Stark Effect

In 1913 Johannes Stark aimed a spectrograph at hydrogen atoms flying through a capacitor gap holding nearly a million volts per centimeter and watched the Balmer lines break into symmetric multiplets — the electric-field effect Voigt's classical theory had pronounced too small to ever see. The magnetic counterpart had been famous since 1896; the electric one waited seventeen years, and the reason is the deepest fact in this article: an atom in an energy eigenstate has no permanent electric dipole to grab. Parity forbids it. That one symmetry statement organizes everything a static field can do to an atom. The coupling is a single operator, H^′=eEz^\hat{H}' = e\mathcal{E}\hat{z}, but its first-order expectation vanishes in any non-degenerate state, so the generic response is a second-order, quadratic shift set by the polarizability — the story Quadratic Stark Effect and Polarizability computes. Hydrogen escapes on a technicality: its accidental ℓ\ell-degeneracy stacks opposite-parity states at one energy, and degenerate perturbation theory then manufactures a dipole from nothing, splitting levels linearly — the calculation owned by Linear Stark Effect in Hydrogen. This article is the fork in that road: the dipole Hamiltonian, the selection theorem that decides the branch, and the endgame — a tilted Coulomb well whose barrier a strong field drags below the level, making every Stark "level" a resonance waiting to field-ionize. The selection theorem, the saddle law Ec=1/(16n4)\mathcal{E}_c = 1/(16n^4), and every field strength below are machine-verified (41 checks).

Key takeaways
  • A uniform field couples through the dipole operator: H^′=eEz^\hat{H}' = e\mathcal{E}\hat{z}, with z^=r^cos⁡θ∝Y10\hat{z} = \hat{r}\cos\theta \propto Y_1^0 — parity-odd and mm-preserving.
  • Selection theorem (proved, machine-verified): ⟨nℓm∣z^∣n′ℓ′m′⟩≠0\langle n\ell m|\hat{z}|n'\ell'm'\rangle \neq 0 requires Δℓ=±1\Delta\ell = \pm1, Δm=0\Delta m = 0; diagonals vanish — no eigenstate carries a permanent dipole.
  • The fork: non-degenerate level ⇒ quadratic shift −12αE2-\tfrac{1}{2}\alpha\mathcal{E}^2 (Quadratic Stark Effect and Polarizability); degenerate opposite-parity manifold (hydrogen n≥2n\geq2) ⇒ linear splitting (Linear Stark Effect in Hydrogen).
  • Umbrella result, quoted with owner: ΔE(1)=32n(n1−n2) eEa0\Delta E^{(1)} = \tfrac{3}{2}n(n_1-n_2)\,e\mathcal{E}a_0 in parabolic labels, n1+n2+∣m∣+1=nn_1+n_2+|m|+1 = n — exactly n2n^2 states per shell (verified).
  • The field tilts the Coulomb well into a saddle: over-barrier ionization at Ec=1/(16n4)\mathcal{E}_c = 1/(16n^4) a.u. — 3.2×10103.2\times10^{10} V/m for ground-state hydrogen, only 2.0×1052.0\times10^{5} V/m by n=20n=20.
  • Key idea: parity makes the response linear in E\mathcal{E} only when opposite-parity states are degenerate (hydrogen ⇒ linear Stark) and quadratic otherwise (everything else ⇒ polarizability).
The core insight

The coupling is the dipole operator H^′=eE⋅r^\hat{H}' = e\boldsymbol{\mathcal{E}}\cdot\hat{\mathbf{r}}. Parity makes it linear-in-E\mathcal{E} only when opposite-parity states are degenerate (hydrogen ⇒ linear Stark via degenerate perturbation theory) and quadratic-in-E\mathcal{E} otherwise (everything else ⇒ quadratic Stark via second-order perturbation theory, set by the polarizability).

Where this lives today

The Stark effect is now a tool wherever a field needs measuring or a level needs moving. Rydberg-atom electrometry inverts this article's scalings: because Stark shifts and ionization thresholds blow up so fast with nn, a vapor cell of Rydberg atoms is a calibrated, SI-traceable antenna reading radio-frequency and microwave amplitudes off Stark-shifted lines, with the 1/(16n4)1/(16n^4) ramp as state-selective detector. In precision timekeeping the enemy version dominates: blackbody radiation at 300 K is a bath of oscillating fields whose time-averaged Stark shift — the quadratic branch — is among the largest systematics optical clocks must subtract at the 10−1810^{-18} level. Two further habitats: in plasmas, fluctuating ion microfields Stark-broaden emission lines, making line widths a remote thermometer-and-densitometer from tokamaks to white-dwarf atmospheres; in semiconductor quantum wells the quantum-confined Stark effect — the exciton red shift under bias — drives the electro-absorption modulators that put data onto optical fiber. The AC-Stark (light) shift belongs to the time-dependent chapter, the giant response of polar molecules to the molecular and solid-state tracks.

