Transcript Hartree-Fock Theory

Hartree-Fock Theory
1
The orbitalar approach
Slater
Mulliken
Y(1,….N) = Pi fi(i)
Pauli principle
spin
Combination
of Slater
determinants
Solutions for S2
2
MO approach
A MO is a wavefunction associated with a
single electron. The use of the term "orbital"
was first used by Mulliken in 1925.
We are looking for wavefunctions for a
system that contains several particules. We
assume that H can be written as a sum of
one-particle Hi operator acting on one
particle each.
H = Si Hi
Y(1,….N) = Pi fi(i)
E = Si Ei
Robert Sanderson
Mulliken
1996-1986
Nobel 1966
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MO approach
Friedrich Hund
1896-1997
Charles Alfred Coulson
1910-1974
MO theory was developed, in the years after valence bond
theory (1927) had been established, primarily through the
efforts of Friedrich Hund, Robert Mulliken, John C. Slater,
and John Lennard-Jones. The word orbital was introduced
by Mulliken in 1932. According to Hückel, the first
quantitative use of MO theory was the 1929 paper of
Lennard-Jones. The first accurate calculation of a molecular
orbital wavefunction was that made by Charles Coulson in
1938 on the hydrogen molecule. By 1950, MO were
completely defined as eigenfunctions of the self-consistent4
field
Generalizing the
LCAO approach:
A linear combination of
atomic orbitals or LCAO
Sir John Lennard-Jones
1894-1954
Linus Carl Pauling
1901-1994 Nobel 1962
It was introduced in 1929 by Lennard-Jones
with the description of bonding in the
diatomic molecules of the first main row of
the periodic table, but had been used earlier
by Pauling for H2+.
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Schrödinger equation for LCAO
HY = EY
This is the set of linear equation solved for
| Hij-E S ij |= 0
Secular determinant
Erwin Rudolf
Josef Alexander
Schrödinger
Austrian
1887 –1961
In an ab-initio calculation, all the Hij and Sij
Integrals are rigorously calculated.
To do that, one need defining a set of AOs
(basis set) and the geometry.
There is no parameterization.
A subjective choice concerns that the basis
set (always incomplete). The best calculation
(SCF level or after IC) is that providing the6
lowest energy.
Particles are
electrons
Born-Oppenheimer
approximation
1927
Max Born German
(1882-1970)
}
Operator acting on e
Julius Robert Oppenheimer
Berkeley- Los alamos
1904 –1967
VN-N
H = [Te +VN-e + Vee] + TN +
He = [Te +VN-e + Vee] VN-e also acts on N; when N is
Should be a
coupling term
considered as fixed: VN-e then
becomes an operator acting on e.
N
e
Y = PY(e) PY(N) Separation of Y(e) and Y(N)
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Born-Oppenheimer
approximation
The forces on both electrons and nuclei due to
their electric charge are of the same order of
magnitude; since the nuclei are so much more
massive than the electrons, they must
accordingly have much smaller velocities.
.
Nucleon mass / Electron mass = 1835
For the same kinetic energy, when an electron travels meter a
nucleus of hydrogen travels 2.3 cm, that of carbon 6.7
millimeters and that of gold 1.7 millimeters. It is then
considered as the movement of nuclei and electrons are
independent, that electrons adapt instantaneously to the
movement of nuclei. It is said that the electrons follow
adiabatically the motion of the nuclei.
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The adiabatic theorem 1928
Max Born German
(1882-1970)
A physical system remains in its
instantaneous eigenstate if a given
perturbation is acting on it slowly
enough and if there is a gap
between the eigenvalue and the
rest of the Hamiltonian's spectrum.
Vladimir
Aleksandrovich
Fock russian
1898–1974
This adiabatic principle is crucial because it allows
us to separate the nuclear and electronic motion,
leaving a residual electron-phonon interaction.
