University Quantum Mechanics II · The Variational Principle · 10.5

Helium with an effective nuclear charge

This is where the variational method starts earning its keep. First-order theory kept both orbitals the size a bare nucleus gives them; letting that size float costs one parameter and one derivative, recovers nearly two thirds of the missing energy, and forces the question every later method must answer: a bound on what, exactly?

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

First-order perturbation theory averaged helium's electron–electron repulsion over two orbitals built for the bare charge Z = 2 and got −74.83 eV; as ⟨H⟩ in a normalised state, that was already a variational bound, one with nothing left to adjust. The model here frees the one thing the repulsion ought to change, the orbitals' size, by putting both electrons in the 1s orbital of a floating charge ζ. Scaling does most of the work: every length in the state goes as a/ζ, so the kinetic energy goes as ζ², the nuclear attraction as Zζ and the repulsion, already integrated at charge Z, as (5/8)ζ. ⟨H⟩ is then a quadratic whose minimum at ζ = Z − 5/16 = 27/16 gives −(27/16)²Eₕ = −77.49 eV (derived), 1.9 per cent above the −79.005 eV measured by stripping both electrons.

What the theorem guarantees it exceeds is the lowest eigenvalue of the Hamiltonian it was computed with, fixed nucleus and no relativity, −79.014 eV (quoted), which lies 0.009 eV below the measurement. Reading 27/16 as each electron hiding 5/16 of a proton from the other is permitted by the result but not demanded by it, since ζ is set by an energy balance, not measured. And the model stops where correlation starts: a product of two orbitals lets the electrons meet as often as independent clouds would, no ζ changes that, and the 1.5 eV it leaves does not shrink as Z grows.

Simple definition: An effective nuclear charge is the variational parameter ζ of a trial that puts both electrons in the hydrogenic 1s orbital of charge ζ instead of Z; minimising ⟨H⟩ over ζ picks the orbital size that best balances kinetic energy against attraction and repulsion.
Example: For helium the minimum is at ζ = 2 − 5/16 = 1.6875, where ⟨H⟩ = −(1.6875)² × 27.211 eV = −77.49 eV and each electron's mean radius 3a/(2ζ) is 47.0 pm instead of the 39.7 pm it has at ζ = 2.

Trial state with a floating chargeψ(r₁, r₂) = φ_ζ(r₁)φ_ζ(r₂), with φ_ζ(r) = (ζ³/πa³)¹⁄² e^(−ζr/a)

Normalised at every ζ, and ζ = Z is first-order theory's unscreened state, so the family's minimum cannot lie above −74.83 eV.

a = 4πε₀ħ²/(me²) = 52.92 pm; ζ is dimensionless; φ_ζ is in m^(−3/2); the spatial part is symmetric, so the spins form the singlet

The energy as a function of ζ (derived)E(ζ) = ⟨T⟩ + ⟨Vₙₑ⟩ + ⟨Vₑₑ⟩ = (ζ² − 2Zζ + 5ζ/8)Eₕ

Lengths scale as a/ζ, so kinetic energy goes as ζ² and both Coulomb terms as ζ; the repulsion is the previous topic's 5Z/8 with Z replaced by ζ.

Eₕ = e²/(4πε₀a) = α²mc² = 27.211 eV; Z = 2 for helium; the three terms are ζ²Eₕ, −2ZζEₕ and 5ζEₕ/8

The optimum charge (derived)dE/dζ = (2ζ − 2Z + 5/8)Eₕ = 0 ⇒ ζ = Z − 5/16 = 27/16

The shift is half the repulsion coefficient: the two electrons' kinetic and attraction terms supply 2ζ − 2Z, the one shared repulsion only 5/8.

Pure numbers; d²E/dζ² = 2Eₕ is positive, so the stationary point is a minimum

The bound, and what it boundsE(27/16) = −(27/16)²Eₕ = −77.49 eV ≥ E₀ = −79.014 eV

The inequality is strict but refers to E₀. The measured energy includes nuclear recoil, relativity and QED, and sits 0.009 eV above E₀.

E₀: lowest eigenvalue of the fixed-nucleus, non-relativistic H (quoted); measured −79.005 eV = −(24.587 + 54.418) eV

Virial check at the optimum⟨Vₙₑ⟩ + ⟨Vₑₑ⟩ = −2⟨T⟩ at ζ = Z − 5/16

Any Aζ² − Bζ has Aζ² = Bζ/2 at its minimum, so optimising a length scale enforces the Coulomb virial theorem: a free check on the algebra.

