Hartree Method

The Hartree method replaces the instantaneous interaction among atomic electrons by self-consistent direct Coulomb fields. Each orbital is solved in a potential generated by the densities of the other occupied orbitals; the resulting orbitals generate new densities, so the procedure must be iterated to a fixed point.

For an electron in orbital $i$, the central idea is

$$\begin{aligned} \left[ -\frac12\nabla^2 -\frac{Z}{r} +V_i^{\mathrm H}(\mathbf r) \right]\phi_i(\mathbf r) = \epsilon_i\phi_i(\mathbf r), \end{aligned}$$

with

$$V_i^{\mathrm H}(\mathbf r) = \sum_{j\ne i} \int \frac{|\phi_j(\mathbf r')|^2} {|\mathbf r-\mathbf r'|} \,d^3r'.$$

The subscript $i$ matters: the electron must not be included in its own direct field.

Hartree’s construction introduced the self-consistent-field way of thinking that still underlies modern atomic calculations. Raw Hartree theory is not, however, a complete fermionic theory. A product of labeled electron orbitals is generally not antisymmetric, so it omits exchange and does not provide a valid variational family for identical electrons. Hartree–Fock repairs that structural defect by using a Slater determinant.

Required background. Central fields provide radial orbitals, exchange and correlation distinguish missing effects, and helium supplies a two-electron example. The variational principle explains stationary energy equations. The general product-state construction is developed in Hartree approximation; here we specialize it to atomic radial fields.

Atomic Hamiltonian

In atomic units, for a fixed point nucleus of charge $Z$,

$$\begin{aligned} H ={}& \sum_{i=1}^{N}h(i) +\sum_{i<j}\frac{1}{r_{ij}},\\ h(i) ={}& -\frac12\nabla_i^2-\frac{Z}{r_i}. \end{aligned}$$

The exact electronic wavefunction depends jointly on all electron coordinates. The Coulomb term prevents a separation into independent one-electron eigenproblems.

Hartree theory chooses a separable ansatz,

$$\Psi_{\mathrm H} (\mathbf r_1,\ldots,\mathbf r_N) = \prod_{i=1}^{N}\phi_i(\mathbf r_i),$$

with

$$\langle\phi_i|\phi_i\rangle=1.$$

The index $i$ labels a factor in the product, not a permanently identifiable physical electron trajectory. Factorization asserts that the joint spatial probability is a product:

$$|\Psi_{\mathrm H}|^2 = \prod_i|\phi_i(\mathbf r_i)|^2.$$

Its spatial position distribution therefore factorizes. This statement concerns the labeled product; a symmetrized or antisymmetrized state has additional exchange correlations.

Particle symmetry

Under exchange of two electron coordinates,

$$\Psi_{\mathrm H}(\ldots,\mathbf r_i,\ldots,\mathbf r_j,\ldots)$$

does not generally acquire the required minus sign. Imposing orbital occupations or orthogonality by hand does not antisymmetrize the product. Consequently:

• the raw Hartree product is not an admissible many-electron wavefunction for generic electronic states;
• its optimized energy is not guaranteed to be an upper bound to the physical fermionic ground-state energy;
• Pauli exclusion and exchange splitting do not follow from the ansatz.

The direct Hartree field remains a meaningful component of Hartree–Fock and density-functional equations. The defect lies in using the direct-only product as the full electronic state.

Hartree Energy Functional

For the product ansatz, define the direct integral

$$J_{ij} ={} \iint \frac{ |\phi_i(\mathbf r)|^2 |\phi_j(\mathbf r')|^2 }{ |\mathbf r-\mathbf r'| } \times d^3r\,d^3r'.$$

The Hartree energy is

$$\begin{aligned} E_{\mathrm H} ={}& \sum_i h_{ii} +\frac12 \sum_{i\ne j}J_{ij},\\ h_{ii} ={}& \langle\phi_i|h|\phi_i\rangle. \end{aligned}$$

The restriction $i\ne j$ removes one-electron self-interaction. The factor $1/2$ then prevents double counting because $J_{ij}=J_{ji}$.

