dr
GGA
x
F(s) GGA functionals
X3LYP
2.0
1.5
PBE
PW91
1.0
0.5
s=
F
B88
0.0
ρ
0.0
(24 π ) ρ
1
2 3
5.0
s
10.0
S
4
3
9
1 sa 1sinh 1 (sa 2 ) a 3s 2
Becke
b = 0.0042 a4 and a5 zero
(s ) =
1
1
1 sa 1sinh (sa 2 )
2
1 sa 1sinh (sa 2 ) (a 3 a 4 e 100s
PW91
(s ) =
F
1
da
(
)
1
sa
sinh
sa
a
s
a
=
6
βa
Here a = (48 π ) ,1
,2 a = 5
1
2
1
2 3
2
2
a
10
2
3 , a = 6 βa , a =
(
)
a
=
a
=
48
π
Here
β
,
1
2
4
2
3
1/ 3
2
81
2 Ax
s
)
a 42 10 6
10
β , a 4 = a 3 , a 5 = 1/ 3
, and d = 4.
2
3
81 Goddard III, all
21 / 32012
A x William A.
2 rights
A x reserved\
© copyright
2
2
1
Lecture 1Ch121a-Goddard-L01
49
adiabatic connection formalism
The adiabatic connection formalism provides a rigorous way to define Exc.
It assumes an adiabatic path between the fictitious non-interacting KS system (λ =
0) and the physical system (λ = 1) while holding the electron density r fixed at its
physical λ = 1 value for all λ of a family of partially interacting N-electron systems:
Exc ] = U xc,l ]d l
1
0
is the exchange-correlation energy at intermediate coupling strength λ.
The only problem is that the exact integrand is unknown.
Becke, A.D. J. Chem. Phys. (1993), 98, 5648-5652.
Langreth, D.C. and Perdew, J. P. Phys. Rev. (1977), B 15, 2884-2902.
Gunnarsson, O. and Lundqvist, B. Phys. Rev. (1976), B 13, 4274-4298.
Kurth, S. and Perdew, J. P. Phys. Rev. (1999), B 59, 10461-10468.
Becke, A.D. J. Chem. Phys. (1993), 98, 1372-1377.
Perdew, J.P. Ernzerhof, M. and Burke, K. J. Chem. Phys. (1996), 105, 99829985.
Mori-Sanchez, P., Cohen, A.J. and Yang, W.T. J. Chem. Phys. (2006), 124,
091102-1-4.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
50
Becke half and half functional
assume a linear model
take
U xc,l =0 = Exexact
partition
the exact exchange of the KS orbitals
U xc,l =1 U
approximate
set
U xc ,l = a bl
LDA
xc ,l =1
ExcLDA = ExLDA EcLDA
Exexact
=xcLDA
ExcLDA
xexact
Exexact
a =a E=xexact
; b ;=b E
E
Get half-and-half functional
1 exact
1 LDA
LDA
Exc ] = ( Ex Ex ) Ec
2
2
Becke, A.D. J. Chem. Phys.
(1993), 98, 1372-1377
© copyright 2012 William A. Goddard III, all rights reserved\
Lecture 1Ch121a-Goddard-L01
51
Becke 3 parameter functional
Empirically modify half-and-half
ExcB3 ] = ExcLDA c1 ( Exexact ExLDA ) c2 ExGGA c3EcGGA
where E x
GGA
E
GGA
is
c
is the gradient-containing correction terms to the LDA exchange
the gradient-containing correction to the LDA correlation,
c1 , c2 , c3 are constants fitted against selected experimental thermochemical data.
The success of B3LYP in achieving high accuracy demonstrates that errors of E xcDFT for
covalent bonding arise principally from the λ 0 or exchange limit, making it important
to introduce some portion of exact exchange
Becke, A.D. J. Chem. Phys. (1993), 98, 5648-5652.
Becke, A.D. J. Chem. Phys. (1993), 98, 1372-1377.
Perdew, J.P. Ernzerhof, M. and Burke, K. J. Chem. Phys. (1996), 105, 99829985.
Mori-Sanchez, P., Cohen, A.J. and Yang, W.T. J. Chem. Phys. (2006), 124,
091102-1-4.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
52
Some popular DFT functionals
LDA: Slater exchange
Vosko-Wilk-Nusair correlation, etc
GGA: Exchange: B88, PW91, PBE, OPTX, HCTH, etc
Correlations: LYP, P86, PW91, PBE, HCTH, etc
Hybrid GGA: B3LYP, B3PW91, B3P86, PBE0,
B97-1, B97-2, B98, O3LYP, etc
Meta-GGA: VSXC, PKZB, TPSS, etc
Hybrid meta-GGA: tHCTHh, TPSSh, BMK, etc
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
53
Truhlar’s DFT functionals
MPW3LYP, X1B95, MPW1B95, PW6B95,
TPSS1KCIS, PBE1KCIS, MPW1KCIS,
BB1K, MPW1K, XB1K, MPWB1K, PWB6K,
MPWKCIS1K
MPWLYP1w,PBE1w,PBELYP1w, TPSSLYP1w
G96HLYP, MPWLYP1M , MOHLYP
M05, M05-2x
M06, M06-2x, M06-l, M06-HF
M06 = HF tPBE + VSXC
Lecture 1Ch121a-Goddard-L01
Hybrid meta-GGA
© copyright 2012 William A. Goddard III, all rights reserved\
54
Mu-Jeng Cheng
Reductive Elimination Thermochemistry
H/D exchange was measured from 153-173K by Girolami (J . Am. Chem. Soc.,
Vol. 120, 1998 6605) by NMR to have a barrier of G‡ = 8.1 kcal/mol.
G(173K)
B3LYP
0.0
M06
0.0
8.7
9.5
(reductive
elimination)
4.6
5.3
(s-bound
complex)
6.4
5.2
(site-exchange)
M06 and B3LYP functionals both consistent with experimental barrier site exchange.
These calculations use extended basis sets and PBF solvation
QM allows first principles predictions on new ligands, oxidation states, and
solvents. But there are error bars in the QM having to do with details of the
caculations (flavor of DFT, basis set). We use the best available methods and
compare to any available experimental data on known systems to assess the
accuracy for new systems. Some examples here and on the next slides
Typical validation: Metal-oxo Oxidations
Phosphine oxidation by (Tp)Re(O)Cl2 and (Tpm)Re(O)Cl2+ was observed from 1550˚C in 1,2-dichlorobenzene by Seymore and Brown (Inorg. Chem., Vol. 39,
2000, 325):
H‡(25C)
Experiment:
17.1 kcal/mol
M06: 16.6
B3LYP: 24.1
H‡(25C)
Experiment: 13.4 kcal/mol
M06: 11.8
B3LYP: 17.1
• M06 performs well
• B3LYP overestimates bimolecular barriers involving bulky or
polarizable species
Fundamental problem in standard DFT methods
bad description London Dispersion (vdW attraction)
Use QM calculations on small systems ~100 atoms get accurate energies,
geometries, stiffness
Fit QM to force field to describe big systems (104 -107 atoms)
Fit to obtain parameters for continuum systems
macroscopic properties based on first principles (QM)
Can predict novel materials where no empirical data available.
General Problem with DFT: bad
description of vdw attraction (London
dispersion)
Invalidates multiscale
paradigm
57
Solution: XYGJ-OS method
include excitations to virtual orbitals in order to describe
London Dispersion in DFT
this goes beyond using just density (occupied orbtials)
E
XYGJ-l OS
xc
] = e E
x
HF
x
(
(1 ex ) E eVWN E
S
x
VWN
c
eLYP E
LYP
c
)e
PT 2
E
PT 2
c , os
include only opposite spin and only local contributions
Get {ex, eVWN, eLYP, ePT2} ={0.7731,0.2309, 0.2754, 0.4364}.
A fast doubly hybrid density functional method close
to chemical accuracy: XYGJ-OS
Igor Ying Zhang, Xin Xu, Yousung Jung,
William A. Goddard III
PNAS 108 : 19896 (2011)
58
Density Functional Theory errors kcal/mol)
atomize
barrier
Popular with physicists
LDA
130.88
15.2
Include density gradient (GGA)
BLYP
10.16
7.9
PW91
22.04
9.3
Popular with physicists
PBE
20.71
9.1
Hybrid: include HF exchange
Popular with chemists
B3LYP
6.08
4.5
PBE0
5.64
3.9
Include KE functional fit to barriers and complexes
M06-L
5.20
4.1
M06
3.37
2.2
M06-2X
2.26
1.3
Include excitations to virtuals
XYGJ-OS
1.81
1.0
The level needed for
G3 (cc)
1.06
0.9
reliable predictions
59
Problem cannot do XYGJ-OS for crystals
Strategy: use XYGJ-OS to get accurate London
Dispersion on small cluster use to obtain
parameter for doing crystals (PBE-ulg)
Sublimation energy (kcal/mol/molecule) PBE-lg 3 to 5% too high
(zero point energy)
Molecules
PBE
PBE-ℓg
Exp.
Benzene
1.051
12.808
11.295
Naphthalene
2.723
20.755
20.095
Anthracene
4.308
28.356
27.042
Cell volume (angstrom3/cell)
PBE-lg 0 to 2% too small,
thermal expansion
Molecules
PBE
PBE-ℓg
Exp.
Benzene
511.81
452.09
461.11
Naphthalene
380.23
344.41
338.79
Anthracene
515.49
451.55
451.59
60
Problem cannot do XYGJ-OS for crystals
Strategy: use XYGJ-OS to get accurate London
Dispersion on small cluster use to obtain
parameter for doing crystals (PBE-ulg)
Sublimation energy (kcal/mol/molecule) PBE-lg 3 to 5% too high
(zero point energy)
Molecules
PBE
PBE-ℓg
Exp.
Benzene
1.051
12.808
11.295
Naphthalene
2.723
20.755
20.095
Anthracene
4.308
28.356
27.042
Cell volume (angstrom3/cell)
PBE-lg 0 to 2% too small,
thermal expansion
Molecules
PBE
PBE-ℓg
Exp.
Benzene
511.81
452.09
461.11
Naphthalene
380.23
344.41
338.79
Anthracene
515.49
451.55
451.59
61
Equation of States of Benzene Crystal
PBE-ulg predicts the
correct coldcompression curve.
Fundamental problem in standard DFT methods
Use QM calculations on small systems ~100 atoms get
accurate energies, geometries, stiffness
Fit QM to force field to describe big systems (104 -107 atoms)
Fit to obtain parameters for continuum systems
macroscopic properties based on first principles (QM)
Can predict novel materials where no empirical data available.
General Problem with DFT: bad
Invalidates multiscale
description of vdw attraction
paradigm
(London dispersion)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
63
XYG3 approach to include London Dispersion in DFT
Görling-Levy coupling-constant perturbation expansion
Exc ] = U xc,l ]d l
1
Take initial slope as the
2nd
0
order correlation energy:
Sum over virtual orbtials
i j ˆee a b
U xc ,l =0 =
2
U xc ,l
l
= 2 EcGL 2
l =0
i ˆx fˆ a
2
1
E =
4 ij ab e i e j e a e b
e i ea
i
a
where ˆee is the electron-electron repulsion operator, ˆx is the local exchange operator,
and fˆ is the Fock-like, non-local exchange operator.
LDA
exact
GL 2 exact
Substitute into
with
or
a
=
E
;
b
=
E
E
b
=
2
E
U xc ,l = a bl
x
xc
x
c
where
GL 2
c
Combine both approaches (2 choices for b)
b = b1 EcGL 2 b2 ( ExcDFT Exexact )
ExcR5 ] = ExcLDA c1 ( Exexact ExLDA ) c2 ExGGA c3 ( EcPT 2 EcLDA ) c4 EcGGA
DFT
a double hybrid DFT that mixes some exact exchange into E x
while also introducing a 2
DFT
PT 2
certain portion of Ec
into Ec
GL 2
PT 2 contains the double-excitation parts of E
E
c
c
i j ˆee a b
1
=
4 ij ab e i e j e a e b
This is a fifth-rung functional (R5) using information from both occupied and virtual KS
Lecture 1Ch121a-Goddard-L01
© copyrightdispersion
2012 William A. Goddard III, all rights reserved\
orbitals.
In principle can now describe
64
Final form of XYG3 DFT
ExcR5 ] = ExcLDA c1 ( Exexact ExLDA ) c2 ExGGA c3 ( EcPT 2 EcLDA ) c4 EcGGA
we adopt the LYP correlation functional but constrain c4 = (1 – c3) to exclude
compensation from the LDA correlation term.
This constraint is not necessary, but it eliminates one fitting parameter.
Determine the final three parameters {c1, c2, c3} empirically by fitting only to the
thermochemical experimental data in the G3/99 set of 223 molecules:
Get {c1 = 0.8033, c2 = 0.2107, c3 = 0.3211} and c4 = (1 – c3) = 0.6789
Use 6-311+G(3df,2p) basis set
XYG3 leads to mean absolute deviation (MAD) =1.81 kcal/mol,
B3LYP: MAD = 4.74 kcal/mol.
M06: MAD = 4.17 kcal/mol
M06-2x: MAD = 2.93 kcal/mol
M06-L: MAD = 5.82 kcal/mol .
G3 ab initio (with one empirical parameter): MAD = 1.05
G2 ab initio (with one empirical parameter): MAD = 1.88 kcal/mol
but G2 and G3 involve far higher computational cost.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
65
Thermochemical accuracy with size
G3/99 set has 223 molecules:
G2-1: 56 molecules having up to 3 heavy atoms,
G2-2: 92 additional molecules up to 6 heavy atoms
G3-3: 75 additional molecules up to 10 heavy atoms.
