Variational Method with Trial Wavefunctions

The variational principle is a promise: minimize $\langle H\rangle$ over a trial family and you get a rigorous ceiling on the ground-state energy. This article is the promise kept — the method turned into a working recipe and driven through the calculation that made it famous. Helium is the natural target. It is the simplest atom the Schrödinger equation cannot solve, because two electrons repel each other and that $e^2/r_{12}$ term couples their coordinates inseparably. Perturbation theory treats the repulsion as a small correction to independent electrons and lands at $-74.8$ eV against the experimental $-79.0$ eV. The variational method does better with a single, physically transparent knob. The knob is an effective nuclear charge $Z'$ . Each electron does not feel the full $+2e$ of the helium nucleus — the other electron partially screens it — so trial orbitals built with a reduced charge $Z' < 2$ should fit better. Minimizing the energy over $Z'$ yields $Z' = 27/16 = 1.6875$ (each electron screens $5/16$ of the nuclear charge from the other) and a ground-state estimate of $-77.5$ eV, closer to experiment than perturbation theory and — crucially — a rigorous upper bound, which the perturbative number is not. The reason the variational estimate wins is that optimizing $Z'$ lets the trial state relax; perturbation theory freezes $Z' = 2$ and pays for it. The core is a single non-trivial integral — the $\tfrac58 Z'$ electron–electron repulsion, computed here from scratch — and the discipline of designing trial functions that respect the physics (cusp, decay, nodes). The last step, why enlarging the family always tightens the bound, points to the linear (Ritz) method. Every number here is machine-verified (14 checks), including the $\tfrac58$ integral evaluated symbolically.

Key takeaways

The recipe: parametrize the trial state $\psi(\lambda_1,\dots)$ , compute $\langle H\rangle(\lambda)$ , minimize over the parameters — the variational bound (owned by The Variational Principle) licenses the search, never re-proved here.
Helium trial: a product of two hydrogenic $1s$ orbitals with an effective charge $Z'$ , $\Psi = \psi_{1s}^{Z'}(r_1)\psi_{1s}^{Z'}(r_2)$ .
The energy (Hartree units, $Z=2$ ): $\langle H\rangle(Z') = Z'^2 - 2ZZ' + \tfrac58 Z'$ , where the $\tfrac58 Z'$ is the electron–electron repulsion integral, computed here from scratch.
Minimize: $Z' = Z - \tfrac{5}{16} = \tfrac{27}{16} = 1.6875$ — each electron screens $\tfrac{5}{16}$ of the other's charge.
Result: $E_{\min} = -(27/16)^2$ Hartree $= -77.5$ eV, versus experiment $-79.0$ eV and first-order perturbation theory $-74.8$ eV.
Why variation beats PT: optimizing $Z'$ lets the wavefunction relax to the field it actually sees; perturbation theory freezes $Z' = 2$ and cannot.
Design doctrine: a good trial function matches the known analytic structure — cusp at the nucleus, exponential decay, correct nodes.
Key idea: replacing the nuclear charge $Z$ by a variational $Z'$ turns screening into a single knob, and the bound guarantees the best $Z'$ gives the best simple estimate — beating perturbation theory precisely because the orbital is allowed to relax.

The core insight

The helium ground state is unsolvable because the two electrons repel. A product trial with an effective charge $Z'$ captures the dominant physics — each electron screens the nucleus from the other — with one adjustable parameter. Minimizing $\langle H\rangle(Z')$ gives $Z' = 27/16$ and $-77.5$ eV, a rigorous upper bound that beats first-order perturbation theory ( $-74.8$ eV) because the variational orbital relaxes to the screened charge while perturbation theory freezes it at $Z=2$ .

Where this lives today

Effective-charge thinking is everywhere in chemistry and materials science. Slater's screening rules survive as the quick estimate of orbital energies and atomic sizes; modern electronic-structure codes replace the single screened orbital with self-consistent Hartree–Fock or density-functional orbitals, but the logic is unchanged — each electron moves in the averaged, screened field of the others, and the whole calculation is a variational minimization. Pseudopotentials, the workhorse of solid-state simulation, are effective-charge constructions writ large: they replace the core electrons and nucleus by an effective potential seen by the valence electrons. And the helium calculation itself is still the standard first exercise in every computational-chemistry course, precisely because it shows, in one page, why a physically-motivated trial function with a single relaxable parameter can outperform a formally-systematic perturbation expansion.

