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Hartree method (Read 2407 times)
Gerrit-Jan Linker
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Posts: 75
Hartree method
29.08.09 at 14:42:01
 
Hartree method
 
Using the molecular Hamiltonian in the Born-Oppenheimer approximation the nuclear coordinates appear as parameters and the nuclear repulsion is a constant independent of the electron coordinates.
 
The electron Hamiltonian then becomes:
Ĥ = ∑ie(i) + ∑i,KeN(i,K) + ∑i>jee(i,j)  
 
This hamitonian describes the motion of electrons in the field of the nuclei and the other electrons.
 
Using the independent particle approximation the wave function can be written als a Hartree product:
ψHP(r1,r2,...,rN)=Ф1(r12(r2)...ФN(rN)  
 
The energy of a system is then obtained by:
E = <ψ|Ĥ|ψ> / <ψ|ψ> or assuming the wave function is normalised: E = <ψ|Ĥ|ψ>
 
Example: one electron, one nucleus (e.g. H)
E = <ψ|Ĥ|ψ>
 = <Ф1(1) | e(1) + eN1(1) |Ф1(1)>
 = <Ф1(1) | ĥ(1) |Ф1(1)>
 = h11 = The electron in the field of the nucleus.
 
Example: two electrons, one nucleus (e.g. H-)
E = <ψ|Ĥ|ψ>
 = <Ф1(1)Ф2(2) | e(1) + e(1) + eN1(1) + eN1(2) + ee(1,2) | Ф1(1)Ф2(2)>
 = <Ф1(1) | e(1) + eN1(1) |Ф1(1)> <Ф2(2)|Ф2(2)> +
  <Ф2(2) | e(2) + eN1(2) |Ф2(2)> <Ф1(1)|Ф1(1)> +
  <Ф1(1)Ф2(2) | ee(1,2) | Ф1(1)Ф2(2)>
 = <Ф1(1) | ĥ(1) |Ф1(1)> +
  <Ф2(2) | ĥ(2) |Ф2(2)> +
  <Ф1(1)Ф2(2) | ĝ(1,2) | Ф1(1)Ф2(2)>
 = h11 + h22 + J12
 
h11 : Energy of electron 1 in the field of the nuclei
h22 : Energy of electron 2 in the field of the nuclei
J12 : Classical Coulomb energy for electrons 1 and 2.
 
See also:
Molecular Hamiltonian
http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1251548847/0#0
Born Oppenheimer approximation
http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1199304972
Hartree product
http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1211044354
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« Last Edit: 29.08.09 at 15:05:18 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
Linker IT Software
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Gerrit-Jan Linker
YaBB Administrator
*****




Posts: 75
Hartree method: Energy of a many electron system
Reply #1 - 10.09.09 at 07:49:21
 
Hartree method: Energy of a many electron system
 
For an arbitrary system with N electrons the energy within the Hartree method is given by:
 
E = <ψ|Ĥ|ψ>  
 = ∑Nii | ĥ |Фi> + ∑Ni≠ji Фj | Фi Фj>  
 
 
We can rewrite the 2 electron integrals changing the indexes in the sum:
 
Ni≠ji Фj | Фi Фj> = ∑Ni,ji Фj | Фi Фj> - ∑Nii Фi | Фi Фi>  
 
where  
Ni≠j in which i is unequal to j is excluded from the sum
Ni,j in which i equal to j is inclusive
Ni in which i is equal to j
 
 
Using this we can rewrite the expression for the energy as:
 
E = ∑Nii | ĥ |Фi> + ( ∑Ni,ji Фj | Фi Фj> - ∑Nii Фi | Фi Фi> )  
  = ∑Nii | ĥ |Фi> + ∑Ni,ji Фj | Фi Фj> - ∑Nii Фi | Фi Фi>  
 
The 2 electron integrals can be rewritten in terms of electron densities. See:
Coulomb integrals
http://www.oraxcel.com/cgi-bin/yabb2/YaBB.pl?num=1251370990/0#0
 
Jij = <ΦiΦjiΦj> = (ΦiΦijΦj) = ∫ ρi(r1) r12-1 ρj(r2) dr1dr2
 
Using this we rewrite the energy expression as:
 
E = ∑Nii | ĥ |Фi> + ∑Ni,j ∫ ρi(r1) r12-1 ρj(r2) dr1dr2 - ∑Nii Фi | Фi Фi>  
  = ∑Ni hii + ∑Ni,j ∫ ρi(r1) r12-1 ρj(r2) dr1dr2 - ∑Nii Фi | Фi Фi>  
  = ∑Ni hii + ∫ [∑Ni ρi(r1)] r12-1 [∑Nj ρj(r2)] dr1dr2 - ∑Nii Фi | Фi Фi>  
 
Using total electron density ρ(r) = ∑Ni ρi(r),
 
E = ∑Ni hii + ∫ ρ(r) r12-1 ρ(r) dr1dr2 - ∑Nii Фi | Фi Фi>  
 
 
where
Ni hii is the sum of one electron energies in the field of all nuclei for all N electrons  
∫ ρ(r) r12-1 ρ(r) dr1dr2 is the energy of classical Coulomb repulsion of the electron cloud with itself  
Nii Фi | Фi Фi> is the self-interaction energy. It compensates for the self-repulsion of the individual electrons.
 
 
To summarise:
The total energy in the Hartree method of an arbitrary many electron system:
The sum of the energies of all the electrons in the field of all nuclei corrected for the classical repulsion between the electrons.
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« Last Edit: 10.09.09 at 08:52:26 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
Linker IT Software
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