Post on 04-Jun-2020
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 1
Ch121a Atomic Level Simulations of Materials and Molecules
William A. Goddard III, wag@caltech.edu316 Beckman Institute, x3093Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology
Lecture 1, January 7, 2019Quantum Mechanics-1: wavefunctions
Yalu Chen <ychen3@caltech.edu>
Room BI 115Lecture: Monday, Wednesday 2-3pm
Lab Session: Friday 2-3pm
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 2
CH121a Atomic Level Simulations of Materials and Molecules
Instructor: William A. Goddard IIIPrerequisites: some knowledge of quantum mechanics, classical mechanics, thermodynamics, chemistry, Unix. At least at the Ch21a levelCh121a is meant to be a practical hands-on introduction to expose students to the tools of modern computational chemistry, computational materials science, and computational biochemistry 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.
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ARTIFICIAL PHOTOSYNTHESIS (JCAP) H2O +h H2+O2,: HER, OER
ARTIFICIAL PHOTOSYNTHESIS (JCAP):CO2 +h fuels: (MO/Cu); NP,
FUEL CELL CATALYST: Oxygen Reduction Reaction; Alkane fuels; Dealloyed
BATTERIES: Li air-CO2, LiS, anionic electrolytes, solid electrolytes
PEROVSKITES: MAPbI3, photoanodes, BaTiO3, Ferroelectrics,
CERAMICS: Ductile ceramics, FC electrodes, FC membranes, HiTc
POLYMERS: Batteries (PEO) ; Fuel Cells electrolytes (Nafion, Anionic)
Proteins-pharma: GPCR Membrane Proteins, Pharma, GP activation
DNA-RNA: Origami, cond-siRNA
2D MATERIALS: MBE-graphene; MoS2, CVD, ALD
CATALYSIS: NH3 synthesis, selective ammoxidation and oxidation alkanes
ENERGETIC MATERIALS: PETN, RDX, HMX, TATB, TATP, Propellants
COMBUSITION: Kinetics from full reaction reactive simulations
THERMOELECTRICS: (mechanical properties (brittleness))
SOLAR ENERGY: dye sensitized solar cells, CuInGaSe (CIGS/CdS) cells, Ionic Liquids
GAS STORAGE (H2, CH4, CO2) : MOFs, COFs, metal alloys, nanoclusters
SEMICONDUCTORS: damage free etching
To solve the most challenging problems, we develop methods and software simultaneously. Current Focus
MultiParadigm Strategy: apply 1st principles to complex systems
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Motivation: Design Materials, Catalysts, Pharma from 1st Principles so can do design prior to experiment
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Big breakthrough making FC simulations practical:
reactive force fields based on QMDescribes: chemistry,charge transfer, etc. For metals, oxides, organics.
Accurate calculations for bulk phases and molecules (EOS, bond dissociation)Chemical Reactions (P-450 oxidation)
time
distance
hours
millisec
nanosec
picosec
femtosec
Å nm micron mm yards
MESO
Continuum(FEM)
QM
MD
ELECTRONS ATOMS GRAINS GRIDS
Deformation and FailureProtein Structure and Function
Micromechanical modelingProtein clusters
simulations real devices full cell (systems biology)
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
Ch121a
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1:Quantum Mechanics Challenge: increased accuracyNew Functionals DFT (dispersion)Meta Dynamics QM (G† 298K)Accurate implicit solvation (CANDLE)Grand Canonical QM (constant Potentia)Accurate Band Gaps2:Force FieldsChallenge: chemical reactionsReaxFF- Describe Chemical Reaction processes, Mixed Metals, Ceramics, PolymersAccelerated Reactive DynamicsNonEquil QM Dynamics (eFF)Hybrid QM-ReaxFF
4:Molecular DynamicsChallenge: Extract properties essential to materials designNon-Equilibrium Dynamics
Viscosity, rheologyThermal ConductivityPlasticity, Dislocations, CrackInterfacial Energies
• surface tension, contact anglesReaction KineticsEntropies, Free energies• surface tension, contact angles5: Coarse Grain Reactive MD
3:Biological Predictions1st principles structures GPCRs1st principles Ligand BindingConditional siRNA therapeutics
6: Integration: Computational Materials Design Facility (CMDF)•Seamless across the hierarchies of simulations using Python-based scripts
Materials Design Requires Improvements in Methods to Achieve Required Accuracy. Our Focus:
Need new theory methods to solve key problems in energy and environment
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 6
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 first 5 weeks homework applies these principles to practical problems making use of modern generally available software.
