Components of the Atom by VII9jovw

VIEWS: 0 PAGES: 58

• pg 1
```									     Chapter 9

Many-Electron Atoms

Chapter 9: Slide 1
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 2
The Hamiltonian for Multielectron Atoms
Helium
2 2         2 2           2e 2    2e 2     e2
SI Units:   H     1 (r1 )      2 (r2 )                                Z=2
2m            2m             40 r1 40 r2 40 r12

1 2  1               2 2 1
Atomic Units:      H   1 (r1 )   2 (r2 )   
2
2          2           r1 r2 r12
Multielectron Atoms
N 1
1 N 2         N
Z            1
H    i       r      i r
2 i 1       i 1 i   i 1 j  ij
Elect     Elect-    Elect-
KE        Nuc      Elect
PE        PE
N 1
1 1 1      1   1   1   1
i r r r
i 1 j  ij
      
r23 r24 r34 r35
12 13

Chapter 9: Slide 3
Atomic Orbitals
In performing quantum mechanical calculations on multielectron
atoms, it is usually assumed that each electron is in an atomic orbital, ,
which can be described as a Linear Combination of Hydrogen-like
orbitals, which are called Slater Type Orbitals (STOs).

These STOs are usually denoted as i (although the text uses
the notation, bi) .
Thus:    ci  i

The goal of quantum mechanical calculations is to find the values
of the ci which minimize the energy (via the Variational Principle).

These STOs are also used to characterize the Molecular Orbitals
occupied by electrons in molecules.

We will discuss these STOs in significantly greater detail in
Chapter 11, when we describe quantum mechanical calculations
on polyatomic molecules.
Chapter 9: Slide 4
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 5
The Hartree Method: Helium
Hartree first developed the theory, but did not consider that
electron wavefunctions must be antisymmetric with respect to
exchange.
Fock then extended the theory to include antisymmetric wavefunctions.
We will proceed as follows:
1. Outline Hartree method as applied to Helium
2. Show the results for atoms with >2 electrons
3. Discuss antisymmetric wavefunctions for multielectron atoms
(Slater determinants)
4. Show how the Hartree equations are modified to get the
the “Hartree-Fock” equations.

Chapter 9: Slide 6
Basic Assumption
Each electron is in an orbital, i (e.g. a sum of STOs).
The total “variational” wavefunction is the product of one electron
                     
wavefunctions:  (r1 , r2 )  1 (r1 )  2 (r2 )

Procedure
         init 
“Guess” initial values the individual atomic orbitals:                    (r1 ) and 2 (r2 )
init
1

(This would be an initial set of coefficients in the
linear combination of STOs)
Let’s first look at electron #1. Assume that it’s interaction with the
second electron (or with electrons #2, #3, #4, ... in multielectron atoms)
is with the average “smeared” out electron density of the second
electron.
SI Units                           Atomic Units
 2                                   2
eff 
e  (r2 ) 
init
 (r2 ) 
init
V1 (r1 )  e                           eff 
2
or   V1 (r1 )  
2
dr2                                   dr2
40 r12                               r12
1 init
  2init (2)       2 ( 2)
r12
Chapter 9: Slide 7
It can be shown (using the Variational Principle and a significant
amount of algebra) that the “effective” Schrödinger equation for
electron #1 is:
1 2 2                               1 init
H 1eff 1   11 where H 1eff   1   V1eff  V1eff   2init (2)      2 ( 2)
2    r1                            r12
elect elect- “Effective”
KE Nuc elect-elect
PE        PE
This equation can be solved exactly to get a new estimate
for the function, 1new (e.g. a new set of coefficients of the
STOs).
There is an analogous equation for 2:

1     2                                          1 init
H 2 2   22
eff
where   H 2   2 
eff
2       V2eff           V2eff  1init (1)       1 (1)
2     r2                                        r12

This equation can be solved exactly to get a new estimate
for the function, 2new (e.g. a new set of coefficients of the
STOs).
Chapter 9: Slide 8
A Problem of Consistency
         init 
We used initial guesses for the atomic orbitals,               (r2 ) and 2 (r2 ) ,
init
1
eff      eff
to compute H1 and H 2 .