B3LYP: MAD = 2.12 kcal/mol (G2-1), 3.69 (G2-2), and 8.97 (G3-3) leads to
errors that increase dramatically with size
B2PLYP MAD = 1.85 kcal/mol (G2-1), 3.70 (G2-2) and 7.83 (G3-3) does not
improve over B3LYP
M06-L
MAD = 3.76 kcal/mol (G2-1), 5.71 (G2-2) and 7.50 (G3-3).
M06-2x MAD = 1.89 kcal/mol (G2-1), 3.22 (G2-2), and 3.36 (G3-3).
XYG3, MAD = 1.52 kcal/mol (G2-1), 1.79 (G2-2), and 2.06 (G3-3), leading to
the best description for larger molecules.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
66
Accuracy (kcal/mol) of various QM methods for
predicting standard enthalpies of formation
Functional
MAD
Max(+)
Max(-)
XYG3 a
1.81
16.67 (SF6)
-6.28 (BCl3)
M06-2x a
2.93
20.77 (O3)
-17.39 (P4)
M06 a
4.17
11.25 (O3)
-25.89 (C2F6)
B2PLYP a
4.63
20.37(n-octane)
-8.01(C2F4)
B3LYP a
4.74
19.22 (SF6)
-8.03 (BeH)
M06-L a
5.82
14.75 (PF5)
-27.13 (C2Cl4)
BLYP b
9.49
41.0 (C8H18)
-28.1 (NO2)
PBE b
22.22
10.8 (Si2H6)
-79.7 (azulene)
LDA b
121.85
0.4 (Li2)
-347.5 (azulene)
HFa
211.48
582.72(n-octane)
-0.46 (BeH)
MP2a
10.93
29.21(Si(CH3)4)
-48.34 (C2F6)
QCISD(T) c
15.22
42.78(n-octane)
-1.44 (Na2)
7.2 (SiF4)
-9.4 (C2F6)
DFT
Ab initio
G2(1 empirical parm) 1.88
Lecture
1Ch121a-Goddard-L01
G3(1
empirical
parm) 1.05
© copyright
2012) William A. Goddard III, all-4.9
rights
7.1 (PF
(Creserved\
F)
67
Comparison of QM methods for reaction surface of
H + CH4 H2 + CH3
30.00
H + CH4 H2 + CH3
HF
25.00
Energy (kcal/mol)
(kcal/mol)
Energy
HF
HF_PT2
XYG3
20.00
CCSD(T)
B3LYP
15.00
CCSD(T)
XYG3
BLYP
SVWN
HF_PT2 SVWN
B3LYP
10.00
BLYP
5.00
SVWN
0.00
-2.00
-1.50
-5.00
Lecture 1Ch121a-Goddard-L01
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
2.50
ReactionR(CH)-R(HH)
coordinate (in Å)
Reaction Coordinate:
© copyright 2012 William A. Goddard III, all rights reserved\
68
All (76)
HT38
HAT12
NS16
UM10
XYG3
1.02
0.75
1.38
1.42
0.98
M06-2x a
1.20
1.13
1.61
1.22
0.92
B2PLYP
1.94
1.81
3.06
2.16
0.73
M06 a
2.13
2.00
3.38
1.78
1.69
M06-La
3.88
4.16
5.93
3.58
1.86
B3LYP
4.28
4.23
8.49
3.25
2.02
BLYP a
8.23
7.52
14.66
8.40
3.51
PBEa
8.71
9.32
14.93
6.97
3.35
LDAb
14.88
17.72
23.38
8.50
Zhao and Truhlar
compiled benchmarks
Ab initio
of accurate barrier
HFb
11.28
13.66
16.87
6.67
heights in 2004
MP2 b
4.57
4.14
11.76
0.74
includes forward and
reverse barrier heights
QCISD(T) b
1.10
1.24
1.21
1.08
for
19 hydrogen transfer (HT) reactions,
6 heavy-atom transfer (HAT) reactions,
8 nucleophilic substitution (NS) reactions and
5Lecture
unimolecular
and association© (UM)
reactions.
1Ch121a-Goddard-L01
copyright
2012 William A. Goddard III, all rights reserved\
5.90
Reaction
barrier
heights
Note: no reaction
barrier heights used
in fitting the 3
parameters in
XYG3)
Functional
DFT
3.82
5.44
0.53
69
A. Total Energy (kcal/mol)
30.00
Test for
London
Dispersion
B. Exchange Energy (kcal/mol)
25.00
B
20.00
15.00
30.00
Ex_B
Ex_B3LYP
Ex_XYG3
Ex_HF
Ex_S
10.00 B3LYP
25.00
5.00
BLYP
Energy (kcal/mol)
20.00
15.00
10.00
5.00
B3LYP
0.00
BLYP
B3LYP
XYG3
CCSD(T)
SVWN
HF_PT2
-5.00 3.0
0.00
-5.00
-10.00
-15.00
4.0
HF_PT2
5.0
S 4.0
HF XYG3
5.0
6.0
Distance (A)
VWN
-3.00 B3LYP
0.00
3.0
(B)
6.0
LDA
(SVWN)
Intermolecular distance
CCSD(T)
(A)
XYG3
Distance (A)
Ec_VWN
Ec_B3LYP
LYP
CCSD(T)
Ec_LYP
-6.00
XYG3
Ec_XYG3
Ec_CCSD(T)
-9.00
PT2
Ec_PT2
(C)
C.
Correlation
Energy
(kcal/mol)
-12.00
Conclusion: XYG3 provides excellent accuracy for London dispersion, as good as
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
CCSD(T)
70
Accuracy of QM methods for noncovalent interactions.
Functional
Note: no
noncovalent
complexes used
in fitting the 3
parameters in
XYG3)
Total
HB6/04
CT7/04
DI6/04
WI7/05
PPS5/05
M06-2x b
0.30
0.45
0.36
0.25
0.17
0.26
XYG3 a
0.32
0.38
0.64
0.19
0.12
0.25
M06 b
0.43
0.26
1.11
0.26
0.20
0.21
M06-L b
0.58
0.21
1.80
0.32
0.19
0.17
B2PLYP
0.75
0.35
0.75
0.30
0.12
2.68
B3LYP
0.97
0.60
0.71
0.78
0.31
2.95
1.17
0.45
2.95
0.46
0.13
1.86
1.48
1.18
1.67
1.00
0.45
3.58
3.12
4.64
6.78
2.93
0.30
0.35
2.08
2.25
3.61
2.17
0.29
2.11
0.64
0.99
0.47
0.29
0.08
1.69
0.57
0.90
0.62
0.47
0.07
0.95
DFT
HB: 6 hydrogen bond PBE c
complexes,
BLYP c
CT 7 charge-transfer
LDA c
complexes
Ab initio
DI: 6 dipole
interaction complexes, HF
WI:7 weak interaction MP2c
complexes,
QCISD(T) c
PPS: 5 pp stacking
complexes.
Lecture 1Ch121a-Goddard-L01
WI and PPS dominated
by London
© copyright 2012 William A. Goddard III, all rights
reserved\dispersion. 71
Problem
XYG3 approach to include London Dispersion in DFT
Görling-Levy coupling-constant perturbation expansion
Exc ] = U xc,l ]d l
1
Take initial slope as the
2nd
0
order correlation energy:
Sum over virtual orbtials
i j ˆee a b
U xc ,l =0 =
2
U xc ,l
l
= 2 EcGL 2
l =0
i ˆx fˆ a
2
1
E =
4 ij ab e i e j e a e b
e i ea
i
a
where ˆee is the electron-electron repulsion operator, ˆx is the local exchange operator,
and fˆ is the Fock-like, non-local exchange operator.
where
GL 2
c
EGL2 involves double excitations to virtuals, scales as N5 with size
MP2 has same critical step
Yousung Jung (KAIST) has figured out how to get linear scaling for MP2
Lecture
1Ch121a-Goddard-L01
XYGJ-OS
and XYGJ-OS
© copyright 2012 William A. Goddard III, all rights reserved\
72
A solution: XYGJ-OS: include excitations to virtual orbitals in order
to describe London Dispersion in DFT
Goes beyond using just density (occupied orbitals)
Scales as (size)**3 just as B3LYP (CCSD scales as (size)**7
E
XYGJ-l OS
xc
] = e E
x
HF
x
(
(1 ex ) E eVWN E
S
x
VWN
c
eLYP E
LYP
c
)e
PT 2
E
PT 2
c , os
include only opposite spin & only local contributions N**3 scaling
Get {ex, eVWN, eLYP, ePT2} ={0.7731,0.2309, 0.2754, 0.4364}.
A fast doubly hybrid density functional method close
to chemical accuracy: XYGJ-OS
Igor Ying Zhang, Xin Xu,
Yousung Jung, WAG
PNAS (2011) in press
Lecture 1Ch121a-Goddard-L01
Xin Xu
© copyright 2012 William A. Goddard III, all rights reserved\
73
XYG4-OS and XYG4-LOS timings
CPU (hours)
Timings XYGJ-OS and XYGJ-LOS for long XYG4-LOS
alkanes
200.0
160.0
80.0
200.0
40.0
XYGJ-OS
XYG4-OS
120.0
B3LYP
200.0
B3LYP
160.0
200.0
0.0
0
160.0
XYG3
80.0
U (hours)
CPU (hours)
CPU (hours)
CPU (hours)
XYG4-LOS
XYGJ-LOS
120.0
160.0
40.0
120.0
80.0
120.0
0.0
0
20
Lecture 1Ch121a-Goddard-L01
XYG4-OS
120.0
XYG3
XYG4-OS and XYG4-LO
XYG4-OS and XYG4-
XYG4-OS and XYG4-LO
20
40
60
XYG4-LOS
alkane chain l
XYGJ-OS
XYG4-OS
XYG4-LOS
B3LYP
XYG4-OS
XYG4-LOS
XYGJ-LOS
XYG3
B3LYP
XYG4-OS
80
100
120
XYG3
80.0
B3LYP
© alkane
copyright 2012
William
A. Goddard III, all rights reserved\
chain
length
40
60
74
Accuracy of Methods (Mean absolute deviations MAD, in eV)
Methods
HOF
(223)
IP
(38)
DFT methods
SPL (LDA)
5.484 0.255
BLYP
0.412 0.200
PBE
0.987 0.161
TPSS
0.276 0.173
B3LYP
0.206 0.162
PBE0
0.300 0.165
M06-2X
0.127 0.130
XYG3
0.078 0.057
0.072 0.055
XYGJ-lOS
MC3BB
0.165 0.120
B2PLYP
0.201 0.109
Wavefunction based methods
HF
9.171 1.005
MP2
0.474 0.163
G2
0.082 0.042
G3
0.046 0.055
EA
(25)
PA
(8)
BDE
(92)
NHTBH
(38)
HTBH
(38)
NCIE
(31)
All
(493)
0.311
0.105
0.102
0.104
0.106
0.128
0.103
0.080
0.084
0.175
0.090
0.276
0.080
0.072
0.071
0.061
0.057
0.092
0.070
0.067
0.046
0.067
0.754
0.292
0.177
0.245
0.226
0.155
0.069
0.068
0.033
0.111
0.124
0.542
0.376
0.371
0.391
0.202
0.154
0.056
0.056
0.049
0.062
0.090
0.775
0.337
0.413
0.344
0.192
0.193
0.055
0.033
0.038
0.036
0.078
0.140
0.063
0.052
0.049
0.041
0.031
0.013
0.014
0.015
0.023
0.023
2.771
0.322
0.562
0.250
0.187
0.213
0.096
0.065
0.056
0.123
0.143
1.148
0.166
0.057
0.049
0.133
0.084
0.058
0.046
0.104
0.363
0.078
0.047
0.397
0.249
0.042
0.042
0.582
0.166
0.054
0.054
0.098
0.028
0.025
0.025
4.387
0.338
0.068
0.046
HOF = heat of formation; IP = ionization potential,
EA = electron affinity, PA = proton affinity,
BDE = bond dissociation energy,
NHTBH, HTBH = barrier heights for reactions,
Lecture =
1Ch121a-Goddard-L01
© copyright
2012 William A. Goddard III, all rights reserved\
NCIE
the binding in molecular
clusters
75
Comparison of speeds
HOF
IP TimeEA
P
HTBH
NCIE
All
Methods
(223) C100
(38)
(38)
(31)
(493)
H202 (25)
C100H100 (8
DFT methods
SPL (LDA)
5.484 0.255
0.311
0.2
0.542
0.775
0.140
2.771
0.376
0.337
BLYP 0.063 0.322
0.412 0.200 0.105 0.0
0.371
0.413
0.052
0.562
PBE
0.987 0.161 0.102 0.0
0.391
0.344
TPSS 0.049 0.250
0.276 0.173 0.104 0.0
0.202
0.192
2.8
12.3 0.0
B3LYP 0.041 0.187
0.206 0.162
0.106
0.154
0.193
PBE0 0.031 0.213
0.300 0.165 0.128 0.0
0.056
0.055
M06-2X0.013 0.096
0.127 0.130 0.103 0.0
0.056
0.033
81.4 0.0
XYG3 0.014 0.065
0.078 200.0
0.057 0.080
7.8
46.4 0.0
0.049
0.038
0.015
0.056
0.072 0.055
0.084
XYGJ-lOS
0.062
0.036
MC3BB 0.023 0.123
0.165 0.120 0.175 0.0
0.090
0.078
0.143
76
Lecture 1Ch121a-Goddard-L01
copyright 2012 William
A. Goddard III,
all rights reserved\
B2PLYP©0.023
0.201
0.109
0.090 0.0
NHTBH
(38)
Density Functional Theory errors kcal/mol)
atomize
barrier
Popular with physicists
LDA
130.88
15.2
Include density gradient (GGA)
BLYP
10.16
7.9
PW91
22.04
9.3
Popular with physicists
PBE
20.71
9.1
Hybrid: include HF exchange
Popular with chemists
B3LYP
6.08
4.5
More rigorous foundation
PBE0
5.64
3.9
Include KE functional fit to barriers and complexes
M06-L
5.20
4.1
No exact exchange, fast
wag uses for catalysis
M06
3.37
2.2
M06-2X
2.26
1.3
Does well not well founded
Include excitations to virtuals
Accuracy needed for predictions
XYGJ-OS
1.81
1.0
G3 CC (4 semiempirical parameters)
77
G3
(cc)
1.06
0.92012 William
cannot
be used
potential
Lecture
1Ch121a-Goddard-L01
© copyright
A. Goddard
III, allfor
rights
reserved\curves
Lecture 1Ch121a-Goddard-L01
G3
G2
XYG4-OS
XYG3
B2PLYP
M06-L
M06-2x
G2-1 small molecules
G2-2
G3-3 Large molecules
M06
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
B3LYP
MAD (kcal/mol)
Heats of formation (kcal/mol)
© copyright 2012 William A. Goddard III, all rights reserved\
78
Truhlar NHTBH38/04 set and HTBH38/04 set
HAT12
NS16
UM10
HT38
20.0
15.0
10.0
Reaction barrier heights (kcal/mol)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
XYG4-OS
XYG3
QCISD(T)
MP2
HF
LDA
PBE
0.0
BLYP
5.0
B3LYP
MAD (kcal/mol)
25.0
79
Truhlar NCIE31/05 set
Nonbonded interaction (kcal/mol)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
XYG4-OS
XYG3
QCISD(T)
MP2
HF
LDA
PBE
HB6
CT7
DI6
WI7
PPS5
BLYP
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
B3LYP
MAD (kcal/mol)
8.0
80
Comparison of QM methods for reaction surface of
H + CH4 H2 + CH3
30.00
H + CH4 H2 + CH3
HF
25.00
Energy (kcal/mol)
(kcal/mol)
Energy
HF
HF_PT2
XYG3
20.00
CCSD(T)
B3LYP
15.00
CCSD(T)
XYG3
BLYP
SVWN
HF_PT2 SVWN
B3LYP
10.00
BLYP
5.00
SVWN
0.00
-2.00
-1.50
-5.00
Lecture 1Ch121a-Goddard-L01
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
2.50
ReactionR(CH)-R(HH)
coordinate (in Å)
Reaction Coordinate:
© copyright 2012 William A. Goddard III, all rights reserved\
81
DFT-ℓg for accurate Dispersive Interactions for Full
Periodic Table
Hyungjun Kim, Jeong-Mo Choi, William A. Goddard, III
1Materials and Process Simulation Center, Caltech
2Center for Materials Simulations and Design, KAIST
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
82
Current challenge in DFT calculation for energetic
materials
• Current implementations of DFT describe well strongly bound
geometries and energies, but fail to describe the long range van der
Waals (vdW) interactions.
