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Introduction Time dependant Schrödinger equation: For atoms and molecules N=number of electrons n=number of nuclei i H t ψ ψ = = 1 ψψ = 2 2 1 1 1 1 () 2 2 N N N i i i i i j i j e H r m r r υ = = =− ∆+ + = K K K 2 () n j j j Z e r r R υ =− K K K
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### Transcript of Introduction - ESIernzerhm/cours/Graduate course/cours1.pdf · Introduction Time dependant...

• Introduction

Time dependant Schrödinger equation:

For atoms and molecules

N=number of electrons

n=number of nuclei

i Htψ ψ∂ =

∂1ψ ψ =

2 2

1 1

1 1( )2 2

N N N

i ii i i j i j

eH rm r r

υ= = ≠

= − ∆ + +−∑ ∑ ∑

2

( )n

j

j j

Z er

r Rυ = −

−∑

• In the following, we use atomic units

The wave function depends on spatial and spin coordinates and on t.

2 1e m= = =

1 1 2 2( , ,... , )N Nr s r s r s tψ ψ= or s α β=

• Time independent Schrödinger equation

We focus on solutions of the form

All the proprieties calculated from such a wave function are independent of time

1 1 1 1( ,..., , ) ( ,..., )iEt

N N N Nr s r s t r s r s eψ ψ−=

• Inserting

into the time-dependent Schrödinger equation yields an equation for ,

1 1( ,..., )iEt

N Nr s r s eψ−

1 1( ,..., )N Nr s r sψ

H Eψ ψ=

• The variational method

We define a function of a function

[ ] HE ϕ ϕϕϕ ϕ

=

[ ]E ϕA function of a function such as is called a functional

We show that the stationary points of are solutions of the time-independent Schrödinger equation.

[ ]E ϕ

• Stationary point means that the variation of E, ( ) upon variationsof is zero to first order, i.e.,

Eδϕ

3[ ]( )

( ) 0E d rE rr

δ δδ ϕδϕ

ϕ= =∫

ϕ

We need to calculate the variation of the energy with respect to a variation of the wave function. This can be done by adding a small change to the wave functionδϕ

ϕ ϕ ϕ→ +∆

• [ ] HE ϕ ϕ ϕ ϕϕ ϕϕ ϕ ϕ ϕ+ ∆ + ∆

+ ∆ =+ ∆ + ∆

Multiplying on both sides with

ϕ ϕ ϕ ϕ+ ∆ + ∆

[ ]EH ϕ ϕ ϕ ϕ ϕ ϕϕ ϕ ϕ ϕ = ++ ∆ + ∆ + ∆ + ∆∆yields:

H H H H

H

ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕ

+ ∆ + ∆ = + ∆ + ∆

+ ∆ ∆

Since we use real trial functions and since H is hermitian,

2H H Hϕ ϕ ϕ ϕ ϕ ϕ∆ + ∆ = ∆

• Now we look at the right-hand-side

[ ]EH ϕ ϕ ϕ ϕϕ ϕ ϕ ϕϕ ϕ+ ∆ + ∆ + ∆+ ∆ + ∆ =

[ ] ( ) 2( )EE rrδϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

δϕ⎧ ⎫

+ ∆ + ∆ + ∆ ∆⎨ ⎬⎩ ⎭

…… { }

Again, we neglect terms that are second order in and we use thefact that

ϕ∆0

( )Erδ

δϕ=

to obtain: [ ] [ ]2 2H H E Eϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ+ ∆ = + ∆

Eliminating zeroth-order contributions yields:

H Eϕ ϕ ϕ ϕ∆ = ∆

• Because is completely arbitrary, we can set ϕ∆

This is the Euler equationcorresdonding to H Eϕ ϕ⇒ = [ ]E ϕ

Hϕ ϕϕ ϕ

The stationary points of

are the solutions of the Schrödinger equation.