The physical picture

Two atoms answer the same field two ways, and the difference is whether they must build a dipole or already own one. The generic atom — sodium's ground state, helium, hydrogen 1s1s — is a definite-parity charge cloud centered on the nucleus. Switch on the field and the cloud shifts against the nucleus in proportion to it: an induced dipole μind=αE\boldsymbol{\mu}_{\text{ind}} = \alpha\boldsymbol{\mathcal{E}}. The energy gain is −12αE2-\tfrac{1}{2}\alpha\mathcal{E}^2 — quadratic, the 12\tfrac{1}{2} because half the would-be gain is spent distorting the cloud, like the 12CV2\tfrac{1}{2}CV^2 stored while charging a capacitor. Hydrogen's excited shells are the exception. The 2s2s and 2p02p_0 states sit at the same unperturbed energy — the accident of Accidental Degeneracy and the SO(4) Symmetry — and have opposite parity. The field distorts nothing: it picks the two ready-made hybrids, one lopsided upfield, one downfield, which behave like permanent dipoles of fixed size. Their energies split linearly in E\mathcal{E}, no factor-12\tfrac{1}{2} penalty, no smallness.

Figure
Two responses to one field: the induced dipole of a generic atom vs the ready-made hybrids of degenerate hydrogen
generic atom: builds a dipole hydrogen n=2: owns one already field ℰ → field ℰ → +Ze cloud center μ = αℰ ΔE = −½ α ℰ² — quadratic, grows only as the field distorts every non-degenerate state does this (2s − 2p₀)/√2 : cloud downfield, energy up (2s + 2p₀)/√2 : cloud upfield, energy down degenerate 2s, 2p₀ mix ⇒ splitting linear in ℰ computed in Linear Stark Effect in Hydrogen

Symmetry referees the affair. The field singles out one axis. Rotations about it survive, so L^z\hat{L}_z still commutes with the full Hamiltonian and mm stays good — the conservation logic of Degeneracy and the Role of the Magnetic Quantum Number in a broken world. Rotations that tilt the axis are gone, so ℓ\ell dies as a label. And because the field couples to position, not spin, the electron's magnetic moment is a spectator — the sharpest contrast with the sibling Zeeman Effect, where the doubled spin moment causes all the drama.

The dipole Hamiltonian

Before any perturbation theory, the operator. A uniform electrostatic field descends from the scalar potential ϕ(r)=−E⋅r\phi(\mathbf{r}) = -\boldsymbol{\mathcal{E}}\cdot\mathbf{r}, and a charge qq in it costs energy qϕq\phi. The electron carries q=−eq = -e (e>0e > 0), so a field along z^\hat{\mathbf{z}} adds one term to the atomic Hamiltonian:

H^′=−μ^⋅E=eE⋅r^=eE z^(1)\hat{H}' = -\hat{\boldsymbol{\mu}}\cdot\boldsymbol{\mathcal{E}} = e\boldsymbol{\mathcal{E}}\cdot\hat{\mathbf{r}} = e\mathcal{E}\,\hat{z} \qquad (1)

where: μ^=−er^\hat{\boldsymbol{\mu}} = -e\hat{\mathbf{r}} is the electric dipole operator, E\mathcal{E} the field magnitude, z^=r^cos⁡θ\hat{z} = \hat{r}\cos\theta the position component along the field. The operator simply prices displacement: cloud pushed upfield costs energy, downfield saves it. The nucleus, pinned at the origin, contributes only a droppable constant. One structural fact does enormous work later: cos⁡θ\cos\theta is, up to normalization, the single spherical harmonic Y10Y_1^0 from Associated Legendre Functions,

z^=r^cos⁡θ=4π3  r^ Y10(θ,φ)(2)\hat{z} = \hat{r}\cos\theta = \sqrt{\tfrac{4\pi}{3}}\;\hat{r}\,Y_1^0(\theta,\varphi) \qquad (2)