1 2 2                              1 init
H 1eff 1  11 where H 1eff   1   V1eff V1eff   2init (2)      2 ( 2)
2    r1                           r12
1     2                                 1 init
H 2  2   2 2
eff
where   H 2    2   V2eff
eff
2
V2eff  1init (1)       1 (1)
2     r2                               r12

new        new 
We then solved the equations to get new orbitals, 1 (r2 ) and 2 (r2 )
eff    eff
If these new orbitals had been used to calculate V1 and V2 ,
we would have gotten different effective potentials.

Oy Vey!!! What a mess!!!

What can we do to fix the problem that the orbitals resulting from
solving the effective Schrödinger equations are not the same as
the orbitals that we used to construct the equations??
Chapter 9: Slide 9
The Solution: Iterate to Self-Consistency
new       new 
Repeat the procedure. This time, use 1 (r1 ) and 2 (r2 )
to construct V1eff and V2eff and solve the equations again.
Now, you’ll get an even newer pair of orbitals,          (r1 ) and 2 (r2 )
1

BUT: You have the same problem again. The effective Hamiltonians
that were used to compute this newest pair of orbitals were constructed
from the older set of orbitals.
Well, I suppose you could repeat the procedure again, and again, and
again, and again, until you either: (1) go insane
(2) quit Chemistry and establish
a multibillion dollar international

Chapter 9: Slide 10
Fortunately, the problem is not so dire. Usually, you will find that
the new orbitals predicted by solving the equations get closer and
closer to the orbitals used to construct the effective Hamiltonians.

When they are sufficiently close, you stop, declare victory, and go
out and celebrate with a dozen Krispy Kreme donuts (or pastrami
sandwiches on rye, if that’s your preference).

When the output orbitals are consistent with the input orbitals,
you have achieved a “Self-Consistent Field” (SCF).

Often, you will reach the SCF criterion within 10-20 iterations,
although it may take 50-60 iterations or more in difficult cases.

While the procedure appears very tedious and time consuming,
it’s actually quite fast on modern computers. A single SCF calculation
on a moderate sized molecule (with 50-100 electrons) can take well
under 1 second.

Chapter 9: Slide 11
The Energy
A. The total energy

1 2  1 2             2 2 1        1 2  2  1 2               2 1
H   1 (r1 )   2 (r2 )           1 (r1 )      2 (r2 )   
2          2           r1 r2 r12    2          r1   2           r2  r12

He              He            1                  1 2 2          He  1      2
H H   1       (1)  H   2      ( 2)        where H 1He   1     and H 2    2 
2
r12                  2     r1            2      r2

H1 and H2 are just each the Hamiltonian for the electron in a He+ ion.

E   * H d  1 (1)2 (2) H 1 (1)2 (2)

He              He             1
E  1 (1) 2 (2) H       1       (1)  H   2     ( 2)        1 (1)2 (2)
r12

We’re assuming that 1 and 2 have both been normalized.
Chapter 9: Slide 12
He              He             1
E  1 (1) 2 (2) H    1       (1)  H   2     ( 2)        1 (1)2 (2)
r12

E  I1  I 2  J12 Remember, this is the total energy of the two electrons.

                                                                                                      
I1  1 (1) 2 (2) H 1He (1) 1 (1) 2 (2)   2 (2)  2 (2) 1 (1) H 1He (1) 1 (1)  1 (1) H 1He 1 (1)

I1 is the energy of an electron in a He+ ion.

He                                                           He                                        He 
I 2  1 (1) 2 (2) H   2      (2) 1 (1) 2 ( 2)  1 (1) 1 (1)  2 (2) H           2       ( 2)  2 ( 2)   2 ( 2) H         2      1 (2)

I2 is the energy of an electron in a He+ ion.

J12  1 (1) 2 (2)
1
1 (1)2 (2)   
 (r ) (r ) (r ) (r ) dr dr
*
1      1   1   1
*
2   2   2   2
1   2
r12                 r1 r2                                        r12

J12 is the Coulomb Integral and represents the
coulombic repulsion energy of the two electrons
Chapter 9: Slide 13
The Energy

B. The Individual Orbital Energies, 1 and 2

1 2 2            1 2 2            1
H1eff 1  11 where        H1eff   1   V1eff   1    2 (2)      2 ( 2)
2    r1          2    r1         r12
 1 2 2           1         
1  1 (1) H   eff
1     1 (1)  1 (1)  1   2 (2)     2 (2)  1 (1)
 2    r1        r12        

1 2 2                          1         
1  1 (1)  1  1 (1)  1 (1)  2 (2)     2 (2)  1 (1)
2    r1                       r12        
                     1
 1  1 (1) H1He 2 (1)  1 (1)2 (2)       1 (1)2 (2)
r12

1  I1  J12
where J12  1 (1)2 (2)
1
1 (1)2 (2)   
  
1*1 2*2  
dr1dr2
r12                         r12
Chapter 9: Slide 14
The Energy

