University Physics V · Atomic Spectroscopy · 13.8

The Stark Effect, Polarisability & Level Mixing

Switch on a uniform field and ask what the energy does. For an isolated level the answer is nothing at first order and −½αF² at second; for a hydrogen manifold carrying parity degeneracy it is a straight line. Knowing which case you are in is the whole skill, and what decides it is a ratio, not a formula.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put an atom in a uniform static field F ẑ and the perturbation is H_S = −d⋅F = eFz, a one-body operator odd under parity. Every eigenket of a central-field H₀ carries definite parity, so ⟨0|z|0⟩ = 0 identically and the first-order shift vanishes at any field strength: an isolated atomic level has no permanent electric dipole. The leading effect is second order, a sum over the states z can reach, ΔE = −½αF² with α = 2e²Σ|⟨k|z|0⟩|²/(Eₖ − E₀).

Read that sum as level repulsion: the field mixes opposite-parity states into |0⟩, the mixture carries an induced dipole ⟨d_z⟩ = αF, and a ground state always moves down because every denominator is positive. Hydrogen breaks the argument without breaking the parity rule. Its accidental l degeneracy puts 2s and 2p₀ at one energy, a denominator is zero, the sum is meaningless, and you must diagonalise eFz inside the manifold instead, where |⟨2s|z|2p₀⟩| = 3a₀ returns eigenvalues ±3ea₀F on equal mixtures of the two parities.

What it all costs is boundedness: the added potential runs to −∞ downhill, so strictly no level is bound at all. Every one is a Siegert resonance with a tunnelling width — invisible to a finite-basis diagonalisation, utterly negligible for the ground state, and decisive for Rydberg states.

Simple definition: The Stark effect is the shift and splitting of atomic levels caused by the odd-parity operator eFz, and the static polarisability α is the coefficient in the second-order shift ΔE = −½αF² it produces whenever no degeneracy blocks the sum.
Example: Hydrogen's ground state has α = 4.5 × 4πε₀a₀³ = 7.42 × 10⁻⁴¹ C m² V⁻¹, so at F = 1.0 × 10⁷ V m⁻¹ it drops by 3.7 × 10⁻²⁷ J, or 23 neV — one part in 4 × 10⁸ of the 10.2 eV gap up to n = 2.

Stark operator and its parityH_S = −d⋅F = eFz, E⁽¹⁾ = eF⟨0|z|0⟩ = 0

Names the operator and settles the leading term in one line: z is odd, |0⟩ has definite parity, so the diagonal element dies.

F in V m⁻¹, z in m, e = 1.602 × 10⁻¹⁹ C; d = −er, so H_S is an energy in J

Second order and the static polarisabilityΔE = −½αF², α = 2e² Σ_(k≠0) |⟨k|z|0⟩|² / (Eₖ − E₀)

Compresses a whole spectrum into one response number, and shows a ground state must move down: every denominator is positive.

α in C² m² J⁻¹; α/4πε₀ is the polarisability volume in m³, 0.667 Å³ for H 1s

Induced dipole and the factor of one half⟨d_z⟩ = αF, ΔE = −∫₀^F ⟨d_z⟩ dF′ = −½αF²

The dipole grows with the field that pays for it, so the shift is half of −dF. A permanent dipole would deliver the full −dF.

⟨d_z⟩ in C m: 7.4 × 10⁻³⁴ C m for H 1s at 1.0 × 10⁷ V m⁻¹, or 8.8 × 10⁻⁵ ea₀

Hydrogen 1s, exactlyα(1s) = (9/2) · 4πε₀a₀³ = 7.42 × 10⁻⁴¹ C m² V⁻¹

Fixes the benchmark and exposes the continuum's 0.84 a₀³, 19% of the total — a bound-state sum on its own always comes up short.

Dalgarno–Lewis, not a truncated sum: 2p alone gives 2.96 a₀³, all discrete np 3.66 a₀³

Degenerate manifold: a linear shift⟨2s|z|2p₀⟩ = −3a₀ ⟹ E = E₂ ± 3ea₀FΔEₙ = (3/2) n (n₁ − n₂) ea₀F

Gives three components from four states, since z cannot reach |2p_(±1)⟩, and a genuine permanent dipole in the field's own basis.

3ea₀ = 2.54 × 10⁻²⁹ C m = 7.62 D; linear only above ΔE_fs/3ea₀ ≈ 2.9 kV cm⁻¹ at n = 2

Saddle point and field ionisationFᵢₒₙ = Eₕ/(16 n⁴ e a₀) = 1/(16n⁴) in atomic units

Sets the field above which no barrier is left, and reminds you that the eigenvalue below it is a resonance, not a bound state.

Atomic field unit Eₕ/ea₀ = 5.14 × 10¹¹ V m⁻¹; n = 30 gives 4.0 × 10⁴ V m⁻¹ = 397 V cm⁻¹

Parity, not smallness, kills the first-order term

H_S = eFz is odd under r → −r. Every eigenket 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 first volt per metre. The whole asymmetry between a linear Zeeman effect and a quadratic Stark effect is a parity statement and nothing else.

