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 Ch2a levelCh121a 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.
Motivation: Design Materials, Catalysts, Pharma from 1st Principles so can do design priorprior to experiment
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
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.
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.
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 dynamicsTight Binding for electronic propertiessolvation, diffusion, mesoscale force fields
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 locations1st term: kinetic energy operator HOW MANY 2nd term: attraction of electrons to nuclei: HOW MANY3rd term: electron-electron repulsion HOW MANY4th term: what is missing?
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 locations1st term: kinetic energy operator: 42 terms 2nd term: attraction of electrons to nuclei: HOW MANY3rd term: electron-electron repulsion HOW MANY4th term: what is missing?
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 locations1st term: kinetic energy operator: 42 terms 2nd term: attraction of electrons to nuclei: 42*12= 5043rd term: electron-electron repulsion HOW MANY4th term: what is missing?
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 locations1st term: kinetic energy operator: 42 terms 2nd term: attraction of electrons to nuclei: 42*12= 5043rd term: electron-electron repulsion 42*41/2= 861 terms4th term: what is missing?
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
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
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.
We have assumed a normalized wavefunction, <Φ|Φ> = 1
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
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 for spin up and for spin down. Thus the ground states of H atom are φ(xyz)(spin) and φ(xyz)(spin)
This it is invariant under 3 kinds of permutations
Space only:
Spin only:
Space and spin simultaneously: )
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
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 HartreeNote h0 = e2/a0 = 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/molThus 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
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 – Z/r is independent of θ and φ
x
z
yθ
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
The ground and excited states of a system can all be written as hφk = ekφk, where <φk |φj> = kj (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
Consider excited states with Znl = 1; these are ns states with l=0The 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.
The cross section is plotted along the z axis, but it would look exactly the same along any other axis. Here
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(-r)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 planerZ11 = x = r sinθ cosφ, which is zero in the yz planerZ1,-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 atomφnlm = 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 planeAlso 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
So far we have the s angular function Z00 = 1 with no angular nodal planesAnd three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with one angular nodal planeCan we form any more angular functions with one nodal plane orthogonal to all 4 of the above functions?
x
z
+
-
px’
x
z
+-
pzpx’
+ -
For example we might rotate px by an angle 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)
So far we have the s angular function Z00 = 1 with no angular nodal planesAnd three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with one angular nodal planeNext in energy will be the d functions with two angular nodal planes. We can easily construct three functions
x
z
+
-
dxz-
+
dxy = xy =r2 (sinθ)2 cosφ sinφ
dyz = yz =r2 (sinθ)(cosθ) sinφ
dzx = zx =r2 (sinθ)(cosθ) cosφ
where dxz is shown here
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,
Z20 = dz2 = [3 z2 – r2 ] m=0, dZ21 = dzx = zx =r2 (sinθ)(cosθ) cosφZ2-1 = dyz = yz =r2 (sinθ)(cosθ) sinφZ22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]Z22 = dxy = xy =r2 (sinθ)2 cosφ sinφWe find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2 = 360ºWhen this number, m=0 we call it a functionWhen m=1 we call it a functionWhen m=2 we call it a functionWhen m=3 we call it a function
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.
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 (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 Φ1sexp(-r)We find that the optimum = (Z – 5/16) = Zeff = 1.6875With this value of , the total energy is E= - 2 = -2.8477 h0
This wavefunction can be interpreted as two electrons moving independently in the orbital Φ1sexp(-r) 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=1.6875The total energy is just the sum of the individual energies, E = -2 (Zeff
2/2) = -2.8477 h0
Ionizing the 2nd electron, the 1st electron readjusts to = Z = 2with E(He+) = -Z2/2 = - 2 h0. Thus the ionization potential (IP) is 0.8477 h0 = 23.1 eV (exact value = 24.6 eV)
Consider adding the 3rd electron to the 2s orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(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
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
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
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
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
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> = jk= 1 if j=k
= 0 if j≠k
Note < Φj| Ĥ|Φk> = Ek < Φj|Φk> = Ek jk
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
= 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> = ij = 1 if i=j
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 PrincipleThus we will take determinants of our wavefunctions.
