I am new in computational chemistry. I'm studying the Hartree–Fock–Roothaan (HFR) method and have some of questions about the HFR basis function coefficients.

  1. Do we have the condition on HFR coefficients like in the Hückel method?

In the Hückel method, we have a normalization condition on the coefficients $\{c_i\}$:

$$c_1^2 + c_2^2 + \dots + c_i^2 = 1 \tag{1}\label{1}$$

In the HF literature I didn't see that condition, so my guess is that we don't have it.

  1. Do we have similar to Hückel method condition on HFR coefficients of bf?

I'm wondering because going deeper in understanding HF I find out that we have certain conditions that looks like $\eqref{1}$. For example, a normalization condition for the Gaussian primitives that are not normalized:

$$ S = \left< \Phi_{\textrm{1s}}^{\text{STO-3G}} | \Phi_{\textrm{1s}}^{\text{STO-3G}} \right> = 1 $$ $$ \Phi_{\textrm{1s}}^{\text{STO-3G}} = N_1C_1e^{-a_1r^2} + N_2C_2e^{-a_3r^2} + N_3C_3e^{-a_3r^2} $$

This is not the same as $\eqref{1}$ but we can represent it as

$$f_1(c_1^2) + f_2(c_2^2) + \dots + f_i(c_i^2) = 1 \tag{2}\label{2}$$

Because HF is more complex than Hückel, $\eqref{2}$ can make a sense. So can I think about $\eqref{2}$ as complex version of $\eqref{1}$? I was hoping to find $\eqref{2}$ or similar equation in HF literature intro part, but those sources that I have do not mention $\eqref{2}$ in intro part. Maybe my attention was scattered or it is just a mess in my head, anyway I will be grateful for any help.

  1. What are the conditions imposed on the coefficients of the basis functions that use in the calculation of the HFR equations?

For example, the electronic charge distribution is

$$ \sum_{mn} P_{mn}S_{mn} = N =\sum_{mn} 2\sum_{a}^{N/2} c_{ma}c_{na}^*\int{f_m(r)f_n^*(r)\mathop{dr}}, $$


$$ S_{mn} =\int{f_m(r)f_n^*(r)\mathop{dr}} = \delta_{mn}, $$

and N is the total number of electrons. Can we consider this as a condition?

  • 4
    $\begingroup$ I would highly recommend getting Szabo and Ostlund's Modern Quantum Chemistry if you are trying to learn about Hartree Fock and the Roothaan-Hall method. @AlexMal $\endgroup$
    – Tyberius
    May 31, 2018 at 14:11

1 Answer 1


Let start by definition the Hartree-Fock-Roothaan equation:

$$\sum_\nu F_{\mu\nu}C_{\nu i}=\varepsilon_i\sum_\nu S_{\mu\nu}C_{\nu i}$$

Where the Fock matrix elements are giving as:

$$F_{\mu \nu} = \left< \phi_\mu \left| \hat{f} \right|\phi_\nu\right>$$

And the overlap matrix is given as:

$$S_{\mu \nu} = \left< \left. \phi_\mu \right|\phi_\nu\right>$$

Here $\phi_\nu$ is our atomic orbitals (i.e. what you get from your basis set).

When we do a Hartree-Fock calculation we want to find a set of orthonormal orbitals, also known as molecular orbitals. The molecular orbitals are constructed as:

$$\psi_i=\sum_{\mu=1}^KC_{\nu i}\phi_\nu$$

For the molecular orbitals to be orthonormal we can write the following condition:

$$\left< \left. \psi_i \right|\psi_j\right>=\delta_{ij}$$

This means that the orbitals have zero overlap with one-another and an overlap of $1$ with them self. If we insert the expansion of the molecular orbitals in atomic orbitals in the above equation we get:

$$\sum_{\mu=1}^K\sum_{\nu=1}^LC_{\nu i}^\dagger C_{\mu j}\left< \left.\phi_\nu \right|\phi_\mu\right>=\delta_{ij}$$

I.e. we can see that when we do a Hartree-Fock calculation we have the constraint on our molecular orbital coefficients that have to transform our atomic orbital overlap to a matrix with ones on the diagonal and zero elsewhere. This can also be written more compact in matrix form:

$$\mathbb{C^\dagger SC=1}$$

Which is a condition equivalent to the one you know from Hückel theory. At last let me make a breif comment about basis set, since you also talked about those in your question. In general we use contracted Gaussian type basis sets, just like STO3G, that you mentioned. Our atomic orbitals can thus be written as:

$$\phi_\nu = \sum_{k=1}^M N_k d_k P(r)\cdot e^{-\alpha_k\cdot r^2}$$

Here $P(r)$ is a function related to the angular moment quantum number.The way the contraction coefficients $d_k$ are choose implies that we normalize our Gaussian functions, i.e:

$$N_k^2 \int_{-\infty}^\infty \left( P(r)\cdot e^{-\alpha_k\cdot r^2} \right)^2 dr = 1$$

We can now write $\left< \left.\phi_\nu \right|\phi_\nu\right> = 1$ as:

$$\overline{N}\sum_{k=1}^M N_k d_k\sum_{l=1}^T N_l d_l\left< \left.P(r)\cdot e^{-\alpha_k\cdot r^2} \right| P(r)\cdot e^{-\alpha_l\cdot r^2}\right> = 1$$

Where $\overline{N}$ is the normalization we need. As we can see there is no easy condition we can put on the contraction coefficients, in order to avoid evaluation $\overline{N}$. It can be shown (I will not show it here) that $\overline{N}$ only depends on the total angular moment and on the Gaussian exponents, thus it is possible to choose the contraction coefficients in such a way, that we do not need to evaluate $\overline{N}$.

  • $\begingroup$ Thank you very much. I appreciate this. But one thing is confuse me. The coefficient C^2 should be less than 1 to hold the equality. But this resource phys.ubbcluj.ro/~vchis/cursuri/cspm/part3.pdf (page 17) gives for some coefficients value more than one. What is my problem? Maybe they are not normalized? $\endgroup$
    – Alex Mal
    May 31, 2018 at 19:59
  • $\begingroup$ No, the pdf is right. I just made a mistake. The final condition is the CSC = 1. The other part was a brain fart from my side. $\endgroup$ May 31, 2018 at 20:28

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