From this point on it is assumed that the electrons
respond instantaneously to the nuclear motion and
always occupy the ground-state of that nuclear
configuration.
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Vibrations, nuclear motion
Developing U around the equilibrium position (r = req)
gives U(Dr) = U° + U’’(req)Dreq2: The angular frequency is
w = (req/m) ½ depending on the reduced mass m=m1m2(m1+m2)
3N-6 degree of freedom + 3 translations and 3 rotations for the general case.
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The surface energy obtained
under the Born-Oppenheimer
approximation: adiabatic
surfaces.
These are surfaces that can be achieved step by step from
calculations developed. They may be significant changes in the
characters of orbital obtained. According to an internuclear
distance ionic or covalent change significantly.
In the case of rapidly changing systems (collision dissociations
from excited states), the nature of the orbital can not change and
the Born-Oppenheimer approximation is no longer valid. The state
surface is therefore a surface potential where the same dominant
character is always found. Such surface, called diabatic surface,
is in some region the ground state and for some region an excited
state. Then, there is a jump from an adiabatic surface to another!
The diabatic surfaces are inherently difficult to obtain since 11
calculations lead to the system that is most stable.
PES: Potential Energy Surfaces.
U(d)
Covalent
Na+ + ClDiabatic
Adiabatic
Ionic
Na + Cl
Diabatic
dNa-Cl
An ionic curve “naturally” correlates with ions whereas a
covalent one dissociates to generate radicals. This leads to an
avoided crossing.
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Finding diabatic surfaces
The diabatic surface should in principle be obtained by
calculating the coupling terms <f1|TN|f2> allowing possible
jumps from one adiabatic surface to another.
This is uneasy since the basis set gives an avoided crossing
without the coupling. The usual procedure is generally the
reverse: impose “natural orbitals” and minimize the coupling
terms to allow the surfaces crossing.
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Diabatic- Adiabatic surfaces.
adiabatic
diabatic
Process velocity
slow
fast
Born-Oppenheimer
valid
Not valid
electrons
adaptation to the
nuclei motion
Adapt
(follow it)
Natural MO
character
May switch
Do not
(delay to
follow nuclei)
Remains
constant
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Approximation of
independent particles
Douglas Rayner
Hartree English
1897 1958
HHartree= Si Fi
Vladimir
Aleksandrovich
1930
Fock russian
1898–1974
The hamiltonian, HHartree , is the sum of Fock operators
only operating on a single electron i.
The wave function is the determinant of orbitals, f(i) to
satisfy the Pauli principle. Fi f(i)=εi f(i)
EHartree= Si εi
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Approximation of
independent particles
Hartree
Fock
1927
Fi = Ti + Sk Zik/dik+ Sj Rij
Each one electron operator is the sum of one
electron terms + bielectronic repulsions
Rij is the average repulsion of the electron j upon i.
The self consistency consists in iterating up to
convergence to find agreement between the postulated
repulsion and that calculated from the electron density.
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This ignores the real position of j vs. i at any given time.
Self-consistency
Given a set of orbitals Yi, we calculate the
electronic distribution of j and its repulsion
with i.
This allows expressing Fi = Ti + Sk Zik/dik+ Sj Rij
and solving the equation to find new Y1i
allowing to recalculate Rij. The process is
iterated up to convergence. Since we get
closer to a real solution, the energy
decreases.
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Assuming Y= Y1Y2
Jij, Coulombic integral for 2 e
Let consider 2 electrons, one in orbital Y1, the other in orbital Y2, and
calculate the repulsion <1/r12>.
Assuming Y= Y1Y2
This may be written
Jij =
Dirac notation used in physics
= < Y1Y2IY1Y2>
= (Y1Y1IY2Y2)
Notation for 18
Quantum chemists
Assuming Y= Y1Y2
Jij, Coulombic integral for 2 e
Jij = < Y1Y2IY1Y2> = (Y1Y1IY2Y2) means the product of
two electronic density  Coulombic integral.
