# What will be the form of coulomb and exchange operators for electron m in an n-electron molecule?

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?

• 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. – Verktaj May 14 at 3:47
• 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? – Raghav Arora May 28 at 5:29
• 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. – Tyberius Sep 29 at 3:17