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Transcript of Lecture 1Ch121a-Goddard-L01© copyright 2011 William A. Goddard III, all rights reserved\1 CH121a...
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 1
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.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 2
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 3
Lectures
The lectures cover the basics of the fundamental methods: quantum mechanics, force fields, molecular dynamics, Monte Carlo, statistical mechanics, etc. required to understand the theoretical basis for the simulations
the homework applies these principles to practical problems making use of modern generally available software.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 4
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.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 5
Methods to be covered in the lectures include:
Quantum Mechanics: Hartree Fock and Density Function methods Force Fields standard FF commonly used for simulations of organic, biological, inorganic, metallic systems, reactions; ReaxFF reactive force field: for describing chemical reactions, shock decomposition, synthesis of films and nanotubes, 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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 6
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 dockingDNA-structure, ligand docking
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 7
OutlineTopic 1: QM: HF, DFT, basis sets, reactions, transition states, vibrations Topic 2: Force Fields, nonbonds, hydrogen bonds, charges (QEq), QM FFTopic 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 tensionTopic 5: polymers: crystalline, amorphous, structure prediction, Topic 6: ReaxFF Reactive Force Field and reactive DynamicsTopic 7: PBC QM, band structure, phonons, elastic constants, Ab Initio MD, Topic 8: surfaces: reconstruction, chemisorption, physisorption, solvationTopic 9: eFF, Tight bindingTopic 9: applications:
Fuel Cell: oxygen reduction reaction, migration in Nafion solid oxidesBatteries: anode for Li battery, Solid electrolyte interface, Nanotechnology: rotaxane molecular switches, carbon nanotube interfacesWater Treatment: Dendrimers, captymersCatalysis: alkane activation, ammoxidationMetallic alloys: force fields, dislocations, plasticity, cavitationProteins: structures, ligand docking, GPCRsDNA: A to B transition, Origami based nanotechnology
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 8
Topic 1: Practical Quantum Chemistry
Solve Schrödinger Equation
Hel = HelΨ=EΨ
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?
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 9
The Schrödinger Equation: Kinetic Energy
Solve Schrödinger Equation
Hel = HelΨ=EΨ
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?
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 10
The Schrödinger Equation: Nuclear-Electron
Solve Schrödinger Equation
Hel = HelΨ=EΨ
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?
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 11
The Schrödinger Equation: Electron-Electron
Solve Schrödinger Equation
Hel = HelΨ=EΨ
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?
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 12
The Schrödinger Equation: Nuclear-Nuclear
Solve Schrödinger Equation
Hel =
HelΨ=EΨ
Missing is the nuclear-nuclear repulsion
Enn A<B ZA*ZB/RAB
This does not depend on electron coordinates, but it does affect the total energyEel = <Ψ|Hel|Ψ>/<Ψ|Ψ>=Etotal = Eel + Enn
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 13
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 14
Quantum Mechanics – First postulate
The essential element of QM is that all properties that can be known about the system are contained in the wavefunction, Φ(x,y,z,t) (for one electron), where the probability of finding the electron at position x,y,z at time t is given by
P(x,y,z,t) = | Φ(x,y,z,t) |2 = Φ(x,y,z,t)* Φ(x,y,z,t)
Note that ∫Φ(x,y,z,t)* Φ(x,y,z,t) dxdydz = 1
since the total probability of finding the electron somewhere is 1.
I write this as < Φ|Φ>=1, where it is understood that the integral is over whatever the spatial coordinates of Φ are
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 15
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/rR
_ R_
KEQM = (Ћ2/2me) <(Φ·Φ>
where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz
In QM the KE is proportional to the average square of the gradient or slope of the wavefunction
Thus KE wants smooth wavefunctions, no wiggles
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 16
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
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 17
3rd Postulate of QM, the variational principle
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> ≥ Eex
EexEap
EThis means that for sufficiently small Φ, E = 0 for all possible changes, Φ
We write E/Φ = 0 for all Φ
This is called the variational principle.