• In general for any trial wave function

where is the ground-state energy.If the equal sign holds then is the exact ground-state wave function.

o

HE

ϕ ϕϕ ϕ

ϕ

The variational principle allows us to obtain simplified Schrödingerequations (Euler equations) by employing simple trial functions

0Eϕ

• The linear variation methodWe introduce an orthonormal basis set

withAnd expand the trial wave function in thebasis set:

ϕ

{ } 1i iχ =… i j ijχ χ δ=

i ii

cϕ χ=∑

• then we make the energy stationary

with respect to variations of the expansioncoefficients i.e. ,

Hϕ ϕϕ ϕ

ic

0c H ci i i i

i i

c cj i i i ii i

ddc

χ χ

χ χ

∑ ∑

∑ ∑=

• This yields the basis set representation of the time-independent Schrödinger equation:

1 1

2 2ij

c cH c E c

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

ij i jH Hχ χ=

• Hartree-Fock approximation

1 1 2 2

1 1 1 2 2 2

( , )( ) ( )r s r sr s r s

ψ ψϕ ϕ

=≈

We consider a two-electron wave function and approximate it bya simple trial wave function

( ) ( ) spin partspin orbital spatial orbital

ii i i

i

i rα

ϕ ϕβ

− −⎫

= × ⎬⎭

• A simple trial functionEach electron occupies an orbital:

Disadvantage: this trial wave function is notantisymmetric

i.e. We will see that not satisfying this conditionintroduces large errors, since we are neglecting exchange effects.

1 2(1) (2).... ( )N Nφ ϕ ϕ ϕ= ⋅

(1,2,3... ) (1,3,2... )N Nφ φ≠ −

• Thus, we antisymmetrise the trial function by using the concept of a Slater determinant.

1 1 1

2 2

(1) (2) ( )(1) (2)1(1,2, , )

!(1) ( )N N

N

NN

N

ϕ ϕ ϕϕ ϕ

φ

ϕ ϕ

=…

{ }1 21(1,2, , ) det (1) (2).... ( )

! NN N

Nφ ϕ ϕ ϕ= ⋅…

{ } { }1 11 1det (1).... ( ) ( ) (1).... ( )

! !N NPN Sign P P N

N Nϕ ϕ ϕ ϕ= ∑

• Hartree-Fock equationsNow we use the trial function and the variationalmethod to derive the Hartree-Fock equations

With out resticting the trial function , we can assume that the orbitals are orthonormal, i.e.,

The energy functional becomes

{ }11(1,.. ) det (1).... ( )

! NN N

Nφ ϕ ϕ=

i j ijϕ ϕ δ=

[ ]E Hϕ ϕ ϕ=

φ

φ

• Calculation of where is a Slaterdeterminant

Hφ φ φ

1

1 1 1( )2 2

( , )

N N

i ii i j i j

H rr r

h V r ri i j

υ= ≠

⎧ ⎫= − ∆ + +⎨ ⎬−⎩ ⎭

= =

∑ ∑

1one-body two-bodyoperator operator

( , )N N

i i ji i j

H h V r r= ≠

= +∑ ∑

11

N

ii

h N hφ φ φ φ=

=∑

• 11

N

ii

h N hφ φ φ φ=

=∑

{ } { }

{ } { }

{ } { }

1 1 1

1 1 1

1 1 1

0

( ) ( ) ( ) ( )!

( ) ( ) ( ) ( )!

( ) ( )!

for

N NP P

N NP P

N NP

N sign P P N h sign P P NNN sign P sign P P N h P NN

N P N h P NN

P P

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

=

=

=

= ≠

∑ ∑

∑ ∑

• We obtain

{ } { }1 1 1 1(1)... ( ) (1)... ( )! N NPNN h P N h P NN

φ φ ϕ ϕ ϕ ϕ= ∑

1 11 1

( 1)!!