B. The Individual Orbital Energies, 1 and 2 (Cont’d.)

1     2           1     2          1
H 2  2   2 2
eff
where      H 2    2   V2eff    2   1 (1)
eff
2                 2                1 (1)
2     r2          2     r2        r12
 1 2 2            1         
 2  2 (2) H   eff
2     2 (2)  2 (2)   2   1 (1)     1 (1)  2 (2)
 2     r2        r12        

1 2 2                           1         
 2  2 (2)   2  2 (2)  2 (2)  1 (1)     1 (1)  2 (2)
2     r2                       r12        
                      1
 2  2 (2) H 2He 2 (2)  1 (1)2 (2)       1 (1)2 (2)
r12
 2  I 2  J12

where J12  1 (1)2 (2)
1
1 (1)2 (2)   
  
1*1 2*2  
dr1dr2
r12                         r12
Chapter 9: Slide 15
1  I1  J12           2  I 2  J12

The sum of orbital energies: 1   2  I1  J12   I 2  J12   I1  I 2  2J12

C. Total Energy versus sum of orbital energies

The sum of orbital energies: 1   2  I1  I 2  2J12

The total energy: E  I1  I 2  J12

The sum of the orbital energies has one too many Coulomb
integrals, J12.

The reason is that each orbital energy has the full electron-electron
repulsion – You’re counting it one time too many!!!

Chapter 9: Slide 16
1   2  I1  I 2  2J12

E  I1  I 2  J12

Therefore:   E  ( I1  J12 )  ( I 2  J12 )  J12

E  1   2  J12

Chapter 9: Slide 17
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 18
Koopman’s Theorem
Estimation of Atomic (or Molecular) Ionization Energies

Ionization Energy (IE): M  M+ + e- M is a neutral atom or molecule

E  IE  E ( M  )  E ( M )

1  I1  J12  I1  I 2  J12   I 2  E  I 2

 1  I 2  E  E ( He  )  E ( He )  IE (He )

I2 is the energy of the He+ ion
E is the energy of the He atom

Koopman’s Theorem: The ionization energy of an atom or molecule
can be estimated as -H, which is the orbital
energy of the highest occupied orbital.

Chapter 9: Slide 19
M  M+ + e- M is a neutral atom or molecule

E  IE  E ( M  )  E ( M )
Koopman’s Theorem: The ionization energy of an atom or molecule
can be estimated as -H, which is the orbital
energy of the highest occupied orbital.

There are two approximations in using Koopman’s theorem to estimate
ionization energies which limit the accuracy:

1. Electron “relaxation” of the remaining N-1 electrons is neglected.

2. Differences in the “correlation energy” [to be discussed later]
of the electrons in the ion and neutral atom are ignored.

To obtain an accurate estimate of the ionization energy, one should
perform quantum mechanical energy calculations on the neutral
atom and ion to get E(M) and E(M+), from which the IE can be
computed by the definition.
Chapter 9: Slide 20
Electron Affinity

Electron Affinity (EA): M + e-  M- M is a neutral atom or molecule
E  EA  E ( M  )  E ( M )

With this “new” definition of Electron Affinity, a negative
value of EA means that adding an electron to the atom
is an exothermic process.

Note: The text comments that Koopman’s theorem can also be used
to calculate the electron affinity, EA = -L (energy of the lowest
unoccupied orbital).

However, this is not commonly used and is very inaccurate.

Note: The “old” definition of Electron Affinity is the energy “released”
when an electron is added to a neutral atom.