This integral is positive (it is a repulsion). It is large when dij
is small.
When Y1 are developed on atomic orbitals f1, bilectronic
integrals appear involving 4 AOs (pqIrs)
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Assuming Y= Y1Y2
Jij, Coulombic integral involved in
two electron pairs
electron 1: Y1 or Y1 interacting with electron 2: Y2 or Y2
When electrons 1 have different spins (or electrons 2 have different spins)
the integral = 0. <aIb>=dij. Only 4 terms are  0 and equal to Jij.
The total repulsion between two electron pairs is equal to 4Jij.
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Particles are electrons!
Pauli Principle
Wolfgang Ernst Pauli
Austrian 1900 1950
electrons are indistinguishable:
|Y(1,2,...)|2 does not depend on
the ordering of particles
1,2...:
| Y (1,2,...)|2 = | Y (2,1,...)|2
Thus either Y (1,2,...)= Y (2,1,...) S bosons
or Y (1,2,...)= -Y (2,1,...)
A fermions
The Pauli principle states that electrons are fermions.
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Particles are fermions!
Pauli Principle
The antisymmetric function is:
YA = Y (1,2,3,...)-Y (2,1,3...)-Y (3,2,1,...)+Y (2,3,1,...)+...
which is the determinant
Such expression does not allow two electrons to be in the same state
the determinant is nil when two lines (or columns) are equal;
"No two electrons can have the same set of quantum numbers".
One electron per spinorbital; two electrons per orbital.
“Exclusion principle”
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Two-particle case
The simplest way to approximate the wave function of a many-particle
system is to take the product of properly chosen wave functions of the
individual particles. For the two-particle case, we have
This expression is a Hartree product. This function is not antisymmetric.
An antisymmetric wave function can be mathematically described as
follows:
Therefore the Hartree product does not satisfy the Pauli principle. This
problem can be overcome by taking a linear combination of both Hartree
products
where the coefficient is the normalization factor. This wave function
is antisymmetric and no longer distinguishes between fermions.
Moreover, it also goes to zero if any two wave functions or two
fermions are the same. This is equivalent to satisfying the Pauli 23
exclusion principle.
Generalization: Slater determinant
.
The expression can be generalized to any number of fermions by writing it as a
determinant.
John Clarke Slater
1900-1976
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Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant
Kij, Exchange integral for 2 e
Let consider 2 electrons, one in orbital Y1, the other in orbital Y2, and
calculate the repulsion <1/r12>.
Assuming Y= 1/√ IY1Y2I
Kij is a direct consequence of the Pauli principle
Dirac notation used in physics
Kij =
= < Y1Y2IY2Y1A>
= (Y1Y2IY1Y2)
Notation for 25
Quantum chemists
Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant
Kij, Exchange integral
It is also a positive integral (-K12 is negative and corresponds to a
decrease of repulsion overestimated when exchange is ignored).
two electron pairs
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Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant
Interactions of 2 electrons
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Unpaired electrons: sgsu
Singlet and triplet states
bielectronic terms
│YiYj│+ │YiYj│
Jij+Kij
│YiYj│
Jij
Kij
│YiYj│
Jij-Kij
│YiYj│ │YiYj│
│YiYj│- │YiYj│
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Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant
Jj and Kj operators
We need to express the ij bielectronic repulsion as one electron
operator Rj acting on Yj and representing the repulsion with an
average distribution of j.
Rj is the sum of 2 operators Jj and Kj.
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Slater rules
Hartree-Fock Operators are 1 or 2 electron operators ( i or ij).