N N

i i i ii i

N N h hN

ϕ ϕ ϕ ϕ= =

−= =∑ ∑

• Now we calculate

gives non-zero contributions only for

1 2( , ) ( 1) ( , )i ji j

V r r N N V r rφ φ ϕ ϕ≠

= −∑

12(1 )P P P= +

{ } { }1 1 2 1( 1) ( ) ( ) ( , ) ( ) ( )

!(*)

N NP P

N N sign P P N V r r sign P P NN

ϕ ϕ ϕ ϕ−=

=

∑ ∑

{ } { }

{ }

1 1 2 12 1

1 2 1 2

( 1)(*) (1)... ( ) ( , ) (1 ) (1)... ( )!

( 1)( 2)! ( , ) ( , )!

N NP

i j i j i j j iij

N N P N V r r P P NN

N N N V r r V r rN

ϕ ϕ ϕ ϕ

ϕϕ ϕϕ ϕϕ ϕ ϕ

−= −

− −== −

• { } { }

11

1 2 1 2

direct or coulomb term exchange term

( , ) ( , )

N

i ii

N N

i j i j i j j ii j i j

H h

V r r V r r

φ φ ϕ ϕ

ϕϕ ϕϕ ϕϕ ϕ ϕ

=

≠ ≠

=

+ −

∑ ∑

Now we have the energy expression corresponding to the trial function.

• Derivation of the Hartree-Fock equation

{ }1 2 1,

[ , , ]N

N i i i j i j i j j ii i j

E h V Vϕ ϕ ϕ ϕ ϕ ϕϕ ϕϕ ϕϕ ϕ ϕ= + −∑ ∑

Now we are going to vary the orbitals and we require that a smallvariation of the orbitals does not change the energy

αϕ eαϕ ϕ+We replace the orbital by

and calculate the first order change in the energy

{ }11

2 4 0N

e j e j j j ej

E h V Vα α αδ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ=

= + − =∑

• One defines the Fock matrix F by the equation

{ }11

2 4 0

2

N

e j e j j j ej

E h V V

F e

α α αδ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕα

=

= + − =

=

It follows that2

11 2

(2)2

Nj

xj

F h d Fr rϕ

= + +−∑∫

with

1 1 2

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

Nj e

x e jj

F dr r

ϕ ϕϕ ϕ

=

⎛ ⎞= ⎜ ⎟⎜ ⎟−⎝ ⎠∑ ∫

• The elements and can bearbitrary modified without changing theHartree-Fock energy and the HF wavefunction!

i jFϕ ϕ i F αϕ ϕ

jFαϕ ϕ Fα βϕ ϕ

, ,....,i j Nϕ ϕ ϕi

j

ϕϕ

α

β

ϕϕ

,..., ...α βϕ ϕ

F =

i jFϕ ϕ Fα βφ ϕ

occ

occ

inocc

inocc

• Diagonalizing the Fock matrix means that the matrix elementscoupling the occupied and the unoccupied orbitals are zero!Zero coupling matrix elements means that the Hartree-Fockenergy is minimized

( )

1

2

0 0 00 0 00 0 00 0 0

abF

εε

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

occ

occ inocc

inocc

• The HF equations are usually written aseigenvalue problem:

i i iFϕ ε ϕ=

[ ]1 , . . . , NH

Eφ φ

ϕ ϕφ φ

=

this is the Euler equation corresponding to the minimization problem

• Note that the Fock operator F depends already on the occupied orbitals, thus we cannot construct the exact Fock operator unless we know already the HF orbitals.

In order to deal with this problem, we use an iterative procedure to solve the HF equations.

• Iterative proceduce:

Guess for the occupied orbitals0 01 .... Nϕ ϕ

Construct Fock operator

Diagonalize F to get1 1

1 ....n n

Nϕ ϕ+ +

1 ?n ni iϕ ϕ+ =

Stop,calculation converged

yesno

• Details of the Hartree-Fock methodSlater determinant

2 electron-2 orbitals:

{ }1 21(1,2,.. ) (1) (2)... ( )