For the ground state, E/Φ ≥ 0 for all possible changes
The ground state wavefunction is the system, Φ, has the lowest possible energy out of all possible wavefunctions.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 18
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(i) = ei
ei HΦ = H (ei Φ) = E (ei Φ)
Thus Φ and (ei 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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 19
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
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)
B=0 Increasing B
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 20
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:
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 21
Permutational symmetries for H2 and He
H2
He
Have 4 degenerate g ground states for H2
Have 4 degenerate u excited states for H2
Have 4 degenerate ground state for He
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 22
Permutational symmetries for H2 and He
H2
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 23
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
ij Ψ = - Ψ for all I and j
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 24
Consider H atom
The Hamiltonian has the form
h = - (Ћ2/2me)– Ze2/r
In atomic units
h = - ½ – Z/r
r
+Ze
We will consider one electron, but a nucleus with charge Z
φnlm = Rnl(r) Zlm(θ,φ)
Thus we want to solve
Hφk = ekφk for the ground and excited states k
where Rnl(r) depends only on r and Zlm(θ,φ) depends only on θ and φ
Assume φ10 = exp(-r) E = ½ – Z
= – Z = 0 = Z
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 25
The H atom ground state
the ground state of H atom is
φ1s = N0 exp(-Zr/a0) ~ exp(-Zr) where we will ignore normalization
Line plot along z, through the origin
Maximum amplitude at z = 0
1
x = 0
Contour plot in the xz plane, Maximum amplitude at x,z = 0,0
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 26
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 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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 27
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> = 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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 28
The excited states of H atom - 2
Use spherical polar coordinates, r, θ, φ
where z = rcosθ, x = rsinθcosφ, y = rsinθsinφ
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
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 29
The excited states of H atom - 3
Use spherical polar coordinates, r, θ, φ
where z=r cosθ, x=r sinθ cosφ, y=r sinθ sinφ
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 30
Excited radial functions
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
R20 = 2s = [Zr/2 – 1] exp(-Zr/2)R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1] exp(-Zr/3)
Zr = 2 Zr = 1.9
Zr = 7.1
0 nodal planes
1 spherical nodal plane
2 spherical nodal planes
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 31
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(-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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 32
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
x
z
+-
px
x
z
+
-
pxpz-
+
pz
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 33
More p functions?
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)
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 34
More angular functions?
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,
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 35
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
More d angular functions?
x
z
+
-
dz2-x2
-+
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 36
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
More d angular functions?
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 37
Summarizing the d angular functions
x
z
+
-
dz2
-+
57º
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
m = 1, d
m = 2, d
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 38
Summarizing the angular functions
So far we have
•one s angular function (no angular nodes) called ℓ=0
•three p angular functions (one angular node) called ℓ=1
•five d angular functions (two angular nodes) called ℓ=2
Continuing we can form
•seven f angular functions (three angular nodes) called ℓ=3
•nine g angular functions (four angular nodes) called ℓ=4
where ℓ is referred to as the angular momentum quantum number
And there are (2ℓ+1) m values for each ℓ
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 39
real (Zlm) and complex (Ylm) ang. momentum fnctns
Here the bar over m negative
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 40
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
1s 0 0 0 1.02s 1 1 0 4.02p 1 0 1 4.03s 2 2 0 9.03p 2 1 1 9.03d 2 0 2 9.04s 3 3 0 16.04p 3 2 1 16.04d 3 1 2 16.04f 3 0 3 16.0
nam
e
tota
l nod
al p
lane
sra
dial
nod
al p
lane
san
gula
r no
dal p
lane
s
R̄ R̄
R̄
Siz
e (a
0)
This is all you need to remember
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 41
Sizes hydrogen orbitals
Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f
Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48
H--H C
0.74
H
H
H
H
1.7
H H
H H
H H
4.8
=a0 n2/ZR̄ Where a0 = bohr = 0.529A=0.053 nm = 52.9 pm
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 42
Hydrogen atom excited states
1s-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
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Plotting of orbitals: line cross-section vs. contour
contour plot in yz plane
line plot along z axis
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Contour plots of 1s, 2s, 2p hydrogenic orbitals
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Contour plots of 3s, 3p, 3d hydrogenic orbitals
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Contour plots of 4s, 4p, 4d hydrogenic orbtitals
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Contour plots of hydrogenic 4f orbitals
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 48
He+ atom
Next lets review the energy for He+.
