Variational Treatment of Helium

Helium is the first atom quantum mechanics cannot solve exactly. Two electrons, one nucleus, and a mutual repulsion $e^2/r_{12}$ that couples their coordinates so tightly that no separation of variables survives. The variational method turns this defeat into a triumph of estimation. The simplest honest trial — put both electrons in a hydrogen-like $1s$ orbital but let the nuclear charge they feel be an adjustable $Z'$ rather than the true $Z=2$ — is developed in Variational Method with Trial Wavefunctions, which minimizes the energy to $Z'=27/16=1.6875$ and lands at $-77.5$ eV against the experimental $-79.0$ eV. This article picks up exactly where that number stops. We diagnose why the simple product can never reach $-79.0$ : it is a product of one-electron functions, so it builds in no correlation — no tendency for the electrons to stay on opposite sides of the atom, dodging each other's repulsion. We then show the fix that the variational bound guarantees will help: multiply the product by an explicit function of $r_{12}$ (a Jastrow factor), and the two-parameter trial's energy drops below $-77.5$ eV toward the exact answer — the correlated bound sitting between the product estimate and the true value, exactly as the interlacing logic of Ritz Variational Method and Excited States demands. Finally we read off a physical observable the variational energy delivers directly: the first ionization energy, $E(\text{He}) - E(\text{He}^+)$ , with the hydrogenic $\text{He}^+$ energy consumed from Hydrogenic Ions and Scaling with Nuclear Charge. The full explicit- $r_{12}$ chase to chemical accuracy belongs to the many-body track; here we own the structure — why correlation matters and that a correlation parameter provably lowers the energy.

Key takeaways

The simple product trial $\Psi=\psi^{Z'}_{1s}(r_1)\psi^{Z'}_{1s}(r_2)$ , minimized over $Z'$ , gives $Z'=27/16$ and $E=-77.5$ eV (owned by Variational Method with Trial Wavefunctions) — $1.9\%$ above the exact $-79.0$ eV.
The gap is correlation: a product wavefunction cannot let the electrons avoid each other, so it always overestimates the repulsion and lands high.
Fix: multiply by a Jastrow factor $(1+b\,r_{12})$ . The two-parameter trial's energy drops to about $-78.7$ eV — provably below the product and still above the exact, as the variational bound requires.
First ionization energy $=E(\text{He}^+)-E(\text{He})$ ; with the hydrogenic $E(\text{He}^+)=-54.4$ eV, the variational $E(\text{He})=-77.5$ eV gives $23.1$ eV vs the experimental $24.6$ eV; the exact $E(\text{He})$ gives $24.6$ eV.
Key idea: a product wavefunction can never reach the true helium energy because it omits electron correlation; a correlation factor provably lowers the variational bound toward $-79.0$ eV.

The core insight

Helium's missing $1.5$ eV is the correlation energy — the piece a product of one-electron orbitals structurally cannot capture. Because every trial energy is an upper bound, adding a factor that lets the electrons stay apart can only lower the estimate, and it demonstrably does: the correlated bound falls below the effective-charge product and toward the exact value.

Where this lives today

The helium calculation is the training ground for every method that treats electron correlation. Hartree–Fock is the product (mean-field) wavefunction taken to self-consistency; its systematic error is the correlation energy, and recovering it is the entire business of post-Hartree–Fock quantum chemistry — configuration interaction, coupled cluster, and the explicitly-correlated (R12/F12) methods that are the modern descendants of Hylleraas's $r_{12}$ trick. Density-functional theory folds correlation into an exchange–correlation functional. Quantum Monte-Carlo uses Jastrow-correlated trial functions exactly like the two-parameter one here, at enormous scale. The full explicit- $r_{12}$ chase to helium's $-79.005$ eV, the para/ortho ( $1s^2$ vs $1s2s$ ) configurations, and multi-electron screening rules all belong to the many-body track.

Origins · 1 minElectron Correlation and how this was discovered

Electron Correlation

1929–1934 · Oslo, Ithaca

The concept that carries this article is electron correlation — the piece of the energy a product of one-electron orbitals structurally misses. Egil Hylleraas made it quantitative in 1929 in Oslo: by writing helium trial functions that depend explicitly on the inter-electron distance $r_{12}$ , he drove the computed ground-state energy onto the experimental value, proving that the "missing" energy was correlation and that it could be captured by breaking the product form. Around the same time Georg Kellner and others pushed the effective-charge product calculations that give the $Z'=27/16$ result. The distinction Hylleraas made sharp — mean-field versus correlated — organizes all of quantum chemistry to this day.

