questions exchange and correlation

I also have similar questions.

To be specific, there should be some logic to arrive at a guess of exchange correlation function in DFT. Is it so or not?

Hi

I think you can see the possible solutions for solving the problem at hand, in three levels: 1) Hartree 2) Hartree Fock and 3) DFT –> to have increasing accuracy.

(1) Hartree method tries to break down the many body wave function purely as a product of individual wave functions, which is overly simplified and doesn’t necessarily have the anti-symmetric property as well.
(2) The HF method is an improvisation to (1) as it does incorporate the anti-symmetric nature. It is relatively better than (1)
(3) Lastly, the DFT method (provided that we know the exact XC) is likely to find the exact solution. So, it can be identified as the best method.

Also, note that the energy difference between solutions of (2)-(1) is defined as the ‘Exchange energy’, and the energy difference (3)-(2) is defined as the correlation energy.

–> So, ‘exact exchange by definition for the HF method’ means HF method includes the exchange energy.
ie. like I previously mentioned,
(2)-(1)=exchange energy
(2)=(1)+exchange energy
where (2) is HF, (1) is Hartree method.
–> Along the same lines, you’d understand the line: ‘there is no correlation by definition for the HF method’. HF accounts for only the exchange energy difference.

Now, previously I mentioned DFT to yield the exact solution, but that was under the assumption that we know the exact XC. In reality, it is hard to identify the exact XC, and we adopt several approximations (like LDA), which reduces the accuracy of the solution nevertheless.
Ideally, DFT by definition is supposed to be account for both the exchange and the correlation energies.
ie.
(3)-(2)=correlation energy
(3)=(2)+correlation energy
(3)=(1)+exchange+correlation energy

However, since we used an approximation (LDA), it can be interpreted to have only ‘some’ exchange and ‘some’ correlation energy contributions’, and NOT the exact.

Hope this helps.

regards,
Arc