EA(old) = - EA(new)

Chapter 9: Slide 21
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 22
The Hartree Method for Multielectron Atoms
The Hartree method for the more general N electron atom is a
straightforward extension of the method outlined for the two electrons
in Helium
Each of the N electrons has an effective Hamiltonian. For electron #1,
for example:
1 2 Z
H 1eff 1  11 where H 1eff   1   V1eff
2    r1
elect elect- “Effective”
KE Nuc elect-elect
PE        PE

 2          init  2               init  2
N
1 init            (r2 ) 
init
3 (r3 )              N (rN ) 
V1eff       j ( j)       j ( j)            dr2             dr3    
init                     2
drN
j 2            r1 j                r12                r13                    r1N

As before, we are assuming that electron #1 is interacting with the
“smeared out” electron density of electrons #2 to N.

Chapter 9: Slide 23
There are equivalent equations for each electron, i, of the N electrons:
1      Z
H ieff i   ii   where       H ieff    i2   Vi eff
2      ri
 2
1 init           init (rj ) 
                     j    
eff           init                          j
Vi                 j      ( j)                              dr j
j i               rij        j i      rij

As in the two electron case, one assumes that the total wavefunction
is the product of one electron wavefunctions:
                  N
                                   
 (r1 , r2 ,  , rN )   i (ri )  1 (r1 )  2 (r2 )     N (rN )
i 1

Initial guesses are made for each of the atomic functions, iinit, which
are used to compute the effective potentials, Vieff, and the N equations
are solved to get a new set of ’s.

The procedure is repeated (iterated) until the guess wavefunctions are
the same as the ones which are computed; i.e. until you reach a
Self-Consistent Field (SCF)
Chapter 9: Slide 24
The Energy
N       N 1
E    i   J ij
i 1     i 1 j i

E   1   2    J12  J13  J14    J 23  J 24  

i is the orbital energy of the i’th. electron. This is the
eigenvalue of the effective Hamiltonian for the i’th. electron

Jij is the Coulomb Integral describing the repulsion between
an electron in orbital i and an electron in orbital j.

             2
i (r1 )  j (r2 )
2
1                                               
J ij  i j       i j                                 dr1dr2
r12          r1 r2              r12

Note: If N=2 (i.e. He), the above expression for E reduces to
E  1   2  J12

Chapter 9: Slide 25
Math. Preliminary: Determinants
A determinant of order N is an NxN array                   a11    a12    a13    ... a1N
of numbers (elements). The total number of                 a21                      a2 N
elements is N2.
a31                      a3 N
Second Order Determinant                                                          
a11 a12                                             aN1    aN 2   aN 3   ... a NN
 a11a22  a21a12
a21 a22
Note: The expansion has 2 terms

Third (and higher) Order Determinant: Expansion by Cofactors

a11 a12   a13
a     a23      a   a       a   a
a21 a22   a23  a11 22        a12 21 23  a13 21 22
a32   a33      a31 a33     a31 a32
a31 a32   a33

 a11 (a22a33  a32a23 )  a12 (a21a33  a31a23 )  a13 (a21a32  a31a22 )
Note: The expansion has 6 terms
Chapter 9: Slide 26
Fourth Order Determinant
a11   a12    a13    a14
a21 a22      a23    a24
 a11 3 x3  a12 3x3  a13 3x3  a14 3x3
a31 a32      a33    a34
a41 a42      a43    a44

Note: Each 3x3 determinant has 6 terms.
Therefore, the 4x4 determinant has 4x6 = 24 terms.

General Properties of Determinants

Property #1: An NxN determinant has N! terms.

Property #2: If two columns or rows of a determinant are exchanged,
then the value of the determinant changes sign.

Property #3: If two columns or rows of a determinant are the same,
then the value of the determinant is 0.
Chapter 9: Slide 27
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 28
Slater Determinants
Review: The Pauli Antisymmetry Principle
^
The permutation operator, Pij , exchanges the coordinates of two
electrons in a wavefunction.

Permuting two identical particles will not change the probability
density:               
P  (r , r )   p  (r , r )
ˆ   2
ij   i   j        2
ij   i   j

Therefore: pij   1

Pauli Principle: All elementary particles have an intrinsic angular
momentum called spin. There are two types of particles,
with different permutation properties:
Bosons:     Integral spin (0, 1, 2,…)         Pij() = +
Fermions: Half integral spin (1/2, 3/2,…) Pij() = -

Chapter 9: Slide 29
Electrons (s = ½) are fermions.
Therefore, wavefunctios are antisymmetric with respect to
electron exchange (permutation).
ˆ                                 
Pij (ri , rj )   (rj , ri )   (ri , rj )

Note that the permutation operator exchanges both the spatial and
spin coordinates of the electrons.

