CHEM 580Week 1: From Schrodinger to Hartree-FockAUG 25TH –AUG 27TH

EMILIE GUIDEZ: EGUIDEZ@IASTATE.EDU

201 SPEDDING HALL

Intro to the Schrodinger equation (SE)

The objective of any quantum chemistry problem is to solve the SE. The time-independent SE (no time dependent external forces) has the form:

is the Hamiltonian operator. is the energy of the system (eigenvalue) is the wavefunction that contains all the information you need about the system.(eigenfunction)

Operator An operator is a mathematical construct that performs some operation on a function.

Examples:1) Let and . What is ?

2) Let . What is ?

The Hamiltonian The Hamiltonian in the SE has the following form:

Kinetic energy operator of electron i

Potential energy operator between proton A and electron i

Potential energy operator between electrons i and j

Atomic units The SE is often solved in atomic units. Atomic units definition:

SI units Atomic unitsMass of the electron (me)

9.109382*10-31 kg 1

Proton charge e 1.6021765*10-19 C 1

ћ 1

4πε0 1

Energy unit:

Length unit: Å

Problem What is the C-H distance in methane in Bohr? (d(C-H)=1.09Å)?

What is the bond dissociation energy? (E=105 kcal/mol)

The Hamiltonian The Hamiltonian in the SE has the following form:

The SE cannot be solved exactly because of the term. For what system can we solve the SE exactly?

The hydrogen atom

The wavefunction solution to the SE for the hydrogen atom defining the orbitals is given by:

Radial function Spherical harmonics

n=1,2,3… Principal quantum numberl=0,1,2…n-1 Angular quantum number|m|≤l Magnetic quantum number

The radial function and spherical harmonics

Real hydrogen-like wavefunctions

Spherical harmonics are complex. In order to work with real functions, we take linear combinations of the complex spherical harmonics.

Energy levels of the Hydrogen atom

Z= Number of protons in the nucleusµ= reduced mass (kg): memp/(me+mp)e= charge of the proton (C)n= principal quantum numberh= Plank’s constant (J.s)ε0=permittivity (C2.N-1.m-2)

In SI units, The energy levels of the hydrogen atom are given by:

In atomic units, the energy levels of the hydrogen atom are given by:

Note: The energy depends on n only. Not on l and m. Therefore, states with the same n and different l and m are degenerate.

Problem 1) What is ground state energy (n=1) of the hydrogen atom in atomic units?

2) Convert the ground state energy of hydrogen to kcal/mol and eV. (1hartree=627.51 kcal/mol, 1hartree=27.21 eV)

The energy to remove the electron from the hydrogen atom (ionization potential) is 13.6 eV.

The Helium atom

The SE equation cannot be solved exactly due to the electron repulsion in the Hamiltonian. (there is no analytical solution for this 3-body problem, this equation is non-separable). What can we do?

1) Ignore the electron repulsion (independent particle model)

The Hamiltonian is the sum of two Hydrogen-like Hamiltonians that we can solve exactly:

Problem What is the ground state energy (n=1) of the He atom in the independent particle model?

The experimental value is -2.9 Hartree, which is a difference of 690 kcal/mol!!!Electron correlation is important!!

The Helium atom 2) Use independent particle wavefunction and correct the Hamiltonian

Unperturbed Hamiltonian. We know the exact solution to the SE: (E0=-4hartree)

Small perturbation

Using perturbation theory we can derive a correction to E0 due to electron repulsion:

The energy is now too positive since we did not take into account the change in the wavefunction due to the presence of the other electron (We used the H-like wavefunction).

The Helium atom 3) Variational method.

Expectation values (average values)

Quantum mechanics postulates that the only values that can be measured for the energy of a system are the eigenvalues of the Hamiltonian. If the wavefunction happens to be an eigenfunction Ψi of the Hamiltonian with eigenvalue Ei, we are sure to get Ei when we measure the energy. Otherwise, we will get one of the eigenvalues of the Hamiltonian when we measure the energy but we can’t predict which.

The expectation value of the energy is the average of the measured energy values E1, E2, E3… of identical systems 1,2,3 in the same state Ψ.