• Get volumes ~ 10% too large
• XYGJ-lOS solves this problem but much slower than standard
methods
• DFT-low gradient (DFT-lg) model accurate description of the longrange1/R6 attraction of the London dispersion but at same cost as
standard DFT
EDFT D = EDFT Edisp
N
Elg = -
Clg,ij
6
6
r
dR
ij , i j ij
eij
C6 single parameter from QM-CC
d =1
R
Rei + Rek (UFF vdW
radii)
Lecture
© copyright
2012 William A. Goddard III, all rights reserved\
eik =1Ch121a-Goddard-L01
83
PBE-lg for benzene dimer
T-shaped
Sandwich
PBE-lg parameters
Elg = -
Clg-CC=586.8, Clg-HH=31.14, Clg-HH=8.691
Clg,ij
N
r dR
6
ij , i j ij
Parallel-displaced
6
eij
RC = 1.925 (UFF), RH = 1.44 (UFF)
First-Principles-Based Dispersion Augmented Density Functional Theory: From
Molecules to Crystals’ Yi Liu and wag; J. Phys. Chem. Lett., 2010, 1 (17), pp
84
2550–2555
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
DFT-lg description for benzene
PBE-lg predicted the EOS of benzene crystal (orthorhombic phase I) in good agreement with
corrected experimental EOS at 0 K (dashed line).
Pressure at zero K geometry: PBE: 1.43 Gpa; PBE-lg: 0.11 Gpa
Zero pressure volume change: PBE: 35.0%; PBE-lg: 2.8%
Heat
sublimation at 0 K: Exp:11.295
kcal/mol;
PBE: A.
0.913;
PBE-lg:
6.762reserved\
Lecture of
1Ch121a-Goddard-L01
© copyright
2012 William
Goddard
III, all rights
85
Binding energy (kcal/mol)
DFT-lg description for graphite
PBE
PBE-lg
Exper E
0.8, 1.0, 1.2
c lattice constant (A)
Exper c 6.556
graphite has AB stacking
(also show AA eclipsed graphite)
© copyright 2012 William A. Goddard III, all rights reserved\
Lecture 1Ch121a-Goddard-L01
86
Universal PBE-ℓg Method
UFF, a Full Periodic Table Force Field for Molecular Mechanics and Molecular
Dynamics Simulations; A. K. Rappé, C. J. Casewit, K. S. Colwell, W. A. Goddard
III, and W. M. Skiff; J. Am. Chem. Soc. 114, 10024 (1992)
Derived C6/R6 parameters from scaled atomic polarizabilities for Z=1-103 (HLr) and derived Dvdw from combining atomic IP and C6
Universal PBE-lg: use same Re, C6, and De as UFF, add a single new
parameter slg
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
87
blg Parameter Modifies Short-range Interactions
12-6 LJ potential (UFF parameter)
blg =1.0
lg potential
lg potential
blg =0.7
When blg =0.6966,
ELJ(r=1.1R0) = Elg(r=1.1R0)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
88
Problem cannot yet do XYGJ-OS for crystals
Solution: use XYGJ-OS or CCSD to get accurate London Dispersion on
small vdW clusters.
Use to modify PBE for doing crystals by adding low gradient correction
(PBE-lg) (also B3LYP-lg) for accurate description of the long-range
1/R6 attraction of the London dispersion
EDFT D = EDFT Edisp
Reik = Rei + Rek (UFF vdW radii)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Universal low gradient (ulg) method for DFT-ulg
Problem with DFT-lg: need a C6 parameter for every pair of atoms.
Can get from XYGJ-OS or CCSD calculation on small <100 atom
complexes, but for atoms up to Lr (Z=103) would need 5356 parameters,
far too tedious
Universal force field (UFF): Rappé, Goddard JACS 114, 10024 (1992)
Generic approach to force fields for whole periodic table (to Z=103 Lr)
For each atom: 6 rule based parameters 618 to describe all molecules
for all atoms up to Z=103
UFF has two vdw parameters: D0 and R0 per atom based on
• atomic polarizability from HF QM
• ionization potential from experiment
• atom size from experiment
ulg strategy: base C6 term in DFT-ulg on the C6 from UFF
wag962. Universal Correction of Density Functional Theory to Include London Dispersion
(up1Ch121a-Goddard-L01
to Lr, Element 103); HJ Kim,
JM Choi,
J.A.Phys.
Chem.
Lett.
2012,
3, 360−363
Lecture
© copyright
2012 wag;
William
Goddard
III, all
rights
reserved\
Universal low gradient (ulg) method
ulg method: use van der Waal’s parameters from Universal Force-Field
DFT-ulg
UFF vdw terms (up to Lr, Z=103)
1. Match long R
2. Match at mid-range regime (r = 1.1R0):
3. Then introduce a single general scaling parameter for
whole periodic table (slg),
With 1 parameter, DFT-ulg
defined for Z=1 to 103
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Determine the single parameter in DFT-ulg from
Benzene dimer interactions
J-M
Choi,
HJ Kim, WAG
Lecture
1Ch121a-Goddard-L01
DFT-ulg fit a single
parameter slg to benzene
dimer CCSD(T)
Get Slg = 0.7012
© copyright 2012 William A. Goddard III, all rights reserved\
Validation: C6 parameters for lg fit to PBE for benzene dimer
does excellent job on crystals
Sublimation energy (kcal/mol/molecule) PBE-lg 3 to 5% too high
(zero point energy)
Molecules
PBE
PBE-ℓg
Exp.
Benzene
1.051
12.808
11.295
Naphthalene
2.723
20.755
20.095
Anthracene
4.308
28.356
27.042
Cell volume (angstrom3/cell)
PBE-lg 0 to 2% too small,
thermal expansion
Molecules
PBE
PBE-ℓg
Exp.
Benzene
511.81
452.09
461.11
Naphthalene
380.23
344.41
338.79
Anthracene
515.49
451.55
451.59
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
93
Crystals: Polyaromatic Hydrocarbons
V0 (Å3)
Benzene crystal:
PBE
PBE-ulg
PBE-Grimme
Exp.
Lecture 1Ch121a-Goddard-L01
511.8
452.1
420.3
461.8
Sublimation Compres.
E (kcal/mol) B0 (GPa)
1.05
1.3
12.81
8.8
13.33
10
11.3
~8
Volume
Naphthalene
Anthracene
Phenantracene
PBE
380.2
515.5
524.5
PBE-ulg
344.4
451.6
461.7
Exp.
342.3
455.2
459.5
Heat Vapor.
Naphthalene
Anthracene
Phenantracene
PBE
0.89
1.75
1.52
PBE-ulg
18.93
25.80
24.39
Exp.
18.4-23.5
24.6-30.0
23.6-26.5
© copyright 2012 William A. Goddard III, all rights reserved\
Equation of States of Benzene Crystal
PBE-ulg predicts the
correct coldcompression curve.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Hobza S22 database
Twenty-two prototypical small molecular complexes for non-covalent
interactions in biological molecules (h-bonded, dispersion dominated,
and mixed)
7 hydrogen bonded Mean average error (MAE)
PBE-ulg:
0.53 kcal/mol
PBE-Grimme: 1.01 kcal/mol
vdw-DF:
0.59 kcal/mol (lundqvist, PRL 2004)
8 dispersion dominated MAE
PBE-ulg:
1.26 kcal/mol
PBE-Grimme: 0.58 kcal/mol
vdw-DF:
1.86 kcal/mol
Overall:
Mean average error (MAE)
PBE-ulg: 0.70 kcal/mol
PBE-ulg: 0.22 kcal/mol
PBE-Grimme: 0.65 kcal/mol
PBE-Grimme: 0.38 kcal/mol
vdw-DF: 1.20 kcal/mol
vdw-DF: 1.06 kcal/mol© copyright 2012 WilliamXYGJ-OS:
kcal/mol
Lecture 1Ch121a-Goddard-L01
A. Goddard III, all0.46
rights reserved\
Big Challenge for DFT
Proper description of spin states
Organometallic reaction barriers depend strongly on spin
Antiferromagnets
Cuprate superconductors
Ground states of Mn, Fe, Co, Ni metals
Current optimization of DFT methods focus mainly on 1st
and 2nd row compounds (H-Ar) but applications involve
transition metals, lanthanides, actinides where local d and f
orbitals can lead to magnetically complex systems
Example of the challenge:
Group 10:
s2d8 (3F) vs. s1d9 (3D) vs. s0d10 (1S)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
97
Ground state configurations for group 10
Ni
Exper GVB-CI HF
Pd
Exper GVB-CI HF
Pt
Exper GVB-CI HF
wag206-Theoretical Studies of Oxidative Addition and Reductive Elimination. II.
Reductive Coupling of H-H, H-C, and C-C Bonds from Pd and Pt Complexes
98
J.