History

1927

— Kellner computes the helium ground state with screened hydrogenic orbitals.

1929

— Hylleraas introduces explicit

r_{12}

dependence and reaches the experimental helium energy.

1930

— Slater's screening rules generalize the effective-charge idea across the periodic table.

1963

— Kolos and Wolniewicz push explicitly-correlated methods to benchmark accuracy for H₂.

The physical picture

Two electrons in helium do not move independently. Each carries a negative charge, and their mutual repulsion creates electron correlation: the probability of finding the electrons close together is reduced compared with an uncorrelated product state. This mutual dodging is called electron correlation, and it is the single feature the simplest quantum-chemical wavefunction throws away. The effective-charge product trial, $\Psi=\psi_{1s}(r_1)\psi_{1s}(r_2)$ , is a product of one-electron functions. That factorization is precisely the mathematical statement that electron 1's position tells you nothing about electron 2's — they are statistically independent. But real electrons are not independent; they are anti-correlated by their repulsion. A product wavefunction therefore lets the two electrons sit on top of each other more often than they really would, overcounting the repulsion energy, and — because the variational energy is always an upper bound — landing above the true ground state. The effective charge $Z'=27/16$ already captures the average screening (each electron shields $5/16$ of the nuclear charge from the other), but averaging is not dodging.

Figure

The correlation hole: real electrons avoid each other; a product wavefunction lets them overlap

The remedy is to break the factorization. If the trial wavefunction is allowed to depend explicitly on the electron–electron distance, $r_{12}=|\mathbf{r}_1-\mathbf{r}_2|$ , it can reduce the probability of configurations in which the electrons are close together — carving out a correlation hole. Egil Hylleraas discovered in 1929 that including $r_{12}$ explicitly drives the helium energy essentially onto the experimental value. The cleanest illustration for us is a minimal Jastrow factor: multiply the product by $(1+b\,r_{12})$ and let $b$ be a second variational parameter alongside $Z'$ . Since any trial energy is an upper bound, and the enlarged family contains the old product (at $b=0$ ), the best correlated energy can only be lower — and it is.

Figure

The energy ladder: product bound, correlated bound, and the exact value it can never cross

From the effective charge to the correlated bound

Variational Method with Trial Wavefunctions computes the product-trial energy as a function of the effective charge, minimizes it, and reports the result. We consume that number rather than re-derive it: the product energy is

E(Z') = \left(Z'^2 - 2ZZ' + \tfrac{5}{8}Z'\right)E_h,

where: $Z=2$ is the true helium nuclear charge, $Z'$ the variational effective charge, $E_h=27.211$ eV the Hartree, and the $\tfrac{5}{8}Z'E_h$ term is the electron–electron repulsion integral $\langle 1/r_{12}\rangle$ evaluated on the product of two $1s$ orbitals of exponent $Z'$ — the celebrated Hylleraas integral, whose coefficient $5/8$ is the one genuinely nontrivial number in the calculation. Minimizing gives $Z'=Z-5/16=27/16=1.6875$ and $E=-(27/16)^2E_h=-77.49$ eV. That is the ceiling a product wavefunction can reach. Everything below it must come from correlation.

Derivation AWhy the product energy sits above the exact, and the screening reading of Z prime

The goal is to make precise two claims the physical picture asserts: that the product-trial energy is a rigorous upper bound (so it lies above the exact ground energy), and that the optimal effective charge $Z'=27/16$ has a clean screening interpretation. We consume the product energy $E(Z')$ from Variational Method with Trial Wavefunctions and analyze it; the $5/8$ integral itself is verified symbolically here.

Step 1. The variational principle (from the Variational Principle) guarantees $E(Z') \ge E_{\rm exact}$ for every $Z'$ , so the minimum over $Z'$ still lies above the exact energy. For this one-parameter product trial, the minimum is $E_{\rm var}\approx -77.5\,\mathrm{eV}$ , so it cannot reach the exact energy of about $-79.0\,\mathrm{eV}$ .

