University Quantum Mechanics II · The Variational Principle · 10.7

Does H-minus bind? A variational existence proof

H⁻ is the cleanest case in which a variational number stops being an estimate and becomes a proof. The lesson teaches the one comparison that decides it, a trial energy against the energy at which the extra electron can leave, and why helium's trial fails that test while Chandrasekhar's passes it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Fix the proton, drop relativity, and H⁻ is helium with Z = 1: two hydrogen Hamiltonians plus the repulsion e²/(4πε₀r₁₂). The question is not how much energy the ion has but whether it has a normalisable state at all, and the variational theorem answers it only if the bound is compared with the right number. Put one electron in hydrogen's 1s and send the other far away, slowly: that costs as little above −13.606 eV as you like, so the spectrum always reaches down to −13.606 eV, the edge of a continuum.

A trial with ⟨H⟩ below that edge forces something discrete beneath it, a bound state; one above it says nothing, because the continuum was there already. Helium's single effective charge, 11/16 at Z = 1, stops at −12.86 eV. The far electron sees a neutral atom with no Coulomb tail, so binding has to come from how the two electrons share space, and one shared orbital cannot express that.

Give each electron its own exponent and symmetrise, as Chandrasekhar did in 1944, and the minimum is −13.968 eV, with one orbital close to hydrogen's 1s and one nearly four times larger: 0.36 eV below the edge, which proves H⁻ binds by at least that much, against 0.754 eV measured. The method stops where its logic does: it proves existence and a floor on the binding, never absence and never a ceiling, and the edge itself is an input, argued physically here and proved by the HVZ theorem, quoted.

Simple definition: A variational existence proof shows that a system has a bound state by exhibiting one admissible trial state whose ⟨H⟩ lies below the continuum edge, the lowest energy at which the system can come apart.
Example: For H⁻ the edge is a hydrogen atom plus an electron at rest, −13.606 eV. Chandrasekhar's trial gives ⟨H⟩ = −13.968 eV, so H⁻ has a bound state, bound by at least 0.362 eV; 0.754 eV is measured.

Helium's single-exponent trial at Z = 1Eₘᵢₙ = −(Z − 5/16)² Eₕ = −(11/16)² Eₕ = −12.86 eV

Derived for any Z in the effective-charge lesson and evaluated here. It lies 0.74 eV above the H + e⁻ edge: a true upper bound that proves nothing about binding.

Eₕ = e²/(4πε₀a₀) = 27.211 eV, a₀ = 52.9 pm; the shared orbital e^{(−ζr/a₀)} has ζ = Z − 5/16 = 11/16

Continuum edge of H⁻Eₜₕ = E(H, 1s) + 0 = −Eₕ/2 = −13.606 eV

Argued here, proved by the HVZ theorem, quoted: every energy from Eₜₕ up belongs to the continuum, so a trial with ⟨H⟩ < Eₜₕ forces a discrete level below the edge.

A hydrogen atom in its ground state plus an electron at rest far away; fixed proton, no relativity

Chandrasekhar's two-exponent trialψ(r₁, r₂) = e^{(−k₁r₁ − k₂r₂)} + e^{(−k₂r₁ − k₁r₂)}

A choice, not a derivation: each electron gets its own radius. k₁ = k₂ returns the single-exponent trial, and k₁ = 1/a₀ with k₂ → 0 returns the edge Eₜₕ exactly.

k₁, k₂ in m⁻¹, quoted as multiples of 1/a₀; ψ is unnormalised and symmetric, so it pairs with the spin singlet

Its energy, from five integrals⟨H⟩ = (h₁₁ + h₂₂ + 2S h₁₂ + J + K)/(1 + S²)

Derived. In Eₕ with k in 1/a₀: hᵢᵢ = kᵢ²/2 − kᵢ, h₁₂ = S(k₁k₂ − k₁ − k₂)/2, K = (5/16)(k₁ + k₂)S². Minimum −0.51330 Eₕ = −13.968 eV at k₁ = 1.039, k₂ = 0.283.