Second order is level repulsion, and it defines α

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.

Dalgarno–Lewis: solve one equation instead of summing

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.

Degeneracy changes the question: diagonalise first

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 a 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.

Which term is bigger? The crossover is a ratio

Real n = 2 hydrogen is not degenerate: fine structure lifts 2p_{3/2} by 10.97 GHz above 2p_{1/2}, and the Lamb shift lifts 2s_{1/2} by 1.06 GHz above it. So the honest model is two levels split by Δ and coupled by V = 3ea₀F, with eigenvalues ±√((Δ/2)² + V²). Expand for V ≪ Δ/2 and the quadratic regime reappears with α = 2μ²/Δ; take V ≫ Δ/2 and the eigenvalues are ±V, linear. The crossover is therefore a ratio, not a fixed field: V = Δ at 275 V cm⁻¹ against the Lamb shift and at 2.9 kV cm⁻¹ against the fine-structure interval. Small denominators make enormous polarisabilities — the two-level α for 2s against the Lamb shift is 1.85 × 10⁻³³ C m² V⁻¹, some 2.5 × 10⁷ times the ground state's — which is why a quadratic description of an excited level expires at fields a bench supply reaches with a couple of hundred volts.

In a uniform field, no level is strictly bound

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Matrix element ⟨2s|z|2p₀⟩3.00 a₀

Zero-field splitting Δ1.05 GHz

Field marker F200 V cm⁻¹

Leave Δ at the 1.05 GHz Lamb shift and walk the marker out: the dashed curve peels away once μF reaches Δ/2, near 140 V cm⁻¹, and by 1200 V cm⁻¹ the mixing angle is a few degrees short of 45°. Raise Δ to the 11 GHz fine-structure interval and the quadratic curve survives the whole axis.

COUPLING μF0.768 GHz

EXACT SHIFT-0.405 GHz

QUADRATIC −½αF²-0.561 GHz

MIXING ANGLE θ27.8 °

Live interpretationCOUPLING μF: 0.768 GHz. EXACT SHIFT: −0.405 GHz. QUADRATIC −½αF²: −0.561 GHz. MIXING ANGLE θ: 27.8 °

Catch the common trap

Explain before calculating.

Misconception check

The error is reading the quadratic shift as a small-field approximation that turns linear once the field is strong enough. It never does. ⟨0|z|0⟩ is a number computed in the zero-field basis, and parity makes it zero whatever F is; raising F does not make that matrix element non-zero. What changes with F is whether perturbation theory converges at all. Hydrogen's linear Stark effect is not the first-order term switching on — it is degenerate perturbation theory answering a different question, because 2s and 2p₀ were already degenerate at F = 0. The mirror-image error is then to say hydrogen 2s has a permanent dipole. The bare |2s⟩ does not; the field's own eigenstates (|2s⟩ ± |2p₀⟩)/√2 do, and it was the field that selected that basis. And the ½ in −½αF² is not a convention: the dipole is induced by the very field it then interacts with, so half the energy goes into making it.

Hydrogen's ground state shifts as −½αF² in a static field, but its n = 2 level splits linearly, into three components with the outer two at ±3ea₀F. Why does a term linear in F appear for n = 2 and not for n = 1?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen's ground state has polarisability α = 4.5 in atomic units, where the atomic unit is 4πε₀a₀³ = 1.649 × 10⁻⁴¹ C² m² J⁻¹. Find the Stark shift and the induced dipole at F = 1.0 × 10⁷ V m⁻¹ (100 kV cm⁻¹), and compare the shift with the 10.2 eV gap to n = 2.

Convert the polarisability: α = 4.5 × 1.649 × 10⁻⁴¹ = 7.42 × 10⁻⁴¹ C m² V⁻¹.
First order is zero by parity, so the leading shift is ΔE = −½αF² = −0.5 × 7.42 × 10⁻⁴¹ × (1.0 × 10⁷)² = −0.5 × 7.42 × 10⁻⁴¹ × 1.0 × 10¹⁴ = −3.71 × 10⁻²⁷ J.
In electronvolts: −3.71 × 10⁻²⁷ ÷ 1.602 × 10⁻¹⁹ = −2.32 × 10⁻⁸ eV, that is −23.2 neV.
Induced dipole: ⟨d_z⟩ = αF = 7.42 × 10⁻⁴¹ × 1.0 × 10⁷ = 7.42 × 10⁻³⁴ C m, which is 7.42 × 10⁻³⁴ ÷ 8.478 × 10⁻³⁰ = 8.8 × 10⁻⁵ ea₀ — the electron cloud has barely moved.
Compare with the gap: 2.32 × 10⁻⁸ ÷ 10.2 = 2.3 × 10⁻⁹, so the perturbation is one part in 4 × 10⁸ of the level spacing and second-order theory is in no danger.

