The energy in the Hartree-Fock approximation is given as:
EHF=⟨ψHF|ˆH|ψHF⟩=∑i,jPi,jHcorei,j+12∑i,j,k,lPi,jPl,k(ij||kl)+VNN
The geometrical derivative of the Hartree-Fock energy can be shown to be [1]:
∂EHF∂XA=∑i,jPi,j∂Hcorei,j∂XA+12∑i,j,k,lPi,jPl,k∂(ij||kl)∂XA−∑i,jQi,j∂Si,j∂XA+∂VNN∂XA
The Hellmann-Feynman Theorem states:
dEdXA=⟨ψ|dˆHdXA|ψ⟩
The Hellmann-Feynman Theorem implies that the energy derivative only depends on the parts of the Hamiltonian that have a dependency on the derivative. This leads to the geometrical derivative being equal to:
dEdXA=⟨ψ|dˆVNNdXA+dˆVNedXA|ψ⟩=⟨ψ|dˆVNNdXA|ψ⟩+⟨ψ|dˆVNedXA|ψ⟩=⟨ψ|ZA∑BZB(XB−XA)R3AB|ψ⟩+⟨ψ|−ZA∑iXi−XAr3iA|ψ⟩
Now the Hellmann-Feynman theorem can be applied to find the derivative geometrical of the Hartree-Fock energy:
∂EHF,Hellmann−Feynman∂XA=⟨ψHF|∂ˆH∂XA|ψHF⟩=∑i,jPi,j∂VNe,ij∂XA+∂VNN∂XA
I tried to implement equation (2) and equation (3), and used them to calculate, the force between H and Li for different distances, with cc-pVDZ and cc-pVTZ. and got the following:
Hartree-Fock refers to equation (2) and Hellmann-Feynman refers to equation (3). I have read that the Hellmann-Feynman theorem is only valid for exact solutions, but should become a better and better approximation, when going towards the Hartree-Fock limit. As can be seen in the picture, equation (3) performs worse with the larger basisset cc-pVTZ. This now leads me to my question, were am I wrong in my understanding of the application of the Hellmann-Feynman theorem to the Hartree-Fock approximation?
[1] Attila Szabo, Neil S. Ostlund; Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory; equation C. 12
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