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, JamalS Mar 19 '15 at 13:22

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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. $\endgroup$ – hft Mar 19 '15 at 0:48
  • $\begingroup$ 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. $\endgroup$ – Another.Chemist Mar 19 '15 at 1:37
  • 1
    $\begingroup$ @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. $\endgroup$ – Wildcat Mar 19 '15 at 8:00
  • 1
    $\begingroup$ 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. $\endgroup$ – ACuriousMind Mar 19 '15 at 9:43

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 $$ \iint \mathrm{d} \vec{x}_{1} \mathrm{d} \vec{x}_{2} \chi_{1}^{*}(1) \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \chi_{1}(1) \, , \\ \iint \mathrm{d} \vec{x}_{1} \mathrm{d} \vec{x}_{2} \chi_{2}^{*}(1) \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \chi_{2}(1) \, . $$

  • $\begingroup$ 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? $\endgroup$ – Another.Chemist Mar 19 '15 at 14:36
  • 1
    $\begingroup$ @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. $\endgroup$ – Wildcat Mar 19 '15 at 15:15
  • $\begingroup$ @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. $\endgroup$ – Wildcat Mar 19 '15 at 15:19
  • 1
    $\begingroup$ @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 $\int \mathrm{d} \vec{x}_{1} \chi_{1}(1) J_{2}(1) \chi_{1}(1)$. $\endgroup$ – Wildcat Mar 19 '15 at 15:29
  • 1
    $\begingroup$ @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. $\endgroup$ – Wildcat Mar 19 '15 at 15:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.