J. Low
and W. A. Goddard
III; Organometallics
5, 609
(1986)
Lecture
1Ch121a-Goddard-L01
© copyright
2012 William A. Goddard
III, all
rights reserved\
Ab initio methods Ni atom (all electron)
method
s1d9 (3D) s0d10 (1s)
exper
-0.69
39.43
HF(wag 1986)
15.30
114.80
1.72
55.17
HF(G3 basis, Yu 2012)
Ni atom
Basis set issues
HF (numerical nonrelativistic)
29.29
126.14 Cowan-Griffin
HF (numerical relativistic)
37.59
139.29 Cowan-Griffin
GVB-CI (wag 1986)
-14.20
MP2(G3 basis, Yu 2012)
-30.92
-44.60 using s2d8 state
not conv
47.07 using s2d8 state
CCSD(G3 basis, Yu 2012)
26.20
Basis set issues:
G3 basis: contraction of 3s and 3p core functions (overlaps 3d)
Reference state issues for MP2 and CCSD: used s2d8
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
99
Compare DFT methods Ni atom (all electron)
Highlight
method s1d9 < 3 kcal/mol
exper
s0d10 <10 kcal/mol
HF(G3 basis, Yu 2012)
s1d9 (3D)
s0d10 (1s)
-0.69
39.43
1.72
55.17
GVB-CI (wag 1986)
-14.20
26.20
PBE(G3 basis, Yu 2012)
-12.30
29.37
PBE0(G3 basis, Yu 2012)
-9.18
85.06
B3LYP(G3 basis, Yu 2012)
-9.11
22.65
M06-L(G3 basis, Yu 2012)
36.92
-51.31
M06(G3 basis, Yu 2012)
-10.33
19.43
M06-HF(G3 basis, Yu 2012)
-14.01
49.11
M06-2X(G3 basis, Yu 2012)
XYGJ-OS (G3 basis, Yu 2012)
-3.59
0.03
48.36
-2.12
Need
to use multiple spin
states
DFT
optimization
Lecture 1Ch121a-Goddard-L01
© copyright
2012 in
William
A. Goddard
III, all rights reserved\
100
Pt atom using LANL Core Effective Potential
method
s0d10 (1s) s2d8 (3F)
Pt atom
exper
11.07
14.76
HF(wag 1986)
31.40
8.40
HF(Yu 2012)
25.41
5.95
HF (numerical nonrelativistic) -32.52
75.64
Cowan-Griffin
HF (numerical relativistic)
20.75
9.22
Cowan-Griffin
GVB-CI (wag 1986)
12.20
14.20
PBE(Yu 2012)
14.39
-0.05
PBE0(Yu 2012)
15.03
9.08 Highlight
B3LYP(Yu 2012)
14.67
6.84 s2d8 < 2 kcal/mol
M06-L(Yu 2012)
14.01
0.86 s0d10 <2 kcal/mol
M06(Yu 2012)
0.40
19.17
M06-HF(Yu 2012)
24.95
21.07
M06-2X(Yu 2012)
11.72
15.24
XYGJ-OS
na
na
101
Ground
state s1d9 (3D)
YuA. basis:
Lecture
1Ch121a-Goddard-L01
© copyright 2012 William
Goddard III,LACV3P**++f,
all rights reserved\
method
s1d9 (3D) s2d8 (3F)
Pd atom
exper
21.91
77.94
HF(wag 1986)
-12.70
41.80
HF(Yu 2012)
1.72
55.17
HF (numerical nonrelativistic) -17.29
86.71
Cowan-Griffin
HF (numerical relativistic)
2.30
50.50
Cowan-Griffin
GVB-CI (wag 1986)
19.60
82.20
Pd
atom
MP2
13.41
5.43
CCSD
16.83
0.92 using LANL
PBE( Yu 2012)
9.43
67.17
Core
PBE0(Yu 2012)
19.99
88.40
Effective
B3LYP(Yu 2012)
19.98
84.79
Potential
M06-L(Yu 2012)
30.66
97.18
M06( Yu 2012)
38.63
114.45 Highlight
M06-HF(Yu 2012)
10.09
89.50 s1d9 < 3 kcal/mol
M06-2X(Yu 2012)
26.07
97.61 s2d8 <5 kcal/mol
XYGJ-OS
na
na
102
Lecture
1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Yu
basis:LACVP
in Qchem
States for Ni d8 atom (real orbitals)
Hole type
exp
0
z2, x2-y2 σδ
0
xy, z2
σδ
HF
PBE PBE0 B3LYP M06-L M06 M06-HF M06-2X
0
0
0
0
0
0
0
0
0
11.3
11.6
11.6
11.6
11.6
0.08
10.2
9.39
9.39
10.4
10.4
0.06
10.4
9.79
9.79
10.5
10.5
0.05
10.25
9.8
9.8
10.42
10.42
-0.42
16.22
13.17
13.17
16.38
16.38
-0.27
10.4
7.5
7.5
10.5
10.5
-0.53
-7.93
0.52
0.52
-7.7
-7.7
-0.83
4.13
6.05
6.05
4.29
4.29
xz, z2
σπ 27.01 34.3 26.6 28.2 28.33 35.85 18.1
10.28
19.9
yz, z2
σπ 27.01 34.3 26.6 28.2 28.33 35.85 18.1
10.28
19.9
δδ
14.84
26.92
xz, yz
ππ 9.004
xz, x2-y2 πδ 9.004
yz, x2-y2 πδ 9.004
xz, xy
πδ 9.004
yz, xy
πδ 9.004
xy, x2-y2
36.02 45.4 36.7 38.3 38.44 48.36 23.8
ok exc exc exc
ok
Bottom line: for transition metal systems, current levels of DFT
based
on foundation of©sand:
must address in next generation DFT
103
Lecture 1Ch121a-Goddard-L01
copyright 2012 William A. Goddard III, all rights reserved\
Parameter Optimization
Implemented in VASP 5.2.11
0.7012
0.6966
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
104
Method:
·
Semi-Empirical, used for very big systems, or for rough approximations of
geometry (extended Huckel theory, CNDO/INDO, AM1, MNDO)
·
HF (Hartree Fock). Simplest Ab Initio method. Very cheap, fairly inaccurate
·
MP2 (Moeller-Plasset 2). Advanced version of HF. Usually not as cheap or as
accurate as B3LYP, but can function as a complement.
·
CASSCF (Complete Active Space, Self Consisting Field). Advanced version
of HF, incorporating excited states. Mainly used for jobs where photochemistry is
important. Medium cost, Medium Accuracy. Quite complicated to run…
·
QCISD (Quadratic Configuration Interaction Singles Doubles). Very
advanced version of HF. Very Expensive, Very accurate. Can only be used on
systems smaller than 10 heavy atoms.
·
CCSD (Coupled Cluster Singles Doubles). Very much like QCISD.
Density Functional Theory
LDA (local density approximation)
PW91, PBE
·
B3LYP (density functional theory). Cheap, Accurate.
Generally, B3LYP is the method of choice. If the system allows it, QCISD or CCSD
can be used. HF and/or MP2 can be used to verify the B3LYP results.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
105
Basis Set: What mathematical expressions are used to describe orbitals. In
general, the more advanced the mathematical expression, the more accurate
the wavefunction, but also more expensive calculation.
·
STO-3G - The ‘minimal basis set’. Not particularly accurate, but cheap and
robust.
·
3-21G - Smallest practical Basis Set.
·
6-31G - More advanced, i.e. more functions for both core and valence.
·
6-31G** - As above, but with ‘polarized functions’ added. Essentially
makes the orbitals look more like ‘real’ ones. This is the standard basis set
used, as it gives fairly good results with low cost.
·
6-31++G - As above, but with ‘diffuse functions’ added. Makes the orbitals
stretch out in space. Important to add if there is hydrogen bonding, pi-pi
interactions, anions etc present.
·
6-311++G** - As above, with even more functions added on… The more
stuff, the more accurate… But also more expensive. Seldom used, as the
increase in accuracy usually is very small, while the cost increases drastically.
·
Frozen Core: Basis sets used for higher row elements, where all the core
electrons are treated as one big frozen chunk. Only the valence electrons are
treated explicitly
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
106
• Software packages
–
–
–
–
–
–
–
Jaguar
GAMESS
TurboMol
Gaussian
Spartan/Titan
HyperChem
ADF
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
107
Running an actual calculation
– Determine the starting geometry of the
molecule you wish to study
– Determine what you’d like to find out
– Determine what methods are suitable and/or
affordable for the above calculation
– Prepare input file
– Run job
– Evaluate result
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
108
Example: Good ol’ water
Starting geometry: water is bent, (~104º), a normal
O-H bond is ~0.96 Å. For illustration, however, we’ll
start with a pretty bad guess.
Simple Z-matrix:
O1
1.00 Å
H2 O1 1.00
H3 O1 1.00 H2 110.00
Lecture 1Ch121a-Goddard-L01
1.00 Å
110º
© copyright 2012 William A. Goddard III, all rights reserved\
109
What do we wish to find out?
How about the IR spectra?
What is a suitable method for this calculation?
Well, any, really, since it is so small. But 99% of the
time the answer to this question is “B3LYP/631G**” – a variant of density functional theory that
is the main workhorse of applied quantum
chemistry, with a standard basis set. Let’s go with
that.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
110
Actual jaguar input:
&gen
igeopt=1
ifreq=1
dftname=b3lyp
basis=6-31g**
&
&zmat
O1
H2 O1 0.95
H3 O1 0.95 H2 120.00
&
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
111
Running time!
Jaguar calculates the wave function for the
atomic coordinates we provided
From the wave function it determines the
energy and the forces on the current geometry
Based on this, it determines in what direction it
should move the atoms to reach a better
geometry, i.e. a geometry with a lower energy
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
112
Forces
Our horrible guess
1.00 Å
1.00 Å
110º
Target geometry
0.96 Å
0.96 Å
104º
Think elastic springs:
The bonds are too long,
so there will be a force
towards shorter bonds
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
113
Optimization –
minimization of the
forces. When all forces
are zero the energy will
not change and we
have the resting
geometry
Lecture 1Ch121a-Goddard-L01
O1
H2 O1 0.9500000000
H3 O1 0.9500000000 H2 120.0000000000
SCF energy: -76.41367730925
-O1
H2 O1 0.9566666804
H3 O1 0.9566666820 H2 106.8986301461
SCF energy: -76.41937497895
-O1
H2 O1 0.9653619358
H3 O1 0.9653619375 H2 103.0739287925
SCF energy: -76.41969584939
-O1
H2 O1 0.9653155294
H3 O1 0.9653155310 H2 103.6688074046
SCF energy: -76.41970381840
--
© copyright 2012 William A. Goddard III, all rights reserved\
114
Accuracy
Computer accuracy
0.9653155294 Å
“actual” accuracy
0.96 Å
0.96 Å
0.9653155294 Å
103.6688074046º
103.7º
Accuracy is a relative concept
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
115
frequencies
1666.01 3801.19 3912.97
No negative frequencies!
(Compare IR spectra for gas-phase water)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
116
Zero Point Energies
Vibrational levels
Zero Point Energy (ZPE)
“zero” level
Optimized energy is at the zero level, but in reality the molecule has a higher
energy due to populated vibrational levels.
At 0 K, all molecules populate the lowest vibrational level, and so the
difference between the “zero” level and the first vibrational level is the Zero
Point Energy (ZPE)
From our calculation:
The zero point energy (ZPE):
Lecture 1Ch121a-Goddard-L01
13.410 kcal/mol
© copyright 2012 William A. Goddard III, all rights reserved\
117
Thermodynamic data at higher temperatures
T = 298.15 K
trans.
rot.
vib.
elec.
total
U
--------0.889
0.889
0.002
0.000
1.779
Cv
--------2.981
2.981
0.041
0.000
6.003
S
--------34.609
10.503
0.006
0.000
45.117
H
--------1.481
0.889
0.002
0.000
2.371
G
---------8.837
-2.243
0.000
0.000
-11.080
Most thermodynamic data can be computed with very good
accuracy in the gas phase. Temperature dependant
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
118
Transition states
Transition State (TS)
Stationary points:
points on the surface
where the derivative
of the energy = 0
Line represents the
reacting coordinate, in this
case the forming C-Cl and
breaking C-Br bonds
Product
Reactant
Reaction coordinate
CH3Br +
Lecture 1Ch121a-Goddard-L01
Cl-
TS
CH3Cl + Br-
© copyright 2012 William A. Goddard III, all rights reserved\
119
Transition state =
stationary point where all forces
except one is at a minimum.
Not a hill, but a
mountain pass
The exception is at its maximum
Reaction coordinate
CH3Br +
Lecture 1Ch121a-Goddard-L01
Cl-
TS
CH3Cl + Br-
© copyright 2012 William A. Goddard III, all rights reserved\
120
Derivative of the energy = 0
TS
Second derivative:
For a minimum > 0
For a maximum < 0
So a TS should have a
negative second derivative of
the energy
Second derivative of the
energy = force
Lecture 1Ch121a-Goddard-L01
Reactant
Product
© copyright 2012 William A. Goddard III, all rights reserved\
121
A transition state should have one
negative (imaginary) frequency!!!
(and ONLY one)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
122
Inflection points
Optimizing transition states:
Simultaneously optimize all
modes (forces) towards their
minimum, except the reacting
mode
But for the computer to know
which mode is the reacting
mode, you must have one
imaginary frequency in your
starting point
TS
Reactant
Product
Region with
imaginary frequency
Must start with a good guess!!!
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
123
Example:
CH3Br + Cl-
CH3Cl + Br-
What do we know about this reaction? It’s an SN2
reaction, so the Cl- must come in from the backside of
the CH3Br. The C-Cl forms at the same time as the CBr forms. The transition state should be five
coordinate
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
124
H
H
Br
C
Cl
2.2
2.0
H
Initial guess: C-Cl = 2.0 Å, C-Br = 2.2 Å
Single point frequency on the above geometry:
frequencies
frequencies
98.64 99.58 109.11 310.66 1339.10 1348.64
1349.46 1428.45 1428.73 2838.52 3017.70 3017.93
No negative frequencies! Bad initial guess
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
125
Refinement :
Initial guess most likely wrong because of erronous CBr and C-Cl bond lengths
Let the computer optimize the five-coordinate structure
Frozen optimizations:
Just like a normal optimization, but with one or more
geometry parameters frozen
In this case, we optimize the structure with all the H-CCl angles frozen at 90º
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
126
Result:
Cl
Br
2.32
2.62
C-Cl and C-Br bonds quite a bit longer in the new structure
Frequency calculation:
frequencies
frequencies
-286.26 168.54 173.32 173.43 874.16 874.76
976.23 1413.99 1414.65 3220.91 3420.84 3421.80
One negative frequency! Good initial guess
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
127
Time for the actual optimization:
Jaguar follows the negative frequency towards the maximum
Geometry optimization 1: SCF Energy = -513.35042353681
Geometry optimization 2: SCF Energy = -513.34995058422
Geometry optimization 3: SCF Energy = -513.35001640704
Geometry optimization 4: SCF Energy = -513.34970196448
Geometry optimization 5: SCF Energy = -513.34968682825
Geometry optimization 6: SCF Energy = -513.34968118535
Final energy higher than starting energy (although only 0.5 kcal/mol)
Frequency calculation
frequencies
-268.67 162.64 174.22 174.31 848.15 848.24
frequencies
960.97 1415.75 1415.96 3220.77 3420.80 3421.15
One negative frequency! We found a true transition state
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
128
Cl
Br
2.46
Final geometry:
2.51
C-Cl = 2.46 Å
C-Br = 2.51 Å
Cl-C-H = 88.7º
Br-C-H = 91.3º
Structure not quite symmetric, the
hydrogens are bending a little bit away
from the Br.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
129
Solvation calculations
Explicit solvents:
Calculations where solvent molecules
are added as part of the calculation
Implicit solvents:
Calculations where solvation effects
are added as electrostatic interactions
between the molecule and a virtual
continuum of “solvent”.