Equivalently,

$$E_{\mathrm H} = \sum_i h_{ii} +\sum_{i<j}J_{ij}.$$

There is no crossed exchange integral $K_{ij}$ in this functional.

Relationship to the Variational Principle

Introduce one Lagrange multiplier $\epsilon_i$ for each normalization condition:

$$\mathcal L = E_{\mathrm H} -\sum_i\epsilon_i \left( \langle\phi_i|\phi_i\rangle-1 \right).$$

Varying with respect to $\phi_i^*(\mathbf r)$ gives

$$\left[ h +\sum_{j\ne i}V_j^{\mathrm H} \right]\phi_i = \epsilon_i\phi_i,$$

where

$$V_j^{\mathrm H}(\mathbf r) = \int \frac{|\phi_j(\mathbf r')|^2} {|\mathbf r-\mathbf r'|} \,d^3r'.$$

Each orbital is therefore an eigenfunction of an operator built from all the other orbitals. The equations are nonlinear as a coupled system.

What the multipliers mean

The $\epsilon_i$ enforce normalization and become one-electron eigenvalues of the converged Hartree operators. They are not separately protected by the many-body variational theorem and are not exact ionization energies.

No automatic orthogonality

The raw product variation imposes normalization but not

$$\langle\phi_i|\phi_j\rangle=0 \qquad (i\ne j).$$

Practical atomic constructions may impose orthogonality to maintain a shell basis, or obtain it when several orbitals solve the same Hermitian central operator. In self-excluded Hartree theory, different orbitals generally see different operators, so orthogonality is not automatic. Adding it is an extra modeling constraint and still does not replace antisymmetry.

Variational status

For distinguishable particles, or for a bosonic problem where the product has the correct symmetry, minimizing this functional gives an upper bound within an admissible trial family. For generic electrons, the labeled product lies outside the antisymmetric Hilbert space. Its stationary energy must not be advertised as a fermionic variational upper bound.

A converged self-consistent solution is only a stationary point of the chosen constrained functional. It can be a local minimum, an excited fixed point, or a saddle.

Spherical Atomic Form

For a central-field description, write a normalized spatial orbital as

$$\phi_{n\ell m}(\mathbf r) = \frac{P_{n\ell}(r)}{r} Y_{\ell m}(\hat{\mathbf r}),$$

with

$$\int_0^\infty |P_{n\ell}(r)|^2\,dr = 1.$$

For a spherically averaged configuration with $q_{n\ell}$ electrons in subshell $n\ell$,

$$n(r) = \frac{1}{4\pi r^2} \sum_{n\ell} q_{n\ell} |P_{n\ell}(r)|^2.$$

The normalization is

$$4\pi \int_0^\infty n(r)r^2\,dr = N.$$

For a specific electron $i$ in subshell $a=(n\ell)$, the configuration-average density of the other electrons is

$$n_i^{(-)}(r) = n(r) -\frac{|P_a(r)|^2}{4\pi r^2}.$$

If $q_a$ electrons occupy that subshell, this subtraction changes its contribution from $q_a$ to $q_a-1$.

Configuration average

A partially filled open shell need not have a spherical density in a state with specified magnetic quantum numbers. The formula above averages occupations over magnetic substates. It gives a central potential and preserves radial shell language, but it erases term-dependent angular structure.

This is a model choice, not a theorem that every oriented atomic state has a spherical instantaneous density. Averaging over a complete degenerate magnetic multiplet does produce a spherical density.

Effective Hartree Potential

For a spherical density, the angular integral can be performed exactly. The self-excluded Hartree potential is

$$V_i^{\mathrm H}(r) ={} 4\pi \left[ \frac{1}{r} \int_0^r n_i^{(-)}(s)s^2\,ds \qquad+ \int_r^\infty n_i^{(-)}(s)s\,ds \right].$$

The first term is the contribution from charge inside radius $r$; the second is the constant potential inside the outer spherical shells.

