lecture 1, january 7, 2019 quantum mechanics-1: wavefunctions · 5 1:quantum mechanics challenge:...

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, [email protected] 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 <[email protected]>

Room BI 115Lecture: Monday, Wednesday 2-3pm

Lab Session: Friday 2-3pm

mailto:[email protected]


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


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.


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


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


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


The stratospheric review of QM

10

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


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

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

AA

A rR

Z

R

ZZ

M

1

2

1

2

1 22


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

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

AA

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

AA

A rR

Z

R

ZZ

M

1

2

1

2

1 22

ji iji i A Ai

A

BA AB

BAi

AA

A rR

Z

R

ZZ

M

1

2

1

2

1 22


Ignore electron-electron interactionsIndependent Particle Approximation

13

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)



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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)---


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)



15

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)---


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


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)


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

17



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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3 electrons


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


The closed shell Hartree Fock Equations

20

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)]


Simple derivation of energyclosed shell wavefunction

21

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


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


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


The Koopmans orbital energy

24

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


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


Bond dissociation for H2

26

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



27

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)]



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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


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


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


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**


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


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


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


Pi orbitals of benzene

Top view


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


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


Software packages

Jaguar: Good for organometallicsQChem: very fast for organicsGaussian: slow, many analysis toolsSpartan easiest to useGAMESSHyperChemADF


HF wavefunctions

39

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.


Electron correlation

40

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


Configuration interaction

41

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


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


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


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!!!


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


The following slides are supplementary materialThey may give you more insight


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


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


Plotting of orbitals: line cross-section vs. contour

49

contour plot in yz plane

line plot along z axis


Contour plots of 1s, 2s, 2p hydrogenic orbitals

50


Contour plots of 3s, 3p, 3d hydrogenic orbitals

51


Contour plots of 4s, 4p, 4d hydrogenic orbtitals

52


Contour plots of hydrogenic 4f orbitals

53


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


He, 2

Ne, 10

Ar, 18

Zn, 30

Kr, 36


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


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


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


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


The Exchange energy

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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


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


Interpret the Exchange term

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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


He atom one slater orbital

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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.


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


Many-electron configurations

67

General aufbau

ordering

Particularly stable


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


General trends along a column of the periodic table

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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


Plot of rφ(r) for the outer s

valence orbital

70


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


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


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


3d = 2.59;


4p = 3.20;


3d = 4.05; 3d<4s<4p

IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;


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


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


Implications for transition metals

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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


Transtion metal valence ns and (n-1)d orbitals

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lecture 1, january 7, 2019 quantum mechanics-1: wavefunctions · 5 1:quantum mechanics challenge:...

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