hᵢⱼ one-electron, J direct and K exchange integrals, all in joules; S = 8(k₁k₂)³⁄²/(k₁ + k₂)³ is the overlap

Direct repulsion of two 1s cloudsJ = [k₁k₂(k₁² + 3k₁k₂ + k₂²)/(k₁ + k₂)³] e²/(4πε₀)

Derived with helium's l = 0 multipole cut. It is 5k/8 when k₁ = k₂, and tends to k₂ when k₂ ≪ k₁: far out, the inner electron screens the proton completely.

k in m⁻¹ and e²/(4πε₀) in J m, so J comes out in joules; with k in 1/a₀ the bracket is J in units of Eₕ

What the bound certifiesB = Eₜₕ − E_gs ≥ Eₜₕ − ⟨H⟩ = 0.362 eV

Derived from E_gs ≤ ⟨H⟩: a ceiling on the energy is a floor on the binding. No trial ever gives a ceiling on B, and one above Eₜₕ gives no floor either.

B is the binding energy of the extra electron, hydrogen's electron affinity, in eV; measured 0.754 eV, quoted

The right zero is a hydrogen atom and a free electron

With the proton fixed and relativity dropped, H = h₁ + h₂ + e²/(4πε₀r₁₂), where hᵢ = pᵢ²/2mₑ − e²/(4πε₀rᵢ) is hydrogen's Hamiltonian for electron i. Bound means a normalisable eigenstate, and zero energy is the wrong yardstick for it. Put electron 1 in hydrogen's 1s and electron 2 in a broad wave packet a distance d away with momentum p: the energy is −13.606 eV + p²/2mₑ plus an interaction that dies away as d grows, because the atom is neutral. Small p and large d bring this as close to −13.606 eV as you like, and adding momentum reaches every energy above it, so from Eₜₕ = −13.606 eV upward the spectrum is continuous whether or not H⁻ binds. The variational theorem bounds the bottom of the whole spectrum, E₀ ≤ ⟨H⟩, and E₀ ≤ Eₜₕ holds before any trial is tried. A discrete level below the edge, the kind of bound state a variational number can prove, exists exactly when E₀ < Eₜₕ, strictly. That the continuum begins at the lowest break-up energy and nowhere lower is the HVZ theorem of Hunziker, van Winter and Zhislin, quoted: this lesson argues it and does not prove it.

Helium's trial at Z = 1 lands above the edge

Helium's effective-charge energy, E(ζ) = (ζ² − 2Zζ + 5ζ/8)Eₕ with Eₕ = e²/(4πε₀a₀) = 27.211 eV, is least at ζ = Z − 5/16. At Z = 1 that is ζ = 11/16 and E = −(11/16)²Eₕ = −0.4727 Eₕ = −12.86 eV, 0.74 eV above Eₜₕ. First-order perturbation theory, ζ = 1, does worse: −(3/8)Eₕ = −10.20 eV. Both are true upper bounds on E₀ and both are useless, since E₀ ≤ −13.606 eV was known before any trial; each is as consistent with a bound state at −14 eV as with none. The trial fails because one orbital must serve two electrons that want different sizes. At ζ = 11/16 its radius a₀/ζ = 1.45a₀ is too loose for an electron that should look like hydrogen's 1s, which alone costs (1 − ζ)²Eₕ/2 = 1.33 eV, and too tight for one that should sit far out. Reshaping the shared orbital cannot rescue it: the best single shared orbital of any shape, the Hartree–Fock limit, reaches −13.28 eV, quoted, still 0.33 eV short. In helium the same flaw cost 1.5 eV out of 79 eV; here the whole question lives in the last 0.75 eV.