Define

$$V_{\mathrm{eff},i}(r) = -\frac{Z}{r} +V_i^{\mathrm H}(r),$$

and the radial kinetic operator

$$T_\ell^{\mathrm{rad}} = -\frac12\frac{d^2}{dr^2} +\frac{\ell(\ell+1)}{2r^2}.$$

The radial Hartree equation is

$$\left[ T_{\ell_i}^{\mathrm{rad}} +V_{\mathrm{eff},i}(r) \right]P_i(r) = \epsilon_iP_i(r).$$

Regular bound solutions obey

$$P_i(r)\propto r^{\ell_i+1} \qquad (r\to0)$$

and decay at large radius when $\epsilon_i<0$.

Poisson equation

The direct potential satisfies

$$\nabla^2V_i^{\mathrm H}(\mathbf r) = -4\pi n_i^{(-)}(\mathbf r).$$

For a spherical field,

$$\frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{dV_i^{\mathrm H}}{dr} \right) = -4\pi n_i^{(-)}(r).$$

Either the integral formula or this boundary-value problem can be used numerically. Their agreement is a useful implementation check.

Screening and the Asymptotic Tail

The self-excluded density contains $N-1$ electrons, so

$$V_i^{\mathrm H}(r) \sim \frac{N-1}{r} \qquad (r\to\infty).$$

The combined effective potential behaves as

$$-\frac{Z}{r} +V_i^{\mathrm H}(r) \sim -\frac{Z-N+1}{r}.$$

For a neutral atom, $N=Z$, and therefore

$$V_{\mathrm{eff},i}(r) \sim -\frac{1}{r}.$$

This is the residual attraction of a singly charged ionic core seen by a distant electron.

If one instead builds the direct field from the full $N$-electron density without subtracting orbital $i$, the neutral-atom tail approaches zero:

$$-\frac{Z}{r} +\frac{N}{r} =0.$$

That is a clear self-interaction failure. It distorts weakly bound and Rydberg orbitals especially strongly.

Near the nucleus, the Hartree potential remains finite for a regular density, while the nuclear term retains the singular behavior $-Z/r$. Core penetration and nodal structure therefore remain controlled by both the nuclear cusp region and the self-consistent screening profile.

Self-Consistency

The orbitals determine the fields that determine the orbitals. A practical atomic iteration is:

1. choose a configuration and trial radial orbitals;
2. normalize them and form the spherical occupation density;
3. subtract the appropriate self-density for each orbital;
4. build each direct potential;
5. solve the radial eigenvalue equations with the required node counts and boundary conditions;
6. mix new and old orbitals, densities, or potentials;
7. repeat until energy, density, orbital residuals, and boundary behavior have converged.

Mixing is numerical, not physical

A simple linear density update is

$$\begin{gathered} n^{(k+1)} = (1-\alpha)n^{(k)} +\alpha n_{\mathrm{out}}^{(k)},\\ 0<\alpha\le1. \end{gathered}$$

For orbital-specific self-exclusion, mix the individual orbital densities (or individual potentials) consistently, not only their total; the latter does not determine which density must be subtracted. Small $\alpha$ can damp oscillations, but it does not change the target stationary equations. More elaborate acceleration methods likewise alter the route to a fixed point rather than the defining Hartree functional.

Multiple fixed points

Different initial orbitals can converge to different stationary configurations. An iterative solver can also cycle or diverge. A small change between two successive iterates is not enough if:

• the residual of the radial equation remains large;
• the density is drifting slowly;
• a symmetry or occupation changed unintentionally;
• a lower stationary solution exists;
• the grid or radial box is not converged.

Total Energy and Orbital Energies

Multiply each Hartree equation by $\phi_i^*$ and integrate:

$$\epsilon_i = h_{ii} +\sum_{j\ne i}J_{ij}.$$

Summing over occupied orbitals gives

$$\sum_i\epsilon_i = \sum_i h_{ii} +\sum_{i\ne j}J_{ij}.$$

The interaction appears twice because each pair contributes to both orbital equations. The total energy is

$$\begin{aligned} E_{\mathrm H} ={}& \sum_i\epsilon_i -\frac12 \sum_{i\ne j}J_{ij}\\ ={}& \sum_i\epsilon_i -\sum_{i<j}J_{ij}. \end{aligned}$$

Adding occupied orbital eigenvalues without the subtraction double counts the direct interaction.

