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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?

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2 Answers 2

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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.)

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  • $\begingroup$ 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. $\endgroup$
    – Lockhart
    Jan 20, 2023 at 14:23
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    $\begingroup$ @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). $\endgroup$
    – Roger V.
    Jan 20, 2023 at 14:29
  • $\begingroup$ 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. $\endgroup$
    – Lockhart
    Jan 20, 2023 at 14:34
  • $\begingroup$ @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. $\endgroup$
    – Roger V.
    Jan 20, 2023 at 14:40
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    $\begingroup$ @Lockhart yes, this is what I think was the difficulty. $\endgroup$
    – Roger V.
    Jan 20, 2023 at 15:57
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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 $M$.

Answer to the remark in comments:

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.

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  • $\begingroup$ Thank you. So is the following right: I assumed a N-electron wavefunction but the HF equation I get in the end nevertheless has infinitely many solutions. If I would solve the HF equation by guessing N wavefunctions and iterating, I would just get N solutions. But if I use the expansion into basis functions I can get more solutions. In this sense I can’t conclude that 2M orbitals mean a 2M-electron wavefunction (which was my mistake)? $\endgroup$
    – Lockhart
    Jan 20, 2023 at 15:55
  • $\begingroup$ @Lockhart I edited my answer. $\endgroup$
    – Hans Wurst
    Jan 20, 2023 at 16:35
  • $\begingroup$ So if we we would just put N trial functions into the HF equation and iterate we would get (after convergence) N molecular orbitals. On the other hand we could use Roothan equations (or solve the HF equation analytically) and get M orbitals. The first N of these M orbitals should the same as we get with the ‚direct‘ iteration i mentions first? $\endgroup$
    – Lockhart
    Jan 20, 2023 at 16:44
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    $\begingroup$ @Lockhart If the space spanned by the M orbitals contains the space spanned by the N orbitals, and if the initial guess is appropriate, then yes. $\endgroup$
    – Hans Wurst
    Jan 20, 2023 at 17:01
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    $\begingroup$ @Lockhart The Fock operator is a nonlinear operator that depends on the set of functions that you use, which is why we have to solve the equations iteratively. So it does depend on M. But if the conditions that I mentioned are met, you should find the same eigenfunctions and eigenvalues for the first 1 to N eigenfunctions. There would be no point in using more than N functions if you knew that they are the best set, but typically you don't know how to pick this set which is why you use more basis functions that cover a larger function space in the hope that it contains the good solution space. $\endgroup$
    – Hans Wurst
    Jan 20, 2023 at 18:21

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