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.

Thus all simulations are first-principles based 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 distance hours millisec nanosec picosec femtosec Å nm micron mm yards MESO Continuum (FEM) QM MD ELECTRONS ATOMS GRAINS GRIDS Deformation and Failure Protein Structure and Function Micromechanical modeling Protein clusters simulations real devices full cell (systems biology) Big breakthrough making FC simulations practical: reactive force fields based on QM Describes: chemistry,charge transfer, etc. For metals, oxides, organics. Accurate calculations for bulk phases and molecules (EOS, bond dissociation) Chemical Reactions (P-450 oxidation)

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.

Homework First 6 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 Ch120a to solve a research problem that can be completed in 4 weeks. The homework for the last 3 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.

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

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

Outline Topic 1: QM: HF, DFT, basis sets, reactions, transition states, vibrations Topic 2: Force Fields, nonbonds, hydrogen bonds, charges (QEq), QM  FF Topic 3: Molecular Dynamics: Verlet, NVE, NVT, NPT, Periodic Systems Topic 4: Statistical mechanics: liquid simulations, entropy, nonequilibrium MD, Green-Kubo, Monte Carlo, Grand Canonical MC, gas storage, surface tension Topic 5: polymers: crystalline, amorphous, structure prediction, Topic 6: ReaxFF Reactive Force Field and reactive Dynamics Topic 7: PBC QM, band structure, phonons, elastic constants, Ab Initio MD, Topic 8: surfaces: reconstruction, chemisorption, physisorption, solvation Topic 9: eFF, Tight binding Topic 9: applications: Fuel Cell: oxygen reduction reaction, migration in Nafion solid oxides Batteries: anode for Li battery, Solid electrolyte interface, Nanotechnology: rotaxane molecular switches, carbon nanotube interfaces Water Treatment: Dendrimers, captymers Catalysis: alkane activation, ammoxidation Metallic alloys: force fields, dislocations, plasticity, cavitation Proteins: structures, ligand docking, GPCRs DNA: A to B transition, Origami based nanotechnology

Topic 1: Practical Quantum Chemistry Solve Schrödinger Equation HelΨ=EΨ Hel = For benzene we have 12 nuclear degrees of freedom (dof) and 42 electronic 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 4th term: what is missing?

The Schrödinger Equation: Kinetic Energy Solve Schrödinger Equation HelΨ=EΨ Hel = For benzene we have 12 nuclear degrees of freedom (dof) and 42 electronic 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 4th term: what is missing?

The Schrödinger Equation: Nuclear-Electron Solve Schrödinger Equation HelΨ=EΨ Hel = For benzene we have 12 nuclear degrees of freedom (dof) and 42 electronic 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 4th term: what is missing?

The Schrödinger Equation: Electron-Electron Solve Schrödinger Equation HelΨ=EΨ Hel = For benzene we have 12 nuclear degrees of freedom (dof) and 42 electronic 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 4th term: what is missing?

The Schrödinger Equation: Nuclear-Nuclear Solve Schrödinger Equation HelΨ=EΨ Hel = 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

The supersonic 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

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

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 PEQM=< Φ| (-e2/r) |Φ> = -e2/ where is the average value of 1/r 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

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). Also we used the preference of KE for smooth wavefunctions to explain the bonding in H2+ and H2. So far we have been able to use the above expressions to reason qualitatively. However to actually solve for the wavefunctions requires the Schrodinger Eqn., which we derive next.

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 Eex Eap E This means that for sufficiently small dΦ, dE = 0 for all possible changes, dΦ We write dE/dΦ = 0 for all dΦ This is called the variational principle. For the ground state, d2E/dΦ ≥ 0 for all possible changes

4th postulate of QM 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. 4th Postulate: wavefunctions differing only by a constant phase factor describe the same state

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=0 Increasing B a b 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)

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: r1  r2 Spin only: s1  s2 Space and spin simultaneously: (r1,s1)  (r2,s2) Since doing any of these interchanges twice leads to the identity, we know from previous arguments 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

Permutational symmetries for H2 and He Have 4 degenerate g ground states for H2 Have 4 degenerate u excited states for H2 He Have 4 degenerate ground state for He

Permutational symmetries for H2 and He 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 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

Consider H atom We will consider one electron, but a nucleus with charge Z r +Ze The Hamiltonian has the form h = - (Ћ2/2me)2 – Ze2/r In atomic units h = - ½ 2 – Z/r 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 = Z

