# Why does Hartree-Fock (HF) theory even work?

Let’s say we have $$N$$ electrons and we want to derive the Hartree-Fock (HF) equations. The first step would be to define a Slater determinant of $$N$$ electrons:

$$\psi(x_1,x_2,… x_N) = \frac{1}{\sqrt{N!}}\begin{vmatrix}\phi_{1}(x_1) & \phi_{2}(x_1) & … & \phi_{N}(x_1)\\ .\\ .\\ .\\ \phi_{1}(x_N) & \phi_{2}(x_N) & … & \phi_{N}(x_N)\end{vmatrix}$$

then we would use the Lagrange minimisation principle to get our HF equations, which are:

$$f\phi_k = \varepsilon_k\phi_k \qquad \forall k=1,\dots,N.$$

$$f := h + \sum_{n=1}^N J_n - K_n,$$

We note the following: We have $$N$$ electrons in our system so we get a $$N\times N$$ slater determinant and $$N$$ molecular wave functions $$\phi_k$$.

If we would try to solve this HF equation we could simply put $$N$$ trial orbitals into the HF equation and solve it iteratively. In practice however, one approximates the molecular orbitals by a linear combination of e.g. $$M>N$$ basis functions: $$\phi_k=\sum_{m=1}^M{C_{mk} \xi_m}$$

If we put this into the HF equation above, we eventually get a matrix equation (Roothan equations).$$FC=\epsilon SC$$ which can be solved on the computer.

My question is:

There are $$M>N$$ expansion coefficients $$C_m$$. By using these $$M$$ coefficients we eventually get $$2M$$ molecular orbitals $$\phi_k$$ with $$k=1,\cdots,2M$$. Or put in words: we get two molecular orbitals for every basis function we use (due to spin). The lowest $$N$$ orbitals are the occupied orbitals the highest $$2M–N$$ are the virtual orbitals. These $$2M$$ molecular orbitals now correspond to a slater determinant with $$2M$$ electrons

$$\psi(x_1,x_2,\cdots, x_{2M}) = \frac{1}{\sqrt{2M!}}\begin{vmatrix}\phi_{1}(x_1) & \phi_{2}(x_1) & … & \phi_{2M}(x_1)\\ .\\ .\\ .\\ \phi_{1}(x_{2M}) & \phi_{2}(x_{2M}) & … & \phi_{2M}(x_{2M})\end{vmatrix}$$

But now we have a problem: We wanted to describe a $$N$$-electron system and stared out by assuming a $$N$$-electron slater determinant. Now we used this expansion into basis functions and ended up with a slater determinant describing $$2M>N$$ electrons. Isn’t this somehow unphysical? How do we know that our result does even describe a $$N$$-electron system in the right way? Since our Slater determinant has a size of $$2M$$ how is it possible to just fill the lowest orbitals and ignore all other?

I think the confusion here is about the number of orbitals vs. the number of electrons in the Slater determinant. In the first equation the indices of $$x_i$$ enumerate the electrons, but the indices of $$\phi_j$$ are actually shortcuts for the orbitals: $$N$$ electrons occupy $$N$$ different states, but it doesn't mean that there are only $$N$$ states - we simply use only these $$N$$ state for constructing the particular $$N$$-electron wave function.

A clearer but more cumbersome notation would be $$\psi_{j_1, j_2,..., j_N}(x_1,x_2,… x_N) = \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_{j_1}(x_1) & \phi_{j_2}(x_1) & … & \phi_{j_N}(x_1)\\ .\\ .\\ .\\ \phi_{j_1}(x_N) & \phi_{j_2}(x_N) & … & \phi_{j_N}(x_N)\end{vmatrix},$$ where $$j_1,...,j_N$$ is a selection of $$N$$ orbitals out of $$M$$ single-particle eigenstates of the one-particle Hamiltonian (i.e., any of the $$j_i$$ can take values in the range $$1...M$$, but of course they should be all different to give a non-zero Slater determinant.)

• Tanks a lot. Nevertheless I am not sure if or how this resolves the problem. If we built a slater determinant with N orbitals in it, we assume that there are N electrons in the system. So if I use the expansion into basis functions I have a slater determinant with 2M orbitals and therefore I have intrinsically assumed 2M electrons. This also effects J and K, which are dependent on the number of electrons. Jan 20, 2023 at 14:23
• @Lockhart I am not familiar with the method that you discuss in the OP, but I think the idea is that we assume that the wave function is a Slater determinant (Hartree-Fock approximation), but we do not know the exact form of functions $\phi_{j_i}$, which we then try to express in terms of non-interacting orbitals (i.e., single-particle solutions.) Hartree-Fock/Slater is needed to reduce the many-particle problem to a single-particle one (or nearly single particle). Jan 20, 2023 at 14:29
• Yes that’s the idea, it’s like a mean field theory. Nevertheless, I am not sure why we can use a larger slater determinant and then just ignore the orbitals we don’t need but at the same time assume the solution to be equivalent to a system with a smaller slater determinant. Jan 20, 2023 at 14:34
• @Lockhart Imagine that you are solving a one-particle problem for Hamiltonian $H=H_0+V$: you start with exact wave-functions of Hamiltonian $H_0$ (which are $M$ in number), and then look for the eigenfunctions of $H$ as an expansion in terms of these unperturbed eigenfucntions. Jan 20, 2023 at 14:40
• @Lockhart yes, this is what I think was the difficulty. Jan 20, 2023 at 15:57

Your expansion coefficient matrix $$C$$ tells you how to transform your original set of unoptimized basis functions $$\xi$$ to a set of optimized orbitals $$\phi$$. The sets have the same size, i.e. if you start out with $$M$$ basis functions you will also end up with $$M$$ orbitals.

I think the point that you are missing is that the number $$M$$ may be larger than the actual number of orbitals that we need to construct our Slater determinant, let that be $$N$$. We will only use the $$N$$ lowest optimized orbitals (lowest with respect to the orbital energy i.e. the associated eigenvalue) to construct the Slater determinant. These orbitals will be the $$N$$ occupied orbitals, which leaves you with $$M-N$$ unoccupied orbitals that are not used. The number $$M$$ can be anything as long as it is equal or greater than $$N$$. The size of the determinant is independent of $$M$$.

There are in principle $$N$$ 1-particle functions that minimize the energy of the $$N$$-particle wavefunction that has the form of a Slaterdeterminant. To find these $$N$$ 1-particle functions we expand the Hartree-Fock equations into a basis set. This leads us to the Roothan equations, which we can solve by iteration. The Roothan equations system can have size $$M \times M$$. And it will yield $$M$$ 1-particle functions as solutions. The lowest $$N$$ solutions are the best approximations to the "exact" $$N$$ 1-particles functions that the Hartree-Fock equations define, to construct the best Slater determinant.
So there are exactly $$N$$ 1-particle functions which we do not know and I would consider these $$N$$ functions to be the solutions of the Hartree Fock equations. We can find good approximations to these $$N$$ 1-particle functions by using a basis set expansion which yields the Roothan equations. And once we have these we can construct the $$N$$-particle wavefunction in Slater determinant form.