Energy is not enough

At a well-resolved stationary solution of a purely Coulombic problem, uniform coordinate scaling gives the virial relation

$$2T+V=0,$$

where $V$ includes nuclear attraction and electron–electron repulsion. A poor virial ratio can expose basis, grid, boundary, or self-consistency errors.

The converse is not guaranteed: a plausible total energy and virial ratio do not prove that exchange, correlation, or term-dependent angular structure has been captured.

Two-Electron Closed-Shell Exception

The helium ground-state symmetry exposes an important nuance. Let two electrons occupy the same spatial orbital $\phi$ and combine into the antisymmetric spin singlet $\chi_{00}$. The total state is

$$\Psi(1,2) = \phi(\mathbf r_1) \phi(\mathbf r_2) \chi_{00}(s_1,s_2).$$

Although its spatial factor is a product, the full spin-space state is antisymmetric and is exactly one Slater determinant built from $\phi\alpha$ and $\phi\beta$.

Its energy functional is

$$E[\phi] = 2\langle\phi|h|\phi\rangle +J[\phi],$$

where

$$J[\phi] = \iint \frac{ |\phi(\mathbf r)|^2 |\phi(\mathbf r')|^2 }{ |\mathbf r-\mathbf r'| } \,d^3r\,d^3r'.$$

Variation gives

$$\left[ h+V^{\mathrm H}[|\phi|^2] \right]\phi = \epsilon\phi.$$

Each electron sees the density of the one other electron. In this special two-electron closed-shell case, the spatial Hartree equation coincides with restricted Hartree–Fock: opposite orthogonal spins have no mutual Fock exchange integral, while each orbital’s self-interaction is canceled.

The approximation still misses electron correlation. The product spatial factor cannot respond explicitly to $r_{12}$ and does not satisfy the exact opposite-spin Coulomb cusp. Helium Atom develops the resulting energy and spectroscopy benchmark.

Missing Exchange and Correlation

Two spin-adapted states with the same orbital densities have the same direct Hartree energy. Their $J_{ab}\pm K_{ab}$ singlet–triplet splitting requires exchange, and their conditional Coulomb avoidance requires correlation beyond a product. Spherical averaging also removes open-shell angular multipoles and term-dependent energies. Thus converging the radial field improves screening and relaxation without recovering the atomic term spectrum.

Hartree, Central Field, and Hartree–Fock

Construction	Effective operator	State interpretation	Main omission
fitted central field	chosen local $V(r)$	model orbitals	transferability and no unique many-body state
Hartree	self-consistent local direct field	labeled product, except special admissible cases	antisymmetry, exchange, and correlation
Hartree–Fock	direct plus nonlocal exchange	optimized Slater determinant	correlation beyond one determinant

Every Hartree field is an effective central field after spherical averaging, but not every central field is Hartree: a model or fitted potential need not be generated by its own orbitals.

Hartree–Fock is also not “Hartree plus an empirical exchange force.” Replacing the product trial family by determinants changes the variational manifold, and the exchange operator follows from evaluating the same Coulomb Hamiltonian in that antisymmetric state.

Exercises

Exercise 1: Derive the orbital equation

Vary the Hartree energy with respect to $\phi_i^*$ while preserving the normalization of every orbital. Explain why no factor of $1/2$ remains in the field.

Solution

The terms containing $\phi_i^*$ are $h_{ii}$ and the pair contributions with either index equal to $i$. Since $J_{ij}=J_{ji}$, differentiating

$$\frac12\sum_{k\ne l}J_{kl}$$

produces two equal halves. The result is

$$\frac{\delta E_{\mathrm H}} {\delta\phi_i^*(\mathbf r)} = \left[ h+\sum_{j\ne i}V_j^{\mathrm H} \right]\phi_i(\mathbf r).$$

Including the normalization multiplier gives

$$\left[ h+\sum_{j\ne i}V_j^{\mathrm H} \right]\phi_i = \epsilon_i\phi_i.$$

The pair factor $1/2$ prevents double counting in the energy; the orbital feels the full field of every other density.