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 7
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 each of the last 5 weeks is to turn in a one page report on progress with the project
The final is a research report ~ 5 page describing the calculations and conclusions
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 8
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, catalysisMolecular Dynamics: structure optimization, vibrations, phonons, elastic moduli, Verlet, microcanonical, Nose, GibbsMonte Carlo and Statistical thermodynamics Growth amorphous structures, Kubo relations, correlation functions, RIS, CCBB, FH methods growth chains, Gauss coil, theta tempCoarse grain approaches eFF for electron dynamicssolvation, diffusion, mesoscale force fields
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 9
Applications will include prototype examples involving such materials as:
ElectrocatalysisHeterogeneous CatalysisHomogeneous CatalysisSemiconductors (IV, III-V, surface reconstruction)Polymers (building amorphous and crystalline);Protein structure (focus GPCR), ligand dockingCeramics (B4C, BaTiO3, LaSrCuOx)Metal alloys (crystalline, amorphous, plasticity)DNA-structure, ligand docking
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The stratospheric review of QM
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You should have already been exposed to much of this material
This overview to remind you of the key points
Overview of Quantum Mechanics, Hydrogen Atom, etcPlease review again to make sure that you are comfortable with the concepts, which you should have seen before
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 11
Classical Mechanics
Energy = Kinetic energy + Potential energy
Kinetic energy =
Potential energy =
Nucleus-Nucleusrepulsion
Nucleus-Electronattraction
Electron-Electronrepulsion
atoms electrons
p p+ +
Classical Mechanics
Can optimize electron coordinates and momenta separately, thus lowest energy: all p=0 KE =0
All electrons on nuclei: PE = - infinity
Therefore electrons collapse into nucleus, no atoms, no molecules, no life
Not consistent with real world. Solution? Quantum mechanics
ji iji i A Ai
A
BA AB
BAi
AA
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ji iji i A Ai
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Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 12
Ab Initio, quantum mechanics
Quantum mechanics
Energy = < Ψ|KE operator|Ψ> + < Ψ|PE operator|Ψ>
Kinetic energy op =
Potential energy =
atoms electrons
Optimize Ψ, get HelΨ=EΨ
Hel =
The wavefunction Ψ(r1,r2,…,rN) contains all information of system determine KE and PE
Schrodinger Equation
Too complicated to solve exactly. What do we do?
ji iji i A Ai
A
BA AB
BAi
AA
A rR
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ji iji i A Ai
A
BA AB
BAi
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Z
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ji iji i A Ai
A
BA AB
BAi
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ji iji i A Ai
A
BA AB
BAi
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Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Ignore electron-electron interactionsIndependent Particle Approximation
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Solve N different 1-electron problems: h(1) ψa(1) = ea ψa(1)Total wavefunction is the product of 1-e orbitals Ψ(1,2,3,4, ..N-1,N) =ψa(1) ψb(2) ψc(3) ----ψN(N)
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Ignore electron-electron interactionsIndependent Particle Approximation
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Prob(1,2,3,4, ..N-1,N) =|Ψ|2= Ψ*ΨΨ*(1,2,3,4, ..N-1,N)Ψ(1,2,3,4, ..N-1,N)= ψa
*(1) ψb*(2) ψc
*(3) ---ψN*(N)ψa(1) ψb(2) ψc(3) ---ψN(N)
=|ψa*(1) ψa(1)| |ψb
*(2) ψb(2)| ----=P(1) P(2)---
Solve N different 1-electron problems: h(1) ψa(1) = ea ψa(1)Total wavefunction is the product of 1-e orbitals Ψ(1,2,3,4, ..N-1,N) =ψa(1) ψb(2) ψc(3) ----ψN(N)
With the product wavefunction the probability of finding e1 at position (x1,y1,z1) is independent of the probability of finding e2 at position (x2,y2,z2)