! NN N

Nϕ ϕ ϕ ϕ= 1 11 , ...r s etc=

1 1 1 1 2 2( ) , ( )r rϕ α ϕ β

1 1 1 1 1 1

1 2 2 1 2 2

( ) ( )1( ) ( )!r rr rN

ϕ α ϕ βϕ

ϕ α ϕ β=

{ }1 1 1 1 2 2 1 2 2 1 1 11 ( ) ( ) ( ) ( )2

r r r rϕ α ϕ β ϕ α ϕ β= −

{ }1 1 1 2 1 2 2 11 ( ) ( )2

r rϕ ϕ α β α β= −

• Orbital is independent of the position of electron 2 there is no correlation between the electrons. The anti symmetry of the wave function is localized in the spin part.Now:

1 1 1 2 2 2( ) , ( )r rϕ α ϕ α

1 1 1 2 1 1

1 2 2 2 2 2

( ) ( )1( ) ( )!r rr rN

ϕ α ϕ αϕ

ϕ α ϕ α=

{ }1 1 1 2 2 2 1 2 2 2 1 11 ( ) ( ) ( ) ( )2

r r r rϕ α ϕ α ϕ α ϕ α= −

{ }1 1 2 2 1 2 2 1 1 21 ( ) ( ) ( ) ( )2

r r r rϕ ϕ ϕ ϕ α α= −

• Now the antisymmetry is in the spatial part of the wave function.Note what happens for !The wave function vanishes, the electrons cannot be close to each other. In this case we have Fermi correlation.

1 2r r=

• Invariance properties of a Slater determinant

In the derivation of the HF equations, we imposed the constraint

{ }1 21 (1) (2)... ( )

! NN

Nϕ ϕ ϕ ϕ=

(*)i j ijϕ ϕ δ=3 ( ) ( )i j i jd r r r ds S Sϕ ϕ= ∫ ∫

ds 1i jαα =∫0i j i jds dsα β β α= =∫ ∫can bei jds S S∫

1i jdsβ β =∫

• Does condition (*) restrict the trial wave function that we use! Would the energy be lower if we allow ?

If we have a set of spatial orbitalsthat are not orthonormal

i.e.

Then we can always find a transformation T such that

3 ( ) ( )i j ijd r r rϕ ϕ δ≠∫

1 1 2 2( ) ( )... ( )N Nr r rϕ ϕ ϕ3 ( ) ( )i j ijd r r rϕ ϕ δ≠∫

3 ( ) ( )i j ijd r r rϕ ϕ δ=∫

• (provided that are linearindependent)

For three matrices A,B and C withA=BC

we have det(A)=det(B)det(C )

( ) ( )ii j jj

r t rϕ ϕ=∑1 1 1

1 1 1 2 1 1 1 1 2 11 2

2 1 2 1

1 11

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

N NN

N NN N N NN

r r r r r rt t tr r

r r rt t

ϕ ϕ ϕ ϕ ϕ ϕϕ ϕ

ϕ ϕ ϕ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ .. ... ( )N Nrϕ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1 1 2 2( ) ( )... ( )N Nr r rϕ ϕ ϕ

• Thus

These two determinants are identical up to aconstant, therefore and

give the same energy !

11 1 1 1 1( ) ... ... ... ... ( ) ... ...

det ... ... ... det ... ... .det ... ... ...... ... ( ) ... ... ... ( )

const

NN N N N N

r t r

r t r

ϕ ϕ

ϕ ϕ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

{ }1 1det ( )... ( )N Nr rϕ ϕ

{ }1 1det ( )... ( )N Nr rϕ ϕ

Furthermore, we see that mixing the occupied orbitals does not change the energy. Therefore, we didn’t need to consider variations of the

orbitals involving occupied orbitals.

• Physical interpretation of the Hartree-Fock method

x

1 , ,

kinetic and potential energy =U =E(due to interaction with the =Hartree term =Exchange energynuclei) of the electrons.