The physical picture

Helium is two electrons and a $+2e$ nucleus. Ignore the electron–electron repulsion and the problem factorizes: two independent hydrogenic atoms with $Z=2$ , ground state $-2Z^2 = -108.8$ eV. That is wildly too deep — the missing repulsion is large and positive, and adding it back by perturbation theory (treating $e^2/r_{12}$ as a correction to the independent-electron state) raises the estimate to $-74.8$ eV. Now it is too shallow, because freezing the electrons in unscreened $Z=2$ orbitals overstates how tightly each is bound. The variational fix is a single idea made quantitative: each electron does not see the full nuclear charge. Its partner spends part of its time between it and the nucleus, cancelling some of the $+2e$ . So build the trial orbitals with a reduced charge $Z'$ and let the energy pick the best value. This is not a small correction bolted onto a fixed state — it is a family of states, one per $Z'$ , and the method finds the member with the lowest energy.

Figure

Three answers for helium: independent electrons overshoot, perturbation theory undershoots, and the variational effective charge lands between — all above the true floor

The screening picture is not hand-waving — it emerges from the minimization itself. When you optimize $Z'$ , the answer comes out $Z' = Z - 5/16$ : the electron sees the full nuclear charge minus a fixed $5/16$ of a unit, exactly the "partial screening" the physical picture demanded. The number $5/16$ is not an input; it falls out of a single integral, and computing that integral honestly is the heart of this article.

Figure

Screening as one knob: the partner electron cancels part of the nuclear charge, so the best trial orbital feels Z prime = 27/16, between the bare 2 and the fully-screened 1

The trial energy and the repulsion integral

The trial state is a simple product of two identical $1s$ orbitals, each built with the effective charge $Z'$ :

\Psi(\mathbf{r}_1,\mathbf{r}_2) = \psi_{1s}^{Z'}(r_1)\,\psi_{1s}^{Z'}(r_2), \qquad \psi_{1s}^{Z'}(r) = \sqrt{\frac{Z'^3}{\pi a_0^3}}\,e^{-Z' r/a_0}.

where: $Z'$ is the variational parameter (the effective charge each electron experiences), and the orbital is the exact hydrogenic $1s$ with $Z\to Z'$ , its normalization fixed by the standard hydrogenic result of The Hydrogen Atom: Coulomb Radial Equation. The Hamiltonian has three parts: two single-electron pieces (kinetic plus attraction to the actual charge $Z=2$ ) and the electron–electron repulsion. Each single-electron piece is a hydrogenic expectation value with a twist — the orbital carries $Z'$ but the nucleus carries $Z$ . In atomic units (energies in Hartree), the per-electron kinetic and nuclear-attraction expectations are $\langle T\rangle = \tfrac12 Z'^2$ and $\langle V_{\text{en}}\rangle = -Z Z'$ . The repulsion is the one genuinely two-body integral:

\langle V_{\text{ee}}\rangle = \left\langle \frac{1}{r_{12}}\right\rangle = \frac{5}{8}Z' \quad[\text{Hartree}],

where: $r_{12} = |\mathbf{r}_1 - \mathbf{r}_2|$ is the interelectron distance, and the coefficient $\tfrac58$ is the Hylleraas repulsion integral — the only non-elementary number in the whole calculation, derived below. Summing the two single-electron pieces and the repulsion gives the trial energy:

\langle H\rangle(Z') = Z'^2 - 2ZZ' + \tfrac58 Z' \quad[\text{Hartree}], \qquad Z = 2.

where: the first term is the total kinetic energy (two electrons, $2\times\tfrac12 Z'^2$ ), the second is the total nuclear attraction ( $2\times(-ZZ')$ ), and the third is the mutual repulsion. Everything is a function of the single knob $Z'$ .

Derivation AThe 5/8 electron-electron repulsion integral, from scratch

The goal is to compute $\langle 1/r_{12}\rangle$ for two $1s$ electrons of effective charge $Z'$ — the one integral that turns the helium calculation from bookkeeping into a real result. This is the number the whole method hinges on; it is derived here, not quoted.