Helium: ⟨T⟩ = 77.49 eV and ⟨V⟩ = −154.98 eV; at ζ = 2 the ratio −⟨V⟩/⟨T⟩ is 1.69, not 2

First ionisation energyI = E(He⁺) − E(He) = [(Z − 5/16)² − Z²/2]Eₕ = 23.07 eV

The whole 1.52 eV error of E(He) lands on a quantity a third its size, so 1.9 per cent in the energy becomes 6.2 per cent in I.

E(He⁺) = −Z²Eₕ/2 = −54.42 eV, exact for a fixed nucleus; measured I = 24.587 eV

Free the orbital's size, starting from first order

Quote the previous topic: averaging the repulsion over the unscreened product φ_Z(r₁)φ_Z(r₂) gives (−Z² + 5Z/8)Eₕ = −74.83 eV, with Eₕ = e²/(4πε₀a) = 27.211 eV. Being ⟨H⟩ in a normalised state, that is already a variational bound, but nothing in it can move, and it sits 4.17 eV above the measured −79.005 eV. Its orbitals were built for a bare charge 2, as though each electron were alone, while in fact each pushes the other outward. So keep the form and free the size: ψ = φ_ζ(r₁)φ_ζ(r₂) with φ_ζ(r) = (ζ³/πa³)¹⁄² e^(−ζr/a). The family is admissible and cheap. It is normalised for every ζ, so ⟨H⟩ needs no denominator. It is symmetric under exchange of r₁ and r₂, as the spin singlet requires. And it contains the first-order state at ζ = Z, so its minimum cannot lie above −74.83 eV: enlarging a trial family can lower the bound or leave it, never raise it.

Get every term from scaling and one old integral

Split H so that the trial is an eigenstate of most of it: H = h_ζ(1) + h_ζ(2) + (ζ − Z)(e²/4πε₀)(1/r₁ + 1/r₂) + e²/(4πε₀r₁₂), where h_ζ is the hydrogenic Hamiltonian for charge ζ. The first two terms give 2 × (−ζ²Eₕ/2) = −ζ²Eₕ. Unit 2's ⟨1/r⟩ = ζ/a turns the third into 2ζ(ζ − Z)Eₕ. The last needs no new integral: every length in the state scales as a/ζ, so the previous topic's ⟨1/r₁₂⟩ = 5Z/(8a) becomes 5ζ/(8a). Adding gives E(ζ) = (ζ² − 2Zζ + 5ζ/8)Eₕ (derived), which splits as ⟨T⟩ = ζ²Eₕ, ⟨Vₙₑ⟩ = −2ZζEₕ and ⟨Vₑₑ⟩ = 5ζEₕ/8. Kinetic energy has two gradients, so it goes as the inverse square of the length scale; a Coulomb energy goes as its inverse. Dimensions: ζ and Z are pure numbers and Eₕ is an energy. Limit: delete the repulsion and the minimum is ζ = Z with E = −Z²Eₕ = −108.85 eV, two independent hydrogenic electrons, as it must be.

Minimise the quadratic, then check it with the virial theorem

dE/dζ = (2ζ − 2Z + 5/8)Eₕ vanishes at ζ = Z − 5/16, and d²E/dζ² = 2Eₕ, being positive, makes it a minimum. For helium ζ = 27/16 = 1.6875 and E = −(27/16)²Eₕ = −2.8477Eₕ = −77.49 eV, or −77.5 eV with 13.6 eV per rydberg. One parameter recovers 2.66 eV, 64 per cent of the gap first order left, and lands 1.52 eV, or 1.9 per cent, above the measured energy. A check that costs nothing: at the optimum ⟨T⟩ = 77.49 eV and ⟨V⟩ = ⟨Vₙₑ⟩ + ⟨Vₑₑ⟩ = −2ζ(Z − 5/16)Eₕ = −154.98 eV, so ⟨V⟩ = −2⟨T⟩, the virial relation every Coulomb eigenstate obeys. It is no accident: any E(ζ) = Aζ² − Bζ is minimised where Aζ² = Bζ/2, so optimising a length scale enforces the theorem, and the minimum energy equals −⟨T⟩. At ζ = 2 the ratio −⟨V⟩/⟨T⟩ is 1.69, and that shortfall is a diagnosis: the kinetic energy is too large for the potential, so the unscreened orbitals are too small.