The Helium atom The expectation value can be expressed as:

Variation theorem: The expectation value of the energy for any normalized trial wavefunction Ф is an upper bound to the true ground state energy E0.

Trial function for He:Effective atomic number

The Helium atom The value ζ that minimizes the energy is 1.69< Z=2. The resulting minimized energy is -2.848 hartree (much closer to the experimental value of -2.9 hartree).

The smaller value ζ reflects the screening due to electron repulsion. However, this screening concept does not always work. For instance, for the hydrogen atom in H2, ζ actually increases.

Spin and antisymmetry

Postulate 1: The wavefunction of a system of identical interacting particles must not distinguish among the particles.

Postulate 2: The total wavefunction of a system of electrons must be antisymmetric with respect to interchange of any two electrons. For instance, for a two-electron system:

ΨHe satisfies postulate 1 but does not satisfy postulate 2 (it is symmetric). What are we missing here?

Spin and antisymmetry For the Hydrogen atom:

α(ms=1/2); β(ms=-1/2)

For the Helium atom: We need a spin function to multiply by:

What are the possible spin functions?

Spatial orbitalSpin-orbital

Spin and antisymmetry1) Which do not satisfy postulate 1?

2) Are the spin functions left symmetric or antisymmetric?

3) What is the symmetry of the resulting total wavefunction when considering the spin functions from 2)? Is postulate 2 satisfied?

Answer: 1) α(1)β(2) and α(2)β(1) 2) Symmetric3)

Symmetric*Symmetric=Symmetric Postulate 2 not satisfied

Spin and antisymmetry What can we do?

Take linear combinations!

1) Do these spin function satisfy postulate 1?2) Are these spin functions symmetric or antisymmetric?3) What is the total wavefunction and is postulate 2 satisfied?

1) Yes!!2) is symmetric, is antisymmetric

3) Total wavefunction:

Symmetric, postulate 2 not satisfied

Antisymmetric, postulate 2 satisfied!!

Slater determinant The total wavefunction for the ground state of Helium is:

It can be written as a determinant:

A general expression for the Slater determinant:

Slater determinant Write a Slater determinant for the Be atom (Z=4).

Pauli exclusion principle What happens if we put the two electrons in a 1s orbital and give them both an α spin?

The wavefunction becomes equal to 0, which means that two electrons can never occupy the same spin-orbital (can never be assigned the same 4 quantum number n,l,m and ms)

He excited states Let’s consider the excitation of one of the 1s electrons of He to the 2s orbital:

What is the problem with these two wavefunctions?

Electrons are not undistinguishable (postulate 1 not satisfied)!!

He excited states What can we do?

Take linear combinations!!

Are these wavefunctions symmetric or antisymmetric?

Symmetric

antisymmetric

He excited states The 4 possible spin functions that satisfy postulate 1:

Which spin functions do we need to satisfy postulate 2 for the excited state of Helium?

Number 3 only

Number 1,2 or 4

Multiplicity S We have 4 possible excited state wavefunctions for Helium:

Degenerate2S+1=3S=1 (triplet)

2S+1=1S=0 (singlet)

Hartree-Fock theory Let us consider a closed-shell system (all occupied orbitals have 2 electrons):

Doubly occupied space

Virtual space ϕυ

ϕμ

The wavefunction of this system can be written as a Slater determinant:

We want to optimize the orbitals that minimize the total energy of the system according to the variation theorem.

Hartree-Fock theoryThe expression of the energy of a Slater determinant is given by:

1 electron term: Average kinetic energy and nuclear attraction energy.

Coulomb integral: Classical coulomb repulsion between electron clouds and

Exchange integral: Results of electron correlation (2 electrons with the same spin cannot occupy the same region of space).

Hartree-Fock theory Minimization of the energy:

How do we optimize the orbitals of the doubly occupied space?

ϕυ

ϕμ

Mix occupied with unoccupied orbitals!

The minimization of the energy can be done by solving the Hartree-Fock equation:

Average field seen by the electron.

Energy of orbital

Hartree-Fock theory The orbitals solution to the Hartree-Fock equations ( ) are called canonical orbitals.

