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Many Electron AtomsMany Electron Atoms
Prof. Dr. Juan Ignacio Rodríguez Hernández
Escuela Superior de Física y MatemáticasInstituto Politécnico Nacional
Mexico City, Nov 2017
1
Many Electron AtomsMany Electron Atoms
2
{−ℏ
2
2mn
∇R2−∑
i=1
Nℏ
2
2m∇ r i
2−∑
i=1
N1
4 πε0
e2
|R− r j|+∑
i=1
N
∑j>i
N1
4 πε0
e2
|r i− r j|}Ψ( R , r i)=EΨ( R , r i)
¿
H Ψ=EΨ
For gold N=79, so we have 3*79=237 independent variables !!!
Atomic Units (a.u.)Atomic Units (a.u.)
3
{−ℏ
2
2 μ∇ r
2−
e2
4 πε0
1r}ψ ( r )=Eψ ( r )
{−12∇r '
2−
1r '}ψ ( r )=Eψ ( r )
r '=ra0
Considering the scaling:
Many electron atom HamiltonianMany electron atom Hamiltonianin a.u.in a.u.
{−ℏ
2
2mn
∇ R2−∑
i=1
N12∇ r i
2 −∑i=1
Ne2
|R− r i|+∑
i=1
N
∑j>i
Ne2
|r i− r j|}Ψ ( R , ri)=EΨ( R , ri)
¿
4
{∑i=1
N −∇ i2
2−∑
i=1
NZr i+∑
i=1
N−1
∑j> i
N1
|r i−r j|}Ψ(r i)=EΨ(r i)
{−ℏ
2
2mn
∇R2−∑
i=1
Nℏ
2
2m∇ r i
2−∑
i=1
N1
4 πε0
e2
|R− r j|+∑
i=1
N
∑j>i
N1
4 πε0
e2
|r i− r j|}Ψ( R , ri)=EΨ( R , ri)
¿
Since the nucleus is much more massive than electrons, let's suppose the nucleus is at rest at the framework origin:
Many electron atom HamiltonianMany electron atom Hamiltonianin a.u.in a.u.
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{∑i=1
N −∇ i2
2−∑
i=1
NZri+∑
i=1
N−1
∑j> i
N1
|ri−r j|}Ψ(ri)=EΨ(ri)
{∑i=1
N
hi+∑i=1
N−1
∑j=1
N1
|ri− r j|}Ψ (ri)=EΨ(ri)
Where:
hi≡−∇ i
2
2−Zri
Many electron WAVE FUNCTION Many electron WAVE FUNCTION
6
How to guess the total wave function of a N electron atom in the variational method?
Ψ(r1, r2 ,... , rN )
That is, how to guess the total wave function of a system of electrons (fermions)?
Which general properties must satisfy?
Ψ(r1, r2 ,... , rN )
Stern-Gerlach Experiment (1922)Stern-Gerlach Experiment (1922)
Beam of Hydrogen or Silver atoms
The beam of identical particles splits into 2 beams under identical conditions (magnetic field)!!!!!
Duble Lines in Sodium SpectrumDuble Lines in Sodium Spectrum
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s2≡ sx
2+ s y
2+ sz
2
[ sx , s y ]= iℏ sz [ s y , s z ]=i ℏ sx [ sz , sx ]=i ℏ s y
s2 f=ℏ2 s( s+1) f s=12
s z f=ℏm s f ms=−12,12
Spin angular momentum Spin angular momentum
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s2 f=ℏ2 s( s+1) f ; s=12
s z f=ℏm s f ; mM=−12,
12
Spin angular momentum Spin angular momentum
s z f 1/2=ℏ12f 1/ 2
s z f −1/2=−ℏ12f−1/2
s zα=ℏ12α
s z β=−ℏ12β
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s2 f=ℏ2 s( s+1) f ; s=12
s z f=ℏm s f ; m s=−12,12
A postulate: A postulate: Spin is an intrinsic variable Spin is an intrinsic variable
s zα=ℏ12α s z β=−ℏ
12β
s=√32ℏ
ms=−12,12
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Spin Variable Spin Variable
α (m s)=δms 1/2 β(ms )=δms−1 /2
ms=1/2,−1/2
α(m s=1/2)=1 α(m s=−1/2)=0
β(ms=1/2)=0 β(ms=−1/2)=1
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Orthonormality ofOrthonormality of spin funcitons spin funcitons
s zα=12α
s z β=−12β
∑ms=−1/ 2
1/2
α*(ms )α (ms )=1
∑ms=−1/ 2
1/2
β* (ms ) β (ms )=1
∑ms=−1/ 2
1/2
α* (ms ) β (ms )=0
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Spin orbitals Spin orbitals
ψ ( r )=ψ ( x , y , z ) ψ ( r ,m s )=ψ ( x , y , z ,ms )
ψ( x)=ψ ( r ,m s )=ψ ( x , y , z ,ms )
x≡( r ,ms)
That is, the electron now has 4 coordinates (3 spatial and one of spin):
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Orthomalized Spin orbitals Orthomalized Spin orbitals
⟨ψ i|ψ j ⟩=∫ψ i¿( r ,ms )ψ j '( r ,ms )dτ≡∑
ms
∫φi¿( r ) gmsi
¿(ms )φ j¿( r )gmsj¿ (ms )d r
=[∫φi¿( r )φ j( r )d r ]∑
m s
gmsi¿(ms )gmsj
(ms)=δij δmsimsj
If φi form an orthonormalized set then so do the spin orbitals ψi ‘s
∫d x=∫ d τ= ∑ms=−1 /2
1/2
∫ℜ3
d r
Notice that now the integration over the electron coordinates implies a sum over spin and a integral over the whole space:
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Spin Statistical Theorem Spin Statistical Theorem
The wave function of a system of identical fermions (bosons) must be antisymmetric (symmetric) under exchange of the coordinates of any two of such fermions (bosons).
