Many Electrons Atomsw3.esfm.ipn.mx/~juan/MM/presentaciones/juanMany... · Many Electron Atoms Juan...

Many Electron AtomsMany Electron Atoms

Juan Ignacio Rodríguez Hernández

Escuela Superior de Física y MatemáticasInstituto Politécnico Nacional

Mexico City

August 2016

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

1h 1em 02 e=

{−ℏ

2

2 μ∇ r

2−

e2

4 πε0

1r}ψ ( r )=Eψ ( r )

{−12∇ r

2−1r}ψ ( r )=Eψ ( r )

Atomic Units (a.u.)Atomic Units (a.u.)

4

5 1 15 31 27.2 2.20 10 6.58 10 2.63 10 / .)Hartree eV cm Hz kJ mol � � �

{−12∇ r

2−1r}ψ ( r )=Eψ ( r ) E=−0 .5 a .u .≡−0.5 Hartrees

1Bohr=0 . 529 A=5 . 29×10−11m

Energy:

Lenth:

Mass:me=9 .1095×10−31kg

Charge:

e=1 .6022×10−19C

Stern-Gerlach ExperimentStern-Gerlach Experiment

Beam of Hydrogen or Silver atoms

7

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

8

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

9

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

10

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

11

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

12

Spin orbitals Spin orbitals

{−12∇ 2−

1r}ψ ( r )=Eψ ( r )

ψ ( r )=ψ ( x , y , z ) ψ ( r ,ms )=ψ ( x , y , z ,ms )

The Hamiltonian is not dependent of the spin operator (to first approximation):

ψ ( r ,ms )=φ( r )gms(ms )

ψ ( r ,1 /2)=φ ( r )α

ψ ( r ,−1 /2)=φ ( r ) β

13

Orthomalized Spin orbitals Orthomalized Spin orbitals

⟨ψ i|ψ j⟩=∫ψ i¿( r ,ms )ψ j ' ( r ,ms )dτ≡∑

m s

∫φ i¿( r ) gm si

¿(m s)φ j¿( r )gmsj¿ (ms )d r

=[∫φi¿( r )φ j ( r )d r ]∑

m s

gmsi

¿(ms )gmsj(m s)=δij δm

simsj

If φi form an orthonormalized set then so do the spin orbitals ψi ‘s

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

14

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:

15

Hartree-Fock ApproximationHartree-Fock Approximationfor the many-electron atom for the many-electron atom

−∑i=1

N12∇ r i

2−∑

i=1

N Z N

ri+∑

i=1

N

∑j>i

N1

|r i− r j|¿Ψ ( r i )=EΨ ( r i)

¿¿¿

Instead of solving:

One minimizes:

F [Ψ ]=⟨Ψ|H|Ψ ⟩

H=−∑i=1

N12∇ r i

2−∑

i=1

N Z N

ri+∑

i=1

N

∑j>i

N1

|r i− r j|

16

HF wave function: Slater HF wave function: Slater Determinant Determinant

Spin Orbital concept:

α (m s1 ) if s=1/2

β (ms 1 ) if s=−1/2¿

g (m s1 )=¿

¿

ui ( x j )≡ψ i( x j )≡φi( r j )g (m s1 )

17

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 function :

1 1 1 1 / 2 1

1 2 1 2 / 2 2

1 1 / 2

( ) (1) ( ) (1) ... ( ) (1)

( ) (2) ( ) (2) ... ( ) (2)1

!

( ) ( ) ( ) ( ) ... ( ) ( )

N

N

N N N N

r r r

r r r

N

r N r N r N

�

r r r

r r r

M M Mr r r

Function Vector Space

HF (Slater determinant) space 18

HF Aproximation HF Aproximation

−∑i=1

N12∇ r i

2−∑

i=1

N Zn

r i

+∑i=1

N

∑j>i

N1

|r i− r j|¿Ψ ( r i)=EΨ ( r i)

¿¿¿

Instead of solving:

One minimizes:

F [Ψ ]=⟨Ψ|H|Ψ ⟩

H=−∑i=1

N12∇ r i

2−∑

i=1

N Z N

r i+∑

i=1

N

∑j>i

Ne2

|r i− r j|

19

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 ⟩

20

F [Ψ ]=⟨Ψ|H|Ψ ⟩=⟨Ψ|∑i=1

N

f i+∑i=1

N

∑j>i

N

g ij|Ψ ⟩

f i≡−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

21

⟨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

22

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

/ 2 *( ') ( ') 'ˆ( ) 2'