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Ignore electron-electron interactionsIndependent Particle Approximation
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Prob(1,2,3,4, ..N-1,N) =|Ψ|2= Ψ*ΨΨ*(1,2,3,4, ..N-1,N)Ψ(1,2,3,4, ..N-1,N)= ψa
*(1) ψb*(2) ψc
*(3) ---ψN*(N)ψa(1) ψb(2) ψc(3) ---ψN(N)
=|ψa*(1) ψa(1)| |ψb
*(2) ψb(2)| ----=P(1) P(2)---
Solve N different 1-electron problems: h(1) ψa(1) = ea ψa(1)Total wavefunction is the product of 1-e orbitals Ψ(1,2,3,4, ..N-1,N) =ψa(1) ψb(2) ψc(3) ----ψN(N)
With the product wavefunction the probability of finding e1 at position (x1,y1,z1) is independent of the probability of finding e2 at position (x2,y2,z2)
The electrons are independent of each other (no correlation of their motions)This wavefunction is called the Hartree approximation
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Pauli Principle
Ψ(1,2,4,3, ..N-1,N) = - Ψ(1,2,3,4, ..N-1,N) Interchanging any two electrons changes the sign of the total wavefunctionHartree wavefunction not satisfy PP2 electrons: ΨH(2,1) = ψa(2) ψb(1) = ψb(1)ψa(2) -ψǂ a(1) ψb(2)
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Simplest wave function satisfying PP for 2 electrons: Ψ(1,2) = ψa(1) ψb(2) - ψb(1)ψa(2) Thus Ψ(1,2) = ψb(1) ψa(2) - ψa(1)ψb(2) = -[ψa(1) ψb(2) - ψb(1)ψa(2)] We write this as a Slater Determinant Ψ(1,2) = ψb(1) ψa(2) - ψa(1)ψb(2)= =
Pauli Principle
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Ψ(1,2,4,3, ..N-1,N) = - Ψ(1,2,3,4, ..N-1,N) Interchanging any two electrons changes the sign of the total wavefunctionHartree wavefunction not satisfy PP2 electrons: ΨH(2,1) = ψa(2) ψb(1) = ψb(1)ψa(2) -ψǂ a(1) ψb(2)
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Simplest wave function satisfying PP for 2 electrons: Ψ(1,2) = ψa(1) ψb(2) - ψb(1)ψa(2) Thus Ψ(1,2) = ψb(1) ψa(2) - ψa(1)ψb(2) = -[ψa(1) ψb(2) - ψb(1)ψa(2)] We write this as a Slater Determinant Ψ(1,2) = ψb(1) ψa(2) - ψa(1)ψb(2)= =
Pauli Principle
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Ψ(1,2,4,3, ..N-1,N) = - Ψ(1,2,3,4, ..N-1,N) Interchanging any two electrons changes the sign of the total wavefunctionHartree wavefunction not satisfy PP2 electrons: ΨH(2,1) = ψa(2) ψb(1) = ψb(1)ψa(2) -ψǂ a(1) ψb(2)
3 electrons
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
General case – independent electrons – Slater determinant
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Ψ(1,2,3,4, ..N-1,N) = A[ψa(1) ψb(2) ψc(3) ----ψN(N)]
Properties of determinants:
Get zero if any two rows or columns are identical (simple Pauli exclusion principle from old QM)
Interchange any two rows or columns changes the sign
Every row or column can be taken as orthogonal to every other row or column
Adding some amount of one column to any other column leaves determinant unchangedEven if electrons do not interact, PP requires a determinant wavefunction
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The closed shell Hartree Fock Equations
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General concept: there are an infinite number of possible orbitals for the electrons. For a system with 2M electrons we will put the electrons into the M lowest orbitals, with two electrons in each orbital (one up or spin, the other down or spin)
M occ orb2M elect
Often the ground state is a closed shell wavefunction in which there are an even number of electrons, N=2Meach occupied orbital has two electrons, one with up or a spin and the other with down or b spin
Ψ(1,2,3,4, ..N-1,N) = A[a(1)α(1) a(2)β(2) b(3)α(3) b(4)β(4)…]
[a]2 [α(1)β(2)- β(1)α(2)] [b]2 [α(3)β(4)- β(3)α(4)]
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Simple derivation of energyclosed shell wavefunction
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N electrons in N/2 orbitals each doubly occupied with up and down spin
2 haa + Jaa4 Jab - 2Kab
E = j=1toN hjj + (1/2)j,k=toN (2Jjk – Kjk) + 1/R
This would appear to have N2/2 Coulomb interactions, but N/2 are canceled by the self exchange, so only get N(N-1/2 two-e terms