Contour plot in the xz plane, Maximum amplitude at x,z = 0,0 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 Contour plot in the xz plane, Maximum amplitude at x,z = 0,0

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 au 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 Thus e1s = The kinetic energy of the 1s state is KE1s = ½ and the potential energy is PE1s = -1 = 1/ where = 1 a0 is the average radius

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 (0 when j=k, 0 otherwise) We say that they are orthogonal. Nodal Theorem  ground state has no nodes (never changes sign), like the 1s state for H atom

The excited states of H atom - 2 The ground and excited states of a system can all be written as hφk = ekφk, where <φk |φj> = dkj (0 when j=k, 0 otherwise) We say that they are orthogonal. Nodal Theorem  ground state has no nodes (never changes sign), like the 1s state for H atom x z y θ φ 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

The excited states of H atom - 3 The ground and excited states of a system can all be written as hφk = ekφk, where <φk |φj> = dkj (0 when j=k, 0 otherwise) We say that they are orthogonal. Nodal Theorem  ground state has no nodes (never changes sign), like the 1s state for H atom x z y θ φ Use spherical polar coordinates, r, θ, φ where z=r cosθ, x=r sinθ cosφ, y=r sinθ sinφ 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 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

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 planes 1 spherical nodal plane 2 spherical 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) R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1] exp(-Zr/3)

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

The p excited angular states of H atom φnlm = Rnl(r) Zlm(θ,φ) Now lets consider excited angular functions, Zlm. They must have nodal planes to be orthogonal to Z00 x z + - pz 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 x z + - px x z + - pxpz pz

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? x z + - px’ a For example we might rotate px by an angle 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 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 radial nodes) z pzpx’ . a + - - x +

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 dyz = yz =r2 (sinθ)(cosθ) sinφ dzx = zx =r2 (sinθ)(cosθ) cosφ where dxz is shown here x z + - dxz 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,

More d angular functions? In addition we can construct three other functions with two nodal planes dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2] dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2] dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2] 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 x z + - dz2-x2

More d angular functions? In addition we can construct three other functions with two nodal planes dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2] dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2] dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2] 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 x z + - dz2-x2 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] = = [3 z2 – r2 ] which we denote as dz2

Summarizing the d angular functions dz2 + 57º - - Z20 = dz2 = [3 z2 – r2 ] m=0, ds Z21 = dzx = zx =r2 (sinθ)(cosθ) cosφ 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 m = 1, dp m = 2, dd x +

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

real (Zlm) and complex (Ylm) ang. momentum fnctns Here the bar over m  negative

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 2p are related, but it turns out that the total energy depends only on n. Enlm = - Z2/2n2 The potential energy is given by PE = - Z2/2n2 = -Z/ , where =n2/Z Thus Enlm = - Z/2 angular nodal planes Size (a0) radial nodal planes total nodal planes name 1s 0 0 0 1.0 2s 1 1 0 4.0 2p 1 0 1 4.0 3s 2 2 0 9.0 3p 2 1 1 9.0 3d 2 0 2 9.0 4s 3 3 0 16.0 4p 3 2 1 16.0 4d 3 1 2 16.0 4f 3 0 3 16.0 ˉ R ˉ R ˉ R This is all you need to remember

Sizes hydrogen orbitals ˉ R =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, 4.77, 8.48 H--H C 0.74 H 1.7 4.8

Hydrogen atom excited states Energy zero 4s -0.033 h0 = -0.9 eV 4p 4d 4f 3s -0.056 h0 = -1.5 eV 3p 3d 2s -0.125 h0 = -3.4 eV 2p Enlm = - Z/2 ˉ R = - Z2/2n2 1s -0.5 h0 = -13.6 eV

Plotting of orbitals: line cross-section vs. contour line plot along z axis contour plot in yz plane

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

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

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

Contour plots of hydrogenic 4f orbitals

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

Estimate J1s,1s, the electron repulsion energy of 2 electrons in He+ 1s orbitals Now consider He atom: EHe = 2(½ z2) – 2Zz + J1s,1s How can we estimate J1s,1s 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 electron repulsion increases the energy (less attractive) by 15.6%

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.