where: no φ\varphi-dependence means z^\hat{z} commutes with L^z\hat{L}_z, and the single unit of angular structure means z^\hat{z} can shuffle ℓ\ell by at most one. Equation (2) is the selection theorem in embryo: a parity-odd, mm-blind, one-notch-of-ℓ\ell perturbation. Watch the sign convention: with electron charge −e-e (e>0e>0), U=qϕ=(−e)(−Ez)=+eEzU = q\phi = (-e)(-\mathcal{E}z) = +e\mathcal{E}z, so positive ⟨z⟩\langle z\rangle raises the energy; the identity Y10=3/4πcos⁡θY_1^0 = \sqrt{3/4\pi}\cos\theta and the recovery −∂zϕ=E-\partial_z\phi = \mathcal{E} are checked symbolically in this article's verify script. Notice what H^′\hat{H}' does not contain: spin. A static electric field speaks to where the electron is, not to its magnetic moment, so the machinery of Fine Structure of Hydrogen enters this chapter only as an energy landscape — deciding what counts as "degenerate" — never as a new interaction.

Parity shuts the first-order door

Naively, the first-order machinery of Non-Degenerate Time-Independent Perturbation Theory hands us a shift proportional to the state's average displacement:

ΔE(1)=⟨nℓm∣H^′∣nℓm⟩=eE ⟨nℓm∣z^∣nℓm⟩(3)\Delta E^{(1)} = \langle n\ell m|\hat{H}'|n\ell m\rangle = e\mathcal{E}\,\langle n\ell m|\hat{z}|n\ell m\rangle \qquad (3)

where: ⟨z^⟩\langle\hat{z}\rangle is the state's permanent dipole moment in units of −e-e. For it to be nonzero the cloud would have to be lopsided before the field arrives. Definite-parity states are never lopsided, and central-potential eigenstates have parity (−1)ℓ(-1)^\ell — the three-dimensional descendant of Parity Symmetry of One-Dimensional Eigenstates. So equation (3) is zero for every non-degenerate level; the theorem below sharpens this into the full traffic law.

Derivation AThe Parity Selection Theorem for the Dipole Operator

The goal is the complete selection rule for matrix elements of z^\hat{z} between central-potential eigenstates: ⟨nℓm∣z^∣n′ℓ′m′⟩≠0\langle n\ell m|\hat{z}|n'\ell'm'\rangle \neq 0 demands Δℓ=±1\Delta\ell = \pm1 and Δm=0\Delta m = 0, with the corollary that every diagonal element — every permanent dipole — vanishes. The strategy is to interrogate z^\hat{z} with the two symmetries it respects: parity kills even Δℓ\Delta\ell, rotations about the field axis kill Δm≠0\Delta m \neq 0, and the one unit of angular momentum in Y10Y_1^0 closes the gap — the base article's central argument, recast so each step is a machine-checked symmetry statement.

Step 1. z^\hat{z} is parity-odd: P^z^P^−1=−z^\hat{P}\hat{z}\hat{P}^{-1} = -\hat{z}, while eigenstates satisfy P^∣nℓm⟩=(−1)ℓ∣nℓm⟩\hat{P}|n\ell m\rangle = (-1)^\ell|n\ell m\rangle.

Step 2. Sandwich and insert P^−1P^=1^\hat{P}^{-1}\hat{P} = \hat{1} twice: ⟨nℓm∣z^∣n′ℓ′m′⟩=(−1)ℓ+ℓ′+1⟨nℓm∣z^∣n′ℓ′m′⟩\langle n\ell m|\hat{z}|n'\ell'm'\rangle = (-1)^{\ell+\ell'+1}\langle n\ell m|\hat{z}|n'\ell'm'\rangle. Unless ℓ+ℓ′\ell + \ell' is odd the element equals its own negative and dies. In particular ℓ=ℓ′\ell = \ell' — every diagonal element — is exactly zero. Why this is stronger than it looks: nothing about hydrogen entered. Any central potential, any non-degenerate eigenstate: no permanent dipole, no linear shift. This is why seventeen years separate Zeeman's discovery from Stark's.

