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I know that the Fock operator for electron $m$ in $n$-electron molecule would be:

$$\hat{f}\!(m)u_i(m) = \varepsilon_iu_i(m)$$

$$\hat{f}\!(m) \equiv -\frac 1 2 ∇_m^2 - \sum_α\frac{Z_α}{r_{ma}} + \sum_{j = 1}^n\left[\hat{j}_j(m) - \hat{k}_j(m)\right]$$

And the Coulomb operator according to Wikipedia is given by:

$$\hat{J}_{\!j}(1)f(1) = f(1)\int\left|\varphi_j(2)\right|^2\frac{1}{r_{12}}\mathrm dr_2$$

My understanding of this equation is that we are finding the interaction of $f(1)$ with the electrons of $j$-th orbital (we take interaction with electron 2 and twice that result, hence $2J$ is used in the expression for Fock operator), and $r_{12}$ is the separation between our electron in $f(1)$ and 2nd electron in $j$-th orbital.

This is the expression for exchange operator for electron $1$ in $j$-th orbital with an arbitrary function '$f$'. What would be the expression of Coulomb and exchange operators for electron $m$? Will we consider the $1$st electron of $(m/2)$-th orbital? Or will we simply put $m$ in place of $1$?

Or have I understood the whole thing incorrectly?

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    $\begingroup$ Remember that the electrons are indistinguishable. Also, you can find an expression for the exchange operator in terms of a permutation operator in Szabo's Modern Quantum Chemistry. $\endgroup$ – Verktaj May 14 at 3:47
  • $\begingroup$ Electrons in the same orbital are indistinguishable, right? So, that only is my doubt. For mth electron, will we consider the 1st electron of (m/2)th orbital, and use the same notation, or would the expression be different? $\endgroup$ – Raghav Arora May 28 at 5:29
  • $\begingroup$ All electrons are indistinguishable. We write the expressions using a dummy index, usually 1, but we can't really label the electrons or say that 2nd electron is in some particular orbital. $\endgroup$ – Tyberius Sep 29 at 3:17

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