A total wave function is antisymmetric if:
Ψ( x1, ... , x i−1 , x i , x i+1, ... , x j−1 , x j , x j+1 , ... , xN )=−Ψ( x1 , ... , x i−1 , x j , xi+1 , ... , x j−1 , x i , x j+1 , ..., xN )
Variational TheoremVariational TheoremGiven a system whose Hamiltonian operator is time independent and whose lowest-energy eigenvalues is E0, if Ф is any normalized-well behaved function of the coordinates of the system’s particles that satisfies the boundary condiition of the problem, then
⟨φ|H|φ ⟩≥E0
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Variational TheoremVariational Theorem
. .. . . .
{ 1}
ˆ ˆming sg s g sE H H
Ψ g. s The ground state function is the eigenfunction with the lowest eigenvalue (ground state energy)
is the function that minimaze the (energy) functional:
1
Ψ g. s
Minimization Constrain:
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Hartree-Fock ApproximationHartree-Fock Approximationfor the many-electron atom for the many-electron atom
Instead of solving:
One minimizes:
F [Ψ ]=⟨Ψ|H|Ψ ⟩
H=−∑i=1
N12∇ r i
2 −∑i=1
N ZN
r i+∑
i=1
N
∑j>i
N1
|r i− r j|
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{∑i=1
N −∇ i2
2−∑
i=1
NZr i+∑
i=1
N−1
∑j> i
N1
|r i−r j|}Ψ(r i)=EΨ(r i)
Ψ is a Slater determinant
HF wave function: HF wave function: Slater Determinant Slater Determinant
Spin Orbital concept:
ui ( x j )≡ψ i( x j )≡φi( r j )g (m s1 )
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The Slater determinant is an antisymmetric (total) wave function of a system of electrons (N electron atom):
Closed shell restricted Closed shell restricted Hartree-Fock Approximation Hartree-Fock Approximation
N is even and there is always a α and β spin orbitals for each spacial orbital :
Function Vector Space
HF (Slater determinant) space 21
Ψ( x1 , ... , x N)=1
√N ! (ϕ1(1)α(1) ϕ1(1)β(1) ⋯ ϕN /2(1)β(1)ϕ1(2)α(2) ϕ1(2)β(2) ⋯ ϕN /2(2)β(2)⋮ ⋮ ⋱ ⋮
ϕ1(N )α(N ) ϕ1(N )β(N ) ⋯ ϕN /2(N )β(N ))
HF ApproximationHF Approximation
F [Ψ ]=⟨Ψ|H|Ψ ⟩=⟨Ψ|−∑i=1
N12∇ r i
2−∑
i=1
N ZN
r i
+∑i=1
N
∑j>i
N1
|r i− r j||Ψ ⟩
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F [Ψ ]=⟨Ψ|H|Ψ ⟩=⟨Ψ|∑i=1
N
hi+∑i=1
N
∑j>i
N
gij|Ψ ⟩
hi≡−12∇ i
2−Z N
rigij≡
1|r i− r j|
F [Ψ ]=⟨Ψ|H|Ψ ⟩=∑i=1
N
⟨u i(1)|f 1|ui(1 )⟩
+∑i∑j>i
[ ⟨ui(1 )u j(2 )|g12|ui(1 )u j(2)⟩−⟨ui (1)u j(2 )|g12|ui (2)u j(1 )⟩ ]
HF ApproximationHF Approximation
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⟨Ψ|∑i
f i|Ψ ⟩=∑i=1
N
⟨u i(1)|f 1|ui(1 )⟩
⟨Ψ|∑i∑j>i
g ij|Ψ ⟩=∑i
N
∑j>i
N
[ ⟨u i(1)u j (2)|g12|u i(1)u j (2)⟩−⟨ui(1 )u j(2)|g12|u i(2 )u j(1)⟩ ]
⟨Ψ|∑i
f i|Ψ ⟩=2∑i=1