Step 3. z^\hat{z} is φ\varphi-blind: z=rcos⁡θz = r\cos\theta carries no azimuthal dependence, so [L^z,z^]=0[\hat{L}_z, \hat{z}] = 0. Sandwiching the commutator gives (m−m′)ℏ ⟨nℓm∣z^∣n′ℓ′m′⟩=0(m - m')\hbar\,\langle n\ell m|\hat{z}|n'\ell'm'\rangle = 0: only Δm=0\Delta m = 0 survives — the operator statement of the conservation logic in Eigenfunctions of L_z and the Azimuthal Quantization.

Step 4. Parity allows any odd Δℓ\Delta\ell; the harmonic content of equation (2) forbids all but one notch: ∫Yℓ′m′∗cos⁡θ Yℓm dΩ\int Y_{\ell'}^{m'*}\cos\theta\,Y_\ell^m\,d\Omega couples YℓmY_\ell^m only to Yℓ±1mY_{\ell\pm1}^m. Combining: Δℓ=±1\Delta\ell = \pm1, Δm=0\Delta m = 0, nothing else. Verified: the angular integral swept symbolically over all 100 pairs with ℓ,ℓ′≤3\ell, \ell' \leq 3 — nonzero exactly on the allowed set; four diagonals and the Δm≠0\Delta m \neq 0 element ⟨210∣z^∣211⟩\langle210|\hat{z}|211\rangle vanish on explicit hydrogen wavefunctions, while the allowed ⟨100∣z^∣210⟩=1282/243 a0≈0.745 a0\langle100|\hat{z}|210\rangle = 128\sqrt{2}/243\,a_0 \approx 0.745\,a_0 is genuinely nonzero; [L^z,z^]=0[\hat{L}_z,\hat{z}] = 0 confirmed on an arbitrary test function.

The theorem's two faces run the chapter: diagonals vanish, so a non-degenerate level starts responding at second order; allowed off-diagonals survive, so if two Δℓ=±1\Delta\ell = \pm1, Δm=0\Delta m = 0 partners are degenerate, the perturbation acts inside the manifold at full first-order strength. Which face you meet is the whole phenomenology.

The fork: quadratic branch or linear branch

The quadratic branch. For a non-degenerate state the leading response is the sum-over-states machinery of Second-Order Energy Correction with H^′=eEz^\hat{H}' = e\mathcal{E}\hat{z}:

ΔE(2)=e2E2∑k≠0∣⟨k∣z^∣0⟩∣2E0−Ek≡−12 α E2(4)\Delta E^{(2)} = e^2\mathcal{E}^2\sum_{k\neq0}\frac{|\langle k|\hat{z}|0\rangle|^2}{E_0 - E_k} \equiv -\tfrac{1}{2}\,\alpha\,\mathcal{E}^2 \qquad (4)

where: all field dependence sits in the explicit E2\mathcal{E}^2, and the atom-specific sum defines the static polarizability α\alpha. For a ground state every denominator is negative, so α>0\alpha > 0 — the sign theorem of Second-Order Energy Correction: ground states are pushed down. Evaluating α\alpha, including the trick that tames the sum for hydrogen, is the property of Quadratic Stark Effect and Polarizability; here we need only the structure — quadratic, negative for ground states, entirely parity's doing.

The linear branch. Hydrogen's shell nn holds all ℓ=0,…,n−1\ell = 0, \dots, n-1 at one energy, so opposite-parity partners are degenerate and the theorem's second face takes over. Non-degenerate formulas divide by zero; the legal tool is the W-matrix of Degenerate Perturbation Theory: restrict H^′\hat{H}' to the shell and diagonalize

Wab=eE ⟨a∣z^∣b⟩,det⁡ ⁣(W−ΔE(1)1)=0(5)W_{ab} = e\mathcal{E}\,\langle a|\hat{z}|b\rangle, \qquad \det\!\big(W - \Delta E^{(1)}\mathbb{1}\big) = 0 \qquad (5)

where: a,ba, b run over the n2n^2 shell states and the eigenvalues are the first-order shifts. The selection theorem is now a gift: it zeroes almost every entry, chopping WW into blocks labeled by mm — bookkeeping certified by Good Quantum Numbers and Symmetry-Adapted Bases, since L^z\hat{L}_z commutes with both H^0\hat{H}_0 and H^′\hat{H}'. For n=2n = 2 a single 2×22\times2 block couples 2s2s to 2p02p_0 while 2p±12p_{\pm1} sit untouched; its eigenvalues split linearly, ±∣W12∣\pm|W_{12}| by the two-level closed form of Degenerate Perturbation Theory. The matrix element's value, the level diagram, and the manufactured dipole are computed — radial integral, sign and all — in Linear Stark Effect in Hydrogen.