Say which Hamiltonian the bound belongs to

The theorem compares ⟨H⟩ with the lowest eigenvalue of the H inside the integrals: a point nucleus of infinite mass, Coulomb forces only, non-relativistic kinetic energy. Correlated calculations put that eigenvalue at −2.903724Eₕ = −79.014 eV (quoted), and −77.49 eV is guaranteed to lie above it. The measured −79.005 eV, the sum of helium's ionisation energies 24.587 eV and 54.418 eV, is a different number, 0.009 eV higher. Nuclear recoil replaces m by μ = m/(1 + m/M); with m/M = 1.37 × 10⁻⁴ for ⁴He every energy scales by μ/m and rises by 0.011 eV (derivable in one line), plus 0.0006 eV of mass polarisation (quoted). Relativistic and radiative corrections, of relative order α² = 5.3 × 10⁻⁵, lower it by 0.002 eV (quoted). The orders show why comparing with experiment is still fair: Eₕ = α²mc², so the calculation is of order α²mc², the omitted physics is about 10⁻⁴ of that, and the variational gap is 2 per cent. But nothing forbids an honest trial for this H from reaching −79.010 eV, below the measurement.

Screening is a reading of Z − 5/16, not a derivation of it

It is natural to read ζ = Z − 5/16 as each electron hiding 5/16 of a proton from the other, and the result is consistent with that: ζ is below Z and each orbital swells, its mean radius 3a/(2ζ) growing by 18.5 per cent. But nothing in the calculation counts enclosed charge. Averaged over positions, the other electron is nearer the nucleus exactly half the time, by symmetry, so a charge-counting argument would give ζ = 1.5, and E(1.5) = −76.53 eV is 0.96 eV worse than the optimum. The 5/16 comes from an energy balance: the repulsion 5ζEₕ/8 rewards a large orbital linearly, the kinetic and attraction terms penalise leaving ζ = Z quadratically, and since the pair shares one repulsion while each electron has its own kinetic and attraction terms, the shift is half the coefficient 5/8. The same algebra reads just as well as the repulsion inflating the orbitals. And ζ belongs to this family, not to the atom: give the two electrons different exponents and they return different values, so no single number is the charge an electron sees.

What no value of ζ can do: correlation

In a product state the joint density is |φ_ζ(r₁)|²|φ_ζ(r₂)|²: where one electron is says nothing about where the other is, so they come close as often as two independent clouds would. The exact electrons avoid each other, and ζ cannot make them, because it rescales both clouds together. One ratio shows it. In every member of the family ⟨1/r₁₂⟩/⟨1/r₁⟩ = (5ζ/8)/ζ = 0.625, whatever ζ is; in the exact ground state it is 0.945818/1.688317 = 0.560 (quoted expectation values). Relative to their distance from the nucleus, real electrons stay farther apart than any member of the family allows. The error has a signature along the two-electron ions. Expanding the optimum, −(Z − 5/16)² = −Z² + 5Z/8 − 25/256 in units of Eₕ, while the exact energy is −Z² + 5Z/8 − 0.1577 + O(1/Z) (quoted). The first two terms agree because the family contains the first-order state; at the third they part, so the absolute error tends to 0.060Eₕ = 1.63 eV as Z grows while the relative error falls as 1/Z². Closing it needs r₁₂ in the trial, or more than one product.

Change one variable at a time

Make the relationship visible.

Interactive model

Floating charge ζ (Z_eff)2.0000

Start at ζ = 2, first-order theory, and lower ζ. Near ζ = 2 the ⟨T⟩ and ⟨Vₙₑ⟩ bars shrink by equal amounts that cancel in E, so it is the shrinking ⟨Vₑₑ⟩ bar that lowers the dot; further down ⟨T⟩ shrinks more slowly than the attraction, so the dot bottoms out where −⟨V⟩/⟨T⟩ = 2, at ζ = 1.6875, and climbs below that, never reaching the dashed line.

KINETIC ⟨T⟩ = ζ²Eₕ108.85 eV

ATTRACTION ⟨Vₙₑ⟩ = −2ZζEₕ-217.69 eV

REPULSION ⟨Vₑₑ⟩ = 5ζEₕ/834.01 eV

ABOVE E₀ = −79.01 eV4.18 eV

Live interpretationKINETIC ⟨T⟩ = ζ²Eₕ: 108.85 eV. ATTRACTION ⟨Vₙₑ⟩ = −2ZζEₕ: −217.69 eV. REPULSION ⟨Vₑₑ⟩ = 5ζEₕ/8: 34.01 eV. ABOVE E₀ = −79.01 eV: 4.18 eV

Catch the common trap

Explain before calculating.