N /2
⟨φ i(1)|f 1|φi (1)⟩
⟨Ψ|∑i∑j>i
g ij|Ψ ⟩=∑i
N /2
∑j>i
N /2
[ 2J ij−K ij ]
J ij≡⟨φi (1)φ j (2)|g12|φi(1 )φ j(2 )⟩ J ij≡⟨φi (1)φ j (2)|g12|φi(2 )φ j(1 )⟩
Minimizing the HF functionalMinimizing the HF functional
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F [Ψ ]=⟨Ψ|H|Ψ ⟩=2∑i=1
N /2
⟨φ i(1)|f 1|φi (1)⟩+∑i=1
N /2
∑j=1
N /2
[2 J ij−K ij ]=F [φi ]
A necessary condition for the φi ‘s that minimize F[φi] is
δFδφi
=0 ; i=1, .. . ,N /2
Variational derivatives
Restricted-close shell HF Restricted-close shell HF Equations Equations
(−12∇ 2−
Zr+ J (r )− K (r ))φi( r )=ε iφi( r ) ; i=1, . .. , N /2
J ϕi( r )≡2ϕi( r )∑j=1
N /2
∫ℜ3
|ϕ j( r ' )|2d r '
|r− r '|
K ϕi( r )≡∑j=1
N /2
ϕj( r )∫ℜ3
ϕi( r ')ϕj*( r ')d r '
|r− r '|
Coulomb operator
Exchange operator
Restricted-close shell HF Restricted-close shell HF Equations Equations
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f φi( r )=εi φi( r ) ; i=1, . .. , N /2
f≡−12∇
2−Zr+J (r )−K (r )≡−
12∇
2−Zr+vHF
vHF≡J (r )−K (r )
Fock operator:
Hartree-Fock potenatial (mean field)
ε i Orbital HF energies
Restricted-close shell HF Restricted-close shell HF Equations Equations
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(−12∇2−
Zr+ J (r )−K (r ))φi ( r )=εi φi( r ) ; i=1,. . ., N /2
● ONE-ELECTRON equations!!!!
● Non linear integral-differential equations
● Coupled equations
We have “psedudoseparated“the n-body problem!!!
Mean field: HF potentialMean field: HF potential
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(−12∇ 2−
Zr+ J (r )− K (r ))φi( r )=ε iφi( r ) ; i=1, . .. , N /2
(h+vHF ))φi ( r )=εiφi ( r ) ; i=1, .. . ,N /2
vHF≡J−K
Where
It is the so called Hartree-Fock potential, which is a mean field, that is, a potential that acts over one electron wave function (orbital) but is has the interaction of that electron with all others electrons in an “average” way
Solving the HF equations:Solving the HF equations:Self-Consistent Field (SCF)method Self-Consistent Field (SCF)method
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(−12∇2−
Zr+ J (r )−K (r ))φi ( r )=εi φi( r ) ; i=1,. . ., N /2
I. Guess the initial HF orbitals
II. Construct J and K operators using
III. Solve the HF equations to obtain “new” orbitals
IV. If the new set of orbitals thus obtained are the same than the previous ones under certain criterion (e.g. )the process is said to converge. If not, make
{ϕi0}
{ϕi1}
|E [ϕi1]−E [ϕi
0]|<ϵ≃0
ϕi0=ϕi
1and go to II
{ϕi0} Convergence
criterion
The Hartree-Fock-Roothaan The Hartree-Fock-Roothaan equationsequations