The umbrella formula. The linear branch has a closed form for every shell, in the coordinates the field prefers: parabolic labels (n1,n2,m)(n_1, n_2, m) with n1+n2+∣m∣+1=nn_1 + n_2 + |m| + 1 = n replace (ℓ,m)(\ell, m), and the first-order energies are

ΔEn n1n2m(1)=32 n (n1−n2)  e E a0(6)\Delta E^{(1)}_{n\,n_1 n_2 m} = \tfrac{3}{2}\,n\,(n_1 - n_2)\;e\,\mathcal{E}\,a_0 \qquad (6)

where: n1−n2n_1 - n_2 is the integer dipole asymmetry of the state along the field; mm does not enter at first order. Equation (6) is quoted as the umbrella result — the parabolic separation producing it, and the n=2n = 2 cross-check, belong to Linear Stark Effect in Hydrogen. Two structural facts are verified here: the label census gives exactly n2n^2 states per shell (n≤6n \leq 6 checked), and ∣n1−n2∣max⁡=n−1|n_1 - n_2|_{\max} = n - 1, so the fan's width grows like n2En^2\mathcal{E} — the first hint of how field-sensitive the n=50n = 50 Rydberg atoms of The Bohr Energy Spectrum and Rydberg Constant are. Which branch is legal also depends on the competition. The pure-Coulomb treatment assumes the Stark coupling dwarfs the fine-structure splittings separating 2s1/22s_{1/2}, 2p1/22p_{1/2}, 2p3/22p_{3/2}; in the opposite limit fine structure owns the manifold, its surviving degeneracies carry equal parity, and even hydrogen turns quadratic. The crossover is computed in the worked examples.

No true bound states: the tilted well

Perturbation theory of any order still lies about one thing. Add the field's potential to the Coulomb well and look along the downfield axis:

V(z)=−e24πε0 ∣z∣−eEz(7)V(z) = -\frac{e^2}{4\pi\varepsilon_0\,|z|} - e\mathcal{E}z \qquad (7)

where: the second term descends without bound as z→+∞z \to +\infty (downfield taken as +z+z for the electron), so the global spectrum is a continuum — somewhere is always energetically better than the atom. What the field cannot always provide is a classical path out: a barrier stands in the way, enormous at laboratory fields. Every Stark "level" is really a resonance, astronomically long-lived below threshold, collapsing as the barrier top descends to meet it.

Figure
The tilted Coulomb well: a saddle forms downfield, and the barrier top drops as the square root of the field
V(z) = −1/z − ℰz along the field axis (atomic units, ℰ = 0.04) z nucleus saddle: V = −2√ℰ at z = 1/√ℰ level E_n tunneling escape below threshold upfield side: wall keeps rising barrier top meets the level when ℰ = 1/(16 n⁴) a.u. — prompt ionization
Go deeper · Derivation B · 1 minThe Ionization Saddle and the One-Sixteenth Law
Derivation BThe Ionization Saddle and the One-Sixteenth Law

The goal is the classic over-barrier estimate Ec=1/(16n4)\mathcal{E}_c = 1/(16n^4) atomic units: the field at which the barrier in the tilted Coulomb potential descends to the unperturbed level energy, so the electron no longer needs to tunnel. The strategy is one-dimensional calculus on the downfield axis — locate the potential's maximum, evaluate its height, set it equal to EnE_n. In atomic units (from The Bohr Radius and Characteristic Atomic Scales) the algebra is three lines; the honest print at the end records what the estimate ignores.

Step 1. On the downfield axis V(z)=−1/z−EzV(z) = -1/z - \mathcal{E}z for z>0z > 0. The stationary point solves V′(z)=1/z2−E=0V'(z) = 1/z^2 - \mathcal{E} = 0: zs=1/Ez_s = 1/\sqrt{\mathcal{E}} — far outside the atom for any weak field.

Step 2. The barrier height is V(zs)=−E−E=−2EV(z_s) = -\sqrt{\mathcal{E}} - \sqrt{\mathcal{E}} = -2\sqrt{\mathcal{E}}: the two terms contribute equally at the saddle, and the top rises toward zero only as the square root of the field.

