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RHF dissociation problem (Read 3976 times)
Gerrit-Jan Linker
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RHF dissociation problem
25.12.09 at 22:33:59
 
RHF dissociation problem
 
Consider H2 in a minimal basis: one atomic 1s orbital on each atom. Two AOs χ lead to two MOs Φ.
 
The AOs combine to form a bonding and an anti-bonding MO. The lowest MO, the bonding MO, will be doubly occupied in H2.
 
Bonding MO: Φ1
Φ1 = χA + χB
 
Anti-bonding MO: Φ2
Φ2 = χA - χB
 
The ground state wavefunction ψ0 can be expressed as a Slater determinant with 2 electrons in the bonding MO:
ψ0 = Φ1(1)α(1)Φ1(2)β(2)-Φ1(2)α(2)Φ1(1)β(1)
ψ0 = Φ1(1)Φ1(2) [α(1)β(2)-α(2)β(1)]
 
Expanding the spatial part of the MO in the atomic basis functions we obtain:
ψ0 = χA(1)χA(2) + χB(1)χB(2) +χA(1)χB(2) + χB(1)χA(2)
 
In this equation there are:
 
2 ionic terms, two electrons in one AO:
χA(1)χA(2)
χB(1)χB(2)
 
2 covalent terms, two electrons shared betweed two AOs:
χA(1)χB(2)
χB(1)χA(2)
 
Therefore at the RHF level, the wavefunction ψ is 50% ionic and 50% covalent. At all bond lenghts!
 
When we consider a situation far from equilibrium bond length, at a large bond length where there should actually be just wo separate H neutral atoms this 50-50 description is not correct. H2 does not dissociate correctly at the RHF level of theory. It should be 100% covalent at large internuclar separations.
 
The RHF dissociation problem is overcome by chosing the wavefunction as a linear combinaton of the bonding and anti-bonding MO's.
Ψ = a0 ψ0 + a1 ψ1
Ψ = a0 Φ1(1)Φ1(2) + a1Φ2(1)Φ2(2)
Ψ = (a0 + a1) [χA(1)χA(2) + χB(1)χB(2)] + (a0 - a1) [χA(1)χB(2) + χB(1)χA(2)]  
 
At very large separation a1 = - a0
The ionic terms disappear and the molecule dissociates correctly into two neutral atoms.
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« Last Edit: 25.12.09 at 22:58:53 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
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Gerrit-Jan Linker
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Posts: 75
RHF dissociation problem
Reply #1 - 28.12.09 at 18:29:20
 
RHF dissociation problem
 
The problem is explained here by working out the total energy expression.  
 
(other notation for the AO and MO has no significance)
 
Consider H2 in a minimal basis: one atomic 1s orbital on each atom. Two AOs s lead to two MOs σ.
 
The AOs combine to form a bonding and an anti-bonding MO. The lowest MO, the bonding MO, will be doubly occupied in H2.
 
Bonding MO: σg
σg = (sa + sb) / √2(Sab+1)
 
Anti-bonding MO: σu
σu = (sa - sb) / √2(Sab+1)
 
where Sab is the overlap between the atomic orbitals.
 
In the ground state the two electrons reside in σg and the wavefunction is:
ψ0 = |σgσg>
 
The ground state energy is then:
E0(H2) = <ψ0|Ĥ|ψ0> = <ψ0| ĥ1 + ĥ2 + 1/r120>
= <σgσg | ĥ1 + ĥ2 + 1/r12 | σgσg >
 
ĥ1 is the hamiltonian for Ha and ĥ2 is the hamiltonian for Hb
 
Since ĥ1 only works on electron one and ĥ2 only works on electron two we can write:
 
E0(H2) = <σg | ĥ1 | σg> + <σg | ĥ2 | σg> + <σgσg | 1/r12 | σgσg>
= 2 <σg | ĥ1 | σg> + <σgσg | σgσg>
 
We now substitute σg = (sa + sb) / √2(Sab+1)
 
E0(H2) = 2 <(sa + sb) / √2(Sab+1) | ĥ1 | (sa + sb) / √2(Sab+1)> + <(sa + sb)(sa + sb) / 2(Sab+1) | (sa + sb)(sa + sb) / 2(Sab+1)>
 = (2 / 2(Sab+1)) <(sa + sb) | ĥ1 | (sa + sb)> + (1 / 4(Sab+1)2) <(sa + sb)(sa + sb) | (sa + sb)(sa + sb)>
 
Now the dissociation: r→∞, Sab=0
 
E0(H2, r→∞) = <(sa + sb) | ĥ1 | (sa + sb)> + <(sa + sb)(sa + sb) | (sa + sb)(sa + sb)>
 
At r→∞ all bicentric integrals are zero and cancel:
 
E0(H2, r→∞) = <sa | ĥ1 | sa> + <sb | ĥ1 | sb> + <sasa + sbsb | sasa + sbsb>
 = 2 <sa | ĥ1 | sa> + ( <sasa | sasa> + <sbsb | sbsb>)
 = 2 <sa | ĥ1 | sa> + <sasa | sasa>
 
E0(H2, r→∞) = 2 E0(H) + <sasa | sasa>
 
This means that at r→∞ the dissociation energy is overestimated by <sasa | sasa>, a repulsion integral associated with the ionic configuration energy. This configuration has the same weight in the dissociated molecule than at the equilibrium distance. In a correct description of the dissociated molecule it must be null.
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« Last Edit: 29.12.09 at 10:01:35 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
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