Since Jaa = Kaa
2 haa + 2 Jaa - Kaa
Self coulomb
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The Hartree Fock Equations
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Variational principle: Require that each orbital be the best possible (leading to the lowest energy) leads to
HHF(1)φa(1)= a φa(1)
where we solve for the occupied orbital, φa, to be occupied by both electron 1 and electron 2
Here HHF(1)= h(1) + b [2Jb(1) - Kb(1)]
This looks like the Hamiltonian for a one-electron system in which the Hamiltonian has the form it would have for the average potential due the electron in all other orbital orbitals
Thus the two-electron problem is factored into M=N/2 one-electron problems, which we can solve to get φa, φb, etc
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Self consistency
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However to solve for φa(1) we need to know b [2Jb(1) - Kb(1)]
which depends on all M orbitals
Thus the HHFφa= a φa equation must be solved iteratively until it is self consistent
But after the equations are solved self consistently, we can consider each orbital as the optimum orbital moving in the average field of all the other electrons
In fact the motions between these electrons would tend to be correlated so that the electrons remain farther apart than in this average field
Thus the error in the HF energy is called the correlation energy
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The Koopmans orbital energy
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The next question is the meaning of the one-electron energy, a in the HF equationHHF(1)φa(1)= a φa(1)
Multiplying each side by φa(1) and integrating leads toa <a|a> = <a|HHF|a> = <a|h|a> + 2b<a|Jb|a> - b<a|Kb|a>
= <a|h|a> + Jaa + b≠a<a|2Jb-Kb|a>
Thus in the approximation that the remaining electron does not change shape, a corresponds to the energy to ionize an electron from the a orbital to obtain the N-1 electron system Sometimes this is referred to as the Koopmans theorem (pronounced with a long o). It is not really Koopmans theorem, which we will discuss later, but we will use the term anyway
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The ionization potential
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There are two errors in using the a to approximate the IPIPKT ~ -a First the remaining N-1 electrons should be allowed to relax to the optimum orbital of the positive ion, which would make the Koopmans IP too largeHowever the energy of the HF description is leads to a total energy less negative than the exact energy, Exact = EHF – Ecorr Where Ecorr is called the electron correlation energy (since HF does NOT allow correlation of the electron motions. Each electron sees the average potential of the other)which would make the Koopmans IP too smallThese effects tend to cancel so that the a from the HF wavefunction leads to a reasonable estimate of IP
(N-1)e
exactHF from Ne
Ne
exactHFexact IP
Koopmans IP
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Bond dissociation for H2
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The closed shell HF wavefunction for H2 φHF(1)φHF(2)[]does ok near Re but goes to the wrong limit as R>3 bohr~1.6A(it is nearly as bad as MO)
HF
exact
ΨHF (1,2)=φHF(1) φHF(2)= =χL(1)χL(2) + χR(1) χR(2) + χL(1) χR(2) + χR(1) χL(2)
φHF(1) = χL(1) + χR(1)
Ionic – ok at Re but terrible at large R
covalent – ok at Re and exact at large R
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Bond dissociation for H2
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Consider now the wavefunction the Unrestricted HF wavefunction (UHF) , whereA[Φa(1)Φb(2)][(1)(2)]But we do not require Φa=Φb
HF
exact
UHF
For R< ~2.2 bohr the optimum is Φa=Φb, closed shell HFBut for R> ~2.2 bohr Φa localizes more on the left while Φb(2) localizes more on the right to that it goes to atomic orbitals at R = ∞, but it has the wrong spin since there is net up spin on left and down spin on rightThe energy curve has ~1/5 the correct bonding starting at ~2.2 bohr.