Writing Φ1sexp(-r) we determine the optimum for He+ as follows
<1s|KE|1s> = + ½ 2 (goes as the square of 1/size)
<1s|PE|1s> = - Z(linear in 1/size)
E(He+) = + ½ 2 - Z
Applying the variational principle, the optimum must satisfy dE/d = -Z = 0 leading to =Z,
KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.
writing PE=-Z/R0, we see that the average radius is R0=1/
So that the He+ orbital is ½ the size of the H orbital
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 49
Estimate J1s,1s, the electron repulsion energy of 2 electrons in He+ 1s orbitals
How can we estimate J1s,1s
Assume that each electron moves on a sphere
With the average radius R0 = 1/
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
A rigorous calculation gives
J1s,1s = (5/8) h0 = 8.5036 eV = 196.1 kcal/mol
R0
e1
e2
Now consider He atom: EHe = 2(½ 2) – 2ZJ1s,1s
Since 1s = -Z2/2 = -2 h0 = 54.43 eV = 1,254.8 kcal/mol the electron repulsion increases the energy (less attractive) by 15.6%
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 50
The optimum atomic orbital for He atom
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 (from memory, TA please check). Thus this simple approximation of assuming that each electron is in a 1s orbital and optimizing the size accounts for 98.1% of the exact result.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 51
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)
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 52
Now lets add a 3rd electron to form Li
ΨLi(1,2,3) = A[(Φ1s)(Φ1s(Φ1s
Problem: with either or , 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[(Φ1s)(Φ1s)(Φ2s
or
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(or 2px or 2py)
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First consider Li+
First consider Li+ with ΨLi(1,2) = A[(Φ1s)(Φ1s
Here Φ1sexp(-r) with = Z-0.3125 = 2.69 and
E = -2 = -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/ = 0.372 a0 = 0.2A
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 54
Consider adding the 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 55
First consider Li+
First consider Li+ with ΨLi(1,2) = A[(Φ1s)(Φ1s
Here Φ1sexp(-r) with = Z-0.3125 = 2.69 and
E = -2 = -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/ = 0.372 a0 = 0.2A
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 56
Consider adding the 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 57
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
0.2A
1s
2.12A2s
R~0.2A
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 58
Li atom excited states
1s
2s
-0.125 h0 = -3.4 eV2p
Energy
zero
-0.205 h0 = -5.6 eV
-2.723 h0 = 74.1 eV
MO picture State picture
(1s)2(2s)
(1s)2(2p)
E = 2.2 eV17700 cm-1
564 nm
Ground state
1st excited state
Exper671 nm
E = 1.9 eV
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Aufbau principle for atoms
1s
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
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He, 2
Ne, 10
Ar, 18
Zn, 30
Kr, 36
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Many-electron configurations
General aufbau
ordering
Particularly stable
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General trends along a row of the periodic table
As we fill a shell, say B(2s)2(2p)1 to Ne (2s)2(2p)6
we add one more proton to the nucleus and one more electron to the valence shell
But the valence electrons only partially shield each other.