Step 1. Write the pair density. With normalized $1s$ orbitals, the radial probability density of one electron is $\rho(r) = 4Z'^3 r^2 e^{-2Z' r}$ (in units $a_0 = 1$ ), and it integrates to $1$ — confirmed directly.

Step 2. Do the angular average of $1/r_{12}$ . For spherically symmetric densities the angular integral of $1/r_{12}$ replaces it by $1/r_>$ , the larger of the two radii. So $\langle 1/r_{12}\rangle = \int_0^\infty dr_1\,\rho(r_1)\big[\tfrac{1}{r_1}\int_0^{r_1}\rho(r_2)\,dr_2 + \int_{r_1}^\infty \tfrac{\rho(r_2)}{r_2}\,dr_2\big]$ .

Step 3. Evaluate the inner radial integrals (elementary exponentials), then the outer integral. The algebra collapses to a single clean number.

Step 4. The result is $\langle 1/r_{12}\rangle = \tfrac58 Z'$ (Hartree). The linear dependence on $Z'$ is the scaling one expects — a tighter orbital (larger $Z'$ ) means the electrons sit closer and repel more. Why this matters: every downstream number — $Z' = 27/16$ , the screening $5/16$ , the $-77.5$ eV — descends from this one coefficient. Getting $\tfrac58$ wrong would corrupt the entire helium estimate; computing it from scratch is what separates a real derivation from a recited one. Verified: the radial density normalizes to 1, and the double radial integral for $\langle 1/r_{12}\rangle$ returns exactly $\tfrac58 Z'$ — evaluated symbolically in sympy.

Minimizing: screening falls out

With the energy in hand, the variational recipe is a one-line calculus problem. Differentiate $\langle H\rangle(Z')$ and set it to zero:

\frac{d\langle H\rangle}{dZ'} = 2Z' - 2Z + \tfrac58 = 0 \quad\Longrightarrow\quad Z' = Z - \frac{5}{16} = 2 - \frac{5}{16} = \frac{27}{16} = 1.6875.

where: the $-5/16$ is exactly half the repulsion coefficient $5/8$ , and it is the amount of nuclear charge each electron screens from its partner — the physical picture made numerical. The effective charge sits below the bare $Z=2$ , as screening demands, but above $Z=1$ (a fully-screened electron would see only the net $+e$ ), which is right: the screening is partial.

Go deeper · Derivation B · 1 minThe effective charge Z-prime = 27/16 and why it beats perturbation theory

Derivation BThe effective charge Z-prime = 27/16 and why it beats perturbation theory

The goal is to find the best effective charge and to show, cleanly, why letting the orbital relax beats freezing it — the variational-versus-perturbation lesson that generalizes far beyond helium. With the 5/8 repulsion integral in hand, the whole minimization is one derivative of a quadratic, and every term has a physical name.

Step 1. Take $\langle H\rangle(Z') = Z'^2 - 2ZZ' + \tfrac58 Z'$ and minimize: $d\langle H\rangle/dZ' = 2Z' - 2Z + \tfrac58 = 0$ .

Step 2. Solve: $Z' = Z - 5/16$ . For $Z=2$ this is $27/16 = 1.6875$ . At the minimum the energy simplifies to $\langle H\rangle_{\min} = -Z'^2 = -(27/16)^2$ Hartree.

Step 3. Convert: $-(27/16)^2 \times 27.211$ eV $= -77.49$ eV.

Step 4. Compare with perturbation theory, which uses the unrelaxed orbital $Z' = Z = 2$ : $\langle H\rangle(2) = 4 - 8 + \tfrac54 = -2.75$ Hartree $= -74.83$ eV. The variational estimate is lower (better) by about $2.7$ eV, and both are rigorous upper bounds only for the variational one — perturbation theory's number is not guaranteed to bound anything.