NHF

i i i j i j i j j ii i j i j

E h V Vϕ ϕ ϕϕ ϕϕ ϕϕ ϕ ϕ=

= + −∑ ∑ ∑

1 ( )2

h rυ= − ∆ +

1 2

1 12

Vr r

=−

• The electron density is given by ,so that

This is the electrostatic repulsion energy of the electron density with itself.

i j i jij

U Vϕϕ ϕϕ=∑

1 2

(1) (1) (2) (2)1 1 22

i i j j

ijd d

r rϕ ϕ ϕ γ

=−∑

( )rρ2

1( ) ( )

N

ii

r rρ ϕ=

=∑

3 3 1 21 2

1 2

1 ( ) ( )2

r rU d rd rr r

ρ ρ=

−∫

• This gives non-zero terms only if ,i.e. the electrons 1 and 2 have to have thesame spin.

x i j j iij

E Vϕϕ ϕ ϕ=∑

1 2

(1) (1) (2) (2)1 1 22

i j j i

ijd d

r rϕ ϕ ϕ ϕ

=−∑∫

1 1 2 23 31 2

1 2

( ) ( ) ( ) ( )12

i j j ii j i j

ij

r r r rd rd r S S S S

r rϕ ϕ ϕ ϕ⎧ ⎫⎪ ⎪= ×⎨ ⎬−⎪ ⎪⎩ ⎭

∑ ∫

i jS S=

• 1 1 2 23 31 2

1 2

1 1 2 23 31 2

1 2

( ) ( ) ( ) ( )12

( ) ( ) ( ) ( )12

occi j i j

xij

occi j i j

ij

x x

r r r rE d rd r

r r

r r r rd rd r

r r

E E

α α α αα

β β β ββ

α β

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

= −−

−−

= +

∑ ∫

∑ ∫

• The Fock operator

α βαβ

2

11 2

(2)2

Nj

xj

F h d Fr rϕ

= + +−∑∫

1

1 2

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

N

j ij

x i jF d r r

ϕ ϕϕ ϕ==

∑∫

If there are more than electrons, andorbitals will have different potentials.

We assume that is anup-spin orbital, we obtain

(1)iϕ

2 213

2 11 2

( ) ( )(1) ( )

occ

j jj

x i j

r rF d r r

r r

αα α

α

ϕ ϕϕ ϕ α

==−

∑∫

• What is the signifiance of the orbital energies?

3 3 2 12 1

1 2 1 2

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

i i i

i i

Ni j i j

j

F

h

r rd r d r d dr r r r

ϕ ϕ ε

ϕ ϕϕ ϕ ϕ ϕρ ρ

=

=

+ +− −∑∫ ∫

We consider a Hartree-Fock determinant

{ }11 det (1)... ( )

!HFN N NN

φ ϕ ϕ=

Nϕand empty the orbital to obtain a determinant for N-1 electrons

{ }1 1 11 det (1)... ( 1)

( 1)!N NN

Nφ ϕ ϕ− −= −−

note that this determinantis not optimized

• Then we calculate

Thus we can interpret the orbital energies areapproximations to the ionization potentials(Koopman’s theorem)

1 1HF HF

N NH Hφ φ φ φ− − −

{ }2N N N j N j N j j Nj

h V Vϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ= + −∑

eigenvalue of theFock operator

N N

Nth

N Fε ϕ ϕ= =

• Note that the sum of the orbital energies is not equal to the total Hartree-Fock energy

{ }1 1

2N N N

i i i i j i j i j j ii i ij

h V Vε ϕ ϕ ϕϕ ϕϕ ϕϕ ϕ ϕ= =

= + −∑ ∑ ∑

{ }N

i j i j i jH

j ij

F

iVE Vϕϕ ϕϕ ϕϕ ϕ ϕ−= +∑

IntroductionTime independent Schrödinger equationThe variational methodThe linear variation methodHartree-Fock approximationA simple trial functionHartree-Fock equationsCalculation of where is a SlaterdeterminantDerivation of the Hartree-Fock equationDetails of the Hartree-Fock methodInvariance properties of a Slater determinantPhysical interpretation of the Hartree-Fock methodThe Fock operatorWhat is the signifiance of the orbital energies?