Components of the Atom by VII9jovw

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									     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.
                                                                    newer 
Now, you’ll get an even newer pair of orbitals,          (r1 ) and 2 (r2 )
                                                  newer
                                                 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
                                       trucking conglomerate (please
                                       remember me in your will).



                                                                   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
 form at the top of this page.
                                                                           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

								
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