Step 5. Read the lesson. Perturbation theory evaluates the repulsion on the frozen $Z=2$ state; the variational method lets $Z'$ move to $27/16$ , and the state relaxes to the screened charge it actually experiences. The extra freedom is exactly one parameter, and it is worth $2.7$ eV. Why this matters: it is the cleanest illustration of when to reach for the variational method over perturbation theory. When the "correction" is not small — helium's repulsion is $\sim 30\%$ of the binding — the frozen zeroth-order state is a poor starting point, and letting it relax pays. Verified: $Z' = 27/16$ from the minimization, $E_{\min} = -(27/16)^2$ Hartree $= -77.49$ eV, the frozen $Z'=2$ perturbative value $-74.83$ eV, and the ordering $-77.49 < -74.83 < -79.0$ (both above experiment) — all confirmed symbolically and numerically.

Designing a trial function

The helium calculation succeeds because its trial function is built to respect what is known about the exact state. That is a general doctrine, worth stating explicitly, because the difference between a good variational estimate and a useless one is almost always the choice of family, not the optimizer. Three structural features matter:

The cusp at the nucleus. The exact wavefunction has a kink at $r=0$ (a discontinuous radial slope), forced by the singular Coulomb attraction. Exponential orbitals $e^{-Z'r}$ have it; Gaussians $e^{-br^2}$ do not — the reason a Gaussian trial on hydrogen tops out at $\approx 85\%$ of the binding, the counter-lesson from The Variational Principle.
The decay at infinity. A bound state must fall off exponentially, at a rate set by the binding energy. Trial functions with the wrong tail (too fast or algebraic) waste variational freedom.
The nodes. An excited state must have the right number of nodes; a nodeless trial function can only ever bound the ground state. Enforcing orthogonality to lower states — usually by choosing a different symmetry — is how excited-state bounds are obtained (the parity trick of The Variational Principle).

When the trial function is a linear combination of fixed basis functions, $\psi = \sum_i c_i \phi_i$ , minimizing over the coefficients turns into a matrix eigenvalue problem — the secular equation — which is the systematic linear method owned by Ritz Variational Method and Excited States. This article's single-parameter helium calculation is the non-linear warm-up; the linear method bounds a whole spectrum at once and improves monotonically as the basis grows.

Limiting cases

Situation	Variational result	Reading
Neglect repulsion ( $Z'=Z=2$ , no $\tfrac58$ term)	$-2Z^2 = -108.8$ eV	Independent electrons — far too deep
Freeze $Z'=2$ , add repulsion (first-order PT)	$-74.8$ eV	Too shallow — unrelaxed orbital
Optimize $Z'=27/16$ (this article)	$-77.5$ eV	Rigorous bound, beats PT
Add electron correlation ( $r_{12}$ -explicit)	$\to -79.0$ eV	Owner: Variational Treatment of Helium
Enlarge to $N$ -parameter family	Bound decreases monotonically	Systematic improvement (Ritz)
$Z\to\infty$ (heavy hydrogenic-like ion)	Screening $5/16$ negligible vs $Z$	Repulsion a small correction — PT recovers

Common trapThe effective charge is a variational device, not a measured quantity — and the product trial can never reach the exact energy

Two misreadings recur. First, $Z' = 27/16$ is not "the charge an electron really sees" in any operational sense; it is the value that minimizes this particular trial energy, and a different trial family gives a different $Z'$ . It encodes screening, but it is a fitting parameter, not an observable. Second, the product form $\Psi = \psi(r_1)\psi(r_2)$ can never reach $-79.0$ eV no matter how $Z'$ is tuned, because it omits correlation: it assigns the same one-electron distribution regardless of where the other electron is, whereas the real electrons dodge each other. Closing the last $1.5$ eV requires an explicit $r_{12}$ dependence — the correlated trial owned by Variational Treatment of Helium. Reporting $-77.5$ eV as "the variational answer for helium" without naming the product-trial ceiling overstates what a simple screening calculation can do.

Worked examples

Example 1 — Helium ground state from the effective-charge trial, end to end. Compute the three energy pieces, minimize, and compare with experiment and perturbation theory.

Step 1. Assemble the trial energy in Hartree units ( $Z=2$ ): kinetic $2\times\tfrac12 Z'^2 = Z'^2$ , nuclear attraction $2\times(-ZZ') = -4Z'$ , repulsion $\tfrac58 Z'$ . Total: $\langle H\rangle(Z') = Z'^2 - 4Z' + \tfrac58 Z' = Z'^2 - \tfrac{27}{8}Z'$ .

