Winter 2013
Chem 356: Introductory Quantum Mechanics
Chapter 9 – Many Electron Atoms ............................................................................................................ 121
Many-Electron Atoms ........................................................................................................................... 121
A: Harttree-Fock: Minimize the Energy of a Single Slater Determinant. .............................................. 126
Hartree-Fock Interation Scheme........................................................................................................... 127
Chapter 9 – Many Electron Atoms
Many-Electron Atoms
Hamiltonian for He Z  2
Hˆ  
2
2me
12 
2
2me
22 
2
2m
 2 
Ze2
4 0 r1

Ze2
e2

4 0 r2 4 0 r1,2
 Tˆ  Vˆ He  Vˆ el
1) Simplification: neglect kinetic energy of heavy nucleus. Assume nucleus is fixed in origin
(exact: separate center of mass motion)
Tˆ  Tˆ el  T N ;
neglect TˆN , only consider electronic problem
2) Simplify units. We will use atomic units
Atomic Units
Mass
me  1 a.u.
9.1  1031 kg
Charge
e  1 a.u.
1.6  1019 C
a0  1 a.u.
4 0 2
a0 
 5.3  1011 m
2
me e
Length
Angular Momentum
Energy
 1 a.u.
 1.05  1036 Js
En  1 a.u.
En 
e2
4 0 a0
 4.36  1018 J
In atomic units (a.u.)
e  me 
 a0  4 0  1 a.u.
This simplifies equations, as all the constants are gone.
Chapter 9 – Many Electron Atoms
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Winter 2013
Chem 356: Introductory Quantum Mechanics
In electronic structure theory, everyone uses a.u. and converts to ‘physical’ units as the need arises
1
1
2 2 1
Hˆ   12   2 2   
2
2
r1 r2 r12
Now we wish to solve
Hˆ  (r1 , r2 )  E (r1 , r2 ) (For the moment we neglect spin)
Let us be more general and write down the Hamiltonian for an N -electron atom
N
N
N
1
Z
1
Hˆ    12    
2
i 1
i 1 rL
i 1 j i rij
where rij  ri  rj
The wavefunction  (r1 , r2 ,....rM ) has 3 N coordinates
This is a very complicated problem. However, if we could neglect the V el ,
1
terms, the problem would
rij
be simple.
Hˆ  hˆ(1)  h(2)  .....hˆ( N )
1
Z
hˆ(i)   i 2 
2
r1
Try product wavefunction
  a (r1 )b (r2 ).... z (rM )
Solve h(r1 )a (r1 )   aa (r1 ) one-electron orbitals, orbital energies
Moreover, each problem for each electron is the same: only the name of the variable is changed. This
would simply be Hydrogen like solutions.
 (r1 ....rM )  a (r1 )b (r2 ).... z (rM )
E   a   b   c  .... z Trivial!
Question: what would the lowest energy state be?
Chapter 9 – Many Electron Atoms
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Winter 2013
Chem 356: Introductory Quantum Mechanics
Answer: It is tempting to fill up these orbital levels with 2 electrons, of either spin. That is not
the lowest energy solution to Ĥ  E !
What is? Put all electrons in the same lowest energy orbital
E  1  1  1....N1  N1
We all know this is physically incorrect, and need to invoke the Pauli principle. Let us try to formulate it,
and understand where it originates.
Let us go back to basics: If an operator  commutes with the Hamiltonian  Hˆ , Aˆ   0 and  is an

eigenstate of Ĥ then:
 

 
ˆ ˆ   AE
ˆ   E Aˆ
Hˆ Aˆ  AH
Hence Â is also an eigenstate with the same energy.
Special types of operators that commute with the Hamiltonian and permutations of coordinates.
Eg.

1
Pˆ12  h(1)  h(2) 
r12



 (1, 2) 