A[Φa(1)Φb(2)][(1)(2)]=[Φa(1)(1)][Φb(2)(2)] –[Φb(1)(1)][Φa(2)(2)]
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Bond dissociation for H2
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HF
GVBexact
Generalized Valence bondA[Φa(1)Φb(2)][(1)(2)-(1)(2)]
Orbital productproper spin function
Best possible orbital product wavefunctionMost general one-electron interpretation
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 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) dʃ 3r2 [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) = Σ C Xwhere the basis functions, {XMBF} are chosen as atomic like functions on the various centersAs a result the HF equations HHFφk = k φk
Reduce to a set of Matrix equationsΣjmHjmCmk = ΣjmSjmCmkk
This is still complicated since the Hjm operator includes exchange terms We still refer to this as solving the HF equationsThis was worked out by Clemens Roothaan (pronounced with a long o and short a) in the RS Mulliken group at Chicago in the 1950s. They called them LCAO SCF equations
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 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(-r) 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(-r), leads to huge computational costs for multicenter molecules and we replace these by an expansion in terms of Gaussian basis functions, such as exp(-r2).
The most popular MBS is the STO-3G set of Pople in which 3 gaussian functions are combined to describe each Slater function
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 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 CThis 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 valenceIn addition it is necessary to provide one level higher angular momentum atomic orbitals to describe the polarization involved in bondingThus 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 1-Ch121a-Goddard-L01 © copyright 2018 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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 33
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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 34
Results for Benzene
The energy of the C1s orbital is ~ - Zeff2/2
where Zeff = 6 – 0.3125 = 5.6875Thus 1s ~ -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 MOsThe highest MOs involve the electrons. Here there are 6 electrons and 6 p atomic orbitals leading to 3 doubly occupied and 3 empty orbitals with the pattern
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 35
Pi orbitals of benzene
Top view
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 36
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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 37
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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 38
Software packages
Jaguar: Good for organometallicsQChem: very fast for organicsGaussian: slow, many analysis toolsSpartan easiest to useGAMESSHyperChemADF
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
HF wavefunctions
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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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Electron correlation
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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(-r)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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Configuration interaction
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Consider a set of N-electron wavefunctions: {i; i=1,2, ..M} where < i|j> = ij {=1 if i=j and 0 if i ≠j)
Write approx = i=1 to M) Ci i
Then E = < approx|H|approx>/< approx|approx>
E= < i Ci i |H| k Ck k >/ < i Ci i | i Ck k >How choose optimum Ci?Require E=0 for all Ci get
k <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 notationCi is a column vector for the ith eigenstate
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Configuration interaction upper bound theorm
42
Consider the M solutions of the CI equationsHCi = SCiEi ordered as i=1 lowest to i=M highest
Then the exact ground state energy of the systemSatisfies Eexact ≤ E1
Also the exact first excited state of the system satisfiesE1st excited ≤ E2 etcThis is called the Hylleraas-Unheim-McDonald Theorem
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Transition states
43
ReactantProduct
TS
Derivative of the energy = 0 Second derivative: For a minimum > 0For a maximum < 0So a TS should have a negative second derivative of the energy, which would lead to an imaginary frequency
Transition state is the stationary point, where all forces are zero, but for which the force is at a minimum for all coordinates but one, where it is at a maximum
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 44
ReactantProduct
TS
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
Inflection points
Region with imaginary frequency
Must start with a good guess!!!