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
130
Reaction energetics and barrier heights
Collect the absolute energies of the reactants, products and transition states
CH3Br +
Cl-
CH3Cl + Br-
TS
-53.078938 + -460.248741
-513.349681
-500.108371 + -13.237607
Sum each term
CH3Br +
Cl-
-513.327679
TS
CH3Cl + Br-
-513.349681
-513.345978
Define reactants as “0”, and deduct the reactant energy from all terms
CH3Br +
Cl-
0
TS
-.022002
CH3Cl + Br-.018299
Convert to kcal/mol (1 hartree = 627.51 kcal/mol)
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
131
Reaction energetics and barrier heights
Convert to kcal/mol (1 hartree = 627.51 kcal/mol)
Cl-
CH3Br +
0
TS
-13.8
CH3Cl + Br-11.5
But this doesn’t make sense
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
132
Reaction energetics and barrier heights
Cl-
CH3Br +
0
TS
-13.8
CH3Cl + Br-11.5
Solvation not included!
Include solvation corrections!
Cl-
CH3Br +
0
Lecture 1Ch121a-Goddard-L01
TS
9.2
CH3Cl + Br-6.4
© copyright 2012 William A. Goddard III, all rights reserved\
133
Quantum Mechanics – First postulate
The essential element of QM is that all properties that can
be known about the system are contained in the
wavefunction, Φ(x,y,z,t) (for one electron), where the
probability of finding the electron at position x,y,z at time t
is given by
P(x,y,z,t) = | Φ(x,y,z,t) |2 = Φ(x,y,z,t)* Φ(x,y,z,t)
Note that ∫Φ(x,y,z,t)* Φ(x,y,z,t) dxdydz = 1
since the total probability of finding the electron
somewhere is 1.
I write this as < Φ|Φ>=1, where it is understood that the
integral is over whatever the spatial coordinates of Φ are
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
134
Quantum Mechanics – Second postulate
In QM the total energy can be written as
EQM = KEQM + PEQM
where for a system with a classical potential energy function,
V(x,y,z,t)
PEQM=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz ≡ < Φ| V|Φ>
Just like Classical mechanics except that V is weighted by P=|Φ|2
For the H atom
_
2/r) |Φ> = -e2/ R
PEQM=<
Φ|
(-e
_
where R is the average value of 1/r
KEQM = (Ћ2/2me) <(Φ·Φ>
where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz
In QM the KE is proportional to the average square of the gradient
or slope of the wavefunction
Thus
KE wants smooth© wavefunctions,
no wiggles
Lecture 1Ch121a-Goddard-L01
copyright 2012 William A. Goddard III, all rights reserved\
135
Summary 2nd Postulate QM
EQM = KEQM + PEQM
where for a system with a potential energy function, V(x,y,z,t)
PEQM= < Φ| V|Φ>=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz
Just like Classical mechanics except weight V with P=|Φ|2
KEQM = (Ћ2/2me) <(Φ·Φ>
where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz
We have assumed a normalized wavefunction, <Φ|Φ> = 1
The stability of the H atom was explained by this KE (proportional
to the average square of the gradient of the wavefunction).
We will use the preference of KE for smooth wavefunctions to
explain the bonding in H2+ and H2.
However to actually solve for the wavefunctions requires the
Schrodinger Eqn., which we derive next.
Lecture 1Ch121a-Goddard-L01
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136
3rd Postulate of QM, the variational principle
The ground state wavefunction is the system, Φ, has the lowest
possible energy out of all possible wavefunctions.
Consider that Φex is the exact wavefunction with energy
Eex = <Φ’|Ĥ|Φ’>/<Φ’|Φ’> and that
Φap = Φex + dΦ is some other approximate wavefunction.
Then Eap = <Φap|Ĥ|Φap>/<Φap|Φap> ≥ Eex
This means that for sufficiently small
dΦ, dE = 0 for all possible changes,
dΦ
We write dE/dΦ = 0 for all dΦ
E
Eex
Eap
This is called the variational principle.
For the ground state, d2E/dΦ ≥ 0 for all
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
possible
changes
137
Side comment: the next 4 slides Derive Schrödinger equation
from variational principle. You are not responsible for this
Write the energy of any approximate wavefunction, Φap, as
Eap = <Φap|Ĥ|Φap>/<Φap|Φap>
Ignoring terms 2nd order in dΦap, we obtain
<Φap|Ĥ|Φap> = Eex + <dΦ|Ĥ|Φex> + <Φex|Ĥ|dΦ>
= Eex + 2 Re[<dΦ|Ĥ|Φex>]
<Φap|Φap> = 1 + <dΦ|Φex> + <Φex|dΦ>
= 1 + 2 Re[<dΦ|Φex>]
where Re means the real part. To 1st order:
(a + db)/(1+dd) = [a /(1+dd)] + db = a+ db –a dd = a+ db –a dd
Thus Eap - Eex = 2 Re[<dΦ|Ĥ|Φex>]} - Eex{2 Re[<dΦ|Φex>]}
Eap - Eex = 2 Re[<dΦ|Ĥ-Eex|Φex>]} = 0 for all possible dΦ
But
∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦ [(Ĥ-Eex)Φex] = 0
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
138
Extra: Derivation of Schrodinger Equation
Assume
EQM = {(Ћ2/2me)<(dΦ/dx)| (dΦ/dx)> + <Φ|V|Φ>}/<Φ|Φ>
Variational principle says that ground state Φ0 leads to the lowest
possible E, E0
Then starting with this optimum Φ0 , and making any change, dΦ
will increase E.
The first order change in E is
dE = (Ћ2/2me)<(d dΦ/dx)| (dΦ/dx)> + < dΦ| V|Φ> + CC
Integrate by parts
dE = -(Ћ2/2me)<(dΦ| (d2Φ/dx2)> + < dΦ| V|Φ> + CC
Lecture 1Ch121a-Goddard-L01
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139
Extra: Derivation of Schrodinger Equation
But even though <Φ0|Φ0> = 1, changing Φ0 by dΦ, might change
the normalization.
Thus we get an additional term
E+dE = E0/{<Φ0|Φ0> + <dΦ|Φ0> + CC} = E0 – E0{<dΦ|Φ0> + CC}
Thus
dE ={-(Ћ2/2me)<(dΦ| (d2Φ0/dx2)>+< dΦ|V|Φ0> -E0<dΦ|V|Φ0>} + CC
At a minimum the energy must increase for both +dΦ and –dΦ,
hence dE=0 = <(dΦ| {-(Ћ2/2me)(d2/dx2)+V -E0}|Φ>} + CC
Must get dE=0 for all possible dΦ, hence the coefficient of dΦ,
must be zero. Get
(H - E0)Φ=0 where H= {-(Ћ2/2me)(d2/dx2)+V} or HΦ= E0Φ
Lecture 1Ch121a-Goddard-L01
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140
Extra Summary deriviation of Schrödinger
Equation
^ | Φ> + < Φ| V|Φ> = <Φ| Ĥ | Φ>
EQM = <Φ| KE
^ + V and KE
^ = - (Ћ2/2m )2
where the Hamiltonian is Ĥ ≡ KE
e
And we assume a normalized wavefunction, <Φ|Φ> = 1
V(x,y,z,t) is the (classical) potential energy for the system
Consider arbitrary Φap = Φex + dΦ and require that
dE= Eap – Eex = 0
Get <dΦ|Ĥ-Eex|Φex>] = ∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦ
This [(Ĥ-Eex)Φex] = 0 or the Schrödinger equation
Ĥ Φex = EexΦex
The exact ground state wavefunction is a solution of this equation
Lecture 1Ch121a-Goddard-L01
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141
4th postulate of QM - Excited states
The Schrödinger equation Ĥ Φk = EkΦk
Has an infinite number of solutions or eigenstates (German
for characteristic states), each corresponding to a possible
exact wavefunction for an excited state
For example H atom: 1s, 2s, 2px, 3dxy etc
Also the g and u states of H2+ and H2.
These states are orthogonal: <Φj|Φk> = djk= 1 if j=k
= 0 if j≠k
Note < Φj| Ĥ|Φk> = Ek < Φj|Φk> = Ek djk
We will denote the ground state as k=0
The set of all eigenstates of Ĥ is complete, thus any arbitrary
function Ө can be expanded as
Ө = Sk Ck Φk where <Φj| Ө>=Cj or Ө = Sk Φk <Φk| Ө>
Lecture 1Ch121a-Goddard-L01
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142
Phase factor
Consider the exact eigenstate of a system
HΦ = EΦ
and multiply the Schrödinger equation by some CONSTANT
phase factor (independent of position and time)
exp(ia) = eia
eia HΦ = H (eia Φ) = E (eia Φ)
Thus Φ and (eia Φ) lead to identical properties and we
consider them to describe exactly the same state.
wavefunctions differing only by a constant phase factor
describe the same state
Lecture 1Ch121a-Goddard-L01
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143
Extra: Configuration interaction
Consider a set of N-electron wavefunctions:
{i; i=1,2, ..M}
where < i|j> = dij {=1 if i=j and 0 if i ≠j)
Write approx = S (i=1 to M) Ci i
Then E = < approx|H|approx>/< approx|approx>
E= < Si Ci i |H| Sk Ck k >/ < Si Ci i | Si Ck k >
How choose optimum Ci?
Require dE=0 for all dCi get
Sk <i |H| Ck k > - Ei< i | Ck k > = 0 ,which we
write as ΣikHikCki = ΣikSikCkiEi
where Hjk = <j|H|k > and Sjk = < j|k >
Which we write as HCi = SCiEi in matrix notation
C
a column vector
for the
ith eigenstate
Lecture
© copyright
2012 William
A. Goddard III, all rights reserved\
i is 1Ch121a-Goddard-L01
144
Extra: Configuration interaction upper bound
theorem
Consider the M solutions of the CI equations
HCi = SCiEi ordered as i=1 lowest to i=M highest
Then the exact ground state energy of the system
Satisfies Eexact ≤ E1
Also the exact first excited state of the system satisfies
E1st excited ≤ E2
etc
This is called the Hylleraas-Unheim-McDonald Theorem
Lecture 1Ch121a-Goddard-L01
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145
Electron spin, 5th postulate QM
Consider application of a magnetic field
Our Hamiltonian has no terms dependent on the magnetic field.
Hence no effect.
But experimentally there is a huge effect. Namely
The ground state of H atom splits into two states
b
B=0
Increasing B
a
This leads to the 5th postulate of QM
In addition to the 3 spatial coordinates x,y,z each electron has
internal or spin coordinates that lead to a magnetic dipole aligned
either with the external magnetic field or opposite.