Misconception check

“Each electron sees a charge 27/16, so each is a hydrogenic electron bound by (27/16)² × 13.6 eV = 38.7 eV, and removing one costs 38.7 eV.” The measured first ionisation energy is 24.59 eV. The step that fails is splitting the total −ζ²Eₕ = −77.49 eV into two equal one-electron shares. That total is ⟨T⟩ + ⟨Vₙₑ⟩ + ⟨Vₑₑ⟩, and halving it charges each electron with half of the 28.70 eV repulsion, whereas removing either electron deletes all of it: with the orbitals frozen, the cost is (77.49 − 28.70)/2 = 24.39 eV. Nor do the orbitals stay frozen: the electron left in He⁺ contracts to ζ = 2 and gives back 1.33 eV. Ionisation energies are differences of total energies, never shares of one. What makes the error tempting is the virial theorem: −ζ²Eₕ equals ⟨H⟩ at the optimum only, and at ζ = 2 the same form would claim −108.8 eV where ⟨H⟩ is −74.8 eV.

A group improves on the 27/16 trial and reports ⟨H⟩ = −79.010 eV for helium, computed with the same fixed-nucleus, non-relativistic Hamiltonian. The measured ground-state energy is −79.005 eV. What should you conclude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHe⁺ is hydrogenic, so E(He⁺) = −Z²Eₕ/2 = −54.42 eV exactly for a fixed nucleus, with Eₕ = 27.211 eV. Predict helium's first ionisation energy from independent electrons (−4Eₕ), from first-order theory (−2.75Eₕ) and from the 27/16 trial, compare each with the measured 24.587 eV, and explain why the trial's percentage error is larger here than in the total energy.

Ionisation takes He to He⁺ plus an electron at rest far away, so I = E(He⁺) − E(He), a difference of two total energies, with E(He⁺) = −2 × 27.211 eV = −54.42 eV.
Independent electrons, E = −4Eₕ = −108.85 eV: I = −54.42 + 108.85 = 54.42 eV, 121 per cent too high. With no repulsion the first electron is held exactly as tightly as the second.
First order, E = −2.75Eₕ = −74.83 eV: I = 20.41 eV, 17.0 per cent low, because the repulsion averaged over unscreened orbitals is too large.
Trial, E = −(27/16)²Eₕ = −77.49 eV: I = 23.07 eV, 6.2 per cent low. Those two energies are upper bounds on E(He) and E(He⁺) is exact, so both ionisation energies are lower bounds on the fixed-nucleus value, −54.42 + 79.01 = 24.59 eV. The independent-electron number bounds nothing, because it drops a positive term from H.
The trial misses E(He) by 1.52 eV and E(He⁺) by nothing, so all of that error lands on I, about a third the size of E: 1.52/24.59 = 6.2 per cent, against 1.52/79.0 = 1.9 per cent for the energy.

AnswerI = 54.4 eV, 20.4 eV and 23.1 eV against 24.59 eV measured: +121, −17 and −6.2 per cent. A difference of two energies, one of them exact, inherits the whole absolute error of the other.

MediumTry a Gaussian for helium instead: both electrons in φ_β(r) = (2β/πa²)³⁄⁴ e^{(−βr²/a²)}, β dimensionless. Per electron ⟨T⟩ = (3/2)βEₕ and ⟨1/r⟩ = 2√(2β/π)/a (standard Gaussian integrals). Find ⟨1/r₁₂⟩ without a new integral, minimise E(β), check the virial theorem, and compare the result with the exponential trial's −77.49 eV.