The Fock operator depends on all occupied orbitals ( because of the exchange and Coulomb operators). Therefore, a specific orbital can only be determined if all the others are known.

One must use iterative methods to solve the HF equations (Self-consistent field method)

Koopman’s theorem

ϕυ

ϕμ

If we assume that orbitals are frozen, the energy required to remove an electron from a closed-shell atom or molecule (ionization potential IP) can be approximated by –ε of the orbital from which the electron is removed. The electron affinity (EA) for adding an electron to a virtual orbital is the negative of the energy of that virtual orbital.

εc

εr

Excited statesKoopman’s theorem does not consider optimization of the orbitals when their occupation changes (frozen orbitals). The energy needed to excite an electron from the occupied orbital i to the unoccupied orbital a is given by:

Singlet and triplet excited states are degenerate. Some methods like Huckel theory use Koopman’s theorem (crude results).

εi

εa

The expression for the singlet and triplet excited state energy are given by:

HF open shells We have only considered closed-shell systems so far (doubly occupied orbitals). Not all systems are closed shells, some are open-shell (singly occupied orbital). There are 2 ways to deal with open-shell:

Restricted Unrestricted

Electrons that are paired occupy the same spatial orbital (closed-shell). The constraint of occupying orbitals in pairs raises the variational energy.

The spatial orbital of two paired electrons are different. Problem: Spin contamination (wave function contaminated by higher multiplicity components).

Hartree-Fock theory•Hartree Fock does not contain any electron correlation except the Fermi hole (two electrons of parallel spin cannot occupy the same orbital). This class will present post HF methods (a.k.a correlation methods) to evaluate the correlation energy.

•Hartree-Fock is a single reference method (wavefunction is described by a single determinant). Some problems may require multiple determinants (multireference methods). Example: Bond breaking.

Linear combination of atomic orbitals

• Most molecules do not have a spherically symmetric potential like atoms do. The Fock operator can then be averaged over all angles to remove its angular dependence. As a result, we can use spin-orbitals that are solutions to a spherically symmetric potential (a.k.a. s,p,d … orbitals). This is called the central field approximation.

• One can attempt to solve the HF equations numerically. This can be done for atoms (spherically symmetric systems) but not for molecules. One way to work around this problem is to expend the molecular HF orbitals into a known basis set χ.

i=1,2,…K

Basis functions Number of basis functions

This expression is exact if the basis set is complete (which would basically require an infinite number of basis functions).

Linear combination of atomic orbitals

For practical reasons, one cannot you a complete basis set. A common approximation is to use a linear combination of atomic orbitals (note: these orbitals are generally NOT a solution to the atomic HF equation). There are two main types of basis functions (chapter 5 of textbook):

◦ Slater-type orbitals (STOs)

Spherical harmonics Radial function

Advantage: Fairly rapid convergence as the number of functions increases.Disadvantage: Most integrals required to solve HF equations cannot be solved analytically except for diatomics. No radial nodes.

Linear combination of atomic orbitals

o Gaussian-type orbitals (GTOs)

Advantage: Can compute integrals analytically and efficiently.Disadvantages: Shape of the orbital not as good as STO. No cusp at the nucleus (they are flat), which messes up NMR and ESR properties. GTOs fall off too fast in the valence region ( One needs more GTOs than STOs to describe the molecular orbitals)

STOs vs GTOs

STO

GTO

Basis sets We saw that we can expend molecular orbitals into a linear combinations of atomic orbitals (STOs or GTOs). The number of atomic orbitals in the extension (K) defines your basis set.

Minimum basis set: Only have enough functions to contain all the electrons of the neutral atom. For H2 for instance, one can use two 1s functions of the H atoms. One needs good functions to get decent description of the orbitals. Slater orbitals are preferred for minimal basis set. One can also use a combination of GTOs to fit one STO (STO-NG).

Basis sets

Double zeta basis set: Double the number of basis functions for each electron. For instance, one can use two functions (1s and 1s’) for each H atom (4 functions total).

Triple zeta basis set: Triple the number of basis functions.