Nj j

j i

r r drJ r

r r

�

��

r r rr

r r/ 2 ( ) ( ') 'ˆ ( ) ( )

'

Nj i

ij i

r r drK r r

r r

�

��

r r rr r

r r

23

(−12∇2−

1r+ J (r )−K (r ))φ i( r )=ε iφi ( r ) ; i=1, .. . ,N /2

f HF φ i( r )=ε iφi ( r ) ; i=1, .. . ,N /2

f HF≡−12∇ 2−

1r+ J (r )−K (r )≡−

12∇ 2−

1r+V HF (r )

V HF (r )≡ J (r )− K (r )

Fock operator

Hartree-Fock potenatial

ε i Orbital HF energies


24

(−12∇2−


● ONE-ELECTRON equations!!!!

● Non linear integral-differential equations

● Coupled equations

(−12∇

2−

1r+∑

i=1

N /2

2∫φ j( r ' )φ j ( r ' )d r '

|r− r '|−K (r ))φi ( r )=εi φi ( r ) ; i=1, .. . ,N /2

We have separated the n-body problem!!!


25

(−12∇2−


J (r )=∑j≠i

N /2

2∫φ j( r ' )φ j( r ' )d r '

|r− r '|=∫

ρ' ( r ' )d r|r− r '|

ρ '( r )=∑j≠i

φ j( r )φ j( r )

Coulomb potential due to the other electrons !!!!

/ 2 ( ) ( ') 'ˆ ( ) ( )'

Nj i

ij i

r r drK r r

r r

�

��

r r rr r

r rDoes not have classical analogous


26

(−12∇2−


J (r )=∑j≠i

N /2

2∫φ j( r ' )φ j( r ' )d r '

|r− r '|=∫

ρ' ( r ' )d r|r− r '|

ρ '( r )=∑j≠i

φ j( r )φ j( r )

Coulomb operator

/ 2 ( ) ( ') 'ˆ ( ) ( )'

Nj i

ij i

r r drK r r

r r

�

��

r r rr r

r rExchange operator

Solving the HF equation:Solving the HF equation:Self-Consistent Field (SCF) Self-Consistent Field (SCF)

method method

27

(−12∇2−


I. Guess the φi ‘sII. Construct J and K operators III. Solve the HF equationsIV. If the new set of orbitals thus obtained are the same than the

previous ones under certain criterion, the process is said to converge. If not:

V. the HF equations are solved again using the new orbitals to calculate J and K and repeating the process until convergence.

The Hartree-Fock-Roothaan The Hartree-Fock-Roothaan equationsequations

28

(−12∇2−


( ) ( )K

i ir C r

�r r

Expanding the space orbitals in a basis set:

β={χ 1 , .. . , χK } Basis set of known and well-behaved functions

K > N

The HF-Roothaan equationsThe HF-Roothaan equations

29

(−12∇2−


HFF C CS

HF: A set of DIFFERENTIAL non-linear equations:

HF-Roothann: A set of ALGEBRAIC non-linear equations:


30

1 1 1 2 1

2 1 2 2 2

1 2

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ ˆ ˆ

HF HF HFK

HF HF HFKKS

HF HF HFK K K K

F F F

F F FH

F F F

� ��

K

M M M

K

1 1 1 2 1

2 1 2 2 2

1 2

K

K

K K K K

S

� ��

��

K

M M M

K

Fock operator matrix representation in the basis set:

Overlap matrix:

The “unknown” matricesThe “unknown” matrices

31

Coefficients matrix:

Orbital energy matrix:

11 12 1

21 22 1

1 2 1

K

K

K K K

C C C

C C CC

C C C

� ��

K

M M M

K

1

2

0 0

0 0

0 0 k

� ��

K

M M M

K


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

32

HFF C CS

Solving HF-Roothaan equationsSolving HF-Roothaan equations

33

HFF C CS

1 1D SD Step 1: Find matrix D so that

1'C D C

Step 2: Define matrices C’ and F’HF

Step 2:

F 'HF≡D−1 FHF D

F 'HFC '=C ' ε C '−1 F 'HFC '=ε

Solving HF-Roothaan equationsSolving HF-Roothaan equations

34

HFF C CS

1 1D SD Step 1: Find matrix D so that

1'C D C

Step 2: Define matrices C’ and F’HF

Step 2:

F 'HF≡D−1 FHF D

F 'HFC '=C ' ε C '−1 F 'HFC '=ε

Properties: EnergyProperties: Energy

35

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

36

HFF C CS K > N/2

K atomic orbitals:

