Question

Hartree-Fock orbitals of a periodic crystal -> Bloch waves?

Answer 1

Hello mate here is your answer»«««
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Hamiltonian does not have the form −12∇⃗ 2+V(r⃗ )−12∇→2+V(r→) but also includes the Coulomb and exchange terms:

HHFψi(r⃗ )=−12∇⃗ 2ψi(r⃗ )+V(r⃗ )ψi(r⃗ )+∑j∫d3r′ψ∗j(r⃗ ′)ψj(r⃗ ′)|r⃗ −r⃗ ′|ψi(r⃗ )−∑j∫d3r′ψ∗j(r⃗ ′)ψi(r⃗ ′)|r⃗ −r⃗ ′|ψj(r⃗ )HHFψi(r→)=−12∇→2ψi(r→)+V(r→)ψi(r→)+∑j∫d3r′ψj∗(r→′)ψj(r→′)|r→−r→′|ψi(r→)−∑j∫d3r′ψj∗(r→′)ψi(r→′)|r→−r→′|ψj(r→)

Here V(r⃗ )V(r→) is the periodic potential of the atoms such that V(r⃗ )=V(r⃗ +R⃗ )V(r→)=V(r→+R→) for any lattice vector R⃗ R→ of the crystal.

So my idea is that it should be enough to show that [HHF,TR⃗ ]=0[HHF,TR→]=0 where TR⃗ =eR⃗ ∇⃗ TR→=eR→∇→ is the translation operator. If this is true then HHFHHF and TR⃗ TR→ have the same eigenfunctions which have the form

ψi(r⃗ )=eik⃗ ir⃗ ui(r⃗ )ψi(r→)=eik→ir→ui(r→)

where ui(r⃗ )=ui(r⃗ +R⃗ )ui(r→)=ui(r→+R→) has the same periodicity as the crystal.

However, the problem is that the Coulomb and exchange operator are composed by the orbitals. So I do not really understand how to compute [HHF,TR⃗ ][HHF,TR→] ?
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Hope this will help you.....»»»