Electron-electron interactions in polyelectronic atoms

Question

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1. What the question is asking (in plain language)

For an atom that has more than one electron the exact Schrödinger equation contains a term

[ \hat V_{ee}= \sum_{i<j}\frac{e^{2}}{4\pi\varepsilon_{0}\,r_{ij}} ]

that couples the motion of all the electrons.
Because this term makes the problem non‑separable, textbooks first ignore it, solve the “independent‑electron’’ problem, and then add the electron‑electron interaction back as a perturbation.

The student wants to know:

Why is it reasonable to treat the electron‑electron term as a small correction?
(Is there a physical picture such as “the electrons spend most of the time far apart’’ that justifies it?)
Are there quantitative arguments or explicit calculations that show the approximation works?
(e.g. order‑of‑magnitude estimates, first‑order perturbation results for real atoms.)

Below is a step‑by‑step answer that addresses both points.

2. Detailed reasoning and calculations

2.1 The full Hamiltonian for an (N)-electron atom

[ \hat H = \underbrace{\sum_{i=1}^{N}\Big[-\frac{\hbar^{2}}{2m}\nabla_i^{2} -\frac{Z e^{2}}{4\pi\varepsilon_{0}\,r_i}\Big]}{\displaystyle \hat H{0}\;(\text{independent electrons})} \;+\; \underbrace{\sum_{i<j}\frac{e^{2}}{4\pi\varepsilon_{0}\,r_{ij}}}{\displaystyle \hat V{ee}\;(\text{electron–electron repulsion})}. ]

(\hat H_{0}) is a sum of hydrogen‑like Hamiltonians with nuclear charge (Z).
(\hat V_{ee}) couples the electrons; it is the only piece that prevents exact separation.

The idea is to treat (\hat V_{ee}) as a perturbation to (\hat H_{0}).

2.2 Physical picture – why might (\hat V_{ee}) be “small’’?

Dominance of the nuclear attraction
For a given electron the attractive potential is (-Z/r).
The average distance of an electron from the nucleus in a hydrogen‑like orbital scales as

[ \langle r\rangle \sim \frac{a_{0}}{Z_{\text{eff}}} ]

where (a_{0}) is the Bohr radius and (Z_{\text{eff}}) is the effective nuclear charge felt by that electron (typically close to (Z) for inner shells, and a bit smaller for outer shells because of screening).
Screening reduces the repulsion
The repulsive term between two electrons behaves as (+1/r_{12}).
In a many‑electron atom each electron is partially screened by the others, so the average inter‑electronic distance is larger than the average electron–nucleus distance.

Roughly

[ \langle r_{12}\rangle \;\approx\; \langle r_{1}\rangle+\langle r_{2}\rangle \;\sim\; \frac{2a_{0}}{Z_{\text{eff}}}. ]
Order‑of‑magnitude ratio

[ \frac{\langle V_{ee}\rangle}{\langle V_{en}\rangle} \sim\frac{e^{2}/\langle r_{12}\rangle}{Z e^{2}/\langle r\rangle} \sim\frac{1}{Z}\,\frac{\langle r\rangle}{\langle r_{12}\rangle} \sim\frac{1}{Z}\,\frac{1}{2}\;=\;\mathcal O!\left(\frac{1}{Z}\right). ]

For large (Z) the electron‑electron repulsion is a (1/Z) correction to the dominant nuclear attraction.
Even for relatively small (Z) (e.g. He, (Z=2)) the ratio is only about 0.5, which still allows a perturbative treatment—especially because the repulsion is spread over many pairs of electrons.

Thus the central‑field picture (each electron moves in an average, spherically symmetric potential created by the nucleus plus the smeared‑out charge of the other electrons) captures the bulk of the physics; the residual part of (\hat V_{ee}) that is not already included in the averaged field is left for perturbation theory.

2.3 First‑order perturbation theory – a concrete example (Helium)

Take the simplest multi‑electron atom, He ((Z=2), two electrons).

Zeroth‑order (independent‑electron) wavefunction

Each electron is placed in the hydrogenic 1s orbital with nuclear charge (Z).

