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I am not sure I fully understood your question, but I will give it a try.

The hydrogen molecule in its ground state only contains two electrons of opposite spin, which means that the exchange (Fermi) contribution to the Hamiltonian (and therefore, the wavefunction, which generates the density) of either electron has to exactly cancel the (unphysical) Coulomb contribution to the Hamiltonian generated by that same electron.

That is, you can just take either converged orbital and square it to get the difference in probability density of the Fermi hole, which in turn equals the alpha- or beta- electron density.

As for the Coulomb hole, you can take the density difference between the solution of the correlated method and the Hartree-Fock solution, which only treats electron potentials generated by opposite spin electrons as an average.

Depending on how precise you want to be I would choose Full-CI vs insteadinstead of DFT because in F-CI there is a one to one-to-one correspondence between the kinetic energy and the rest of Hamiltonian terms with HF of each Hamiltonian term, as opposed towhereas DFT, which approximates and then corrects some of these operators in a way that breaksways that exactbreak the mathematical correspondence. Full-CI for a two-electron system only scales as O(N^4) with the size of the basis set, but even sothen most calculations for a linear molecule can be eliminatedcancelled out by symmetry.

You can use practically any QC program, as most of them output the AO basis weights for the solution orbitals or densities, and then use your favorite plotting tool (GNUplot, Mathematica, etc.) to visualize it. You might also need the weights and exponents used for the basis set, which you can often get from the program, but can also be found here:

https://www.basissetexchange.org/

I am not sure I fully understood your question, but I will give it a try.

The hydrogen molecule in its ground state only contains two electrons of opposite spin, which means that the exchange (Fermi) contribution to the Hamiltonian (and therefore, the wavefunction, which generates the density) of either electron has to exactly cancel the (unphysical) Coulomb contribution to the Hamiltonian generated by that same electron.

That is, you can just take either converged orbital and square it to get the difference in probability density of the Fermi hole, which in turn equals the alpha- or beta- electron density.

As for the Coulomb hole, you can take the density difference between the solution of the correlated method and the Hartree-Fock solution, which only treats electron potentials generated by opposite spin electrons as an average.

Depending on how precise you want to be I would choose Full-CI vs instead of DFT because there is a one to one correspondence between the kinetic energy and the rest of Hamiltonian terms with HF, as opposed to DFT, which approximates and then corrects some of these operators in a way that breaks that exact correspondence. Full-CI for a two-electron system only scales as O(N^4) with the size of the basis set, but even so most calculations can be eliminated by symmetry.

You can use practically any QC program, as most of them output the AO basis weights for the solution orbitals or densities, and then use your favorite plotting tool (GNUplot, Mathematica, etc.) to visualize it. You might also need the weights and exponents used for the basis set, which you can often get from the program, but can also be found here:

https://www.basissetexchange.org/

I am not sure I fully understood your question, but I will give it a try.

The hydrogen molecule in its ground state only contains two electrons of opposite spin, which means that the exchange (Fermi) contribution to the Hamiltonian (and therefore, the wavefunction, which generates the density) of either electron has to exactly cancel the (unphysical) Coulomb contribution to the Hamiltonian generated by that same electron.

That is, you can just take either converged orbital and square it to get the difference in probability density of the Fermi hole, which in turn equals the alpha- or beta- electron density.

As for the Coulomb hole, you can take the density difference between the solution of the correlated method and the Hartree-Fock solution, which only treats electron potentials generated by opposite spin electrons as an average.

Depending on how precise you want to be I would choose Full-CI instead of DFT because in F-CI there is a one-to-one correspondence with HF of each Hamiltonian term, whereas DFT approximates and then corrects some of these operators in ways that break the mathematical correspondence. Full-CI for a two-electron system only scales as O(N^4) with the size of the basis set, but even then most calculations for a linear molecule can be cancelled out by symmetry.

You can use practically any QC program, as most of them output the AO basis weights for the solution orbitals or densities, and then use your favorite plotting tool (GNUplot, Mathematica, etc.) to visualize it. You might also need the weights and exponents used for the basis set, which you can often get from the program, but can also be found here:

https://www.basissetexchange.org/

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paperclip optimizer
paperclip optimizer
  • 141
  • 3

I am not sure I fully understood your question, but I will give it a try.

The hydrogen molecule in its ground state only contains two electrons of opposite spin, which means that the exchange (Fermi) contribution to the Hamiltonian (and therefore, the wavefunction, which generates the density) of either electron has to exactly cancel the (unphysical) Coulomb contribution to the Hamiltonian generated by that same electron.

That is, you can just take either converged orbital and square it to get the difference in probability density of the Fermi hole, which in turn equals the alpha- or beta- electron density.

As for the Coulomb hole, you can take the density difference between the solution of the correlated method and the Hartree-Fock solution, which only treats electron potentials generated by opposite spin electrons as an average.

Depending on how precise you want to be I would choose Full-CI vs instead of DFT because there is a one to one correspondence between the kinetic energy and the rest of Hamiltonian terms with HF, as opposed to DFT, which approximates and then corrects some of these operators in a way that breaks that exact correspondence. Full-CI for a two-electron system only scales as O(N^4) with the size of the basis set, but even so most calculations can be eliminated by symmetry.

You can use practically any QC program, as most of them output the AO basis weights for the solution orbitals or densities, and then use your favorite plotting tool (GNUplot, Mathematica, etc.) to visualize it. You might also need the weights and exponents used for the basis set, which you can often get from the program, but can also be found here:

https://www.basissetexchange.org/