We label these as a for spin up and b for spin down. Thus the
ground states of H atom are φ(xyz)a(spin) and φ(xyz)b(spin)
Lecture 1Ch121a-Goddard-L01
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146
Permutational symmetry, summary
Our Hamiltonian for H2,
H(1,2) =h(1) + h(2) + 1/r12 + 1/R
Does not involve spin
This it is invariant under 3 kinds of permutations
Space only: 1 2
Spin only: s1 s2
Space and spin simultaneously: (1,s1) (2,s2)
Since doing any of these interchanges twice leads to the identity,
we know that
Ψ(2,1) = Ψ(1,2) symmetry for transposing spin and space coord
Φ(2,1) = Φ(1,2) symmetry for transposing space coord
Χ(2,1) = Χ(1,2) symmetry for transposing spin coord
Lecture 1Ch121a-Goddard-L01
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147
Permutational symmetries for H2 and He
H2
Have 4 degenerate g
ground states for H2
Have 4 degenerate u
excited states for H2
He
Have 4 degenerate
ground state for He
Lecture 1Ch121a-Goddard-L01
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148
Permutational symmetries for H2 and He
H2
He
Lecture 1Ch121a-Goddard-L01
the only states
observed are
those for
which the
wavefunction
changes sign
upon
transposing all
coordinates of
electron 1 and
2
Leads to the
6th postulate of
149
© copyright 2012 William A. Goddard III, all rights reserved\QM
The 6th postulate of QM: the Pauli Principle
For every eigenstate of an electronic system
H(1,2,…i…j…N)Ψ(1,2,…i…j…N) = EΨ(1,2,…i…j…N)
The electronic wavefunction Ψ(1,2,…i…j…N) changes
sign upon transposing the total (space and spin)
coordinates of any two electrons
Ψ(1,2,…j…i…N) = - Ψ(1,2,…i…j…N)
We can write this as
tij Ψ = - Ψ for all I and j
Lecture 1Ch121a-Goddard-L01
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150
Consider H atom
We will consider one electron, but a nucleus with charge Z
r
The Hamiltonian has the form
h = - (Ћ2/2me)2 – Ze2/r
In atomic units: Ћ=1, me=1, e=1
h = - ½ 2 – Z/r
+Ze
Thus we want to solve
hφk = ekφk for the ground and excited states k
φnlm = Rnl(r) Zlm(θ,φ)
where Rnl(r) depends only on r and
Zlm(θ,φ) depends only on θ and φ
Assume φ10 = exp(-zr) E = ½ z2 – Z z
dE/dz
= z – Z = 0 z =© copyright
Z
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151
The H atom ground state
the ground state of H atom is
φ1s = N0 exp(-Zr/a0) ~ exp(-Zr) where we will ignore normalization
1
x=0
Line plot along z, through the origin
Maximum amplitude at z = 0
Lecture 1Ch121a-Goddard-L01
Contour plot in the xz
plane, Maximum
amplitude at x,z = 0,0 152
© copyright 2012 William A. Goddard III, all rights reserved\
Atomic units
We will use atomic units for which me = 1, e = 1, Ћ = 1
For H atom the average size of the 1s orbital is
a0 = Ћ2/ mee2 = 0.529 A =0.053 nm = 1 bohr is the unit of length
For H atom the energy of the 1s orbital []ionization potential (IP) of
H atom is
e1s = - ½ me e4/ Ћ2 = - ½ h0 = -13.61 eV = -313.75 kcal/mol
In atomic units the unit of energy is me e4/ Ћ2 = h0 = 1, denoted as
the Hartree
Note h0 = e2/a0 = 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/mol
The kinetic energy of the 1s state is KE1s = ½ and
the potential energy is PE1s = -1 = 1/R, where R = 1 a0 is the
average radius
Lecture 1Ch121a-Goddard-L01
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153
The excited states of H atom - 1
The ground and excited states of a system can all be written as
hφk = ekφk, where <φk |φj> = dkj
Here dkj the Kronecker delta function is
1 when j=k, but it is
0 otherwise
We say that different excited states are orthogonal.
Lecture 1Ch121a-Goddard-L01
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154
Nodal theorem
The ground state has no nodes (never changes sign),
like the 1s state for H atom
Lecture 1Ch121a-Goddard-L01
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155
The excited states of H atom - 2
Use spherical polar coordinates, r, θ, φ
where z = rcosθ, x = rsinθcosφ, y = rsinθsinφ
2 = d2/dx2 + d2/dy2 + d2/dy2 transforms like
r2 = x2 + y2 + z2 so that it is independent of θ, φ
Thus h(r,θ,φ) = - ½ 2 – Z/r is independent of θ
and φ
Lecture 1Ch121a-Goddard-L01
z
θ
φ
y
x
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156
The excited states of H atom - 3
z
Use spherical polar coordinates, r, θ, φ
θ
where z=r cosθ, x=r sinθ cosφ, y=r sinθ sinφ
y
2 = d2/dx2 + d2/dy2 + d2/dy2 transforms like
x
r2 = x2 + y2 + z2 so that it is independent of θ, φ φ
Thus h(r,θ,φ) = - ½ 2 – Z/r is independent of θ
and φ
Consequently the eigenfunctions of h can be factored into Rnl(r)
depending only on r and Zlm(θ,φ) depending only on θ and φ
φnlm = Rnl(r) Zlm(θ,φ)
The reason for the numbers nlm will be apparent later
Lecture 1Ch121a-Goddard-L01
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157
Excited radial functions
Consider excited states with Znl = 1; these are ns states with l=0
The lowest is R10 = 1s = exp(-Zr), the ground state.
All other radial functions must be orthogonal to 1s, and hence
must have one or more radial nodes.
0 nodal
1 spherical
2 spherical
planes
nodal plane
nodal planes
Zr = 7.1
Zr = 2
Zr = 1.9
The cross section is plotted along the z axis, but it would look
exactly the same along any other axis. Here
R20 = 2s = [Zr/2 – 1] exp(-Zr/2)
2/27 – 2(Zr)/3 + 1] exp(-Zr/3)
R
=
3s
=
[2(Zr)
30
Lecture 1Ch121a-Goddard-L01
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158
Angularly excited states
Ground state 1s = φ100 = R10(r) Z00(θ,φ), where Z00 = 1 (constant)
Now consider excited states, φnlm = Rnl(r) Zlm(θ,φ), whose angular
functions, Zlm(θ,φ), are not constant, l ≠ 0.
Assume that the radial function is smooth, like R(r) = exp(-ar)
Then for the excited state to be orthogonal to the 1s state, we
must have
<Z00(θ,φ)|Zlm(θ,φ)> = 0
Thus Zlm must have nodal planes with equal positive and negative
regions.
The simplest cases are
rZ10 = z = r cosθ, which is zero in the xy plane
rZ11 = x = r sinθ cosφ, which is zero in the yz plane
rZ1,-1 = y = r sinθ sinφ, which is zero in the xz plane
These are denoted as angular momentum l=1 or p states
Lecture 1Ch121a-Goddard-L01
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159
The p excited angular states of H atom
φnlm = Rnl(r) Zlm(θ,φ)
z
Now lets consider excited angular functions, Zlm.
They must have nodal planes to be orthogonal to Z00 +
The simplest would be Z10=z = r cosθ, which is
zero in the xy plane.
Exactly equivalent are
Z11=x = rsinθcosφ which is zero in the yz plane,
and
Z1-1=y = rsinθsinφ, which is zero in the xz plane
Also it is clear that these 3 angular functions
with one angular nodal plane are orthogonal to
each other. Thus <Z10|Z11> = <pz|px>=0 since
the integrand has nodes in both the xy and xz
planes, leading to a zero integral
pz
Lecture 1Ch121a-Goddard-L01
pz
x
z
px
-
+
x
z
-
+
+
-
© copyright 2012 William A. Goddard III, all rights reserved\
pxpz
x160
More p functions?
So far we have the s angular function Z00 = 1 with no angular
nodal planes
And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each
with one angular nodal plane
Can we form any more angular functions with one nodal plane
orthogonal to all 4 of the above functions?
z px’
For example we might rotate px by an angle a
a +
about the y axis to form px’. However multiplying,
say by pz, leads to the integrand pzpx’ which
clearly does not integrate to zero
x
z pzpx’
a +
-
Thus there are exactly three pi functions, Z1m,
. with m=0,+1,-1, all of which have the same KE.
Since the p functions have nodes, they lead to a
higher KE than the s function (assuming no
x radial nodes)
161
+
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
More angular functions?
So far we have the s angular function Z00 = 1 with no angular
nodal planes
And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with
one angular nodal plane
Next in energy will be the d functions with two angular nodal
planes. We can easily construct three functions
dxy = xy =r2 (sinθ)2 cosφ sinφ
z
dyz = yz =r2 (sinθ)(cosθ) sinφ
dxz
+
2
dzx = zx =r (sinθ)(cosθ) cosφ
where dxz is shown here
+
-
x
Each of these is orthogonal to each other (and to the s and the
three p functions). <dxy|dyz> = ʃ (x z y2) = 0, <px|dxz> = ʃ (z x2) = 0,
Lecture 1Ch121a-Goddard-L01
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162
More d angular functions?
In addition we can construct three other functions with two
nodal planes
z
dz2-x2
2
2
2
2
2
2
dx2-y2 = x – y = r (sinθ) [(cosφ) – (sinφ) ]
+
dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2]
dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2]
x
+
where dz2-x2 is shown here,
Each of these three is orthogonal to the previous three d
functions (and to the s and the three p functions)
This leads to six functions with 2 nodal planes
Lecture 1Ch121a-Goddard-L01
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163
More d angular functions?
In addition we can construct three other functions with two
nodal planes
z
dz2-x2
2
2
2
2
2
2
dx2-y2 = x – y = r (sinθ) [(cosφ) – (sinφ) ]
+
dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2]
dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2]
x
+
where dz2-x2 is shown here,
Each of these three is orthogonal to the previous three d
functions (and to the s and the three p functions)
This leads to six functions with 2 nodal planes
However adding these 3 (x2 – y2) + (y2 – z2) + (z2 – x2) = 0
Which indicates that there are only two independent such
functions. We combine the 2nd two as
(z2 – x2) - (y2 – z2) = [2 z2 – x2 - y2 ] = [3 z2 – x2 - y2 –z2] =
2 – r2 ] which we denote as d
=
[3
z
z2 A. Goddard III, all rights reserved\
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William
164
Summarizing the d angular functions
z
dz2
+
Z20 = dz2 = [3 – ]
m=0, ds
Z21 = dzx = zx =r2 (sinθ)(cosθ) cosφ
m = 1, dp
Z2-1 = dyz = yz =r2 (sinθ)(cosθ) sinφ
Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]
Z22 = dxy = xy =r2 (sinθ)2 cosφ sinφ
We find it useful to keep track of how often the
wavefunction changes sign as the φ coordinate is
increased by 2p = 360º
When this number, m=0 we call it a s function
When m=1 we call it a p function
When m=2 we call it a d function
When m=3 we call it a f function
z2
r2
Lecture 1Ch121a-Goddard-L01
-
-
+
57º
x
m = 2, dd
© copyright 2012 William A. Goddard III, all rights reserved\
165
Summarizing the angular functions
So far we have
•one s angular function (no angular nodes) called ℓ=0
•three p angular functions (one angular node) called ℓ=1
•five d angular functions (two angular nodes) called ℓ=2
Continuing we can form
•seven f angular functions (three angular nodes) called ℓ=3
•nine g angular functions (four angular nodes) called ℓ=4
where ℓ is referred to as the angular momentum quantum number
And there are (2ℓ+1) m values for each ℓ
Lecture 1Ch121a-Goddard-L01
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166
real (Zlm) and complex (Ylm) ang. momentum fnctns
Here the bar over
m negative
Lecture 1Ch121a-Goddard-L01
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167
Size (a0)
name
total nodal planes
radial nodal planes
angular nodal planes
Combination of radial and angular nodal
planes
Combining radial and angular functions gives the
various excited states of the H atom. They are
named as shown where the n quantum number is
the total number of nodal planes plus 1
The nodal theorem does not specify how 2s and
1s 0 0 0 1.0
2p are related, but it turns out that the total
2s 1 1 0 4.0
energy depends only on n.
2p 1 0 1 4.0
3s 2 2 0 9.0 Enlm = - Z2/2n2
3p 2 1 1 9.0
The potential energy is given by
3d 2 0 2 9.0
4s 3 3 0 16.0 PE = - Z2/n2 = -Z/ Rˉ, where Rˉ=n2/Z
4p 3 2 1 16.0
Thus Enlm = - Z/2 Rˉ
4d 3 1 2 16.0
all you
need to remember
168
4f
3 1Ch121a-Goddard-L01
0 3 16.0 This ©iscopyright
Lecture
2012 William A. Goddard III, all rights reserved\
Sizes hydrogen orbitals
Rˉ =a0 n2/Z
Where a0 = bohr = 0.529A=0.053 nm = 52.9 pm
Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f
Angstrom (0.1nm) 0.53, 2.12,
H
H C
H--H
H
H
0.74
1.7
4.77,
8.48
H
H
H
H
H
H
4.8
Lecture 1Ch121a-Goddard-L01
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169
Hydrogen atom excited states
Energy
zero
-0.033 h0 = -0.9 eV
-0.056 h0 = -1.5 eV
-0.125 h0 = -3.4 eV
4s
4p
4d
3s
3p
3d
2s
2p
4f
Enlm = - Z/2 Rˉ = - Z2/2n2
-0.5 h0 = -13.6 eV
Lecture 1Ch121a-Goddard-L01
1s
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170
Plotting of orbitals:
line cross-section vs. contour
line plot
along z axis
Lecture 1Ch121a-Goddard-L01
contour plot
in yz plane
© copyright 2012 William A. Goddard III, all rights reserved\
171
Contour plots of 1s, 2s, 2p hydrogenic orbitals
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
172
Contour plots of 3s, 3p, 3d hydrogenic orbitals
Lecture 1Ch121a-Goddard-L01
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173
Contour plots of 4s, 4p, 4d hydrogenic orbtitals
Lecture 1Ch121a-Goddard-L01
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174
Contour plots of hydrogenic 4f orbitals
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
175
He+ atom
Next lets review the energy for He+.