Developing the polyelectronic wavefunction leads only to the
following contributions
<c1Ic1> <c2Ic2> ….. <ciIOIci> ….. <cNIcN>
or <c1Ic1> <c2Ic2> ….. <cicjIOIcicj> ….. <cNIcN>
Only the terms <c1Ic1>=1 remain, the others being nil: <c1Ic1>=0
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Slater rules
Non zero terms arise from 3 possibilities:
N
• Dl = DR
<DlIOIDR> = S [<kikjIOIlilj> - <kikjIOIlilj>]
i<j
• Dl differ from DR by one orbital i
<DlIOIDR> = <kiIOIlj>
N
<DlIOIDR> = S [<kikjIOIlilj> - <kikjIOIlilj>]
ij
• Dl differ from DR by two orbitals i and j
<DlIOIDR> = <kikjIOIlilj> - <kikjIOIlilj>
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Electronic energy
i is the index for an orbital:
occ
EE = Sεi
occ
εi = <fiIhIfi> + S (2Jij-Kij)
j
Each i electron interacts with 2 j
electrons with opposite spins Jij-Kij with
that of the same spin for and Jij for that
with different spin.
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Electronic energy – total energy
The sum of the energies of all the electrons
contains the ij repulsion twice:
occ
occ
EE = Sεi
i
occ
εi = <fiIhIfi> + S (2Jij-Kij)
EE = S[hii +
i
j
occ
S repij ] = Si [hii +2 S Repij ]
occ
occ
j
ESCF is defined as the energy of YSCF applying Slater rules
Pauli principle
occ
occ occ
ESCF = <YSCFIHIYSCF> = Shii + S
i
i
occ
Sj<i(2Jij-Kij)
ETOTAL = ESCF +EN ESCF = EE - Repij = Shii + Repij
i
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Restricted Hartree-Fock
Closed-shell system: an MO is doubly occupied or
vacant. The spatial functions are independent from the
spin. Assuming k and l with spin a, the expression of
a Fock matrix-element, Fkl, is
occ
Fkl = hkl +
occ
S [(klIjj)-(klIjj)] - S (klIjj)
J of spin b
J of spin a
Fkl = h + 2J – K
occ
and
εi = hii + S (2Jij-Kij)
j
occ
occ occ
ESCF = 2S εi – SS(2Jij-Kij)
i
There is no exchange
term for electrons with
different spin (<aIb>=0)
Altogether there are 2 J
for one K.
i j<i
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Restricted Hartree-Fock
Why the HOMO-LUMO electron gap is overestimated?
occ
εi = hii + Sj (2Jij-Kij)
i
For a virtual i orbital the repulsion
concerns all the electrons
i
For an occupied i orbital the repulsion
concerns all the electrons minus 1
(itself)
The repulsion is larger for an unoccupied level than for
an occupied one!
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Koopmans theorem
εi = - IP
occ
εi = hii + Sj (2Jij-Kij)
This theorem applies for a “vertical transition”:
sudden without time for electronic reorganization.
The orbitals do not relax!
Tjalling C.
Koopmans
Dutch
Nobel Prize in
Economic
Sciences
in 1975
This theorem is obvious for Hückel-type approaches:
The AOs are fixed. The total energy is the sum of the orbital
energies:
j occi
j occ
j occi
j occi
IP = 2S εj - 2S εj = 2S εj - 2S εj j
j
j
(Physica, 1, 104 (1933)
It is less obvious in HF since ESCF
S εi
εi
For DFT, the Janak theorem is a generalization of Koopman’s theorem. J. F. Janak, Phys.
Rev. B
36
18, 7165; J.P. Perdew, R. G. Parr, M. Levy et J. L. Balduz Jr., Phys. Rev. Lett. 23 (1982) 1691.
Koopmans theorem
εk = - IP
Tjalling C.
Koopmans
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What is wrong with Koopmans
theorem?
The AOs are not “fixed” but relax (“breath”); the Slater exponents
depend on the bielectronic repulsion. The subtraction in the previous
slide suppose no variation. The relaxation is important for polarisable
systems (metals, the work function is generally weak).
Calculating with AOs for the neutral species underestimates the
energy for the ionized species. IP is overestimated by Koopmans
theorem.