11 12 1

21 22 1

1 2 1

K

K

K K K

C C C

C C CC

C C C

� ��

K

M M M

K

( ) ( )K

i ir C r

�r r


37

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< .. .<εK

φK ( r )=∑ν=1

K

Cv 1 χK ( r ) → εK

φK−1 ( r )=∑ν=1

K

C vK−1 χ ν ( r ) → εK−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


38

φK ( r )=∑ν=1

K

Cv 1 χK ( r ) → εK

φK−1 ( r )=∑ν=1

K

C vK−1 χ ν ( r ) → εK−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

Properties: EnergyProperties: Energy

39

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 )⟩

Koopman’s TheoremKoopman’s Theorem

40

ε a=⟨φa|f|φa ⟩+∑b=1

N /2

(2J ab−K ab)

N E−N−1 E=⟨ NΨ|H|N Ψ ⟩−⟨N−1 Ψ a|H|

N−1 Ψ a ⟩=−ε a

N−1Ψ a is the N-1 Slater determinant obtained by eliminating

the orbitalφa


41

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 Koopman’s TheoremUsing Koopman’s Theorem

42

Atom HF-631p ExperimentalN 0.469 0.189

Mg 0.253 0.281Al 0.214 0.219Si 0.240 0.299P 0.323 0.403

HF calculations using gaussian program: HF/6-31*

Ionization energies (a.u.)


43


Orbital energies for nitrogen:

IP=ε


44


Orbital energies for silicon:

IP=ε


45

EA=N E−N +1 E=−εr

Given an N-electron Hartree-Fock single determinant with

virtual energy , then :

The Electron affinity potential to produce an (N+1) electron single

Determinant is equal to

N Ψ

ε rN +1 Ψ a

ε r

? Ionization Potential!!

Hartree-Fock ApproximationHartree-Fock Approximationfor molecules for molecules

−∑i=1

N12∇ r i

2−∑

A=1

M12∇ R

A

2−∑

A=1

M

∑i=1

N Z A

|R A− r i|+∑

i=1

N

∑j>i

N1

|r i−r j|¿Ψ ( r i )=EΨ ( r i )

¿¿¿

Schrodinger equation for a molecule:

46

−∑i=1

N12∇ r i

2−∑

i=1

N Z N

ri+∑

i=1

N

∑j>i

N1

|r i− r j|¿Ψ ( r i )=EΨ ( r i)

¿¿¿

ATOMS

Born-Oppenheimer approximation:

−∑i=1

N12∇ r i

2−∑

A=1

M

∑i=1

N Z A

|R A−r i|+∑

i=1

N

∑j> i

N1

|r i−r j|¿Ψ ( r i)=EΨ ( r i )

¿¿¿

MOLECULES


47

(−12∇2−


(−12∇2−∑

A=1

M

∑i=1

N Z A

|R A− r i|+ J (r )−K (r ))φi ( r )=εi φi( r ) ; i=1,. . . ,N /2

Restricted-close shell HF Restricted-close shell HF Equations for moleculesEquations for molecules

48

(−12∇2−∑

A=1

M

∑i=1

N Z A

|R A− r i|+ J (r )−K (r ))φi ( r )=εi φi( r ) ; i=1,. . . ,N /2

HFF C CS


49

φK ( r )=∑ν=1

K

Cv 1 χK ( r ) → εK

φK−1 ( r )=∑ν=1

K

C vK−1 χ ν ( r ) → εK−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

Properties: Electronic EnergyProperties: Electronic Energy

50

Eelectronic≠∑i=1

N

εi

Eelectronic≡⟨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 ]

Eelectronic=⟨Ψ|H|Ψ ⟩=∑i=1

N /2

εa+∑i=1

N / 2

⟨φ i(1)|f 1|φi (1)⟩

Total Energy (Molecular energy)Total Energy (Molecular energy)

51

ETot=Eelectronic+∑A=1

M

∑B>A

M Z A ZB

|RA−RB|

ETot ( R1 ,. . , RM )=Eelectronic( R1 , .. , RM )+∑A=1

M

∑B>A

M Z AZ B

|R A−RB|

The energy of the molecule depends on the nuclear coordinates as paremeters!!!

Molecular ground state energyMolecular ground state energy

52

Eg . s .=min

{R1 , . . , RM }0

ETot( R1 , . ., RM )

{R1 , .. , RM }→ ground state structure

Geometry optimization

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