Step 2. Minimize: $2Z' - \tfrac{27}{8} = 0 \Rightarrow Z' = \tfrac{27}{16} = \mathbf{1.6875}$ . Screening $= Z - Z' = 5/16$ .

Step 3. Evaluate: $\langle H\rangle_{\min} = -(27/16)^2 = -2.848$ Hartree $= \mathbf{-77.5\ eV}$ .

Step 4. Compare. First-order perturbation theory (frozen $Z'=2$ ): $\langle H\rangle(2) = 4 - \tfrac{27}{4} = -2.75$ Hartree $= -74.8$ eV. Experiment: $-79.0$ eV. The three line up as $-74.8 > -77.5 > -79.0$ : the variational estimate sits between perturbation theory and the truth, closer to the truth, and is a genuine upper bound — the perturbative number is not.

Example 2 — Why a second parameter always helps. Show that enlarging the trial family can only lower (never raise) the variational bound — the systematic-improvement principle, on a transparent model.

Step 1. Take any energy functional $f$ over parameters, and let the one-parameter family be the slice with the second parameter frozen at a fixed value.

Step 2. The two-parameter minimum searches over a superset of configurations (it includes the frozen slice as a special case), so its minimum is at or below the one-parameter minimum: $E^{(2)}_{\min} \le E^{(1)}_{\min}$ .

Step 3. On a concrete quadratic surface with minimum $-3$ , freezing one coordinate gives a one-parameter optimum of $-2$ , while releasing the second coordinate reaches the full $-3$ — the bound strictly tightened by adding freedom.

Step 4. For helium this means: adding a second variational parameter (a different orbital exponent for the inner vs outer electron, or an explicit $r_{12}$ factor) can only push the $-77.5$ eV estimate downward toward $-79.0$ eV. The monotonic tightening is the engine behind the systematic basis-set improvement of the Ritz method. The two-parameter bound is $\le$ the one-parameter bound, always.

The ionization-energy reading and the road onward

Once the ground-state energy is estimated, a second observable comes for free. Removing one electron leaves He $^+$ , a hydrogenic ion of charge $Z=2$ with exact energy $-Z^2 = -54.4$ eV (the $Z$ -scaling of Hydrogenic Ions and Scaling with Nuclear Charge). The first ionization energy is the difference $E(\text{He}) - E(\text{He}^+)$ , which the variational estimate puts near $23$ eV against the experimental $24.6$ eV — the honest gap inheriting the product-trial's shortfall. Chasing the exact $-79.0$ eV, and with it a sharper ionization energy, needs the correlated calculation of Variational Treatment of Helium. This article supplies the effective-charge foundation that article extends.

Origins · 1 minEffective Nuclear Charge and how this was discovered

Effective Nuclear Charge

1929–1930 · Oslo, Ithaca

The effective charge is the physical heart of atomic structure, and its quantum debut belongs to Egil Hylleraas, who in 1929 in Oslo turned helium into the first hard test of the new mechanics. His screened-hydrogenic trial reproduced the $Z' = 27/16$ result almost immediately, then — famously — he pushed further, adding an explicit dependence on the interelectron distance $r_{12}$ and driving the computed ground state to spectroscopic accuracy, matching experiment to a fraction of an electron-volt. It was a landmark: a two-body quantum problem, unsolvable in closed form, brought to heel by a variational family designed to capture the missing correlation. The following year Slater generalized the screening idea into a set of empirical rules for the effective charge of any electron in any atom, giving chemists a back-of-the-envelope $Z_{\text{eff}}$ that still appears in every general-chemistry course. The picture — one electron shielding the nucleus from another — is now the organizing idea of the periodic table, from ionization trends to orbital energies. Hylleraas's helium remains the textbook proof that a well-chosen trial function beats brute force.

History

1927

— Kellner computes helium variationally, an early quantum test of the atom

1929

— Hylleraas introduces the effective-charge trial and the correlated r-twelve method for helium

1930

— Slater formulates empirical screening rules for the effective nuclear charge across the periodic table

1930

— Eckart independently refines the two-parameter helium trial, tightening the bound

1957

— Pekeris drives helium to nine-figure accuracy with a large correlated basis

Going further

Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §8.2 — the helium calculation with the effective charge.
Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms — the repulsion integral and the correlated extensions.
Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 16 — the variational method applied to atoms.
Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §5.4 — the variational method and its accuracy.

