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[ Cross-posted to the Computational Science Stack Exchange: https://scicomp.stackexchange.com/questions/1297/why-does-iteratively-solving-the-hartree-fock-equations-result-in-convergence ]

In the Hartree-Fock self-consistent field method of solving the time-independent electronic Schroedinger equation, we seek to minimize the ground state energy, $E_{0}$, of a system of electrons in an external field with respect to the choice of spin orbitals, $\{\chi_{i}\}$.

We do this by iteratively solving the 1-electron Hartree-Fock equations, $$\hat{f}_{i}\chi(\mathbf{x}_{i})=\varepsilon\chi(\mathbf{x}_{i})$$ where $\mathbf{x}_{i}$ is the spin/spatial coordinate of electron $i$, $\varepsilon$ is the orbital eigenvalue and $\hat{f}_{i}$ is the Fock operator (a 1-electron operator), with the form $$\hat{f}_{i} = -\frac{1}{2}\nabla^{2}_{i}-\sum_{A=1}^{M}\frac{Z_{A}}{r_{iA}}+V^{\mathrm{HF}}_{i}$$ (the summation runs over nuclei, here, with $Z_{A}$ being the nuclear charge on nucleus A and $r_{iA}$ being the distance between electron $i$ and nucleus $A$). $V^{\mathrm{HF}}_{i}$ is the average potential felt by electron $i$ due to all the other electrons in the system. Since $V_{i}^{\mathrm{HF}}$ is dependent on the spin orbitals, $\chi_{j}$, of the other electrons, we can say that the Fock operator is dependent on it's eigenfunctions. In "Modern Quantum Chemistry" by A. Szabo and N. Ostlund, pp. 54 (the first edition) they write that "the Hartree-Fock equation (2.52) is nonlinear and must be solved iteratively". I have studied the details of this iterative solution as part of my research, but for this question I think they are unimportant, except to state the basic structure of the method, which is:

  1. Make an initial guess of the spin-orbitals, $\{\chi_{i}\}$ and calculate $V_{i}^{\mathrm{HF}}$.
  2. Solve the eigenvalue equation above for these spin orbitals and obtain new spin-orbitals.
  3. Repeat the process with your new spin orbitals until self-consistency is reached.

In this case, self-consistency is acheived when the spin-orbitals which are used to make $V_{i}^{\mathrm{HF}}$ are the same as those obtained on solving the eigenvalue equation.

My question is this: how can we know that this convergence will occur? Why do the eigenfunctions of the successive iterative solutions in some sense "improve" towards converged case? Is it not possible that the solution could diverge? I don't see how this is prevented.

As a further question, I would be interested to know why the the converged eigenfunctions (spin orbitals) give the best (i.e. lowest) ground state energy. It seems to me that the iterative solution of the equation somehow has convergence and energy minimization "built-in". Perhaps there is some constraint built into the equations which ensures this convergence?

James Womack
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2 Answers2

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I remember doing SCF calculations back in the early 80s, and it was by no means guaranteed that the calculation would converge or that it would give you the ground state. Several of my calculations diverged at the first attempt, though putting a bit more thought into the starting point would usually produce convergence.

I don't think I ever accidentally ended up with an excited state, though I'm sure I remember this happening to colleagues. However it was usually easy to see that you didn't have the ground state.

I can't comment on whether this issues are inherent to the method or whether it was the particular implementation that was at fault. I don't remember the name of the software being used. This was at the Chemistry department in Cambridge in 1982/3.

John Rennie
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There is no guaranteed solution to get convergence on the ground state. But there are really good algorithms used like DIIS. And as always you need a good starting point to not get stuck at a local minima. And this is e.g. Hückel operator or INDO guess.

As the Hartree-Fock method relys on the variational principle you find a lower energy with a better trial wavefunction. A better wavefunction is the Slater determinant and eigenvector of orbitals. The orbitals are made of a finite basis set. So the self consistent field solutions are accurate to the given basis set if and only if the system is given by a single Slater determinant and you have choosen the correct one. Modern quantum chemistry software like Gaussian, is really good in getting the correct ground state by e.g. checking the symmetry, good starting guess, do an annealing process etc. Colleagues told me to get an excited state they had to manually start with a unphysical wrong symmetry so the calculation do not converge instantly to the ground state.

James Womack
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Alex1167623
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