How to proceed more rigorously?
Calculate the IP as the difference of the total energy for two states.
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Cu+
(Koop.)
Cu+
(exp.)
Na
(Koop.)
Na
(exp.)
1s
658.4
662
79.4
81.2
2s
82.3
81.3
5.2
6.0
2p
71.83
69.6
2.8
3.66
3s
10.65
9.6
.378
.372
3p
7.28
6.1
3d
1.61
0.79
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Open-Shell – spin contamination
Let assume that here is one a electron more than b.
For them Fock matrix elements, there is one more exchange term for the
a electrons than for the b’s. The matrix elements are not the same and
solving the secular determinants for a and b independently leads to
different solutions.
fi  fi .
The UHF (unrestricted Hartree-Fock method) solves
the secular equation for a and b independently and
obtains different spatial functions for a and b.
This does not allow to have eigenfunctions of the spin
k
operators.
k
j
j
i
i
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ROHF
The same spatial function is taken whatever the
spin is.
Robert K. Nesbet
2e occ
1e occ
F = h + S [( Ijj)-( jI j)] - S ( Ijj)-1/2 ( jI j)
This allowsj eigenfunctions of spin
j operators.
This allows separating spatial and spin functions.
It is not consistent with variational principle (that
does better without this constraint).
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Spin operator
The spinorbitals should be solution of SZ and
S2. This is not the case for UHF.
42
Correspondence principle 1913/1920
For every physical quantity
one can define an operator.
The definition uses
formulae from classical
physics replacing
quantities involved by the
corresponding operators
Niels Henrik David Bohr
Danish
1885-1962
For an angular momentum
Classical expression
Quantum expression
lZ= xpy-ypx
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Angular momentum
In quantum mechanics angular momentum is defined by an operator
where r and p are the position and momentum operators respectively. In
particular, for a single particle with no electric charge and no spin, the angular
momentum operator can be written in the position basis as
This orbital angular momentum operator is the most commonly encountered
form of the angular momentum operator, though not the only one. It satisfies the
following canonical commutation relations:
,
It follows, for example,
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Relations in Angular momentum
L2 is the norm and Lx, Ly, Lz are the projections.
[Jx, Jy]= i Jz
[Jx, J2]= 0
[Jy, Jz]= i Jx eigenfunctions of J2 are j(j+1)
[Jz, Jx]= i Jy eigenfunctions of Jz are m
J2Ij,m> = j(j+1) Ij,m> and Jz Ij,m> = m Ij,m>
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In atomic units, h =1
Introducing J+ = Jx+iJy and J- = Jx-iJy
J2 = Jx2 + Jy2 + Jz2 = Jz2 + ½ (J+J- + J-J+ )
J+J- and J-J+ operators have |j,m> as eigenfunction:
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J+J- |j,m> = (j+m)(j-m+1) |j,m>
J-J+ |j,m> = (j+m+1)(j-m) |j,m>
spin
• For an electron: s =1/2 → S2=s (s +1) =3/4
and ms=±1/2.
• A spin vector is either a |s,ms> = |1/2,1/2>
or b: |s,ms> = |1/2,-1/2>
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spin
For several electrons: the total spin is the vector sum of the
individual spins.
• For the projection Sz=Ms = Ss ms
• For the norm S2 = S12+S22+2S1S2
S2=[S1z2+S1z+S1-S1+]+[S2z2+S2z+S2-S2+]+[(S1+S2- ++S1S2+)+2S1zS2z]
• For 2 electrons, there are 4 spin functions: a(1) a(),
b(1) b(), a(1) b() and a() b(1) that are solutions of Sz.
Are they solution of S2?