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(−12∇2−
Zr+ J (r )−K (r ))φi ( r )=εi φi( r ) ; i=1,. . ., N /2
Expanding the space orbitals in a “finite” basis set:
B={χ 1 , .. . , χK } Basis set of known and well-behaved functions are the unknown coefficients
K > Nϕi( r )=∑ν=1
K
Cν iχν( r )
Cν i
The HF-Roothaan equationsThe HF-Roothaan equations
31
(−12∇2−
Zr+ J (r )−K (r ))φi ( r )=εi φi( r ) ; i=1,. . ., N /2
HF: A set of DIFFERENTIAL non-linear equations:
HF-Roothann: A set of ALGEBRAIC non-linear equations:
FHFC=SC ϵ
The HF-Roothaan equationsThe HF-Roothaan equations
32
Fock operator matrix representation in the basis set:
Overlap matrix:
FHF=(⟨ χ1|f|χ1 ⟩ ⟨χ1|f|χ2⟩ ⋯ ⟨χ1|f|χK ⟩
⟨ χ2|f|χ1 ⟩ ⟨χ2|f|χ2⟩ ⋯ ⟨χ2|f|χK ⟩⋮ ⋮ ⋱ ⋮
⟨χK|f|χ1⟩ ⟨χK|f|χ2 ⟩ ⋯ ⟨χK|f|χK ⟩)
S=(⟨χ1||χ1⟩ ⟨ χ1||χ2⟩ ⋯ ⟨ χ1||χK ⟩
⟨χ2||χ1⟩ ⟨ χ2||χ2⟩ ⋯ ⟨ χ2||χK ⟩⋮ ⋮ ⋱ ⋮
⟨χK||χ1⟩ ⟨χK||χ2⟩ ⋯ ⟨χK||χK ⟩)
The “unknown” matricesThe “unknown” matrices
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Coefficients matrix:
Orbital energy matrix:
C=(C11 C12 ⋯ C1K
C21 C22 ⋯ C2K
⋮ ⋮ ⋱ ⋮CK 1 C K 2 ⋯ CKK
)
ϵ=(ϵ1 0 ⋯ 00 ϵ2 ⋯ 0⋮ ⋮ ⋱ ⋮0 0 ⋯ ϵK
)
The HF-Roothaan equationsThe HF-Roothaan equations
It represents a system of nonlinear algebraic equations
In general, the basis functions are not orthogonal. So S is not always the identity matrix.
represents a generalized eigenvalue problem. The matrices C and represent the eigenvectors and eigenvalues, respectively
The better the quality of the basis set, the better the solution of HFR equaitons
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FHFC=SC ϵ
Solving HF-Roothaan equationsSolving HF-Roothaan equations
35
1 1D SD Step 1: Find matrix D so that
1'C D C
Step 2: Define matrices C’ and F’HF
Step 3:
F 'HF≡D−1 FHF D
F 'HFC '=C ' ε C '−1 F 'HFC '=ε
FHFC=SC ϵ
Solving HF-Roothaan equationsSolving HF-Roothaan equations
36
Step 4: Diagonalized F’ to obtain C’ and obtain C=DC’
Step 5: Obtain the HF orbitals as
FHFC=SC ϵ
ϕi( r )=∑ν=1
K
C ν iχν( r )
Properties: EnergyProperties: Energy
37
E≠∑i=1
N
εi
E=⟨Ψ|H|Ψ ⟩=2∑i=1
N /2
⟨φ i(1)|f 1|φi (1)⟩+∑i=1
N /2
∑j=1
N /2
[2J ij−K ij ]
E=⟨Ψ|H|Ψ ⟩=∑i=1
N /2
εa+∑i=1
N /2
⟨φi (1)|f 1|φi(1 )⟩
Solution of HF-Roothaan Solution of HF-Roothaan equttions equttions
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K > N/2
K atomic orbitals:
C=(C11 C12 ⋯ C1K
C21 C22 ⋯ C2K
⋮ ⋮ ⋱ ⋮CK 1 C K 2 ⋯ CKK
) ϕi( r )=∑ν=1
K
Cν iχν( r )
FHFC=SC ϵ
Solution of HF-Roothaan Solution of HF-Roothaan equttions equttions
39
We have K orbitals and we need only N/2 atomic orbitals, which ones should we choose????
