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.

5

{∑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

9

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

10

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

11

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

12

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

13

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

14

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):

15

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:

16

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

17

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:

18

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|

19

{∑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 )

20

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 ⟩

22

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

23

⟨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

24

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

26

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

27

(−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

28

(−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

29

(−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

30

(−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

33

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

34

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

38

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