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
Interchanging electrons 1 and 3 leads to
Guaranteeing that the Pauli Principle is always satisfied
<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
The total energy is that of the product plus the exchange term which 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 >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>=0Thus all are zeroThus 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 isE = haa + hbb + (Jab –Kab) + 1/R
Energy for 2 electrons with same spinThe 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 followsKab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><(1)|(1)><(2)|(2)> ≡ Kab
where Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>Involves only spatial coordinates.
Now consider the exchange term for spin orbitals with opposite spinKab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><(1)|(1)><(2)|(2)> = 0Since <(1)|(1)> = 0.
Energy for 2 electrons with opposite spin
Thus the total energy is
E = haa + hbb + Jab + 1/R With no exchange term unless the spins are the same
Since <ψa|ψb>= 0 = < Φa| Φb><|> There is no orthogonality condition of the spatial orbitals for opposite spin electronsIn general we can have <Φa|Φb> =S, where the overlap S ≠ 0
The total energy is that of the product wavefunction plus the new terms arising from exchange term which 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 >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>=0Thus all are zeroThus 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 isE = haa + hbb + (Jab –Kab) + 1/R
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
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
Ψ(1,2,3,4) = A[(a)(a)(b)(b)]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 + φ1 the change in the energy isE = 4<φ1|h|φ1> + 4 <φ1|2J1-K1|φ1> + 4 <φ1|2J2-K2|φ1>
= 4 <φ1|HHF|φ1> Where HHF = h + Σj=1,2 [2Jj-Kj] is called the HF HamiltonianIn the above expression we assume that φ1 was normalized, <φ1|h|φ1> = 1.Imposing this constraint (a Lagrange multiplier) leads to<φ1|HHF – 1|φ1> = 0 and <φ2|HHF – 2|φ2> = 0Thus the optimum orbitals satisfy HHFφk = k φk the HF equations
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 orbitalsFirst consider the componets of the total energyΣ i=1,21< φi|h|φi> from the 21 up spin orbitalsΣ i=1,21< φi|h|φi> from the 21 down spin orbitalsΣ I<j=1,21 [Jij – Kij] interactions between the 21 up spin orbitalsΣ I<j=1,21 [Jij – Kij] interactions between the 21 down spin orbitalsΣ I≠j=1,21 [Jij] interactions of the 21 up spin orbitals with the 21 down spin orbitalsEnn = Σ A<B=1,12 ZAZB/RAB nuclear-nuclear repulsionCombining these terms leads toE = 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
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 writeE = Σ i=1,21 2< φi|h|φi> + Enn + Σ I≠j=1,21 [2Jij-Kij] + Σ I=1,21 [2Jii-Kii]and henceE = 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
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 2Also 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 + φ1 the change in the energy isE = 4<φ1|h|φ1> + 4 <φ1|2J1-K1|φ1> + 4 <φ1|2J2-K2|φ1>
= 4 <φ1|HHF|φ1> Where HHF = h + Σj=1,2 [2Jj-Kj] is called the HF HamiltonianIn the above expression we assume that φ1 was normalized, <φ1|h|φ1> = 1.Imposing this constraint (a Lagrange multiplier) leads to<φ1|HHF – 1|φ1> = 0 and <φ2|HHF – 2|φ2> = 0Thus the optimum orbitals satisfy HHFφk = k φk the HF equations
For the general case the HF closed shell equations areHHFφk = k φk where we solve for k=1,M occupied orbitalsHHF = 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 electronsProblem: sum over 2Jj leads to 2M Coulomb terms, not 2M-1This is because we added the self Coulomb and exchange termsBut (2Jk-Kk) φk = (Jk) φk so that these self terms cancel.
The optimum orbitals for the 4 electron closed shell wavefunctionΨ(1,2,3,4) = A[(a)(a)(b)(b)]Are eigenstate of the HF equationsHHFφk = k φk for k=1,2where 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 energySince (Jk-Kk)φk = 0 there spurious terms cancel.
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) = Σ C Xwhere the basis functions, {XMBF} are chosen as atomic like functions on the various centers
As a result the HF equations HHFφk = k φk
Reduce to a set of Matrix equations
ΣjmHjmCmk = ΣjmSjmCmkEk
This is still complicated since the Hjm operator includes Exchange terms
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
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** 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
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
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