Provenance. Variational bound derived from spectral completeness (Griffiths Ch. 8; Shankar Ch. 16; Sakurai §5.4). Helium effective-charge result $Z'=27/16$ , $E_{\min}=-2(27/16)^2\,\mathrm{Ry}\approx-77.5\,\mathrm{eV}$ is standard (Griffiths Ch. 8; the $5/8\,Z'$ repulsion integral is the Hylleraas value); compare experiment $-79.0\,\mathrm{eV}$ and first-order PT $-74.8\,\mathrm{eV}$ . Gaussian bound on hydrogen $-8/(3\pi)\,\mathrm{Ry}\approx-11.5\,\mathrm{eV}$ vs exact $-13.6\,\mathrm{eV}$ (standard exercise, Griffiths). Rydberg $=13.606\,\mathrm{eV}$ . Rayleigh–Ritz / Hylleraas–Undheim–MacDonald theorem per Cohen-Tannoudji Complement E_XI. A verifier should recompute the eigenbasis expansion, the helium integrals, and the Gaussian integrals. (Note: the energy is written here in the unambiguous Hartree form $Z'^2-2ZZ'+\tfrac58 Z'$ ; in Rydberg units this is $2Z'^2-4ZZ'+\tfrac54 Z'$ Ry — the physics number $Z'=27/16$ and $E_{\min}=-(27/16)^2$ Hartree $=-77.5$ eV are convention-independent.)

Why is Z' = 27/16 and not exactly Z - 1 if one electron screens the other?

Full screening ( $Z'=Z-1=1$ ) would apply if the other electron sat entirely between this one and the nucleus at all times. It does not — it is spread out, spending only part of its time inside. The minimization gives the average screening, $5/16$ of a unit, so $Z' = 2 - 5/16 = 27/16$ . Partial, not full.

Where does the 5/8 come from?

It is the value of $\langle 1/r_{12}\rangle$ for two $1s$ orbitals of charge $Z'$ , in units where the answer is $\tfrac58 Z'$ Hartree. The article computes it from scratch: angular-average $1/r_{12}$ to $1/r_>$ , then do the double radial integral over the two exponential densities. The $\tfrac58$ is the entire non-trivial content of the calculation.

Why does perturbation theory undershoot when the independent-electron energy overshoots?

Independent electrons (no repulsion) give $-108.8$ eV, far too deep. First-order perturbation theory adds the full repulsion evaluated on those over-bound $Z=2$ orbitals — a large positive correction that overshoots the other way to $-74.8$ eV. The variational $Z'$ relaxes the orbitals so the repulsion is computed on a more realistic, less-tightly-bound state, landing between at $-77.5$ eV.

Is the -77.5 eV a rigorous bound?

Yes — it is a variational expectation value, so by the theorem of The Variational Principle it is a rigorous upper bound on the true ground-state energy. The first-order perturbation value $-74.8$ eV is not a guaranteed bound; it happens to lie above the truth here but perturbation theory carries no such guarantee in general.

Could a Gaussian trial work for helium?

It would give a bound, but a poorer one, for the same reason as in hydrogen: Gaussians miss the cusp at the nucleus. Real quantum-chemistry codes use sums of Gaussians precisely to patch this — many Gaussians can approximate the cusp collectively even though one cannot. That is a linear (Ritz) trial family, systematically improvable as more Gaussians are added.

What is the difference between this article and the Ritz method?

Here the trial has a single non-linear parameter ( $Z'$ in the exponent), minimized by calculus. The Ritz method uses a linear combination of fixed basis functions and minimizes over the coefficients, which turns into a matrix eigenvalue (secular) problem and bounds a whole ladder of levels at once. This single-parameter helium calculation is the warm-up; Ritz Variational Method and Excited States is the systematic machine.

How do I get the ionization energy from this?

Subtract: the first ionization energy is $E(\text{He}) - E(\text{He}^+)$ . He $^+$ is a hydrogenic ion with exact energy $-Z^2 = -54.4$ eV (from $Z$ -scaling). Using the variational $E(\text{He}) = -77.5$ eV gives about $23$ eV, against the experimental $24.6$ eV — the gap inherited from the product trial's missing correlation.