Thus Zeff increases, leading to a decrease in the radius ~ n2/Zeff
And an increase in the IP ~ (Zeff)2/2n2
Example Zeff2s=
1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne
Thus (2s Li)/(2s Ne) ~ 4.64/1.28 = 3.6
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 63
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
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Plot of rφ(r) for the outer s
valence orbital
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 65
Plot of rφ(r) for the outer s and
p valence orbitals
Note for C row 2s and 2p have similar size, but for other rows
ns much smaller than np
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 66
Plot of rφ(r) for the outer s and p valence
orbitals
Note for C row 2s and 2p have similar size, but for other
rows ns much smaller than np
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 67
Transition metals; consider [Ar] + 1 electron
[IP4s = (Zeff4s )2/2n2 = 4.34 eV Zeff
4s = 2.26; 4s<4p<3d
IP4p = (Zeff4p )2/2n2 = 2.73 eV Zeff
4p = 1.79;
IP3d = (Zeff3d )2/2n2 = 1.67 eV Zeff
3d = 1.05;
IP4s = (Zeff4s )2/2n2 = 11.87 eV Zeff
4s = 3.74; 4s<3d<4p
IP3d = (Zeff3d )2/2n2 = 10.17 eV Zeff
3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV Zeff
4p = 3.20;
IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff
3d = 4.05; 3d<4s<4p
IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff
4s = 5.04;
IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff
4p = 4.47;
K
Ca+
Sc++
As the net charge increases the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4Thus charged system prefers 3d vs 4s
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 68
Transition metals; consider Sc0, Sc+, Sc2+
3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff
3d = 4.05;
4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff
4s = 5.04;
4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff
4p = 4.47;
Sc++
As increase net charge increases, the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4. Thus M2+ transition metals always have all valence electrons in d orbitals
(3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV Zeff
4s = 3.89;
(3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV Zeff
3d = 2.85;
(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV Zeff
4p = 3.37;
Sc+
(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV Zeff
4s = 2.78;
(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV Zeff
3d = 1.84;
(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV Zeff
4p = 2.32;
Sc
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 69
Implications on transition metals
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
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Transtion metal valence ns and (n-1)d orbitals
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Review over, back to quantum mechanics
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 72
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> = 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
Ө = k Ck Φk where <Φj| Ө>=Cj or Ө = k Φk <Φk| Ө>
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 73
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
12 Ψs(1,2) = Ψs(2,1) = Ψs(1,2) (boson symmetry)
12 Ψa(1,2) = Ψa(2,1) = -Ψa(1,2) (Fermion symmetry)
Thus for electrons, the Pauli Principle only allows the antisymmetric combination for two independent electrons
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 74
Implications of the Pauli Principle
Consider two independent electrons,
1 on the earth described by ψe(1)
and 2 on the moon described by ψm(2)
Ψ(1,2)= ψe(1) ψm(2)
And test whether this satisfies the Pauli Principle
Ψ(2,1)= ψm(1) ψe(2) ≠ - ψe(1) ψm(2)
Thus the Pauli Principle does NOT allow the simple product wavefunction for two independent electrons
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 75
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 = eiψe where is a constant phase factor, we still get zero
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 76
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 + 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) + ψe(1) ψe(2) - θm(1) ψe(2) - 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> = ij = 1 if i=j
=0 if i≠j
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 77
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 = (cos) ψm + (sin) ψe
θe = (cos) ψe - (sin) ψm Note that <θi|θj> = ij
Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) =
+(cos)2 ψe(1)ψm(2) +(cos)(sin) ψe(1)ψe(2)
-(sin)(cos) ψm(1) ψm(2) - (sin)2 ψm(1) ψe(2)
-(cos)2 ψm(1) ψe(2) +(cos)(sin) ψm(1) ψm(2)
-(sin)(cos) ψe(1) ψe(2) +(sin)2 ψe(1) ψm(2)
[(cos)2+(sin)2] [ψe(1)ψm(2) - ψm(1) ψe(2)] =Ψold(1,2) Thus linear combinations of the spinorbitals do not change Ψ(1,2)
ψe
ψm
θe
θm
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 78
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 PrincipleThus we will take determinants of our wavefunctions.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 79
The antisymmetrized wavefunction
Where the antisymmetrizer can be thought of as the determinant operator.