Step 2. Verify the repulsion integral. For two $1s$ orbitals of exponent $a$ , the mutual Coulomb energy $\langle 1/r_{12}\rangle$ reduces (after the angular/monopole integration) to a radial double integral over the two hydrogenic densities. Evaluating it gives exactly $\tfrac{5}{8}a$ in Hartree — the $5/8$ coefficient. This is the term that couples the two electrons in the energy; without it the problem would separate and $Z'$ would be $Z$ .

Step 3. Minimize $E(Z')=(Z'^2-2ZZ'+\tfrac58 Z')E_h$ : setting $dE/dZ'=0$ gives $2Z'-2Z+\tfrac58=0$ , so $Z'=Z-\tfrac{5}{16}=2-\tfrac{5}{16}=\tfrac{27}{16}=1.6875$ .

Step 4. Read the screening. Each electron sees not the bare $Z=2$ but $Z'=27/16$ : the other electron's cloud shields $5/16$ of the nuclear charge. The minimization derived this screening — it was not assumed. The optimal energy is the compact $E=-(Z')^2E_h=-(27/16)^2E_h=-77.49$ eV. Why this matters. The screening constant $5/16$ is not a fudge factor; it drops out of a genuine energy minimization, and it is the reason a variational calculation ( $-77.5$ eV) beats naive first-order perturbation theory ( $-74.8$ eV, which freezes $Z'=2$ ). Optimizing the orbital shape is worth $2.7$ eV. The $5/8$ Hylleraas repulsion integral $J=\tfrac{5}{8}a$ can be confirmed symbolically via the radial double integral; the minimizer gives $Z'=27/16$ and $E=-(27/16)^2E_h=-77.49$ eV; and the first-order PT comparator evaluates to $(-4+\tfrac54)E_h=-74.83$ eV.

The product trial is honest but blind: it treats the two electrons as independent, so it cannot represent the fact that finding one electron here makes the other more likely there. The next derivation shows that repairing this blindness—with a single extra correlation parameter—lowers the variational energy.

Go deeper · Derivation B · 2 minA correlation factor provably lowers the helium bound

Derivation BA correlation factor provably lowers the helium bound

The goal is to prove — not merely assert — that letting the wavefunction depend on the inter-electron distance $r_{12}$ improves the variational estimate, and to quantify the improvement. This is the structural result the article owns: the product's $-77.5$ eV is not the best a compact trial can do, and the reason is correlation, demonstrated by direct computation. We build a two-parameter Jastrow-correlated trial and evaluate its energy.

Step 1. Take the correlated trial $\Psi=e^{-Z'(r_1+r_2)}\,(1+b\,r_{12})$ , with two variational parameters: the orbital exponent $Z'$ and the correlation strength $b\ge0$ . At $b=0$ this is exactly the product trial — so the enlarged family contains the old one, and by the monotone-improvability logic of Ritz Variational Method and Excited States the minimum over $(Z',b)$ can only fall or hold relative to the product.

Step 2. The factor $(1+b\,r_{12})$ with $b>0$ increases the relative weight of configurations with larger electron–electron separation and suppresses the relative weight of configurations where the electrons are close — a correlation hole. This captures the anti-correlation produced by electron repulsion.

Step 3. Compute $\langle H\rangle(Z',b)$ by evaluating the local energy over the six-dimensional coordinate space (Monte-Carlo, fixed seed). At $b=0,\ Z'=27/16$ we recover the product energy $-77.5$ eV (a check on the machinery). Scanning $b>0$ with $Z'$ re-optimized, the energy drops: the best two-parameter value reaches about $-78.7$ eV.

Step 4. The correlated bound $-78.7$ eV sits strictly below the product $-77.5$ eV and strictly above the exact $-79.0$ eV — exactly as the variational bound demands. Correlation has recovered most, but not all, of the missing energy; a single Jastrow parameter is a first correlated step, not the full story. Why this matters. This is the qualitative lesson of quantum chemistry in one calculation: a product (mean-field) wavefunction has a definite error floor set by the neglected correlation, and the only way through it is to make the wavefunction depend on the electrons' relative coordinates. The full explicit- $r_{12}$ Hylleraas expansion pushes the energy to very high accuracy, but the structure — correlation lowers a rigorous variational bound — is already visible with two parameters. The Monte-Carlo product energy reproduces $-77.5$ eV within the sampling band; the correlated two-parameter minimum reaches $-78.66$ eV, lying below $-77.5$ eV and above the exact $-79.005$ eV — consistent with the variational bound at every step.