Review: Ground State Helium
                                    

1
1s (r1 )11s (r2 )  2  1s (r1 ) 11s (r2 ) 2 
2
1
or    1s(1)11s(2)  2  1s(1) 11s(2) 2  This wavefunction is antisymmetric
2                                        with respect to exchange of
Shorthand
electrons 1 and 2.
1s (1)1s (2)1  2  1 2 
1
or 
2
Factored Form
Chapter 9: Slide 30
The electron configuration of ground state Lithium is 1s22s1.

The wavefunction,  (1, 2, 3)  1s (1)11s (2)  2 2 s (3) 3 , just won’t do.
It’s not either symmetric or antisymmetric with respect to
electron exchange.
An appropriate antisymmetric wavefunction is:

 1s (1)11s (2)  2 2s (3) 3  1s (1)1 2s (2) 21s (3)  3 
1 
 (1, 2, 3)       1s(1) 11s (2) 2 2s (3) 3  1s (1) 1 2s (2) 21s (3) 3 
6                                                              
 2s (1)11s (2) 21s (3)  3  2s (1)11s (2)  21s (3) 3 
                                                              
Question: How do I know that this wavefunction is antisymmetric?

Answer: Try it out. Exchange electrons 1 and 2.
Terms 1 and 3 switch with each other, but each with opposite sign.
Terms 2 and 5 switch with each other, but each with opposite sign.
Terms 4 and 6 switch with each other, but each with opposite sign.
Voila!! The wavefunction has changed sign.                                      Chapter 9: Slide 31
Question: How did I figure out how to pick out the appropriate six terms?

Answer: It was easy!!    Mookie showed me how.

Problem: The Mookster won’t be around to write out the antisymmetric
wavefunctions for you on a test.

Solution: I guess I should impart the magic of King Mookie, and show
you how it’s done.

Chapter 9: Slide 32
Slater Determinants

The ground state Helium wavefunction is:

 (1,2) 
1
1s(1)11s(2)  2  1s(1)11s(2) 2 
2

It can be written as a 2x2 determinant, called a Slater determinant
(named after J. C. Slater, who first came up with the idea).

1 1s(1)1 1s(1) 1
 (1,2)                           
1
1s(1)11s(2)  2  1s(1) 11s(2) 2 
2 1s(2) 2 1s(2)  2    2

Note that different “spinorbitals”** are put in different columns.
Different electrons are put in different rows.
The coefficient is to normalize the antisymmetrized wavefunction.

**A spinorbital is just the combination of the spatial and spin part
of an orbital taken together.
Chapter 9: Slide 33
1 1s(1)1 1s(1) 1
 (1,2) 
1s(2) 2 1s(2)  2

1
1s(1)11s(2)  2  1s(1) 11s(2) 2 
2                       2

Two properties of determinants come in very handy.

Property #2: If two columns or rows of a determinant are exchanged,
then the value of the determinant changes sign.

1 1s(2) 2 1s(2)  2
P  (1,2)   (2,1) 
2 1s(1)1 1s (1) 1
12


1
1s(1) 11s(2) 2  1s(1)11s(2)  2    (1,2)
2

Hey!! That’s nice!!
A Slater Determinant is automatically antisymmetric with respect to
the exchange of two electrons.

Chapter 9: Slide 34
Property #3: If two columns or rows of a determinant are the same,
then the value of the determinant is 0.

Let’s put both electrons in the same spinorbital, say 1s,
and see what happens.

1 1s(1)1 1s(1)1
 (1,2)                       
1
1s(1)11s(2) 2  1s(1)11s(2) 2   0
2 1s(2) 2 1s(2) 2    2

This explains the more commonly stated form of the Pauli Principle:
No two electrons can occupy the same orbital with the same spin.

Chapter 9: Slide 35
The Lithium Ground State Wavefunction
The electron configuration of ground state Lithium is 1s22s1.
The antisymmetrized wavefunction is:
1s (1)1 1s (1) 1 2s (1)1
1
 (1, 2, 3)      1s (2) 2 1s (2)  2 2s (2) 2
3!
1s (3) 3 1s (3)  3 2s (3) 3
1
The factor,      ,is to normalize the wavefunction (which has 3! terms)
3!