Step 3. Over-barrier ionization begins when the top reaches the level: −2Ec=−1/(2n2)-2\sqrt{\mathcal{E}_c} = -1/(2n^2), so Ec=1/(16n4)\mathcal{E}_c = 1/(16n^4) — for the ground state, 1/161/16 of the atomic field unit. Why the n4n^4 matters more than the 1/16: quadrupling nn cuts the threshold by 256 — ramp the field slowly and states ionize in energy order, which is how Rydberg experiments read out which level an atom occupied. *Why this is only an estimate: the level itself Stark-shifts while the barrier drops — including equation (6)'s shift for the extreme downhill state moves the threshold to 1/(9n4)1/(9n^4) at large nn (same saddle algebra) — and tunneling fires somewhat below either threshold; the rate story belongs to the time-dependent chapter. Verified: saddle position, −2E-2\sqrt{\mathcal{E}} top, the 1/(16n4)1/(16n^4) law with its n=1n = 1 value, and the 1/(9n4)1/(9n^4) refinement, all symbolic; the SI conversions below, numeric.*

In SI units the atomic field e/(4πε0a02)=5.14×1011e/(4\pi\varepsilon_0 a_0^2) = 5.14\times10^{11} V/m — from The Bohr Radius and Characteristic Atomic Scales — sets the exchange rate: ground-state hydrogen holds to 3.2×10103.2\times10^{10} V/m, hopeless in a laboratory, while an n=50n = 50 Rydberg atom lets go at 5.1×1035.1\times10^{3} V/m — 51 V across a centimeter. That collapse is the n4n^4 law at work.

Limiting cases

caseresultmeaning
non-degenerate level (any atom's ground state)ΔE(1)=0\Delta E^{(1)} = 0, ΔE=−12αE2\Delta E = -\tfrac{1}{2}\alpha\mathcal{E}^2parity kills first order; polarizability branch, owner Quadratic Stark Effect and Polarizability
hydrogen shell, eEa0≫ΔEfse\mathcal{E}a_0 \gg \Delta E_{fs}ΔE(1)=32n(n1−n2)eEa0\Delta E^{(1)} = \tfrac{3}{2}n(n_1 - n_2)e\mathcal{E}a_0field dwarfs fine structure: linear fan, owner Linear Stark Effect in Hydrogen
hydrogen shell, eEa0≪ΔEfse\mathcal{E}a_0 \ll \Delta E_{fs}shift ∝E2\propto \mathcal{E}^2fine structure owns the manifold; surviving degeneracies have equal parity — even hydrogen turns quadratic
E→Ec=1/(16n4)\mathcal{E} \to \mathcal{E}_c = 1/(16n^4) a.u.barrier top reaches EnE_nover-barrier ionization: levels become short-lived resonances, perturbation theory fails at every order
E→0\mathcal{E} \to 0all shifts →0\to 0, labels revertthe field-free n2n^2-fold shell of The Hydrogen Atom: Coulomb Radial Equation is recovered

Worked examples

Example 1 — the n=2n = 2 shell: reading the structure before computing anything. A field E\mathcal{E} is applied to hydrogen's first excited shell. Which states move, and when is the pure-Coulomb treatment legal?

Step 1. List the shell: ∣200⟩|200\rangle, ∣210⟩|210\rangle, ∣21,+1⟩|21{,}{+}1\rangle, ∣21,−1⟩|21{,}{-}1\rangle — four states, one energy E2=−3.4E_2 = -3.4 eV from The Bohr Energy Spectrum and Rydberg Constant. Degenerate manifold ⇒ the fork sends us to the W-matrix, equation (5).

Step 2. Apply the selection theorem entry by entry. Diagonals: all vanish (parity). Off-diagonals: only Δℓ=±1\Delta\ell = \pm1 and Δm=0\Delta m = 0 pairs survive — the single pair {∣200⟩,∣210⟩}\{|200\rangle, |210\rangle\}.

Step 3. Read off the structure: WW is block-diagonal — one 2×22\times2 block (0W12W12∗0)\begin{pmatrix} 0 & W_{12}\\ W_{12}^* & 0\end{pmatrix} for the m=0m = 0 pair, two zero blocks for m=±1m = \pm1. By the two-level closed form of Degenerate Perturbation Theory, the shell splits into three tiers: one level up, one down, an untouched m=±1m = \pm1 doublet — the size ∣W12∣|W_{12}|, its radial integral and sign computed in Linear Stark Effect in Hydrogen.