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)

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 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)

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

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 0.2A 1s 2.12A 2p

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

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 0.2A 1s 2.12A 2p

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 at ~ 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 0.2A 1s 2.12A 2s R~0.2A

Li atom excited states Energy MO picture State picture zero DE = 2.2 eV 17700 cm-1 564 nm 1st excited state -0.125 h0 = -3.4 eV (1s)2(2p) 2p -0.205 h0 = -5.6 eV (1s)2(2s) 2s Ground state Exper 671 nm DE = 1.9 eV -2.723 h0 = 74.1 eV 1s

Aufbau principle for atoms Energy 14 10 4f Kr, 36 4d 6 4p Zn, 30 2 10 4s Ar, 18 3d 6 3p 2 3s Ne, 10 6 2p Get generalized energy spectrum for filling in the electrons to explain the periodic table. Full shells at 2, 10, 18, 30, 36 electrons 2 2s He, 2 2 1s

Kr, 36 Zn, 30 Ar, 18 Ne, 10 He, 2

Many-electron configurations General aufbau ordering Particularly stable

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

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

Plot of rφ(r) for the outer s valence orbital

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 [IP4s = (Zeff4s )2/2n2 = 4.34 eV  Zeff4s = 2.26; 4s<4p<3d 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 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 IP4s = (Zeff4s )2/2n2 = 21.58 eV  Zeff4s = 5.04; IP4p = (Zeff4p )2/2n2 = 17.01 eV  Zeff4p = 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 4 Thus charged system prefers 3d vs 4s

Transition metals; consider Sc0, Sc+, Sc2+ 3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV  Zeff3d = 4.05; 4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV  Zeff4s = 5.04; 4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV  Zeff4p = 4.47; Sc++ (3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV  Zeff4s = 3.89; (3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV  Zeff3d = 2.85; (3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV  Zeff4p = 3.37; Sc+ (3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV  Zeff4s = 2.78; (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; 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

Implications on 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

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

Review over, back to quantum mechanics

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>

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

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

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

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

Consider some simple cases: nonuniqueness Starting with the wavefunction Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2) Consider the new spinorbitals θm and θe where θm = (cosa) ψm + (sina) ψe θe = (cosa) ψe - (sina) ψm Note that <θi|θj> = dij Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) = +(cosa)2 ψe(1)ψm(2) +(cosa)(sina) ψe(1)ψe(2) -(sina)(cosa) ψm(1) ψm(2) - (sina)2 ψm(1) ψe(2) -(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) ψm ψe a θm θe a a Thus linear combinations of the spinorbitals do not change Ψ(1,2)

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.

The antisymmetrized wavefunction Now put the spinorbitals into the matrix and take the determinant Where the antisymmetrizer can be thought of as the determinant operator. Similarly starting with the 3!=6 product wavefunctions of the form The only combination satisfying the Pauil Principle is

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

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 ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2 Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0

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

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 = haa + hbb + Jab + 1/R

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

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

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

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.

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

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)]= 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

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 3Eab = 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

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

Re-examine He atom with spin and the Pauli Principle Ψ(1,2) = A[(φ1s a) (φ1s b)] = 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

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 Jab is the Coulomb repulsion between densities ra=[ψa(1)]2 and rb

The energy for an antisymmetrized product, Aψaψb 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 One new term from the antisymmetrizer

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 ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2 Since the integrand ra(1) rb(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

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

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

Starting point for First Principles QM Energy = Kinetic energy + Potential energy Kinetic energy = Potential energy = Nucleus-Nucleus repulsion Nucleus-Electron attraction Electron-Electron atoms electrons p + Classical Mechanics 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

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 = Potential energy = atoms electrons Optimize Ψ, get HelΨ=EΨ Hel =

Schrodinger Equation Hel = HelΨ=EΨ 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

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

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

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

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.

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.

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

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

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

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**

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

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 organometallics QChem: very fast for organics Gaussian: many analysis tools GAMESS HyperChem ADF Spartan/Titan

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

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

The energy diagram for N2 TAs put energies of 7 occupied orbitals plus lowest 2 unoccupied orbitals, use correct symmetry notation

The HF orbitals of H2O TAs put energies of 5 occupied orbitals plus lowest 2 unoccupied orbitals, use correct symmetry notation Show orbitals

The HF orbitals of ethylene TAs put energies of 8 occupied orbitals plus lowest 2 unoccupied orbitals, use correct symmetry notation Show orbitals

The HF orbitals of benzene TAs put energies of 21 occupied orbitals plus lowest 4 unoccupied orbitals, use correct symmetry notation Show orbitals

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.

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

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