Each Cartesian component of r₁ or r₂ is Gaussian with variance a²/(4β), so each component of r₁ − r₂ has variance a²/(2β): the density of a Gaussian orbital with β replaced by β/2. Hence ⟨1/r₁₂⟩ = 2√(2(β/2)/π)/a = 2√(β/π)/a.
Assemble: E(β) = [2 × (3/2)β − 2Z × 2√(2β/π) + 2√(β/π)]Eₕ = (3β − B√β)Eₕ, with B = (4√2 Z − 2)/√π = 5.2547 for Z = 2. Kinetic energy goes as β and both Coulomb terms as √β, the Gaussian version of ζ² against ζ; β and B are pure numbers, so each term is an energy.
Minimise in u = √β: dE/du = (6u − B)Eₕ = 0 gives u = B/6 = 0.8758, so β = 0.767 and E = −(B²/12)Eₕ = −2.3010Eₕ = −62.61 eV.
Virial: ⟨T⟩ = 3βEₕ = 62.61 eV and ⟨V⟩ = −BuEₕ = −125.23 eV = −2⟨T⟩, as for any optimised length scale. Limit: without the repulsion the same algebra gives −8Z²Eₕ/(3π) = −92.39 eV, 0.849 of the exact −108.85 eV, the same 8/(3π) a Gaussian achieves for hydrogen.
Compare: −62.61 eV is a valid bound, above −79.014 eV, but 14.88 eV above the exponential trial and 12.22 eV above even unscreened first-order theory. The Gaussian's mean radius, 0.911a, is close to the exponential's 0.889a; what it lacks is the cusp at r = 0, where the attraction is strongest, and the slow exponential tail.

Answer⟨1/r₁₂⟩ = 2√(β/π)/a; β = 0.767 and E = −2.301Eₕ = −62.6 eV, with ⟨V⟩ = −2⟨T⟩. A legal bound, but 14.9 eV worse than the exponential: a wrong shape costs more than five times the 2.66 eV that choosing ζ gains.

HardHelium-4's nucleus has M/m = 7294.30. Measured from the nucleus, the non-relativistic Hamiltonian with recoil is H = (p₁² + p₂²)/(2μ) + p₁⋅p₂/M + V, with μ = mM/(m + M). Minimise ⟨H⟩ for the trial φ_ζ(r₁)φ_ζ(r₂), and set the result against the fixed-nucleus eigenvalue −79.0144 eV, the recoil-included eigenvalue −79.0029 eV (both quoted) and the measured −79.0052 eV. What does each variational number bound?

The cross term vanishes for this trial. ⟨p₁⋅p₂⟩ = ⟨φ_ζ|p|φ_ζ⟩⋅⟨φ_ζ|p|φ_ζ⟩, and for a real, normalisable φ, ⟨p⟩ = −iħ∫φ∇φ d³r = −(iħ/2)∫∇(φ²) d³r = 0, a surface term at infinity. Only the mass in the kinetic term changes.
At fixed ψ the kinetic energy goes as 1/mass, so ⟨T⟩ becomes (m/μ)ζ²Eₕ, with Eₕ and a still defined by m; the potential terms do not involve the masses. So E(ζ) = [(m/μ)ζ² − 2Zζ + 5ζ/8]Eₕ.
dE/dζ = 0 gives ζ = (μ/m)(Z − 5/16) and E = −(μ/m)(Z − 5/16)²Eₕ. With μ/m = 1/(1 + 1/7294.30) = 0.999863, ζ = 1.68727 and E = −77.4781 eV, 0.0106 eV above the fixed-nucleus −77.4887 eV. The orbital is slightly larger, as a lighter effective mass requires.
Checks: μ/m and ζ are pure numbers, so E is an energy; M → ∞ sends μ/m → 1 and returns the fixed-nucleus result; and the shift is (m/M)|E| to first order, 1.371 × 10⁻⁴ × 77.49 eV = 0.0106 eV.
From lowest to highest: −79.0144 eV (fixed-nucleus eigenvalue), −79.0052 eV (measured), −79.0029 eV (recoil eigenvalue), −77.4887 eV and −77.4781 eV. The fixed-nucleus trial is guaranteed only to lie above −79.0144 eV, the recoil trial above −79.0029 eV. The measurement sits between the eigenvalues because relativity and QED, in neither H, lower the energy by 0.0022 eV (quoted), a relative 2.8 × 10⁻⁵, of order α².
So no theorem puts either variational number above the measured energy. What makes the 1.9 per cent comparison meaningful is that the physics outside H is about 0.01 eV, more than a hundred times smaller than the 1.5 eV variational gap.

Answerζ = (μ/m)(27/16) = 1.6873 and E = −77.478 eV, 0.011 eV higher. It bounds the recoil eigenvalue −79.0029 eV, as −77.489 eV bounds the fixed-nucleus −79.0144 eV; neither theorem involves the measured −79.0052 eV, which lies between them.