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Homework Lecture 1
45
Homework: wavefunction for butadiene
Draw the structure, minimize with force field in Maestro (Jaguar) then minimize in JaguarCompare cis and trans. Plot orbital energies, Plot orbitals, calculate the vibrational levels, get the free energy and entropy, calculate the rotational barrier
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 46
The following slides are supplementary materialThey may give you more insight
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Sizes hydrogen orbitals
47
Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f
Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48
H--H C
0.74
H
H
H
H
1.7
H H
H H
H H
4.8
=a0 n2/ZR̄ Where a0 = bohr = 0.529A=0.053 nm = 52.9 pm
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Hydrogen atom excited states
481s
-0.5 h0 = -13.6 eV
2s-0.125 h0 = -3.4 eV
2p
3s-0.056 h0 = -1.5 eV
3p 3d
4s-0.033 h0 = -0.9 eV
4p 4d 4f
Energy zero
Enlm = - Z/2 R̄ = - Z2/2n2
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Plotting of orbitals: line cross-section vs. contour
49
contour plot in yz plane
line plot along z axis
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Contour plots of 1s, 2s, 2p hydrogenic orbitals
50
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Contour plots of 3s, 3p, 3d hydrogenic orbitals
51
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Contour plots of 4s, 4p, 4d hydrogenic orbtitals
52
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Contour plots of hydrogenic 4f orbitals
53
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Aufbau principle for atoms
541s
2s2p
3s3p
3d4s
4p 4d 4f
Energy
2
2
6
2
62
6
10
10 14
He, 2
Ne, 10
Ar, 18Zn, 30
Kr, 36
Get generalized energy spectrum for filling in the electrons to explain the periodic table.
Full shells at 2, 10, 18, 30, 36 electrons
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 55
He, 2
Ne, 10
Ar, 18
Zn, 30
Kr, 36
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Consider the product wavefunction
Ψ(1,2) = ψa(1) ψb(2)
And the Hamiltonian
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>
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>= ʃ [ψa(1)]2 [ψb(1)]2/r12
Represent 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
Energy for Hartree Product Wavefunctionconsider 2 electron case
Very simple: the one-electron
energy for ψa(1) and for ψa(2) plus the e-e repulsion
between them (and the 0 electron nuclear energy
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Energy for Hartree wavefunction N electrons
57
ΨH(1,2,3,4, ..N-1,N) =ψa(1) ψb(2) ψc(3) ----ψN(N)
EH= a=1,N haa + a<b=1,N Jab + 1/R
N one-electron terms,
N*(N-1)/2 Coulomb terms
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The Coulomb energy
58
Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>
Jab = ʃ1,2 [a(1)]2 [b(2)]2/r12
Jab =ʃ1 [a(1)]2 Jb(1)
where Jb(1) = ʃ [b(2)]2/r12 is the coulomb potential evaluated at point 1 due to the charge density [b(2)]2 integrated over all space
Thus Jab is the total Coulomb interaction between the electron density a(1)=|a(1)|2 and b(2)=|b(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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 59
The energy for an antisymmetrized product, 2e case
The new terms energy is from the exchange It is negative with 4 partsEex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa > - < ψaψb|1/r12|ψb ψa >
But <ψb|ψa>=0 get zero for < ψa|h(1)|ψb><ψbψa >+ <ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R
Thus the only nonzero term is the 4th term: -< ψaψb|1/r12|ψb ψa >
E = <ψaψb |H|A ψaψb >/ <ψaψb |A ψaψb >
<ψaψb |A ψaψb > = <ψaψb |ψaψb > - <ψaψb |ψbψa >
=<ψa|ψa> <ψb |ψb > - <ψa |ψb> <ψb |ψa >
e1 e2
1 1 0 0
- ∫ ψa(1) ψb (2)|1/r12|ψb(1) ψa (2) = - ∫ψa(1)ψb (1)|1/r12|ψb(2) ψa (2)
Exchange term
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The 2 electron exchange term
60
The 4th term leads to-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 isE = haa + hbb + (Jab –Kab) + 1/R
generalizing from 2 electrons to 4
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The Exchange energy
61
Kab= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>
Kab = ʃ1 [a(1)b(1)] ʃ2 [b(2)a(2)]/r12
Kab = ʃ1 [a(1) {b(2)] ʃ2[b(2)]/r12 } a(1)] = ʃ1 [a(1) Kb(1) a(1)]
No simple classical interpretation, but we have written it in terms of an integral operator Kb(1) so that is looks similar to the Coulomb case
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 62
The sign of the exchange energy
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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Interpret the Exchange term
63
For the gu excited states H2 we found two states:3 = [φgφu - φuφg][] 1 = [φgφu + φuφg][]The energy of these two wavefunctions is 3E=Jgu – Kgu 1E=Jgu + Kgu
In 2-e space this leads to electron correlation asThe biggest contribution is when both electrons are at the same spot, say z1=z2Which is along the upper diagonal. 2Kgu is just the difference of these 2 cases
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ Ch120a-Goddard-L02
64
Fundamental Postulate of QM, the variational principle
To find the best Φap we change parameters in Φap until we get the minimum energy. For this best approximate wavefunction Φbest it must be that Eap = 0 for all possible changes, Φ.