Writing Φ1s = exp(-zr) we determine the optimum z for He+ as
follows
<1s|KE|1s> = + ½ z2 (goes as the square of 1/size)
<1s|PE|1s> = - Zz (linear in 1/size)
E(He+) = + ½ z2 - Zz
Applying the variational principle, the optimum z must satisfy
dE/dz = z - Z = 0 leading to z = Z,
KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.
writing PE=-Z/R0, we see that the average radius is R0=1/z = 1/2
So that the He+ orbital is ½ the size of the H orbital
Lecture 1Ch121a-Goddard-L01
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176
Estimate J1s,1s, the electron repulsion energy of 2
electrons in He+ 1s orbitals
e1
Now consider He atom: EHe = 2(½ z2) – 2Zz J1s,1s
How can we estimate J1s,1s
R0
e2
Assume that each electron moves on a sphere
With the average radius R0 = 1/z =1/2
And assume that e1 at along the z axis (θ=0)
Neglecting correlation in the electron motions, e2 will on the
average have θ=90º so that the average e1-e2 distance is
~sqrt(2)*R0
Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.707 z
Lecture 1Ch121a-Goddard-L01
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177
Estimate J1s,1s, the electron repulsion energy of 2
electrons in He+ 1s orbitals
e1
Now consider He atom: EHe = 2(½ z2) – 2Zz J1s,1s
How can we estimate J1s,1s
R0
e2
Assume that each electron moves on a sphere
With the average radius R0 = 1/z =1/2
And assume that e1 at along the z axis (θ=0)
Neglecting correlation in the electron motions, e2 will on the
average have θ=90º so that the average e1-e2 distance is
~sqrt(2)*R0
Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.707 z
A rigorous calculation gives
J1s,1s = (5/8) z = 0.625 z = (5/16) h0 = 8.5036 eV = 196.1 kcal/mol
Since e1s = -Z2/2 = -2 h0 = 54.43 eV = 1,254.8 kcal/mol the
178
electron
repulsion increases
(lessIII,attractive)
by 15.6%
Lecture 1Ch121a-Goddard-L01
© copyrightthe
2012energy
William A. Goddard
all rights reserved\
The optimum atomic orbital for He atom
He atom: EHe = 2(½ z2) – 2Zz (5/8)z
Applying the variational principle, the optimum z must satisfy
dE/dz = 0 leading to
2z - 2Z + (5/8) = 0
Thus z = (Z – 5/16) = 1.6875
KE = 2(½ z2) = z2
PE = - 2Zz (5/8)z = -2 z2
E= - z2 = -2.8477 h0
Lecture 1Ch121a-Goddard-L01
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179
The optimum atomic orbital for He atom
He atom: EHe = 2(½ z2) – 2Zz (5/8)z
Applying the variational principle, the optimum z must satisfy
dE/dz = 0 leading to
2z - 2Z + (5/8) = 0
Thus z = (Z – 5/16) = 1.6875
KE = 2(½ z2) = z2
PE = - 2Zz (5/8)z = -2 z2
E= - z2 = -2.8477 h0
Ignoring e-e interactions the energy would have been E = -4 h0
The exact energy is E = -2.9037 h0 (from memory, TA please
check).
Thus this simple approximation of assuming that each electron is
in a 1s orbital and optimizing the size accounts for 98.1% of the
exact result.
Lecture 1Ch121a-Goddard-L01
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180
Interpretation: The optimum atomic orbital for He atom
Assume He(1,2) = Φ1s(1)Φ1s(2) with Φ1s = exp(-zr)
We find that the optimum z = (Z – 5/16) = Zeff = 1.6875
With this value of z, the total energy is E= - z2 = -2.8477 h0
This wavefunction can be interpreted as two electrons moving
independently in the orbital Φ1s = exp(-zr) which has been
adjusted to account for the average shielding due to the other
electron in this orbital.
On the average this other electron is closer to the nucleus about
31.25% of the time so that the effective charge seen by each
electron is Zeff = 2 - 0.3125=1.6875
The total energy is just the sum of the individual energies,
E = -2 (Zeff2/2) = -2.8477 h0
Ionizing the 2nd electron, the 1st electron readjusts to z = Z = 2
with E(He+) = -Z2/2 = - 2 h0.
Thus the ionization potential (IP) is 0.8477 h0 = 23.1 eV (exact
value = 24.6 eV)
181
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Now lets add a 3rd electron to form Li
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)(Φ1sg)]
Problem: with either g = a or g = b, we get ΨLi(1,2,3) = 0
Since there are two electrons in the same spinorbital
This is an essential result of the Pauli principle
Thus the 3rd electron must go into an excited orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2sa)]
or
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
Lecture 1Ch121a-Goddard-L01
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182
First consider Li+
First consider Li+ with ΨLi(1,2) = A[(Φ1sa)(Φ1sb)]
Here Φ1s = exp(-zr) with z = Z-0.3125 = 2.69 and
E = -z2 = -7.2226 h0.
For Li2+ we get E =-Z2/2=-4.5 h0
Thus the IP of Li+ is IP = 2.7226 h0 = 74.1 eV
The size of the 1s orbital for Li+ is
R0 = 1/z = 0.372 a0 = 0.2A
Lecture 1Ch121a-Goddard-L01
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183
Consider adding the 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
Since the 2p orbital goes to zero at z=0, there is very
little shielding so that the 2p electron sees an effective
charge of
Zeff = 3 – 2 = 1, leading to
a size of R2p = n2/Zeff = 4 a0 = 2.12A
And an energy of e = -(Zeff)2/2n2 = -1/8 h0 = 3.40 eV
1s
0.2A
2p
2.12A
Lecture 1Ch121a-Goddard-L01
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184
Consider adding the 3rd electron to the 2s orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
The 2s orbital must be orthogonal to the 1s, which means that
it must have a spherical nodal surface below ~ 0.2A, the size
of the 1s orbital. Thus the 2s has a nonzero amplitude at z=0
so that it is not completely shielded by the 1s orbitals.
The result is Zeff2s = 3 – 1.72 = 1.28
This leads to a size of R2s = n2/Zeff = 3.1 a0 = 1.65A
And an energy of e = -(Zeff)2/2n2 = -0.205 h0 = 5.57 eV
1s
0.2A
2s
Lecture 1Ch121a-Goddard-L01
2.12A
R~0.2A
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185
Li atom excited states
Energy
MO picture
State picture
zero
1st
excited
state
-0.125 h0 = -3.4 eV
-0.205 h0 = -5.6 eV
2p
2s
E = 2.2 eV
17700 cm-1
564 nm
(1s)2(2p)
(1s)2(2s)
Ground
state
Exper
-2.723 h0 = -74.1 eV
671 nm
1s
186
E = 1.9 eV
Lecture 1Ch121a-Goddard-L01
© copyright 2012 William A. Goddard III, all rights reserved\
Aufbau principle for atoms
Energy
Kr, 36
Zn, 30
Ar, 18
6
2
2
4s
10
6
6
2s
He, 2
2
Lecture 1Ch121a-Goddard-L01
4p
14
4d
4f
3d
3p
3s
Ne, 10
2
10
1s
2p
Get generalized energy
spectrum for filling in the
electrons to explain the
periodic table.
Full shells at 2, 10, 18, 30,
36 electrons
187
© copyright 2012 William A. Goddard III, all rights reserved\
Kr, 36
Zn, 30
Ar, 18
Ne, 10
He, 2
Lecture 1Ch121a-Goddard-L01
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188
Many-electron configurations
General
aufbau
ordering
Lecture 1Ch121a-Goddard-L01
Particularly stable
© copyright 2012 William A. Goddard III, all rights reserved\
189
General trends along a row of the periodic table
As we fill a shell, say B(2s)2(2p)1 to Ne (2s)2(2p)6
we add one more proton to the nucleus and one more electron to
the valence shell
But the valence electrons only partially shield each other.
Thus Zeff increases, leading to a decrease in the radius ~ n2/Zeff
And an increase in the IP ~ (Zeff)2/2n2
Example Zeff2s=
1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne
Thus (2s Li)/(2s Ne) ~ 4.64/1.28 = 3.6
Lecture 1Ch121a-Goddard-L01
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190
General trends along a column of the periodic
table
As we go down a column
Li [He}(2s) to Na [Ne]3s to K [Ar]4s to Rb [Kr]5s to Cs[Xe]6s
We expect that the radius ~ n2/Zeff
And the IP ~ (Zeff)2/2n2
But the Zeff tends to increase, partially compensating for
the change in n so that the atomic sizes increase only
slowly as we go down the periodic table and
The IP decrease only slowly (in eV):
5.39 Li, 5.14 Na, 4.34 K, 4.18 Rb, 3.89 Cs
(13.6 H), 17.4 F, 13.0 Cl, 11.8 Br, 10.5 I, 9.5 At
24.5 He, 21.6 Ne, 15.8 Ar, 14.0 Kr,12.1 Xe, 10.7 Rn
Lecture 1Ch121a-Goddard-L01
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191
Plot of rφ(r) for
the outer s
valence orbital
Lecture 1Ch121a-Goddard-L01
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192
Plot of rφ(r) for
the outer s and
p valence
orbitals
Note for C row
2s and 2p have
similar size, but
for other rows
ns much
smaller than np
Lecture 1Ch121a-Goddard-L01
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193
Plot of rφ(r) for the
outer s and p valence
orbitals
Note for C row 2s
and 2p have similar
size, but for other
rows ns much
smaller than np
Lecture 1Ch121a-Goddard-L01
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194
Transition metals; consider [Ar] + 1 electron
[IP4s = (Zeff4s )2/2n2 = 4.34 eV Zeff4s = 2.26; 4s<4p<3d
K
IP4p = (Zeff4p )2/2n2 = 2.73 eV Zeff4p = 1.79;
IP3d = (Zeff3d )2/2n2 = 1.67 eV Zeff3d = 1.05;
IP4s = (Zeff4s )2/2n2 = 11.87 eV Zeff4s = 3.74; 4s<3d<4p
Ca+
IP3d = (Zeff3d )2/2n2 = 10.17 eV Zeff3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV Zeff4p = 3.20;
IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff3d = 4.05; 3d<4s<4p
Sc++
IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff4s = 5.04;
IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff4p = 4.47;
As the net charge increases the differential shielding for 4s vs 3d
is less important than the difference in n quantum number 3 vs 4
195
Lecture 1Ch121a-Goddard-L01
© copyright 2012
Thus
charged system prefers
3d William
vs 4sA. Goddard III, all rights reserved\
Transition metals; consider Sc0, Sc+, Sc2+
3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff3d = 4.05;
Sc++
4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff4s = 5.04;
4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff4p = 4.47;
(3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV Zeff4s = 3.89;
Sc+
(3d)2:
IP3d = (Zeff3d )2/2n2 = 12.28 eV Zeff3d = 2.85;
(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV Zeff4p = 3.37;
(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV Zeff4s = 2.78;
Sc
(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV Zeff3d = 1.84;
(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV Zeff4p = 2.32;
As increase net charge increases, the differential shielding for 4s
vs 3d is less important than the difference in n quantum number 3
vs 4. Thus M2+ transition metals always have all valence
electrons in d orbitals © copyright 2012 William A. Goddard III, all rights reserved\
196
Lecture 1Ch121a-Goddard-L01
Implications for transition metals
The simple Aufbau principle puts 4s below 3d
But increasing the charge tends to prefers 3d vs 4s.
Thus Ground state of Sc 2+ , Ti 2+ …..Zn 2+ are all (3d)n
For all neutral elements K through Zn the 4s orbital is
easiest to ionize.
This is because of increase in relative stability of 3d for
higher ions
Lecture 1Ch121a-Goddard-L01
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197
Transtion metal valence ns and (n-1)d orbitals
Lecture 1Ch121a-Goddard-L01
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198
Review over, back to quantum mechanics
Stopped Lecture 1
Lecture 1Ch121a-Goddard-L01
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199
Quick fix to satisfy the Pauli Principle
Combine the product wavefunctions to form a symmetric
combination
Ψs(1,2)= ψe(1) ψm(2) + ψm(1) ψe(2)
And an antisymmetric combination
Ψa(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
We see that
t12 Ψs(1,2) = Ψs(2,1) = Ψs(1,2) (boson symmetry)
t12 Ψa(1,2) = Ψa(2,1) = -Ψa(1,2) (Fermion symmetry)
Thus for electrons, the Pauli Principle only allows the
antisymmetric combination for two independent
electrons
Lecture 1Ch121a-Goddard-L01
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200
Implications of the Pauli Principle
Consider two independent electrons,
1 on the earth described by ψe(1)
and 2 on the moon described by ψm(2)
Ψ(1,2)= ψe(1) ψm(2)
And test whether this satisfies the Pauli Principle
Ψ(2,1)= ψm(1) ψe(2) ≠ - ψe(1) ψm(2)
Thus the Pauli Principle does NOT allow
the simple product wavefunction for two
independent electrons
Lecture 1Ch121a-Goddard-L01
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201
Consider some simple cases: identical spinorbitals
Ψ(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
Identical spinorbitals: assume that ψm = ψe
Then Ψ(1,2)= ψe(1) ψe(2) - ψe(1) ψe(2) = 0
Thus two electrons cannot be in identical spinorbitals
Note that if ψm = eia ψe where a is a constant phase
factor, we still get zero
Lecture 1Ch121a-Goddard-L01
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202
Consider some simple cases: identical spinorbitals
Ψ(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
Identical spinorbitals: assume that ψm = ψe
Then Ψ(1,2)= ψe(1) ψe(2) - ψe(1) ψe(2) = 0
Thus two electrons cannot be in identical spinorbitals
Note that if ψm = eia ψe where a is a constant phase
factor, we still get zero
Lecture 1Ch121a-Goddard-L01
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203
Consider some simple cases: orthogonality
Consider the wavefunction
Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
where the spinorbitals ψm and ψe are orthogonal
hence <ψm|ψe> = 0
Define a new spinorbital θm = ψm + l ψe (ignore normalization)
That is NOT orthogonal to ψe. Then
Ψnew(1,2)= ψe(1) θm(2) - θm(1) ψe(2) =
=ψe(1) θm(2) + l ψe(1) ψe(2) - θm(1) ψe(2) - l ψe(1) ψe(2)
= ψe(1) ψm(2) - ψm(1) ψe(2) =Ψold(1,2)
Thus the Pauli Principle leads to orthogonality of
spinorbitals for different electrons, <ψi|ψj> = dij = 1 if i=j
=0 if i≠j
Lecture 1Ch121a-Goddard-L01
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204
Consider some simple cases: nonuniqueness
Starting with the wavefunction
ψm
Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
Consider the new spinorbitals θm and θe where
θm = (cosa) ψm + (sina) ψe
Note that <θi|θj> = dij
a
Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) =
+(cosa)2 ψ (1)ψ (2) +(cosa)(sina) ψ (1)ψ (2) θm
ψe
θe = (cosa) ψe - (sina) ψm
e
m
e
e
-(sina)(cosa) ψm(1) ψm(2) - (sina)2 ψm(1) ψe(2)
θe
a
-(cosa)2 ψm(1) ψe(2) +(cosa)(sina) ψm(1) ψm(2)
-(sina)(cosa) ψe(1) ψe(2) +(sina)2 ψe(1) ψm(2)
[(cosa)2+(sina)2] [ψe(1)ψm(2) - ψm(1) ψe(2)] =Ψold(1,2)
Thus
linear combinations
of the spinorbitals do not change Ψ(1,2)
205
Lecture 1Ch121a-Goddard-L01
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Determinants
The determinant of a matrix is defined as
The determinant is zero if any two columns (or rows) are identical
Adding some amount of any one column to any other column
leaves the determinant unchanged.