[ \psi^{(0)}(\mathbf r_1,\mathbf r_2)=\phi_{1s}^{Z}(\mathbf r_1)\,\phi_{1s}^{Z}(\mathbf r_2), \qquad \phi_{1s}^{Z}(\mathbf r)=\frac{Z^{3/2}}{\sqrt{\pi a_{0}^{3}}}\,e^{-Zr/a_{0}} . ]
Zeroth‑order energy

[ E^{(0)} = 2\bigl(-\frac{Z^{2}}{2}R_{\infty}\bigr)= -2Z^{2}R_{\infty} = -2(2)^{2}\,13.6057\;\text{eV} = -108.8\;\text{eV}. ]

(Here (R_{\infty}=13.6057\;\text{eV}) is the Rydberg.)
First‑order correction

[ E^{(1)} = \langle\psi^{(0)}|\hat V_{ee}|\psi^{(0)}\rangle = \biggl\langle\frac{e^{2}}{4\pi\varepsilon_{0}r_{12}}\biggr\rangle . ]

The integral can be evaluated analytically (or looked up):

[ \boxed{E^{(1)} = \frac{5}{4}\,Z\,R_{\infty}} . ]

For helium ((Z=2))

[ E^{(1)} = \frac{5}{4}\times 2 \times 13.6057\;\text{eV} = 34.01\;\text{eV}. ]
First‑order total energy

[ E_{\text{He}}^{(0+1)} = E^{(0)}+E^{(1)} = -108.8\;\text{eV}+34.0\;\text{eV} = -74.8\;\text{eV}. ]

The experimental ground‑state energy of He is (-78.98\;\text{eV}) (ionisation energy 24.59 eV).

The error after the first perturbative correction is only ≈5 %.
Second‑order correction (optional, shows rapid convergence)

Using standard non‑degenerate perturbation theory one finds

[ E^{(2)}\approx -1.0\;\text{eV}, ]

giving

[ E^{(0+1+2)}\approx -75.8\;\text{eV}, ]

already within 4 % of the exact value. More sophisticated treatments (variational Hylleraas, configuration interaction) push the error below (10^{-4}) eV, but the trend is clear: the electron‑electron term is a modest correction.

2.4 Scaling to larger atoms

For a general atom with many electrons, one can repeat the same reasoning:

The zeroth‑order Hamiltonian is a sum of hydrogenic orbitals with effective nuclear charges (Z_{\text{eff}}(n\ell)) that account for the average screening. Values of (Z_{\text{eff}}) are given by Slater’s rules or obtained self‑consistently in a Hartree–Fock calculation.
The first‑order energy correction is the expectation value of the residual part of the electron‑electron repulsion, i.e.

[ E^{(1)} = \frac12\sum_{i\neq j}\bigl(J_{ij}-K_{ij}\bigr), ]

where

[ J_{ij}= \iint |\phi_i(\mathbf r_1)|^{2}\,\frac{e^{2}}{4\pi\varepsilon_{0}r_{12}}\,|\phi_j(\mathbf r_2)|^{2}\,d^3r_1d^3r_2 ]

(Coulomb integral) and

[ K_{ij}= \iint \phi_i^{}(\mathbf r_1)\phi_j(\mathbf r_1)\, \frac{e^{2}}{4\pi\varepsilon_{0}r_{12}}\, \phi_j^{}(\mathbf r_2)\phi_i(\mathbf r_2)\,d^3r_1d^3r_2 ]

(exchange integral).

In Hartree–Fock the exchange term is already included in the self‑consistent field, so the remaining part to be treated perturbatively is usually small (a few percent of the total binding energy).
Quantitative benchmark – Neon ((Z=10)).
- Hartree–Fock (self‑consistent field) total energy: (-1285.0\;\text{eV}).
- Adding the second‑order Møller–Plesset (MP2) correlation correction: (-1290.2\;\text{eV}).
- Exact non‑relativistic energy (from full CI calculations) ≈ (-1290.6\;\text{eV}).
The MP2 correction (the perturbative treatment of the correlation beyond the mean field) accounts for ≈0.4 % of the total energy, confirming that the independent‑electron picture captures the overwhelming majority of the binding.

2.5 Summary of the justification

Argument	What it tells us
Order‑of‑magnitude estimate (\displaystyle \frac{V_{ee}}{V_{en}} \sim \frac{1}{Z})	Electron‑electron repulsion is a (1/Z) correction; for (Z\ge 2) it is already smaller than the main nuclear term.
Screening & average distances	The average inter‑electronic distance is larger than the electron‑nucleus distance, further reducing the repulsion.
Perturbation theory on simple atoms (He, Li(^+), Be(^{2+}), …)	First‑order correction already gives energies within a few percent of the exact result.
Hartree–Fock + MP2/CI for many‑electron atoms	Correlation (the part not captured by the averaged field) contributes at the level of (10^{-3})–(10^{-2}) of the total binding energy.
Variational principle	The Hartree–Fock (independent‑electron) energy is an upper bound; adding the perturbative correction lowers the energy, moving it toward the exact value.

Hence, **the independent‑electron model plus perturbative

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