The ground state wavefunction of the system, Φ, has the lowest possible energy out of all possible wavefunctions.
For the ground state, curvature E/Φ positive for all possible changes, E/Φ ≥ 0
EexEap
E
Consider that Φex is the exact wavefunction with energy Eex = <Φ’|Ĥ|Φ’>/<Φ’|Φ’> and that
Φap = Φex + Φ is some other approximate wavefunction.
Then Eap = <Φap|Ĥ|Φap>/<Φap|Φap> ≥ Eexact
Thus the condition for the best Φap is Eap/Φ = 0 for all Φ from Φap
Get Ĥ Φexact = Eexact Φexact Schrodinger Eqn
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
He atom one slater orbital
65
If one approximates each orbital as φ1s = N0 exp(-r) a Slater orbital then it is only necessary to optimize the scale parameter In this case
He atom: EHe = 2(½ 2) – 2Z(5/8)
Applying the variational principle, the optimum must satisfy dE/d = 0 leading to 2- 2Z + (5/8) = 0Thus = (Z – 5/16) = 1.6875KE = 2(½ 2) = 2
PE = - 2Z(5/8) = -2 2 E= - 2 = -2.8477 h0
Ignoring e-e interactions the energy would have been E = -4 h0
The exact energy is E = -2.9037 h0 Thus this simple approximation of assuming that each electron is in a H1s orbital and optimizing the size accounts for 98.1% of the exact result.
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
The Hartree Fock Equations
66
Variational principle: Require that each orbital be the best possible (leading to the lowest energy) leads to
HHF(1)φa(1)= a φa(1)
where we solve for the occupied orbital, φa, to be occupied by both electron 1 and electron 2
Here HHF(1)= h(1) + b [2Jb(1) - Kb(1)]
This looks like the Hamiltonian for a one-electron system in which the Hamiltonian has the form it would have for the average potential due the electron in all other orbital orbitals
Thus the two-electron problem is factored into M=N/2 one-electron problems, which we can solve to get φa, φb, etc
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Many-electron configurations
67
General aufbau
ordering
Particularly stable
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General trends along a row of the periodic table
68
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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
General trends along a column of the periodic table
69
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
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Plot of rφ(r) for the outer s
valence orbital
70
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\ 71
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
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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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Transition metals; consider [Ar] + 1 electron
73
[IP4s = (Zeff4s )2/2n2 = 4.34 eV Zeff
4s = 2.26; 4s<4p<3d
IP4p = (Zeff4p )2/2n2 = 2.73 eV Zeff
4p = 1.79;
IP3d = (Zeff3d )2/2n2 = 1.67 eV Zeff
3d = 1.05;
IP4s = (Zeff4s )2/2n2 = 11.87 eV Zeff
4s = 3.74; 4s<3d<4p
IP3d = (Zeff3d )2/2n2 = 10.17 eV Zeff
3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV Zeff
4p = 3.20;
IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff
3d = 4.05; 3d<4s<4p
IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff
4s = 5.04;
IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff
4p = 4.47;
K
Ca+
Sc++
As the net charge increases the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4Thus charged system prefers 3d vs 4s
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Transition metals; consider Sc0, Sc+, Sc2+
74
3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff
3d = 4.05;
4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff
4s = 5.04;
4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff
4p = 4.47;
Sc++
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
(3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV Zeff
4s = 3.89;
(3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV Zeff
3d = 2.85;
(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV Zeff
4p = 3.37;
Sc+
(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV Zeff
4s = 2.78;
(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV Zeff
3d = 1.84;
(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV Zeff
4p = 2.32;
Sc
Lecture 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Implications for transition metals
75
The simple Aufbau principle puts 4s below 3dBut 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 1-Ch121a-Goddard-L01 © copyright 2018 William A. Goddard III, all rights reserved\
Transtion metal valence ns and (n-1)d orbitals
76