Exercise 2: Derive the spherical potential

Starting from a spherical density $n(r)$, show that

$$\int \frac{n(r')}{|\mathbf r-\mathbf r'|} \,d^3r'$$

has the inside-plus-outside form used above.

Solution

The angular average of the Coulomb kernel is

$$\int d\Omega' \frac{1}{|\mathbf r-\mathbf r'|} = \frac{4\pi}{\max(r,r')}.$$

Splitting the radial integral at $r'=r$ gives

$$V(r) ={} \frac{4\pi}{r} \int_0^r n(s)s^2\,ds +4\pi \int_r^\infty n(s)s\,ds.$$

The interior charge acts as if concentrated at the origin; every exterior spherical shell contributes a spatially constant potential inside that shell.

Exercise 3: An analytic direct field

A normalized $1s$ density is $n(r)=\zeta^3e^{-2\zeta r}/\pi$. Insert it into the spherical-potential formula and determine the finite value at the origin and the far-field charge. Explain its use in a two-electron closed-shell iteration.

Solution

The elementary integrals give

$$V^{\mathrm H}(r)=\frac{1-e^{-2\zeta r}(1+\zeta r)}{r}.$$

Expansion at small $r$ yields $V^{\mathrm H}(0)=\zeta$ and zero first radial derivative. At large $r$, $rV^{\mathrm H}\to1$. Each electron in a helium-like closed shell sees precisely this one-electron direct density for an exponential trial orbital; using twice this potential would retain self-interaction.

Exercise 4: Recover the total energy

Given

$$\epsilon_i=h_{ii}+\sum_{j\ne i}J_{ij},$$

express $E_{\mathrm H}$ in terms of occupied orbital multipliers.

Solution

Summing gives

$$\sum_i\epsilon_i = \sum_i h_{ii} +\sum_{i\ne j}J_{ij}.$$

The Hartree functional contains only half of the ordered pair sum. Therefore

$$E_{\mathrm H} = \sum_i\epsilon_i -\frac12\sum_{i\ne j}J_{ij}.$$

Equivalently, subtract one $J_{ij}$ for every unordered pair.

Exercise 5: One-electron limit

Apply the Hartree equations to a hydrogenic ion with $N=1$. What direct potential and total energy must result?

Solution

The sum over $j\ne i$ is empty:

$$V_1^{\mathrm H}=0.$$

The orbital equation is exactly the one-electron Coulomb equation,

$$\left( -\frac12\nabla^2-\frac{Z}{r} \right)\phi = \epsilon\phi.$$

There is no pair energy, so $E_{\mathrm H}=\epsilon$ for the occupied state. Any nonzero electron–electron contribution signals that the electron has been included in its own field.

Exercise 6: Why helium is special

Explain why the spatial product $\phi(\mathbf r_1)\phi(\mathbf r_2)$ can be part of an admissible fermionic state for the helium ground configuration, yet the corresponding approximation still lacks correlation.

Solution

The spatial product is symmetric. Multiplying it by the antisymmetric spin singlet makes the full spin-space wavefunction antisymmetric:

$$\Psi = \phi(\mathbf r_1) \phi(\mathbf r_2) \chi_{00}.$$

This is the determinant of $\phi\alpha$ and $\phi\beta$. Its optimized equation contains the direct density of the one opposite-spin electron and coincides with restricted Hartree–Fock for a two-electron closed shell.

The spatial factor still separates into one-electron functions and has no explicit $r_{12}$ response. It cannot reproduce the exact Coulomb cusp or conditional radial and angular motion, so correlation remains missing.

Exercise 7: Diagnose false convergence

An SCF run changes its total energy by less than $10^{-10}$ hartree, but the largest radial-equation residual is $10^{-3}$ hartree and doubling the radial box changes the outer orbital substantially. Is the calculation converged?

Solution

No. A small change between successive energies can result from heavy mixing, stagnation, or cancellation. The large orbital residual shows that the reported orbitals do not solve their final equations, and sensitivity to the radial box shows that the boundary representation is unresolved.

The calculation should tighten the radial solve, enlarge the box, check the asymptotic tail, and repeat the SCF cycle. Energy, density, potential, and residual criteria must be satisfied together.

References

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