Answerα = 7.42 × 10⁻⁴¹ C m² V⁻¹, ΔE = −3.71 × 10⁻²⁷ J = −23.2 neV, ⟨d_z⟩ = 7.42 × 10⁻³⁴ C m ≈ 8.8 × 10⁻⁵ ea₀; the shift is 2.3 × 10⁻⁹ of the 10.2 eV gap.

MediumModel hydrogen 2s₁/₂ and 2p₁/₂ as two levels split by the Lamb shift Δ = 1.058 GHz = 4.37 × 10⁻⁶ eV and coupled by V = μF with μ = 3ea₀. Find the two-level polarisability of 2s, the field at which V = Δ/2, and the error the quadratic formula makes there.

Dipole matrix element: μ = 3ea₀ = 3 × 1.602 × 10⁻¹⁹ × 5.292 × 10⁻¹¹ = 2.543 × 10⁻²⁹ C m, or 7.62 D.
Splitting in joules: Δ = 4.37 × 10⁻⁶ × 1.602 × 10⁻¹⁹ = 7.00 × 10⁻²⁵ J.
Two-level polarisability: expanding ±√((Δ/2)² + V²) for V ≪ Δ/2 gives ∓V²/Δ = −½αF², so α = 2μ²/Δ = 2 × (2.543 × 10⁻²⁹)² ÷ 7.00 × 10⁻²⁵ = 1.29 × 10⁻⁵⁷ ÷ 7.00 × 10⁻²⁵ = 1.85 × 10⁻³³ C m² V⁻¹, some 2.5 × 10⁷ times the ground state's 7.42 × 10⁻⁴¹.
Crossover field: V = Δ/2 needs F = Δ/(2μ) = 7.00 × 10⁻²⁵ ÷ (2 × 2.543 × 10⁻²⁹) = 1.38 × 10⁴ V m⁻¹ = 138 V cm⁻¹.
Exact shift there: Δ/2 − √((Δ/2)² + (Δ/2)²) = (Δ/2)(1 − √2) = −0.414 × (Δ/2), while the quadratic formula gives −V²/Δ = −(Δ/2)²/Δ = −0.500 × (Δ/2).
Error: 0.500 ÷ 0.414 = 1.21, so second-order theory overstates the shift by 21% at only 138 V cm⁻¹ — an excited level with a small denominator leaves the quadratic regime almost immediately.

Answerμ = 2.54 × 10⁻²⁹ C m, α(2s) = 1.85 × 10⁻³³ C m² V⁻¹ (2.5 × 10⁷ × the ground state's), V = Δ/2 at 138 V cm⁻¹, where −½αF² already overstates the shift by 21%.

HardFor a hydrogen Rydberg state with n = 30, work in atomic units (Eₕ = 27.211 eV, field unit Eₕ/ea₀ = 5.142 × 10¹¹ V m⁻¹). Find the binding energy, the classical field-ionisation threshold Fᵢₒₙ = 1/(16n⁴), the Stark manifold width 3n²F at that threshold, and the Inglis–Teller field 1/(3n⁵) at which the n and n + 1 manifolds first overlap.

Binding energy: Eₙ = 1/(2n²) = 1/1800 = 5.556 × 10⁻⁴ Eₕ = 5.556 × 10⁻⁴ × 27.211 = 15.1 meV.
Ionisation threshold: n⁴ = 8.10 × 10⁵, so Fᵢₒₙ = 1/(16 × 8.10 × 10⁵) = 7.72 × 10⁻⁸ a.u. = 7.72 × 10⁻⁸ × 5.142 × 10¹¹ = 3.97 × 10⁴ V m⁻¹, that is 397 V cm⁻¹.
Manifold width at threshold: 3n²Fᵢₒₙ = 3 × 900 × 7.72 × 10⁻⁸ = 2.08 × 10⁻⁴ Eₕ = 5.67 meV, already 38% of the 15.1 meV binding — the manifold is a substantial fraction of the well before it ionises.
Inglis–Teller: manifolds touch when 3n²F = 1/n³, so F_IT = 1/(3n⁵) = 1/(3 × 2.43 × 10⁷) = 1.37 × 10⁻⁸ a.u. = 7.05 × 10³ V m⁻¹ = 70.5 V cm⁻¹.
Ratio: Fᵢₒₙ/F_IT = (1/16n⁴)(3n⁵) = 3n/16 = 5.6, so there is a wide window in which neighbouring manifolds already overlap and anticross while the atom is still bound.
Scale check against the ground state: Fᵢₒₙ scales as n⁻⁴, so n = 1 needs 30⁴ = 8.1 × 10⁵ times more field, 3.2 × 10¹⁰ V m⁻¹ — which is why static field ionisation is a Rydberg technique and not a ground-state one.

AnswerE₃₀ = 15.1 meV, Fᵢₒₙ = 3.97 × 10⁴ V m⁻¹ (397 V cm⁻¹) with a manifold width of 5.67 meV there, F_IT = 7.05 × 10³ V m⁻¹ (70.5 V cm⁻¹), and Fᵢₒₙ/F_IT = 3n/16 = 5.6.