The Stark effect, polarisability and level mixing

H_S = eFz is odd under r → −r. Every eigenket |n l m⟩ of a central-field H₀ is a parity eigenket, |n l m⟩ → (−1)ˡ|n l m⟩, so |⟨r|n l m⟩|² is even while z is odd, and ⟨n l m|z|n l m⟩ = 0 exactly — for hydrogen 1s as for sodium 3s. That is a statement about the operator's symmetry, not about F being small: the diagonal element is a fixed number computed in the zero-field basis, and it does not grow when you turn the field up. Contrast the Zeeman case. H_Z = (μ_B/ħ)(L + gₛ S)⋅B is built from axial vectors, even under parity, so its diagonal elements survive and the shift is linear in B from the weakest field. The whole asymmetry between a linear Zeeman effect and a quadratic Stark effect is a parity statement and nothing else.

With first order gone, second order runs: ΔE = Σ_(k≠0) |⟨k|eFz|0⟩|²/(E₀ − Eₖ). For a ground state every Eₖ > E₀, so every term is negative and the level always moves down; write it as −½αF² and α = 2e²Σ|⟨k|z|0⟩|²/(Eₖ − E₀) is positive by construction. The same first-order mixing |ψ⟩ = |0⟩ + Σₖ |k⟩⟨k|eFz|0⟩/(E₀ − Eₖ) supplies the induced dipole: ⟨ψ|d_z|ψ⟩ = αF to first order in F, and only the cross term between |0⟩ and the admixture survives, precisely because ⟨0|z|0⟩ = 0. Shift and dipole are the same physics counted twice. For an excited state a term can flip sign — anything lying below |0⟩ contributes a negative denominator — so α need not be positive once you leave the ground state, and a level squeezed between two neighbours can barely move at all.

The sum over k is not a practical formula. For hydrogen 1s the discrete np states contribute 3.66 a₀³ and the continuum a further 0.84 a₀³, so truncating at 2p returns 2.96 a₀³ — only 66% of the exact 4.5 a₀³. Two escapes. Closure bounds it: α ≤ 2e²⟨0|z²|0⟩/ΔEₘᵢₙ, and with ⟨1s|z²|1s⟩ = ⟨r²⟩/3 = a₀² and ΔEₘᵢₙ = 0.375 Eₕ that gives 5.33 a₀³, 18% high, which at least brackets the answer. Or stop summing and solve. Dalgarno and Lewis replace the sum by one inhomogeneous equation, (H₀ − E₀)|ψ⁽¹⁾⟩ = −(H_S − ⟨H_S⟩)|0⟩, closed in atomic units by ψ⁽¹⁾ = −F(r + r²/2) cosθ ψ⁰; then ⟨ψ⁰|H_S|ψ⁽¹⁾⟩ = −2.25 F² gives α = 4.5 exactly. Numerically it is a linear solve on the l = 1 radial grid, or a scipy.linalg.eigh diagonalisation of H₀ + eFz in a finite basis with the curvature read off at small F.

Hydrogen's Coulomb degeneracy makes Eₖ − E₀ = 0 for a state that z connects, and the second-order sum divides by zero. Degenerate perturbation theory answers instead: diagonalise H_S inside the degenerate subspace. For n = 2 the four states are |2s⟩ and |2p₀⟩, |2p_(±1)⟩; z is an odd-parity, rank-one spherical tensor with q = 0, so it demands Δl = ±1 and Δm = 0 and the only non-zero element is ⟨2s|z|2p₀⟩ = −3a₀. The (|2s⟩, |2p₀⟩) block is eF times [[0, −3a₀], [−3a₀, 0]], with eigenvalues ∓3ea₀F on (|2s⟩ ± |2p₀⟩)/√2, while |2p_(±1)⟩ stay put. So n = 2 splits into three components — up, unmoved and doubly degenerate, down — and the shifted eigenstates are not parity eigenstates: each carries a real permanent dipole of 3ea₀ = 2.54 × 10⁻²⁹ C m, or 7.62 D. At 10 kV cm⁻¹ that puts the outer components at ±0.159 meV.

Real n = 2 hydrogen is not degenerate: fine structure lifts 2p₃/₂ by 10.97 GHz above 2p₁/₂, and the Lamb shift lifts 2s₁/₂ by 1.06 GHz above it. So the minimal model is two levels split by Δ and coupled by V = μF, with eigenvalues ±√((Δ/2)² + V²); ignoring spin, μ = 3ea₀. Expand for V ≪ Δ/2 and the quadratic regime reappears, with α = 2μ²/Δ for the lower level and −2μ²/Δ for the upper; take V ≫ Δ/2 and the eigenvalues are ±V, linear. The crossover is therefore a ratio, not a fixed field: with μ = 3ea₀, V = Δ at 275 V cm⁻¹ against the Lamb shift and at 2.9 kV cm⁻¹ against the fine-structure interval, and the spin-resolved 2s₁/₂–2p₁/₂ element, √3 ea₀, moves the Lamb-shift crossing up to 477 V cm⁻¹. Small denominators make enormous polarisabilities — the spinless two-level α for 2s against the Lamb shift is −1.85 × 10⁻³³ C m² V⁻¹, negative because its partner lies below it, and some 2.5 × 10⁷ times the ground state's in size — which is why a quadratic description of an excited level expires at fields a bench supply reaches with a few hundred volts.

Add eFz to the Coulomb potential and the total runs to −∞ downhill, so the spectrum is continuous and every level is a Siegert resonance E − iΓ/2. The classical saddle vanishes at Fᵢₒₙ = 1/(16n⁴) in atomic units: 3.2 × 10¹⁰ V m⁻¹ at n = 1, but only 4.0 × 10⁴ V m⁻¹, that is 397 V cm⁻¹, at n = 30 — which is how Rydberg populations are counted by field ionisation. Below threshold the width is exponentially small: for hydrogen 1s at F = 0.01 a.u. = 5.1 × 10⁹ V m⁻¹, Γ = (4/F) exp(−2/(3F)) = 1.8 × 10⁻¹⁰ s⁻¹, a lifetime near 170 years, so calling the state bound costs nothing. A finite-basis diagonalisation cannot see any of this — it returns real eigenvalues by construction. Complex scaling r → r e^(iθ) rotates the continuum away and turns the resonance into an ordinary complex eigenvalue, width included. Between manifolds, the Inglis–Teller field 1/(3n⁵) marks where n and n+1 overlap: 70 V cm⁻¹ at n = 30, comfortably below ionisation.