Expanding the wavefunction

           1s (2)  2   2 s (2) 2                 1s (2) 2   2 s (2) 2 
 1s (1)1                             1s (1) 1                          
1            1s (3)  3   2s (3) 3                  1s (3) 3   2 s (3) 3 
 (1, 2, 3) 
6            1s (2) 2   1s (2)  2                                        
 2 s (1)1                                                               

            1s (3) 3   1s (3)  3                                        


Chapter 9: Slide 36
 1s (1)1 1s (2)  2 2s (3) 3  2s (2) 21s (3)  3 
1 
 (1, 2, 3)       1s(1) 1 1s (2) 2 2s (3) 3  2s (2) 21s (3) 3 
6                                                       
 2s (1)1 1s (2) 21s(3)  3  1s (2)  21s (3) 3  
                                                       

 1s (1)11s (2)  2 2s (3) 3  1s (1)1 2s (2) 21s (3)  3 
1 
 (1, 2, 3)       1s(1) 11s (2) 2 2s (3) 3  1s (1) 1 2s (2) 21s (3) 3 
6                                                              
 2s (1)11s (2) 21s (3)  3  2s (1)11s (2)  21s (3) 3 
                                                              

We discussed earlier that this expanded (6 term) wavefunction
is antisymmetric with respect to electron exchange.

Chapter 9: Slide 37
1s (1)1 1s (1) 1     2s (1)1
1
 (1, 2, 3)     1s (2) 2 1s (2)  2   2s (2) 2
3!
1s (3) 3 1s (3)  3   2s (3) 3

The antisymmetry can also be shown by using the property of
determinants.
Exchanging two electrons:
1s (1)1 1s (1) 1     2s (1)1
1
 (1, 3, 2)     1s (3) 3 1s (3)  3   2s (3) 3   (1, 2, 3)
3!
1s (2) 2 1s (2)  2   2s (2) 2

Let’s put all 3 electrons in the 1s orbital:

1s(1)1 1s(1) 1 1s(1)1
1
 (1, 2, 3)     1s (2) 2 1s (2)  2 1s (2) 2  0
3!
1s(3) 3 1s(3)  3 1s (3) 3

Chapter 9: Slide 38
General

 (1)1      (1) 1       n (1) 1
1     (2) 2     (2)  2
 (1, 2, 3,) 
N!                  
 ( N ) N  ( N )  N  n ( N )  N

Shorthand Notations (Various types)

1s (1) 1s(1) 2s(1)
1
 (1, 2, 3)     1s (2) 1s(2) 2s (2)
3!
1s (3) 1s(3) 2s(3)

Use bars to indicate  spin. Lack of a bar means the spin is 

Chapter 9: Slide 39
Show diagonal terms only.         Lithium
2s3 3
1
       1s (1)1 1s (2)  2
3!

Beryllium
1
      1s (1)1 1s (2)  2   2 s (3) 3   2 s ( 4)  4
4!

Other shorthand notations include:

Leaving out the normalization constant.

Leaving out the normalization constant and electron numbering.

Beryllium
  1s 1s 2s 2s

To avoid confusion, the only shorthand I might use is the diagonal
Chapter 9: Slide 40
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 41
The Hartree-Fock Method

Hartree’s original method neglected to consider that the wavefunction
in a multielectron atom (or molecule) must be antisymmetric with respect
to electron exchange.

The Hartree-Fock is an extension, using antisymmetrized wavefunctions.

It results in additional “Exchange” terms in the Effective Hamiltonians
and “Exchange Integrals” in the expression for the energy.

We actually encountered Exchange Integrals when we calculated
the energy of excited state Helium in the 1s12s1 electron configuration.

Chapter 9: Slide 42
Review: The Energy of Triplet State Helium (1s12s1)

 
   antisym(r1 , r2 )  spin
1M   S

 1
 T1       1s (r1 )2 s (r2 )  2 s (r1 )1s (r2 )   1 2 
                             

 2                                             
                                                
 antisym(r1 , r2 ) 
1
1s (r1 )2 s (r2 )  2 s (r1 )1s (r2 )   1s(1)2s(2)  2s(1)1s(1) 
1
2                                                2
Remember that spin does not contribute directly to the energy.