Limiting cases

Situation	Variational energy	Reading
No repulsion ( $1/r_{12}\to0$ )	$E=-2Z^2\,\tfrac12 E_h=-4E_h=-108.8$ eV	Two independent hydrogenic electrons; the repulsion is the whole difficulty
Frozen $Z'=Z=2$ (first-order PT)	$-74.8$ eV	No orbital relaxation; PT overestimates the energy
Optimized product ( $Z'=27/16$ )	$-77.5$ eV	Screening captured on average; the mean-field ceiling
Two-parameter Jastrow	$\approx-78.7$ eV	Correlation hole opened; below the product, above exact
Full Hylleraas $r_{12}$ expansion	$\to-79.0$ eV	Chemical accuracy — owner: the many-body track

Worked examples

Example 1 — Helium's first ionization energy from the variational calculation. The first ionization energy is the energy required to remove one electron from helium: $I_1=E(\mathrm{He}^+)-E(\mathrm{He})$ . We get $E(\text{He})$ from the variational estimate and $E(\text{He}^+)$ from the exact hydrogenic result, consumed from Hydrogenic Ions and Scaling with Nuclear Charge.

Step 1. $\text{He}^+$ is a one-electron ion with $Z=2$ , so its ground energy is exactly hydrogenic: $E(\text{He}^+)=-\tfrac{Z^2}{2}E_h=-2E_h=\mathbf{-54.42}$ eV (the $Z^2$ scaling from the hydrogenic owner).

Step 2. Use the variational $E(\text{He})=-77.49$ eV. Then $I_1=E(\text{He}^+)-E(\text{He})=-54.42-(-77.49)=\mathbf{+23.07}$ eV.

Step 3. Compare to experiment, $24.59$ eV. The variational estimate is $1.5$ eV low — and this is the same $1.5$ eV correlation deficit as the ground-state error, because $\text{He}^+$ (one electron) has no correlation energy, so the entire error in $I_1$ comes from the two-electron helium side. If we instead use the exact $E(\text{He})=-79.005$ eV, we get $I_1=-54.42-(-79.005)=\mathbf{24.58}$ eV, matching experiment. The ionization energy is a direct, honest test of the variational energy.

Example 2 — How far a correlation parameter moves the bound.

Step 1. Evaluate the product trial ( $b=0$ ) at $Z'=27/16$ : the energy is $-77.5$ eV, reproducing Variational Method with Trial Wavefunctions.

Step 2. Turn on the Jastrow factor $(1+b r_{12})$ and re-minimize over $(Z',b)$ . The optimal point moves the exponent up (to $Z'\approx1.85$ , because the correlation factor changes how much screening the orbital must carry) and lands at $b\approx0.4$ , with energy $\mathbf{-78.7}$ eV.

Step 3. Read the ladder. Correlation recovered about $1.2$ eV of the $1.5$ eV product deficit — roughly $80\%$ of the missing correlation energy from a single extra parameter. The remaining $\sim0.3$ eV requires the full explicit- $r_{12}$ expansion (the many-body track). The bound never crosses $-79.0$ eV: the variational principle holds at every step.

Check yourself

Why can a product wavefunction $\psi(r_1)\psi(r_2)$ never reach helium's exact energy?
What does the factor $(1+b\,r_{12})$ do physically, and why must it lower the energy?
Where does the $5/8$ come from, and where does $Z'=27/16$ ?
Why is the error in the first ionization energy the same as the ground-state error?

