# Hartree-Fock: Coulomb integral [closed]

Today I was wondering how to better understand the Coulomb integral in the Hartree-Fock approximation.

Extracted from: Szabo & Ostlund, Modern Quantum Chemistry, p. 112

The Coulomb term has a simple interpretation. In an exact theory, the Coulomb interaction is represented by the two-electron operator $r_{ij}^{-1}$. In the Hartree-Fock approximation, an electron in a state $\chi_{a}$ experiences a one-electron Coulomb potential:

$$v_{a}^\text{coul}(1) = \sum_{b \neq a} \int \mathrm{d} \vec{x}_{2} | \chi_{b}(2) |^{2} r_{12}^{-1} \tag{3.7}$$

Now, the Coulomb integral (CI) for a system of two interacting electrons, using Slater determinants, is:

$$\hat{J}=\frac{1}{2} \left( \int \psi_1^*\left( 1 \right) \psi_2^*\left( 2 \right) \frac{1}{r_{12}} \psi_2\left( 1 \right)\psi_1\left( 2 \right) + \psi_2^*\left( 1 \right) \psi_1^*\left( 2 \right) \frac{1}{r_{12}} \psi_1\left( 1 \right)\psi_2\left( 2 \right) \mathrm{d}x \right)$$

Now, the help I'm looking for how this single Slater determinant definition comes up from the mathematical point of view.

This is how I understand the HF approximation: Let's start only with the Coulomb integral. This integral, as I understand it, should be understood as follows for the first electron:

\begin{align}\text{CI} & =\int \psi_1^*\left( 1 \right) \psi_2^*\left( 2 \right) \frac{1}{r_{12}} \psi_2\left( 1 \right)\psi_1\left( 2 \right) \mathrm{d}x \\ & =\int \psi_1^*\left( 1 \right) \big( \sum_{n=2}^{n=2} \psi_2^*\left( 2 \right) \frac{1}{r_{12}} \psi_1\left( 2 \right) \big) \psi_2\left( 1 \right) \mathrm{d}x\\ & =\int \psi_1^*\left( 1 \right) \big( \sum_{n=2}^{n=2} \hat{J}_n \big) \psi_2\left( 1 \right)\end{align}

And, more general, the Coulomb integral for the first electron in a system of $n$ electrons would be:

$$\text{CI}= \int \psi_1^*\left( 1 \right) \big( \sum_{n \neq 1}^{n} \hat{J}_n \big) \psi_2\left( 1 \right) \mathrm{d}x$$

Is this correct?

If this is not correct, would someone enlighten me?

To ACuriousMind, Thank so much for editing this post.

Specific help I'm looking for:

I'll try to explain myself better through these questions:

i.) This operator $\frac{1}{r_{ij}}$ involves the spatial coordinates of electrons $i$ and $j$. So, taken this CI:

$$J_{2}(1) \chi_{1}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1)$$

Mathematically speaking, the first step is to integrate $\int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2)$ (which only the spatial coordinates of electron 2 are taken in account) and then multiply the result to $\chi_1 \left( 1 \right)$ ?

ii.) This question is more curiosity. For the same electron (Let's say the first one), How would it be with a system of N-electrons?

## closed as unclear what you're asking by ACuriousMind♦, Martin, Kyle Kanos, Danu, JamalSMar 19 '15 at 13:22

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• This question seems too confused and poorly edited to even begin to give an answer. I would recommend you check out the book by Bethe "Intermediate Quantum Mechanics", he goes over this in great detail. – hft Mar 19 '15 at 0:48
• I have already read from page 58 to 64. But in the book you suggested me still is not answering my question. The only difference is that explicitly include the spin. – Another.Chemist Mar 19 '15 at 1:37
• @Alejandro looks like you are badly and totally confused by the HF method. Basically, all you statements and formulas are "not quite right" to put it mildly. Szabo & Ostlund book is quite good at explaining the HF method, but you have to be careful and patient to follow through. – Wildcat Mar 19 '15 at 8:00
• Hi Alejandro! I have edited your question for better formatting and grammar, and I don't think I've changed the meaning of anything, but I still cannot discern what your actual question really is except that you do not really understand the HF approximation. – ACuriousMind Mar 19 '15 at 9:43

## 1 Answer

Now, the Coulomb integral for a system of two interacting electrons, until Slater determinants, is:

