The Hartree-Fock approximation is used in solving many-body quantum mechanical systems. The problems of these type of systems are the e-e repulsion term in the Hamiltonian and the single electron wavefunction which is unkown.

I understand that:

  1. The e-e repusion term contains two positions variable and is not separable, therefore to solve such problems we need an approximated expression of such term.
  2. By using Fock operator, we can determined the single electron wavefunction and solve the problem.

My questions are: how this Fock operator is related with e-e repulsion term and how to proceed from this term to Fock operator? what is the physical interpretation of Fock-operator? Also in particular, which type of systems can be solved from this approximation.

  • 1
    $\begingroup$ "mainly in solid state systems" - that's ultimately a judgement call, but unless you're fully aware of the role of Hartree-Fock methods in atomic physics and molecular quantum chemistry (i.e. aware enough to be able to justify why you're dismissing them) you should be careful with claims like that. $\endgroup$ – Emilio Pisanty Oct 2 '18 at 7:33
  • $\begingroup$ @Emilio for you i have edited my question. $\endgroup$ – Aman pawar Oct 2 '18 at 7:47
  • $\begingroup$ related physics.stackexchange.com/questions/168042/… $\endgroup$ – jkds Oct 2 '18 at 10:23

The Hartree-Fock approximation makes an assumption on the state, namely that your $N$-electron wave function is the antisymmetric product of single-electron states; if you drop the antisymmetry (i. e. you forget that electrons are fermions), you get the Hartree approximation. You can implement the antisymmetrization procedure by plugging the single-particle wave functions into a Slater determinant.

To find an approximate ground state of the system, you can use a variational principle under the constraint that all the single-particle wave functions are orthogonal (that's the Pauli principle again). The Hartree-Fock equations now come out as the Euler-Lagrange equations to that variational principle. Note that the ground state is approximate, because instead of minimizing over all antisymmetric wave functions, you are only admitting antisymmetrized product states.

In summary, you approximate the true ground state of the system by an antisymmetrized product state of single-particle wave that solve a non-linear equation. The true ground state will not be a product state, but in many situations it is energetically close to a product state.

| cite | improve this answer | |
  • $\begingroup$ "asymmetric" is very different to "antisymmetric". $\endgroup$ – Emilio Pisanty Oct 3 '18 at 6:43
  • 1
    $\begingroup$ That must be "Antisymmetric" instead of "Asymmetric". Also your answer is not touching the asked questions! please answer the main questions. Also how this "The Hartree-Fock equations now come out as the Euler-Lagrange equations to that variational principle" thing works. Can you show explicitly? $\endgroup$ – Aman pawar Oct 3 '18 at 8:13
  • $\begingroup$ Yes, I meant to write antisymmetric, sorry. I think I did address your questions: the e-e interaction term becomes the non-linearity in the Hartree-Fock equations, which mixes the different single-electron wave functions. You can make additional, ad hoc approximations to obtain more tractable expressions. The derivation of the Hartree-Fock equations is quite standard: Compute the energy expectation value of a slater determinant. This gives you quadratic and quartic terms in $\psi_j$. Then take the variational derivative under the constraints of normalization and orthonormality. $\endgroup$ – Max Lein Oct 4 '18 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.