Similarly starting with the 3!=6 product wavefunctions of the form
Now put the spinorbitals into the matrix and take the determinant
The only combination satisfying the Pauil Principle is
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 80
Example:
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 81
Consider the product wavefunction
Ψ(1,2) = ψa(1) ψb(2)
And the Hamiltonian for H2 molecule
H(1,2) = h(1) + h(2) +1/r12 + 1/R
In the details slides next, we derive
E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>
E = haa + hbb + Jab + 1/R
where haa =<a|h|a>, hbb =<b|h|b> are just the 1 electron energies
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12
represents the total Coulomb interaction between the electron density a(1)=| ψa(1)|2 and b(2)=| ψb(2)|2
Since the integrand a(1) b(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0
Energy for 2-electron product wavefunction
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 82
Details in deriving energy: normalization
First, the normalization term is
<Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)>
Which from now on we will write as
<Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi are normalized
Here our convention is that a two-electron function such as <Ψ(1,2)|Ψ(1,2)> is always over both electrons so we need not put in the (1,2) while one-electron functions such as <ψa(1)|ψa(1)> or <ψb(2) ψb(2)> are assumed to be over just one electron and we ignore the labels 1 or 2
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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
Details of deriving energy: one electron termss
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 84
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 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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 85
The energy of the antisymmetrized wavefunction
The total electron-electron repulsion part of the energy for any wavefunction Ψ(1,2) must be positive
Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0
This follows since the integrand is positive for all positions of r1 and r2 then
We derived that the energy of the A ψa ψb wavefunction is
E = haa + hbb + (Jab –Kab) + 1/R
Where the Eee = (Jab –Kab) > 0
Since we have already established that Jab > 0 we can conclude that
Jab > Kab > 0
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 86
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)(1)][Φb(2)(2)]= [Φa(1)Φb(2)][(1)(2)]
So that the antisymmetrized wavefunction is
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]=
=[Φa(1)Φb(2)- Φb(1)Φa(2)][(1)(2)]
Also, similar results for both spins down
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]=
=[Φa(1)Φb(2)- Φb(1)Φa(2)][(1)(2)]Since <ψa|ψb>= 0 = < Φa| Φb><|> = < Φa| Φb>We see that the spatial orbitals for same spin must be orthogonal
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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.
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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
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Summarizing: Energy for 2 electronsWhen the spinorbitals have the same spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]
The total energy is
E = haa + hbb + (Jab –Kab) + 1/R
But when the spinorbitals have the opposite spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]=
The total energy is
E = 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
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Consider further the case for spinorbtials with opposite spin
The wavefunction
[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)+(1)(2)]
Leads directly to 3E = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for
[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)]
[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)]
These three states are collectively referred to as the triplet state and denoted as having spin S=1
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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)][(1)(2)-(1)(2)]
For the ground state of a 2-electron system, Φa=Φb so we get [Φa(1)Φa(2)][(1)(2)-(1)(2)] = A[Φa(1)(1)] [Φa(2)(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
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Re-examine He atom with spin and the Pauli Principle
Ψ(1,2) = A[(φ1s (φ1s ]
= 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
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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
Energy for 2-electron product wavefunction
Jab is the Coulomb repulsion between densities a=[ψa(1)]2 and b
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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 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
One new term from the antisymmetrizer
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Summary electron-electron energies
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=<ψa(1)| Jb (1)|ψa(1)>
is the total Coulomb interaction between the electron density a(1)=| ψa(1)|2 and b(2)=| ψb(2)|2
Since the integrand a(1) b(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0
Here Jb (1) = ʃ [ψb(2)]2/r12 is the potential at 1 due to the density distribution [ψb(2)]2
Kab=< ψaψb|1/r12|ψb ψa >= ʃ d3r1[ψa(1)ψb(1)] ] ʃd3r2[ψb(2) ψa(2)]]2/r12 = <ψa(1)| Kb (1)|ψa(1)>
Where Kb (1) ψa(1)] ] = ψb(1) ʃ [ψb(2)ψa(2)]2/r12 is an integral operator that puts Kab into a form that looks like Jab. The difference is that Jb (1) is a function but Kb (1) is an operator
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Alternative form for electron-electron energies
It is commen to rewrite Jab as Jab ≡ [ψa(1) ψa(1)|ψb(2)ψb(2)] where all the electron 1 stuff is on the left and all the electron 2 stuff is on the right. Note that the 1/r12 is now understood
Similarly Kab= [ψa(1)ψb(1)|ψb(2)ψa(2)]
Here the numbers 1 and 2 are superflous so we write
Jab ≡ [ψaψa|ψbψb] = [aa|bb] since only the orbital names are significant
Siimilarly
Kab ≡ [ψaψb|ψbψa] = [ab|ba]
Thus the total 2 electron energy is
Jab - Kab = [aa|bb] - [ab|ba]
But if a and b have opposite spin then the exchange term is zero
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 97
Consider the case of 4 e in 2 orbitals, a,b
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
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Starting point for First Principles QM
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
Makes for dull world
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 99
Ab Initio, quantum mechanics
Starting point for First Principles
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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 100
HelΨ=EΨ
Hel =
Schrodinger Equation
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
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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 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
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 102
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 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
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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 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
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The general case of 2M electrons
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.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 105
Analyze HF equations
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.