Going further

This article closes the variational chapter. It consumes the effective-charge product calculation from Variational Method with Trial Wavefunctions (the $Z'=27/16$ , $-77.5$ eV, and $5/8$ integral), the upper-bound guarantee from The Variational Principle, the monotone-improvability logic from Ritz Variational Method and Excited States, and the hydrogenic $\text{He}^+$ energy from Hydrogenic Ions and Scaling with Nuclear Charge. The explicit- $r_{12}$ Hylleraas chase to $-79.005$ eV, Hartree–Fock and its correlation correction, and the para/ortho-helium configurations belong to the many-body track. Provenance. The key numerical benchmarks are: $Z_{\mathrm{eff}}=27/16=1.6875$ and $E_{\min}=-(27/16)^2 E_h\approx-77.5\,\mathrm{eV}$ , obtained by minimizing $E(Z_{\mathrm{eff}})=(Z_{\mathrm{eff}}^2-2ZZ_{\mathrm{eff}}+\tfrac58 Z_{\mathrm{eff}})E_h$ with $Z=2$ and $E_h=27.211\,\mathrm{eV}$ ; the direct repulsion integral $\tfrac58 Z_{\mathrm{eff}}E_h$ equals $34.0\,\mathrm{eV}$ at $Z_{\mathrm{eff}}=2$ ; and the perturbation result is $-74.8\,\mathrm{eV}$ , against the experimental ground-state energy $-79.0\,\mathrm{eV}$ and ionization energy $24.6\,\mathrm{eV}$ . Handoff note: the simple-product $-77.5$ eV is owned by Variational Method with Trial Wavefunctions; this article owns the correlated improvement and the correlation-energy diagnosis. The full explicit- $r_{12}$ chase to $-79.005$ eV is deferred by name to the many-body track.

Why exactly can't a product wavefunction get the right answer?

Because a product $\psi(r_1)\psi(r_2)$ encodes statistical independence: the probability of finding electron 2 anywhere is unchanged by where electron 1 is. Real electrons are anti-correlated — electron 2 avoids the region near electron 1. No single product can express that avoidance; you need the wavefunction to depend on $r_{12}$ . The energy difference between the best product and the truth is, by definition, the correlation energy.

Is the effective charge Z' = 27/16 a real physical charge?

No — it is a variational parameter, the single orbital exponent that best fits both electrons' average experience of the screened nucleus. The true screening is position-dependent; $27/16$ is the best constant approximation. Reading it as "each electron blocks $5/16$ of the nuclear charge" is a useful average picture, not a literal geometric fact.

Why does adding the Jastrow factor push Z' UP from 27/16?

Because the two parameters trade off. Once the $(1+b\,r_{12})$ factor handles the electrons' mutual avoidance, the orbital no longer has to be as diffuse to keep them apart, so the optimal exponent rises (toward $\approx1.85$ in the two-parameter fit). The parameters co-adapt at the joint minimum — you cannot optimize one while freezing the other and expect the product-trial value of $Z'$ to stay best.

How much of the missing energy does one correlation parameter recover?

About $80\%$ in this minimal case: the product misses $\sim1.5$ eV, and the single Jastrow parameter recovers roughly $1.2$ eV, landing near $-78.7$ eV. The last few tenths of an eV need the full explicit- $r_{12}$ expansion — many terms, many parameters — which is why chemical accuracy is hard and why helium was Hylleraas's tour de force, not a homework problem.

Does the variational bound still hold for the correlated trial?

Yes, always. Any normalizable trial function gives an energy $\ge$ the exact ground state, regardless of how it is built. That is why the correlated bound ( $-78.7$ eV) stays above the exact $-79.0$ eV — it approaches from above and cannot cross. If a calculation ever reports below the exact energy, it is a numerical error (often a badly conditioned integral or normalization), not better physics.

Why is first-order perturbation theory worse than the variational product?

Perturbation theory treats the repulsion $1/r_{12}$ as a small correction to independent hydrogenic electrons and freezes the orbital exponent at $Z=2$ . The variational calculation lets the exponent relax to $Z'=27/16$ , which lowers the energy by $2.7$ eV ( $-74.8$ to $-77.5$ eV). The repulsion is not really "small" — it is comparable to the binding — so freezing the orbitals is a poor approximation, and orbital relaxation is the cheap, big win.

Does helium have a triplet ground state like the exchange story suggests?

No — the ground configuration is $1s^2$ , a spin singlet, because both electrons occupy the same spatial $1s$ orbital and the Pauli principle forces their spins antiparallel. The exchange splitting between singlet and triplet appears in excited configurations like $1s2s$ (para- vs ortho-helium), where the two electrons are in different spatial orbitals. That is a separate story, owned by the many-body track.