Thus each column can be made orthogonal to all other
columns.(and the same for rows)
The above properties are just those of the Pauli Principle
Thus
we will take determinants
of our wavefunctions.
Lecture 1Ch121a-Goddard-L01
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206
The antisymmetrized wavefunction
Now put the spinorbitals into the matrix and take the determinant
Where the antisymmetrizer
determinant operator.
can be thought of as the
Similarly starting with the 3!=6 product wavefunctions of the form
The only combination satisfying the Pauil Principle is
Lecture 1Ch121a-Goddard-L01
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207
Example:
Interchanging electrons 1 and 3 leads to
From the properties of determinants we know that interchanging
any two columns (or rows), that is interchanging any two
spinorbitals, merely changes the sign of the wavefunction
Guaranteeing that the Pauli Principle is always satisfied
Lecture 1Ch121a-Goddard-L01
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208
Energy for 2-electron product wavefunction
Consider the product wavefunction
Ψ(1,2) = ψa(1) ψb(2)
And the Hamiltonian for H2 molecule
H(1,2) = h(1) + h(2) +1/r12 + 1/R
In the details slides next, we derive
E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>
E = haa + hbb + Jab + 1/R
where haa =<a|h|a>, hbb =<b|h|b> are just the 1 electron energies
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12
represents the total Coulomb interaction between the electron
density a(1)=| ψa(1)|2 and b(2)=| ψb(2)|2
Since the integrand a(1) b(2)/r12 is positive for all positions of 1
and 2, the integral is positive, Jab > 0
Lecture 1Ch121a-Goddard-L01
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209
Details in deriving energy: normalization
First, the normalization term is
<Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)>
Which from now on we will write as
<Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi are normalized
Here our convention is that a two-electron function such as
<Ψ(1,2)|Ψ(1,2)> is always over both electrons so we need not put
in the (1,2) while one-electron functions such as <ψa(1)|ψa(1)> or
<ψb(2) ψb(2)> are assumed to be over just one electron and we
ignore the labels 1 or 2
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210
Details of deriving energy: one electron termss
Using H(1,2) = h(1) + h(2) +1/r12 + 1/R
We partition the energy E = <Ψ| H|Ψ> as
E = <Ψ|h(1)|Ψ> + <Ψ|h(2)|Ψ> + <Ψ|1/R|Ψ> + <Ψ|1/r12|Ψ>
Here <Ψ|1/R|Ψ> = <Ψ|Ψ>/R = 1/R since R is a constant
<Ψ|h(1)|Ψ> = <ψa(1)ψb(2) |h(1)|ψa(1)ψb(2)> =
= <ψa(1)|h(1)|ψa(1)><ψb(2)|ψb(2)> = <a|h|a><b|b> =
≡ haa
Where haa≡ <a|h|a> ≡ <ψa|h|ψa>
Similarly <Ψ|h(2)|Ψ> = <ψa(1)ψb(2) |h(2)|ψa(1)ψb(2)> =
= <ψa(1)|ψa(1)><ψb(2)|h(2)|ψb(2)> = <a|a><b|h|b> =
≡ hbb
The remaining term we denote as
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> so that the total energy is
E
= h1Ch121a-Goddard-L01
Lecture
© copyright 2012 William A. Goddard III, all rights reserved\
aa + hbb + Jab + 1/R
211
The energy for an antisymmetrized product, A ψaψb
The total energy is that of the product plus the exchange term
which is negative with 4 parts
Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa >
- < ψaψb|1/r12|ψb ψa >
The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+
<ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R
But <ψb|ψa>=0
Thus all are zero
Thus the only nonzero term is the 4th term:
-Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy
(or the 2-electron exchange) since it arises from the exchange
term due to the antisymmetrizer.
Summarizing, the energy of the Aψaψb wavefunction for H2 is
E = haa + hbb + (Jab –Kab) + 1/R
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212
The energy of the antisymmetrized wavefunction
The total electron-electron repulsion part of the energy for any
wavefunction Ψ(1,2) must be positive
Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0
This follows since the integrand is positive for all positions of r1
and r2 then
We derived that the energy of the A ψa ψb wavefunction is
E = haa + hbb + (Jab –Kab) + 1/R
Where the Eee = (Jab –Kab) > 0
Since we have already established that Jab > 0 we can conclude
that
Jab > Kab > 0
Lecture 1Ch121a-Goddard-L01
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213
Separate the spinorbital into orbital and spin parts
Since the Hamiltonian does not contain spin the spinorbitals can
be factored into spatial and spin terms.
For 2 electrons there are two possibilities:
Both electrons have the same spin
ψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)]
So that the antisymmetrized wavefunction is
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]=
=[Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)]
Also, similar results for both spins down
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][b(1)b(2)]=
=[Φa(1)Φb(2)- Φb(1)Φa(2)][b(1)b(2)]
Since <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb>
We see that the spatial orbitals for same spin must be orthogonal
Lecture 1Ch121a-Goddard-L01
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214
Energy for 2 electrons with same spin
The total energy becomes
E = haa + hbb + (Jab –Kab) + 1/R
where haa ≡ <Φa|h|Φa> and hbb ≡ <Φb|h|Φb>
where Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>
We derived the exchange term for spin orbitals with same spin as
follows
Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)>
`````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|a(1)><a(2)|a(2)>
≡ Kab
where Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>
Involves only spatial coordinates.
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215
Energy for 2 electrons with opposite spin
Now consider the exchange term for spin orbitals with opposite
spin
Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)>
`````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|b(1)><b(2)|a(2)>
=0
Since <a(1)|b(1)> = 0.
Thus the total energy is
Eab = haa + hbb + Jab + 1/R
With no exchange term unless the spins are the same
Since <ψa|ψb>= 0 = < Φa| Φb><a|b>
There is no orthogonality condition of the spatial orbitals for
opposite spin electrons
In general we can have <Φa|Φb> =S, where the overlap S ≠ 0
Lecture 1Ch121a-Goddard-L01
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216
Summarizing: Energy for 2 electrons
When the spinorbitals have the same spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]
The total energy is
Eaa = haa + hbb + (Jab –Kab) + 1/R
But when the spinorbitals have the opposite spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]=
The total energy is
Eab = haa + hbb + Jab + 1/R
With no exchange term
Thus exchange energies arise only for the case in
which both electrons have the same spin
Lecture 1Ch121a-Goddard-L01
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217
Consider further the case for spinorbtials with opposite spin
The wavefunction
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]
Leads directly to
3E
ab = haa + hbb + (Jab –Kab) + 1/R
Exactly the same as for
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)]
[Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)]
These three states are collectively referred to as the triplet state
and denoted as having spin S=1
Lecture 1Ch121a-Goddard-L01
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218
Consider further the case for spinorbtials with opposite spin
The other combination leads to one state, referred to as the
singlet state and denoted as having spin S=0
[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]
For the ground state of a 2-electron system, Φa=Φb so we get
[Φa(1)Φa(2)][a(1)b(2)-b(1)a(2)] = A[Φa(1)a(1)] [Φa(2)b(2)]
Leading directly to
1Eaa = 2haa + Jaa + 1/R
This state is referred to as the closed shell single state and
denoted as having spin S=0
Lecture 1Ch121a-Goddard-L01
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219
Re-examine He atom with spin and the Pauli Principle
Ψ(1,2) = A[(φ1s a) (φ1s b)]
E= 2 <1s|h|1s> + J1s,1s
Which is exactly what we assumed above when we
ignore spin and the Pauli Principle
So for 2 electrons nothing changes
Lecture 1Ch121a-Goddard-L01
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220
Energy for 2-electron product wavefunction
Consider the product wavefunction
Ψ(1,2) = ψa(1) ψb(2)
And the Hamiltonian for H2 molecule
H(1,2) = h(1) + h(2) +1/r12 + 1/R
In the details slides next, we derive
E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>
E = haa + hbb + Jab + 1/R
where haa =<a|h|a>, hbb =<b|h|b> are just the 1 electron energies
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ d3r1[ψa(1)]2 ʃd3r2[ψb(2)]2/r12 =
= ʃ [ψa(1)]2 Jb (1) = <ψa(1)| Jb (1)|ψa(1)>
Where Jb (1) = ʃ [ψb(2)]2/r12 is the Coulomb potential at 1 due to
the density distribution [ψb(2)]2
2
221
© copyrightbetween
2012 William A.
Goddard III, all
rights
reserved\
JLecture
the Coulomb repulsion
densities
ab is 1Ch121a-Goddard-L01
a=[ψ
a(1)] and b
The energy for an antisymmetrized product,
A ψ aψ b
The total energy is that of the product wavefunction plus the new
terms arising from exchange term which is negative with 4 parts
Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa >
- < ψaψb|1/r12|ψb ψa >
The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+
<ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R
But <ψb|ψa>=0
Thus all are zero
Thus the only nonzero term is the 4th term:
-Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy
(or the 2-electron exchange) since it arises from the exchange
term due to the antisymmetrizer.
Summarizing, the energy of the Aψaψb wavefunction for H2 is
E = haa + hbb + (Jab –Kab) + 1/R
Lecture 1Ch121a-Goddard-L01
One© new
term from the antisymmetrizer
copyright 2012 William A. Goddard III, all rights reserved\
222
Summary electron-electron energies
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=<ψa(1)| Jb (1)|ψa(1)>
is the total Coulomb interaction between the electron density
a(1)=| ψa(1)|2 and b(2)=| ψb(2)|2
Since the integrand a(1) b(2)/r12 is positive for all positions of 1
and 2, the integral is positive, Jab > 0
Here Jb (1) = ʃ [ψb(2)]2/r12 is the potential at 1 due to the density
distribution [ψb(2)]2
Kab=< ψaψb|1/r12|ψb ψa >= ʃ d3r1[ψa(1)ψb(1)] ] ʃd3r2[ψb(2) ψa(2)]]2/r12
= <ψa(1)| Kb (1)|ψa(1)>
Where Kb (1) ψa(1)] ] = ψb(1) ʃ [ψb(2)ψa(2)]2/r12 is an integral
operator that puts Kab into a form that looks like Jab. The
difference is that Jb (1) is a function but Kb (1) is an operator
Lecture 1Ch121a-Goddard-L01
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223
Alternative form for electron-electron energies
It is commen to rewrite Jab as
Jab ≡ [ψa(1) ψa(1)|ψb(2)ψb(2)] where all the electron 1 stuff is on
the left and all the electron 2 stuff is on the right. Note that the
1/r12 is now understood
Similarly Kab= [ψa(1)ψb(1)|ψb(2)ψa(2)]
Here the numbers 1 and 2 are superflous so we write
Jab ≡ [ψaψa|ψbψb] = [aa|bb] since only the orbital names are
significant
Siimilarly
Kab ≡ [ψaψb|ψbψa] = [ab|ba]
Thus the total 2 electron energy is
Jab - Kab = [aa|bb] - [ab|ba]
But if a and b have opposite spin then the exchange term is zero
Lecture 1Ch121a-Goddard-L01
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224
Consider the case of 4 e in 2 orbitals, a,b
Ψ(1,2,3,4) = A[(aa)(ab)(ba)(bb)]
E = 2 haa + 2 hbb + Enn + [2Jaa-Kaa] +2[2Jab-Kab] + [2Jbb-Kbb]
= 2 haa + 2 hbb + Enn + 2(aa|aa)-(aa|aa)+4(aa|bb)-2(ab|ba)
+2(bb|bb)-(bb|bb)
Where we see that the self-Coulomb and self-exchange can
cancel.
Now change φ1 to φ1 + dφ1 the change in the energy is
dE = 4<dφ1|h|φ1> + 4 <dφ1|2J1-K1|φ1> + 4 <dφ1|2J2-K2|φ1>
= 4 <dφ1|HHF|φ1>
Where HHF = h + Σj=1,2 [2Jj-Kj] is called the HF Hamiltonian
In the above expression we assume that φ1 was normalized,
<φ1|h|φ1> = 1.
Imposing this constraint (a Lagrange multiplier) leads to
<dφ1|HHF – l1|φ1> = 0 and <dφ2|HHF – l2|φ2> = 0
Thus the optimum orbitals satisfy HHFφk = lk φk the HF equations
Lecture 1Ch121a-Goddard-L01
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225