A neutral atom offers the far electron nothing

Why is the last 0.75 eV so hard to reach? From outside, a hydrogen atom is neutral. The direct repulsion between 1s clouds of exponents k₁ and k₂ is J = k₁k₂(k₁² + 3k₁k₂ + k₂²)/(k₁ + k₂)³, in Eₕ with k in 1/a₀, derived by helium's l = 0 multipole cut; at k₁ = 1 it expands as J = k₂ − 2k₂³ + …. The outer electron's attraction to the proton is −k₂, so the two cancel at first order: there is no Coulomb tail to fall into. For a plain product φ₁(r₁)φ₂(r₂) with the inner 1s unchanged, what is left above Eₜₕ is k₂²/2 − 2k₂³ + …, a kinetic cost at second order against a penetration gain only at third. Minimised over both exponents, the plain product never gets below Eₜₕ at all; its best is the edge itself, reached as k₂ → 0. The real attraction is subtler. The far electron polarises the atom and draws an induced-dipole energy falling as 1/r⁴, and the two electrons keep apart radially, one in while the other is out. Both are correlation, so only a trial whose electrons respond to each other can bind.

Two exponents, symmetrised: one compact orbital, one diffuse

Chandrasekhar's 1944 trial, written to compute H⁻'s absorption in the Sun's atmosphere, gives each electron its own exponent and adds the swapped product: ψ = e^{(−k₁r₁ − k₂r₂)} + e^{(−k₂r₁ − k₁r₂)}, symmetric, the spatial partner of the spin singlet. With normalised orbitals φᵢ ∝ e^(−kᵢr) the overlap is S = 8(k₁k₂)³⁄²/(k₁ + k₂)³ and ⟨H⟩ = (h₁₁ + h₂₂ + 2S h₁₂ + J + K)/(1 + S²), with hᵢᵢ = kᵢ²/2 − kᵢ, h₁₂ = S(k₁k₂ − k₁ − k₂)/2 and exchange K = (5/16)(k₁ + k₂)S², all in Eₕ. Two limits check the algebra: k₁ = k₂ gives S = 1 and helium's single-exponent energy, and k₁ = 1, k₂ → 0 gives exactly −Eₕ/2, the edge. Minimising gives k₁ = 1.039/a₀, k₂ = 0.283/a₀, S = 0.552 and ⟨H⟩ = −0.51330 Eₕ = −13.968 eV, 0.362 eV below Eₜₕ. The mean radii 3/(2k) are 1.44a₀, a slightly squeezed hydrogen 1s, and 5.30a₀ = 280 pm. The swapped term is what binds: weight it by w instead of 1, re-minimise, and the proof survives only for w above about a quarter.

One-way logic: what the number proves and what it cannot

From E_gs ≤ ⟨H⟩ = −13.968 eV, the binding B = Eₜₕ − E_gs is at least 0.362 eV. That is all a variational number certifies: a floor on B, 48% of the 0.754 eV measured by laser photodetachment, quoted. The fixed-nucleus eigenvalue, −0.527751 Eₕ = −14.361 eV from highly correlated calculations, is quoted too, and gives B = 0.7551 eV. The logic runs one way. A lower ⟨H⟩ raises the floor, but no trial, however good, puts a ceiling on B, because the true level can always lie deeper than any trial found. And a trial that stops above Eₜₕ, as helium's did at −12.86 eV, is not evidence against binding: H⁻ binds, and that trial could not show it. Proving that a state does not exist takes a different kind of argument, one that bounds energies from below; Hill's 1977 proof that H⁻ has exactly one bound state, quoted here, is of that kind. The same asymmetry governs the hydrogen molecule ion two lessons on, whose clamped-proton curve proves a bond only where it dips below −13.6 eV.

