Question

Questions 3-5

3. The predecessor to Hartree-Fock was the Hartree method, where the main difference is that the Hartree-Fock method includes an trial wavefunction by writing it as a Slater Determinant, while the Hartree method uses a simple product wavefunction that does not capture anti- symmetry. In particular, for a minimal-basis model of, the Hartree method's trial wavefunction is given in the while the Hartree-Fock trial wav is given by where and are molecular orbitals, and and coordinates of electron 1 and 2, respectively denote the Write an expression for the total values of and in terms of 1- and 2-electron integrals and comment on the differences. In particular, estimate which method will yield the lower total energy. written in terms of 1- and 2-electron operators as where is the 1-electron operator for electron and is the 2-electron operator for electrons and y expectation Recall the Hamiltonian operator can be 4. Use the following function as a trial wavefunction for the particle-in-a-box ground stat Compare the energy expectation value of to where is the true ground state of the particle in a box of length Discuss whether or not this result is consistent with the variational principle 5. The energy expression for Hartree-Fock theory can be written as where is the number of basis fu the computational cost of the Hartree-Fock energy grows with the number of basis functions. Explain how 4. Use the following function as a trial wavefunction for the particle-in-a -box ground state: 2x (L-x Compare the energy expectation value of φ (x) to ψι (x) where ψι (x) is the true ground state of the particle in a box of length L. Discuss whether or not this result is consistent with the variational principle. 5. The energy expression for Hartree-fock theory can be written as where N is the number of basis functions. Explain how the computational cost of the Hartree-Fock energy grows with the number of basis functions

Answer 1

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5) With an increase in the basis functions, the number of integrals (single-particle integrals and two-body integrals) would increase exponentially. The calculation of these integrals is computationally expensive. That's why the computational cost of Hartree-Fock energy grows with the number of basis functions.