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a(1)a() is solution of S2
a(1) (Sz2+Sz+S-S+) a(2)
[1/2]2 + 1/2 + 0
0.75
a(2) (Sz2+Sz+S-S+) a(1)
[1/2]2 + 1/2 + 0
0.75
2½½
0.5
S-a(1)S+a(2)
0
0
S+a(1)S-a(2)
0
0
2Sza(1)Sza(2)
total
S2 a(1) a() = 2 a(1) a()
2
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a(1)b() is not a solution of S2
a(1) (Sz2+Sz+S-S+) b(2)
[1/2]2 - 1/2 + 1
0.75
b(2) (Sz2+Sz+S-S+) a(1)
[1/2]2 + 1/2 + 0
0.75
2 ½ (-½)
-.5
S-a(1)S+b(2)
b(1) a(2)
b(1) a(2)
S+a(1)S-b(2)
0
0
2Sza(1)Szb(2)
total
S2 a(1) b() = a(1) b() + b(1) a()
2
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a(1)b()± b (1) a () are solutions of S2
S2 a(1) b() = a(1) b() + b(1) a()
S2 b(1) a() = b(1) a() + a(1) b()
S2 (a(1) b() + b(1) a()) = 2 (a(1) b() + b(1) a())
S2 (a(1) b() - b(1) a()) = 0 (a(1) b() - b(1) a())
The spinorbitals must then be a product of such expressions by a
constant (a spatial term which is a constant for the spin operator).
We should be able to separate the spatial and the spin contributions.
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Use of indentation operators
upon vector sums
S2=Sz2+Sz+S-S+
S-S+ |1,1> = 0
S-S+ |1,0> = (1+0+1)(1-0) |1,0> = 2|1,0>
S-S+ |1,-1> = (1-1+1)(1+1) |1,-1> = 2 |1,-1>
Singlet S2|0,0> = [0+0+0] =0 |0,0>
Triplet S2|1,1> = [1+1+0] |1,1>
Triplet S2|1,0> = [0+0+2] |1,0>
Triplet S2|1,-1> = [1-1+2] |1,-1>
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S2 as an operator on determinant
S2(D)= Pab(D) + [(na-nb)2 +2na +2nb ] D
Pab is an operator exchanging the spins :a  b
na and nb are the # of electrons of each spin
Application:
S2(IaaI) = IaaI+IaaI = -IaaI+IaaI = 0
S2(IabI) = IabI+IabI
S2(IabI-IbaI) = 2 (IabI+IabI)= 2 (IabI-IbaI)
S2(IabI+IbaI) = IabI+IbaI+IabI+IbaI
= -IbaI-IabI+IabI+IabI = 0
53
A single Slater determinant is not
necessarily an eigenfunction of S2
Separation of spatial and spin functions
IabI= a(1)a(1) b(2)b(2) - b(1)b(1) a(2)a(2)
= a(1)b(2) a(1)b() - b(1)a(2) a()b(1)
is not a eigenfunction of S2 if ab
54
A combination of Slater
determinant then may be an
eigenfunction of S2
IabI± IbaI = a(1)b(2) a(1)b() - b(1)a(2) a()b(1)
± b(1)a(2) a(1)b() +- a(1)b(2) a()b(1)
IabI+ IbaI = [a(1)b(2) + b(1)a(2)] (a(1)b() - a()b(1))
IabI- IbaI = [a(1)b(2) - b(1)a(2)] (a(1)b() + a()b(1))
are a eigenfunction of S2
The separation of spatial and spin functions is possible
If the spatial function (a or b) is associated with opposite spins
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UHF: Variational solutions are not
eigenfunctions of S2
Iab’I+ Iba’I = a(1)b’(2) a(1)b() - b’ (1)a (2) a()b(1)
+ b(1)a’(2) a(1)b() – a’(1)b(2) a()b(1)
=[a(1)b’(2) + b(1)a’(2)](a(1)b() - [a’(1)b(2) + b’(1)a(2)] a()b(1))
Not equal
The separation of spatial and spin functions is not possible
If 2 spatial functions (a  a’; b  b’) are associated with
opposite spins
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