1 
  h(2)  h(1)   (2,1)
r21 

ˆ  (1, 2)
 HP
12
Or
ˆ ˆ This is true for any permutation, true for many electron
Pˆ12 Hˆ  HP
12
systems.
Special permutations are transpositions Tij , which interchange coordinates 1 and 2. These operators are
Hermitian, and TijTij  1 , interchanging twice does nothing.
Ĥ and Tij have a set of common eigenfunctions, since they commute
Tˆij (1,...i, j , M )  
Chapter 9 – Many Electron Atoms
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Chem 356: Introductory Quantum Mechanics
Tˆij (Tˆij )  Tˆij    2  
Hence  2  1    1,   1
From these considerations (plus a bit more solid reasoning) we can deduce there are 2 types of particles:
1) Bosons   1 : The wave function is symmetric under interchange of coordinates (any
transposition)
 (1,...i,... j ,...M )   (1,... j ,...i, N )
2) Fermions   1: The wavefunction changes sign under a transposition
Tij   (1,... j ,...i,...M )   (1,...i,... j , N )
All spin
1
particles, like protons and electrons are Fermions. I do not have a simple explanation for this
2
(spin-statistics theorem in quantum field theory).
All permutations can be written as a product of transpositions p  Tij Thl Tej .... either odd or even.
Hence action of all Tij determines action of P .
For electrons wavefunction is antisymmetric under interchange of 2 electron coordinates (This includes
the spin degree of freedom)
If we return to our orbital product wavefunction   a (1)b (2)... z ( N ) one can design a clever trick
to make ‘antisymmetric’ wavefunctions
Put the orbitals in a determinant:
a (1) b (1)
 a (1)b (2)  b (1)a (2)
a (2) b (2)
Or
a
a (1)
2 a (2)
3  a (3)
1
b
b (1)
b (2)
b (3)
c
c (1)
c (2)
c (3)
Interchanging the coordinates of 2 electrons = interchange 2 rows in a determinant.  automatic
minus sign.
Also: interchange orbitals a and b  introduces minus sign
We can simplify notation and write   a (1)b (2)c (3) indicating the diagonal terms only, Then
a (2)b (1)c (3)   a (1)b (2)c (3)
  b (1)a (2)c (3)
Chapter 9 – Many Electron Atoms
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Chem 356: Introductory Quantum Mechanics
Slater determinants easily fold in antisymmetry. If we consider a (1) as a spin-orbital, then in the
determinant we flip spin-orbitals by permutations, or we should flip space and spin coordinates upon
interchanging.
Finally :
 h(1)  h(2)  ...h( z )  a (1)b (2)... z ( N ) 
  a   b  ... z  a (1)...z ( N )
If
h(1)i (1)   ii (1)
Now we have the Pauli principle too: If two orbitals in a determinant are equal then
 0
a (1)b (2)...a (k )    c (1)....c (k )   0 (2 equal columns)
For this reason we put one electron in each spin-orbital
A simple Slater determinant is the appropriate anti symmetric generalization of a product of orbitals.
These would be eigenfunctions if Hˆ 
 hˆ(i) , a sum of one-electron Hamiltonians. It is not, we have
i
to worry about Vˆ el 
1
 rij .
2 i j
To deal with Vˆ el one applies the variational procedure in 2 steps.
a) Minimize the energy of a single determinant wavefunction. This is the Hartree-Fock selfconsistent field procedure. It leads to an optimal set of orbitals (one-electron
wavefunctions) and orbital energies.
b) Apply linear variational principle and minimize energy of
 c 
x
in which   are Slater
x
determinants. This is called Configuration Interaction. It is important to go behind HartreeFock (single determinant) to get accurate results. Moreover for degenerate states one has
to use CI to get splitting (see atomic term symbols later on.)
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A: Harttree-Fock: Minimize the Energy of a Single Slater Determinant.
If we label orbitals in the determinant a , to indicate spatial index a , and spin index    ,  , then
the following integrals are relevant:
ha  La   dr1a *(r1 )h(r1 )a (r1 )dr1  | 
 '
J ab  J ab


 '
kab

dr1 dr2a * (r1 )b ' (r2 ) *

1 
a (r1 )b ' (r2 )  |   ' |  '
r12
dr1 dr2a (r1 ) *b ' (r2 ) *
1 '
b (r1 )a (r2 )  |  '  ' | 
r12
The J ab integral describes the Coulomb interaction between
a (r1 )  a (r1 )
b (r2 )  b (r2 )
And
2
It is independent of spin.
 '
The kab
integral arises due to the antisymmetry in  , This integral is the Coulomb interaction
between distributions
ab (r1 )  a (r1 ) * b (r2 )
And
ba (r2 )  a * (r2 )b (r2 )
It is called exchange interaction. It only arises for spin-orbitals that have equal spin. In the
energy it enters with a minus sign
EHF   ha 
a
If one has a closed shell with equal
1
1
 '

J ab
  kab

2 a
2 a ,b,
b '
 and  orbitals, the expression reduces to
EHF  2 ha  2 (2 J ab  kab )
a
a ,b
The exchange interaction is the reason that if orbitals are degenerate, it is preferable to align spins
parallel. It maximizes the number of “ - k ab ” like-spin contributions.
We now have an expression (did not derive) for the Hartree-Fock energy. This expression can be
minimized to find an optimal set of orbitals. If one expands the orbitals in a basis set X i (r ) , one obtains
a matrix eigenvalue equation.
Fij c j     ci
(orthonomal basis)
The matrices F depend on the occupied orbitals. For this reason the equations have to be solved “selfconsistently”
Chapter 9 – Many Electron Atoms
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Hartree-Fock Interation Scheme
Guess starting orbitals  a( old )
Construct F  aold 
Diagonalize Fc  c pick up the lowest energy orbitals  F  anew 
Check if F  anew   F  aold 
If not: update orbitals to get new set of improved orbitals
If F  anew   F  aold  convergence reached
This is called the self consistent field (SCF) iteration scheme. It is a powerful numerical scheme to
minimize the energy of a single determinant.
Some Notes:
-
The orbitals in HF are not unique. The only thing that matters is the determinant. In a
determinant I can make linear combinations of rows and columns. Same is true for orbitals
1
1
  a   b   a   b     a   b
2
2
-
The energy EHF is not equal to

a
One needs to use the full expression, given before.
a
-
-
-
HF is surprisingly accurate in many cases. For molecules it often yields ~ 99% of the exact
energy. For purposes in chemistry this is not accurate enough.
1% of energy > 100 kcal mol-1 typically!
HF SCF equations are tricky. It is easy to get trapped in a higher energy solution. Pitfall of nonlinear optimization.
In practice one use a basis set. The accuracy of a HF calculation is limited by the size of the
basis set.
Hartree-Fock results can be qualitatively wrong when you expect degeneracies between ground
and excited states; or when certain orbitals in a degenerate manifold are occupied, and others
not in the HF determinant. Usually one needs to go beyond HF to get qualitatively correct
results then.
One always needs to go beyond HF to get quantitatively accurate results
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