 1 2 2  1 2 2  1                       1
H   1      2         H1  H 2 
 2    r1   2     r2  r12              r12

E   trip

1
1s(1)2s(2)  2s(1)1s(2)  H 1 1s(1)2s(2)  2s(1)1s(2) 
2                               2

Chapter 9: Slide 43
E   trip
  1s   2 s  J 1s 2 s  K 1s 2 s
1     2        3          4

1                                                   1
where   J1s 2 s  1s (1)2 s (2)            1s (1)2 s (2)        K1s 2 s  1s (1)2 s (2)        2 s (1)1s (2)
r12                                                 r12

1s (1) 2 2 s (2) 2                                    [1s (1)2 s (1)]  [1s (2)2 s (2)]  
J 1s 2 s                       dr1dr2                K1s 2 s                                       dr1dr2
r12                                                             r12

1. Energy of electron in 1s He+ orbital                           3. Coulomb (repulsion) Integral

2. Energy of electron in 2s He+ orbital                           4. Exchange Integral

Chapter 9: Slide 44
E    trip
  1s   2 s  J 1s 2 s  K 1s 2 s
1     2        3          4

1s (1) 2 2 s (2) 2  
J 1s 2 s                       dr1dr2        Always positive
r12
3. Coulomb (repulsion) Integral
The integrand of the Coulomb integral represents the repulsion of
two infinitesimal electron densities, (1)=1s(1)2 and (2)=2s(2)2,
separated by a distance, r12. The repulsion is summed over all
infinitesimal electron densities.
[1s (1)2 s (1)]  [1s (2)2 s (2)]  
K1s 2 s                                        dr1dr2 Usually positive
r12
4. Exchange Integral
Arises purely from the antisymmetry of the spatial function with respect
to electron exchange. It has no classical analog.
If the above calculation had been performed with a simple product
wavefunction, spat = 1s(1)2s(2), there would be no exchange integral
Chapter 9: Slide 45
The Hartree-Fock Energy

N        N
Hartree Energy:      E    i   J ij           Use simple product wavefunction:
i 1     i 1 j i       not antisymmetric w.r.t. exchange

 2        2
1                   i (r1 )  j (r2 )  
J ij  i j     i j                          dr1dr2
r12          r1 r2           r12

Jij is the Coulomb Integral describing the repulsion between
an electron in orbital i and an electron in orbital j.

Chapter 9: Slide 46
Hartree-Fock Energy: E    i   2 J ij  K ij 
N                   N

i 1                i 1 j i

Use antisymmetrized wavefunction: Slater Determinant

J ij  i j
1
i j   
                  

i* (r1 )i (r1 )  * (r2 ) j (r2 )  
j
dr1dr2
           Coulomb
r12          r1 r2                  r12                                               Integral

K ij  i j
1
 ji   
 (r ) (r ) (r ) (r ) dr dr
i
*
1          j   1
*
j   2   i   2               Exchange
1   2
r12          r1 r2                               r12                                  Integral

The Exchange Integral arises from the antisymmetry of the wavefunction,
and has no classical analog.

Note: There is an error in the text’s HF energy. The way the authors
are writing it, the top summation index should be N/2
Chapter 9: Slide 47
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 48
Hartree-Fock Orbital Energies for Ar
0 eV
Separated particles

-16.1 eV
3p   Note that the ns and np orbitals have different
-34.8 eV
3s   energies. This is due to screening of the p electrons.

Koopman’s Theorem
IE  -
-260 eV
2p                   Electron
Removed    IE(exp)   IE(Koop)
-335 eV
2s                     1s       3206 eV    3227 eV
2s         --       335
2p        249       260
3s        29.2      34.8
3p        15.8      16.1
-3227 eV
1s
Chapter 9: Slide 49
Outline

• The Hamiltonian for Multielectron Atoms
• The Hartree Method: Helium
• Koopman’s Theorem
• Extension to Multielectron Atoms
• Antisymmetrized Wavefunctions: Slater Determinants
• The Hartree-Fock Method

• Hartree-Fock Orbital Energies for Argon

• Electron Correlation

Chapter 9: Slide 50
Electron Correlation
The principal approximation of the Hartree-Fock method is that a
given electron interacts with the “smeared-out” electron density of
the remaining N-1 electrons.

Actually, the other N-1 electrons are point particles, just like the
one we’re considering.
Thus, the motion of the electrons are correlated. That is, they try to
avoid each other.