Checks: dimensions, orders in α and the real ion

Every energy here is a number times Eₕ = e²/(4πε₀a₀) = α²mₑc² = 27.211 eV, and every exponent a number over a₀, so each formula can be checked by counting: J is k times e²/(4πε₀), inverse metres times joule metres, so joules, and S is k³ over k³, dimensionless. The binding, 0.0277 Eₕ, is small because the neutral atom nearly cancels the far electron's attraction, not because a coupling is weak: it is still of order α²mₑc². Relativistic corrections are of order α² relative to it, about (1/137)² × 0.755 eV = 4 × 10⁻⁵ eV, invisible at the digits quoted. The proton's finite mass matters more. Every energy scales with the reduced mass μ = mₑ/(1 + mₑ/mₚ), so the fixed-nucleus 0.7551 eV becomes 0.7547 eV, derived; most of the rest of the way to the measured 0.7542 eV is mass polarisation, a cross term in the two electrons' momenta that the fixed-nucleus model omits, quoted. None of this touches the proof, which concerns the Hamiltonian it was computed with: its level lies at least 0.36 eV below its own edge, hundreds of times the corrections.

Change one variable at a time

Make the relationship visible.

Interactive model

Compact exponent k₁1.04 1/a₀

Diffuse exponent k₂ (the dot)0.28 1/a₀

At k₁ = 1.04 and k₂ = 0.28 the dot sits 0.36 eV under the −13.61 eV line: a proof. The dashed plain-product curve never gets under that line, whatever k₁ is. The solid dip clears it only for k₁ from 0.87 to 1.20; outside that, the trial proves nothing, though the ion has not changed.

⟨H⟩ WITH SWAP-13.97 eV

⟨H⟩ PLAIN PRODUCT-13.13 eV

OVERLAP S0.547

BINDING PROVED0.36 eV

Live interpretation⟨H⟩ WITH SWAP: −13.97 eV. ⟨H⟩ PLAIN PRODUCT: −13.13 eV. OVERLAP S: 0.547. BINDING PROVED: 0.36 eV

Catch the common trap

Explain before calculating.

Misconception check

“⟨H⟩ is negative, so the ion is bound.” The hydrogen-atom habit of reading negative energy as bound fails in a two-electron problem. Zero is the energy of a proton and two electrons all at rest far apart, but H⁻ does not have to fall apart completely to be unbound: it only has to shed one electron, and a hydrogen atom plus a slow free electron can have any energy from −13.606 eV upward. A state at −13 eV can be negative and still unbound, a hydrogen atom with an electron flying past it. Helium's trial at −12.86 eV is negative and proves nothing; the comparison that decides binding is with Eₜₕ = −13.606 eV, the lowest energy at which the system can break into two pieces. The mirror-image error is as common: reading a trial that stops above the edge as proof that the ion does not exist.

Two trial functions for H⁻ are minimised with the same fixed-nucleus, non-relativistic Hamiltonian. Trial P gives ⟨H⟩ = −13.80 eV and trial Q gives ⟨H⟩ = −13.40 eV; hydrogen's ground state lies at −13.606 eV. What do the two results establish?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyChandrasekhar studied H⁻ because its photodetachment, H⁻ + photon → H + e⁻, dominates the visible and near-infrared opacity of the Sun's photosphere. His trial gives ⟨H⟩ = −0.51330 Eₕ, with Eₕ = 27.2114 eV. What detachment threshold does it certify, as an energy and as a wavelength, and how does that compare with the measured affinity, 0.754 eV?

Convert: ⟨H⟩ = −0.51330 × 27.2114 eV = −13.968 eV. The edge is hydrogen's ground state plus an electron at rest, Eₜₕ = −Eₕ/2 = −13.606 eV.
⟨H⟩ < Eₜₕ, so a bound state exists, and E_gs ≤ ⟨H⟩ gives the binding B = Eₜₕ − E_gs ≥ −13.606 eV + 13.968 eV = 0.362 eV.
A photon detaches the electron only if hc/λ ≥ B, so the threshold wavelength is λₜₕ = hc/B. With hc = 1239.8 eV nm and B ≥ 0.362 eV, λₜₕ ≤ 1239.8/0.362 nm = 3.42 × 10³ nm = 3.42 μm.
Measured: λₜₕ = 1239.8/0.754 nm = 1.64 × 10³ nm = 1.64 μm, shortward of the certified limit as it must be. The trial's floor on B is 48% of the measured binding.

AnswerB ≥ 0.362 eV, so the detachment edge lies at 3.42 μm or shorter; the measured 0.754 eV puts it at 1.64 μm.