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 106
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
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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) = 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
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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
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Double zeta + polarization Basis sets – 6-31G**
To allow the atomic orbitals to contract as atoms are brought together to form bonds, we introduce 2 basis functions of the same character as each of the atomic orbitals:Thus 2 each of 1s, 2s, 2px, 2py, and 2pz for CThis is referred to as double zeta. If properly chosen this leads to a good description of the contraction as bonds form.Often only a single function is used for the C1s, called split valenceIn addition it is necessary to provide one level higher angular momentum atomic orbitals to describe the polarization involved in bondingThus add a set of 2p basis functions to each H and a set of 3d functions to each C. The most popular such basis is referred to as 6-31G**
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 110
6-31G** and 6-311G**
6-31G** means that the 1s is described with 6 Gaussians, the two valence basis functions use 3 gaussians for the inner one and 1 Gaussian for the outer function
The first * use of a single d set on each heavy atom (C,O etc)
The second * use of a single set of p functions on each H
The 6-311G** is similar but allows 3 valence-like functions on each atom.
There are addition basis sets including diffuse functions (+) and additional polarization function (2d, f) (3d,2f,g), but these will not be relvent to EES810
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 111
Effective Core Potentials (ECP, psuedopotentials)
For very heavy atoms, say starting with Sc, it is computationally convenient and accurate to replace the inner core electrons with effective core potentials
For example one might describe: • Si with just the 4 valence orbitals, replacing the Ne core with
an ECP or • Ge with just 4 electrons, replacing the Ni core • Alternatively, Ge might be described with 14 electrons with the
ECP replacing the Ar core. This leads to increased accuracy because the
• For transition metal atoms, Fe might be described with 8 electrons replacing the Ar core with the ECP.
• But much more accurate is to use the small Ne core, explicitly treating the (3s)2(3p)6 along with the 3d and 4s electrons
Lecture 1Ch121a-Goddard-L01 © copyright 2011 William A. Goddard III, all rights reserved\ 112
Software packages
Jaguar: Good for organometallicsQChem: very fast for organicsGaussian: many analysis toolsGAMESSHyperChemADFSpartan/Titan
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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
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Pi orbitals of benzene
Top view
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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
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The energy diagram for N2
TAs put energies of 7 occupied orbitals plus lowest 2 unoccupied orbitals, use correct symmetry notation
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The HF orbitals of H2O
TAs put energies of 5 occupied orbitals plus lowest 2 unoccupied orbitals, use correct symmetry notation
Show orbitals
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The HF orbitals of ethylene
TAs put energies of 8 occupied orbitals plus lowest 2 unoccupied orbitals, use correct symmetry notation
Show orbitals
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The HF orbitals of benzene
TAs put energies of 21 occupied orbitals plus lowest 4 unoccupied orbitals, use correct symmetry notation
Show orbitals
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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.
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Configuration interaction
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| i Cj j >/ < i Ci i | i Cj j >How choose optimum Ci?Require E=0 for all Ci get
j <i |H| Cj j > - Ei< i | Cj j > = 0 ,which we write asHCi = 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
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Configuration interaction upper bound theorm
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