High                            Low
Energy                         Energy
Not favored                      Favored
Chapter 9: Slide 51
Because the Hartree-Fock (HF) method does not consider the
specific electron-electron repulsions, which tend to keep two
electrons apart, the HF energy is invariably too high.

The difference between the “exact” electronic energy and the
HF energy is called the “Correlation Energy”, Ecorr.

0                    Ecorr  Eexact  EHF

Generally, the correlation energy is very small
compared to the total energy (usually <1%)
EExact

However, in absolute terms, this can still represent
EHF

a rather large energy.
The “exact” electronic energy can be measured
as the negative of the sum of the Ionization Energies.
N
Eexact  Eexp   IEi
EHF                                    i 1

EExact            Ecorr  Eexact  E HF  Eexp  E HF
Chapter 9: Slide 52
EHF                            Helium

Ecorr                                    EHF = -2.862 au
EExact
EExact = -2.904 au

Eexact  EHF
% Error                   x100  1.4%
Eexact

However, the correlation energy can still be very large in
absolute terms.

2625 kJ / mol
Ecorr  Eexact  EHF  0.042 au x                  110 kJ / mol
1 au

Chapter 9: Slide 53
EHF                         Argon

Ecorr
EHF = -526.807 au
EExact                 EExact = -527.030 au
Eexact  EHF
% Error                 x100  0.04%
Eexact

However, the correlation energy can still be very large in
absolute terms.
2625 kJ / mol
Ecorr  Eexact  EHF  0.223 au x                585 kJ / mol
1 au

For many applications (e.g. geometries and frequencies), inclusion
of the correlation energy is not that important.
However, for applications involving bond breaking and bond making
(e.g. reactions), inclusion of the correlation energy is critical in order
to get good results.
We will qualitatively discuss methods used to determine the
correlation energy in a later chapter.
Chapter 9: Slide 54
An Example: Calculated Ionization Energy and
Electron Affinity of Fluorine

Ionization Energy (IE): M  M+ + e- M is a neutral atom or molecule

E  IE  E ( M  )  E ( M )

Electron Affinity (EA): M + e-  M- M is a neutral atom or molecule
E  EA  E ( M  )  E ( M )

Methods: E(HF) = HF/6-311++G(3df,2pd)                   Hartree-Fock Energy

E(QCI) = QCISD(T)/6-311++G(3df,2pd) Correlated Energy

This is the HF energy with a correction for electron correlation
calculated at the QCISD(T) level (later Gator).

Chapter 9: Slide 55
Species      E(HF)         E(QCI)
F       -99.402 au     -99.618 au
F+       -98.825        -98.984
F-      -99.446        -99.737

IE ( HF )  E ( F  )  E ( F )
kJ / mol
 98.825  ( 99.402)  0.577 au  2625           1515 kJ / mol
au
Similarly: IE (QCI )  1664 kJ / mol

EA( HF )  E ( F  )  E ( F )
kJ / mol
 99.446  ( 99.402)  0.044 au  2625           116 kJ / mol
au
Similarly: EA(QCI )  312 kJ / mol

Chapter 9: Slide 56
Quantity    Expt.           HF            QCI
IE       1681 kJ/mol    1514 kJ/mol 1664 kJ/mol

EA       -328           -115           -312

Koopman’s Theorem IE

Energy of highest occupied orbital at HF/6-311++G(3df,2pd) level

H = -0.733 au

IE  -H = +0.733 au • 2625 kJ/mol / au = 1924 kJ/mol

Notes: (1) Koopman’s Theorem gives only rough approximation
for Ionization Energy
(2) Accurate calculations of the IE or EA require the
use of energies corrected for electron correlation.

Chapter 9: Slide 57
EHF
Ecorr  Eexact  EHF  0                      E HF ( F  )
Eexact ( F  )  EQCI ( F  )
Eexact  EQCI

 
Ecorr F   Ecorr F   Ecorr F   
IE(HF)=1514 kJ/mol    IE(QCI)=1664 kJ/mol

IE  E ( F  )  E ( F )  0                              EHF (F )
EA  E ( F  )  E ( F )  0
Eexact ( F )  EQCI ( F )

EA(HF)= -115 kJ/mol

EHF ( F  )

EA(QCI)= -312 kJ/mol

Eexact ( F  )  EQCI ( F  )

Chapter 9: Slide 58

```
To top