φ1( r )=∑ν=1
K
Cv1 χ ν ( r ) → ε 1
φ2( r )=∑ν=1
K
Cv 2 χ ν( r ) → ε2
⋮
φN /2 ( r )=∑ν=1
K
C vN /2 χν ( r ) → εN /2
⋮
φK ( r )=∑ν=1
K
C vK χ ν( r ) → εK
If ε1<ε 2<. . .<εN /2< .. .<εKφK ( r )=∑
ν=1
K
C v 1 χK ( r ) → εK
φN /2+1 ( r )=∑ν=1
K
C vN/2+1 χν ( r ) → εN /2+1
⋮
φN /2 ( r )=∑ν=1
K
C vN /2 χν ( r ) → εN /2
⋮
φ1( r )=∑ν=1
K
C v1 χ ν ( r ) → ε 1
Solution of HF-Roothaan Solution of HF-Roothaan equttions equttions
ε1<ε 2<. . .<εN /2< .. .<εK
Virtual orbitals
Occupied orbitals
40
φK ( r )=∑ν=1
K
Cv 1 χK ( r ) → εK
φN /2+1 ( r )=∑ν=1
K
C vN/2+1 χν ( r ) → εN /2+1
⋮
φN /2 ( r )=∑ν=1
K
C vN /2 χν ( r ) → εN /2
⋮
φ1( r )=∑ν=1
K
Cv1 χ ν ( r ) → ε 1Dobleoccupation
Aufbauf occupation Aufbauf occupation
ε1<ε 2<. . .<εN /2< .. .<εK
Virtual orbitals
Occupied orbitals
41
φK ( r )=∑ν=1
K
Cv 1 χK ( r ) → εK
φN /2+1 ( r )=∑ν=1
K
C vN/2+1 χν ( r ) → εN /2+1
⋮
φN /2 ( r )=∑ν=1
K
C vN /2 χν ( r ) → εN /2
⋮
φ1( r )=∑ν=1
K
Cv1 χ ν ( r ) → ε 1Dobleoccupation
Solution of HF-Roothaan Solution of HF-Roothaan equttions equttions
42
φK ( r )=∑ν=1
K
Cv 1 χK ( r ) → εK
φN /2+1 ( r )=∑ν=1
⋮
KCvN/2−1 χ ν( r ) → εN /2+1
φN /2 ( r )=∑ν=1
K
C vN /2 χν ( r ) → εN /2
⋮
φ1( r )=∑ν=1
K
Cv1 χ ν ( r ) → ε 1
ε1<ε 2<. . .<εN /2< .. .<εK
Virtual orbitals
Occupied orbitals
HOMO
LUMO
Properties: EnergyProperties: Energy
43
E=⟨Ψ|H|Ψ ⟩=∑i=1
N /2
εa+∑i=1
N /2
⟨φi (1)|f 1|φi(1 )⟩
E=⟨Ψ|H|Ψ ⟩=2∑i=1
occ
⟨φ i(1)|f 1|φi (1)⟩+∑i=1
occ
∑j=1
occ
[2J ij−K ij ]
E=⟨Ψ|H|Ψ ⟩=2∑i=1
N /2
⟨φ i(1)|f 1|φi (1)⟩+∑i=1
N /2
∑j=1
N /2
[2J ij−K ij ]
E=⟨Ψ|H|Ψ ⟩=∑i=1
occ
εa+∑i=1
occ
⟨φi (1)|f 1|φi(1 )⟩
Koopmans TheoremKoopmans Theorem
44
ε a=⟨φa|f|φa ⟩+∑b=1
N /2
(2J ab−K ab)
IP≡EN−1−EN
The ionization potential, IP, is the minimum energy to get off one electron of an atom (or molecule):
IP≡EN−1−EN=⟨ΨDS
N−1| ^H N−1
|ΨDSN−1⟩−⟨ΨDS
N|H N|ΨDS
N⟩=−ϵHOMO=−ϵN /2
Koopmans' theorem states that :
ΨDSN−1
ΨDSN
is the Slater determinant of the atom with N electrons
is the Slater determinant of the atom with N-1 electrons
Koopmans' TheoremKoopmans' Theorem
45
N E−N−1 E=⟨ NΨ|H|N Ψ ⟩−⟨N−1 Ψ a|H|N−1 Ψ a ⟩=−ε a
Given an N-electron Hartree-Fock single determinant with
occupied energies and virtual energies , then the
ionization potential to produce an (N-1) electron single determinant
is equal to
N Ψ
ε a ε rN−1Ψ a
ε a
? Ionization Potential!!
IP=N E−N−1 E
Using Koopmans' TheoremUsing Koopmans' Theorem
46
Atom IP_ExpN 0.469 0.189Mg 0.253 0.281Al 0.214 0.219Si 0.240 0.299P 0.323 0.403
RHF calculations using gaussian program: HF/6-31*
Ionization energies (a.u.)
ϵHOMO
Using Koopmans' TheoremUsing Koopmans' Theorem
47
HF calculations using gaussian program: HF/6-31*
Orbital energies for nitrogen:
IP=ε
Using Koopmans' TheoremUsing Koopmans' Theorem
48
HF calculations using gaussian program: HF/6-31*
Orbital energies for silicon:
IP=ε