I don't understand what the heck do you mean by "until Slater determinants", but for a system of two electrons you have a system of two HF equations (3.9): each HF equation for each spin-orbital $\chi_{a}$, $$[h(1) + J_{2}(1) - K_{2}(1) ] \, \chi_{1}(1) = \varepsilon_{1} \chi_{1}(1) \, , \\ [h(1) + J_{1}(1) - K_{1}(1) ] \, \chi_{2}(1) = \varepsilon_{2} \chi_{2}(1) \, , \\$$ where $$J_{2}(1) \chi_{1}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1) \, , \\ K_{2}(1) \chi_{1}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{1}(2) \Big] \chi_{2}(1) \, , \\ J_{1}(1) \chi_{2}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \Big] \chi_{2}(1) \, , \\ K_{1}(1) \chi_{2}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{1}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1) \, . \\$$

So, you have two different Coulomb terms $J_{2}(1) \chi_{1}(1)$ and $J_{1}(1) \chi_{2}(1)$, as well as two different exchange terms $K_{2}(1) \chi_{1}(1)$ and $K_{1}(1) \chi_{2}(1)$, entering each and every HF equation.

Coulomb integrals are expression of the following form $$\int \mathrm{d} \vec{x}_{1} \chi_{1}^{*}(1) J_{2}(1) \chi_{1}(1) \, , \\ \int \mathrm{d} \vec{x}_{1} \chi_{2}^{*}(1) J_{1}(1) \chi_{2}(1) \, ,$$ which contribute to the HF electronic energy $\langle \Phi\,|\,H\,|\,\Phi\rangle$. By the definition of Coulomb operator $J_{2}(1)$ above could be written as follows $$\int \mathrm{d} \vec{x}_{1} \chi_{1}^{*}(1) \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1) \, , \\ \int \mathrm{d} \vec{x}_{1} \chi_{2}^{*}(1) \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \Big] \chi_{2}(1) \, ,$$ or as follows $$\iint \mathrm{d} \vec{x}_{1} \mathrm{d} \vec{x}_{2} \chi_{1}^{*}(1) \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \chi_{1}(1) \, , \\ \iint \mathrm{d} \vec{x}_{1} \mathrm{d} \vec{x}_{2} \chi_{2}^{*}(1) \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \chi_{2}(1) \, .$$

• nope, until this point, I guess, I understand it pretty well. I'll try to explain myself better through these questions: i.) This operator $\frac{1}{r_{ij}}$ involves the spatial coordinates of electrons $i$ and $j$. So, taken the first Coulomb expression of your last list. Mathematically speaking, first this integral $\int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2)$ (which only the spatial coordinates of electron 2 are into the integral) must be integrated and then multiplied to $\chi_{1}(1)$ ? and ii.) (with N-e) it would the same with the N-1 remaining electrons? – Another.Chemist Mar 19 '15 at 14:36
• @Alejandro I just wrote expression for Coulomb integrals explicitly. So, for instance, for the first Coulomb integral, you can think of it as follows. You first calculate the average Coulomb potential $J_{2}(1)$ which e-1 "feels" due to e-2 on $\chi_{2}$ by integrating $\chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2)$ over coordinates of e-2. You then calculate the resulting Coulomb repulsion energy between e-1 on $\chi_{1}$ and e-2 on $\chi_{2}$ by integrating $\chi_{1}(1) J_{2}(1) \chi_{1}(1)$ over coordinates of e-1. – Wildcat Mar 19 '15 at 15:15
• @Alejandro, and yes, for an $N$-electron system you indeed do calculate all such Coulomb contributions for e-1 with all other $N-1$ electrons in the system. And you do the same of e-2, and e-3, etc. You then sum up all the Coulomb integrals (minus all the exchange ones) together with the sum of all core Hamiltonian energies to calculate the HF energy. – Wildcat Mar 19 '15 at 15:19
• @Alejandro, now it looks like you are missing the second integration which enters an expression for a Coulomb integral. So $J_{2}(1) \chi_{1}(1)$ is not a Coulomb integral, it just an expression which defines the Coulomb operator $J_{2}(1)$ by its action on spin-orbital $\chi_{1}(1)$. The corresponding Coulomb integral is written as follows $\int \mathrm{d} \vec{x}_{1} \chi_{1}(1) J_{2}(1) \chi_{1}(1)$. – Wildcat Mar 19 '15 at 15:29
• @Alejandro, as I said, you can first calculate $J_{2}(1)$ using its definition and integrating over coordinates of e-2. You than can substitute the resulting potential $J_{2}(1)$ into $\int \mathrm{d} \vec{x}_{1} \chi_{1}^{*}(1) J_{2}(1) \chi_{1}(1)$ and integrate over coordinates of e-1. So, you multiply $J_{2}(1)$ by $\chi_{1}(1)$ from the right and by its complex-conjugate $\chi_{1}^{*}(1)$ from the lefts and then integrate. – Wildcat Mar 19 '15 at 15:33