MediumTreat the nuclear charge Z of a two-electron ion as a continuous parameter. For which Z does helium's single-exponent trial, Eₘᵢₙ = −(Z − 5/16)²Eₕ, first prove that the nucleus holds two electrons? Find the margin at Z = 1 and Z = 2 in eV, and compare with the critical charge, about 0.911, below which no bound state exists, quoted from precise calculations.

The edge is the one-electron ion in its ground state plus a free electron at rest: Eₜₕ = −Z²Eₕ/2, the hydrogenic −Z²Eₕ/(2n²) at n = 1.
Binding is proved when (Z − 5/16)² > Z²/2. For Z > 5/16 take square roots: Z − 5/16 > Z/√2, so Z(1 − 1/√2) > 5/16 and Z > 0.3125/0.29289 = 1.067.
At Z = 1: Eₘᵢₙ = −(0.6875)²Eₕ = −0.47266 Eₕ against Eₜₕ = −0.5 Eₕ, so the trial sits 0.02734 Eₕ = 0.744 eV above the edge and proves nothing.
At Z = 2: Eₘᵢₙ = −(1.6875)²Eₕ = −2.84766 Eₕ against −2 Eₕ, which is 0.84766 Eₕ = 23.07 eV below the edge. Helium holds its second electron by at least 23.07 eV; its measured first ionisation energy is 24.59 eV.
The critical charge, about 0.911, lies below 1, so for 0.911 < Z < 1.067 the nucleus does bind two electrons while this trial cannot show it, and H⁻ at Z = 1 sits inside that window.

AnswerThe trial proves binding only for Z > 1.067: it misses H⁻ by 0.744 eV at Z = 1 and certifies at least 23.07 eV for helium at Z = 2.

HardWithout minimising, test Chandrasekhar's trial at the round values k₁ = 1/a₀ and k₂ = 1/(4a₀). Evaluate S, the one-electron integrals, J and K in Eₕ, find ⟨H⟩ in eV, and decide whether this one evaluation proves H⁻ binds. Then evaluate the plain product e^{(−k₁r₁ − k₂r₂)}, without the swapped term, at the same exponents.

Overlap: S = 8(k₁k₂)³⁄²/(k₁ + k₂)³ = 8(1/4)³⁄²/(5/4)³ = 1/1.953125 = 0.512, so S² = 0.262144.
One-electron integrals, hᵢᵢ = kᵢ²/2 − kᵢ: h₁₁ = 0.5 − 1 = −0.5 and h₂₂ = 0.03125 − 0.25 = −0.21875. Off-diagonal: h₁₂ = S(k₁k₂ − k₁ − k₂)/2 = 0.512 × (0.25 − 1.25)/2 = −0.256, so 2S h₁₂ = −0.262144.
Repulsion: J = k₁k₂(k₁² + 3k₁k₂ + k₂²)/(k₁ + k₂)³ = 0.25 × 1.8125/1.953125 = 0.232, and K = (5/16)(k₁ + k₂)S² = 0.3125 × 1.25 × 0.262144 = 0.1024.
⟨H⟩ = (−0.5 − 0.21875 − 0.262144 + 0.232 + 0.1024)/(1 + 0.262144) = −0.646494/1.262144 = −0.51222 Eₕ = −13.938 eV.
That is 0.332 eV below Eₜₕ = −13.606 eV, so this single evaluation at round numbers already proves a bound state; minimising only deepens it, to −13.968 eV.
Plain product: ⟨H⟩ = h₁₁ + h₂₂ + J = −0.5 − 0.21875 + 0.232 = −0.48675 Eₕ = −13.245 eV, 0.361 eV above the edge. Same orbitals, no proof: the swapped term, the electrons trading inner and outer places, supplies the binding.

Answer⟨H⟩ = −0.51222 Eₕ = −13.938 eV, 0.332 eV below the edge, so H⁻ binds; the plain product gives −13.245 eV, above the edge, and proves nothing.