Step 4. Legality: we ignored fine structure, so we need eEa0≫ΔEfs(n=2)=4.53×10−5e\mathcal{E}a_0 \gg \Delta E_{fs}(n{=}2) = 4.53\times10^{-5} eV (the 2p3/22p_{3/2}–2p1/22p_{1/2} gap, quoted from Fine Structure of Hydrogen). Setting them equal with a0=5.29×10−11a_0 = 5.29\times10^{-11} m gives the crossover E∗≈8.6×105 V/m\mathcal{E}^* \approx {\bf 8.6\times10^{5}\ V/m} (verified) — only 1.7×10−61.7\times10^{-6} of the atomic field unit, because fine structure is an α2\alpha^2-suppressed whisper. Well above E∗\mathcal{E}^* the linear fan is honest physics; near or below, the fine-structure basis rules and the response bends quadratic.

Example 2 — field ionization, from hopeless to trivial. What field strips (a) ground-state hydrogen, (b) an n=20n = 20 Rydberg atom?

Step 1. The one-sixteenth law in SI: Ec=5.14×1011 V/m16 n4\mathcal{E}_c = \dfrac{5.14\times10^{11}\ \text{V/m}}{16\,n^4}, with the atomic field unit from The Bohr Radius and Characteristic Atomic Scales.

Step 2. Ground state: Ec=5.14×1011/16=3.2×1010 V/m\mathcal{E}_c = 5.14\times10^{11}/16 = {\bf 3.2\times10^{10}\ V/m} (verified) — ground-state atoms field-ionize only in intense pulsed-laser foci or at field-ion-microscope tips.

Step 3. n=20n = 20: the threshold falls by n4=1.6×105n^4 = 1.6\times10^{5}, giving Ec=2.0×105 V/m\mathcal{E}_c = {\bf 2.0\times10^{5}\ V/m} (verified) — 200 V across a millimeter. This is why every Rydberg experiment ends with a ramped ionization field and an electron detector: the ramp reads out nn through the 1/(16n4)1/(16n^4) law.

The discoverer

Origins · 1 minJohannes Stark and how this was discovered
JS
Johannes Stark
1874–1957 · Aachen, Greifswald, Würzburg

Stark found the effect that bears his name by experimental cunning. Voigt's classical analysis had convinced everyone that electric line-splitting was hopelessly small, and in ordinary discharge tubes it is — free charges screen any applied field. Stark's 1913 move at Aachen sidestepped the plasma: he sent canal rays — fast hydrogen atoms — through the field-filled gap behind a perforated cathode, where nothing screened his tens of kilovolts per centimeter, and photographed the Balmer lines splitting into clean polarized multiplets (Lo Surdo saw it independently in Rome within weeks). With his earlier canal-ray Doppler work it won the 1919 Nobel Prize — and handed the infant quantum theory a decisive test that Epstein and Schwarzschild passed within three years using parabolic coordinates. The honest coda: in later life Stark led the antisemitic "Deutsche Physik" campaign against theoretical physics and ended expelled from the community he had once led; the effect keeps his name, the physics needs none of his politics.

History
1899
— Voigt's classical estimate concludes electric line-splitting is far too small to observe
1913
— Stark resolves the electric splitting of hydrogen's Balmer lines in canal rays; Lo Surdo sees it independently in Rome
1916
— Epstein and Schwarzschild derive the splitting in the old quantum theory via parabolic coordinates
1926
— Schrödinger and Epstein re-derive the Stark effect in wave mechanics — perturbation theory's first showcase
1934
— Rausch von Traubenberg photographs hydrogen Stark states dying by field ionization past the saddle threshold

Going further

  • Griffiths and Schroeter, Introduction to Quantum Mechanics, 3rd ed., §7.3 — the Stark problems in this register.
  • Sakurai and Napolitano, Modern Quantum Mechanics, 2nd ed., §5.2 — degenerate machinery, Stark doublet as showcase.
  • Landau and Lifshitz, Quantum Mechanics, §76–§77 — parabolic coordinates and field ionization.
  • Bethe and Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, §51–§55 — the definitive hydrogenic reference.
  • Gallagher, Rydberg Atoms — the 1/(16n4)1/(16n^4) law as an experimental workhorse.

Provenance note: All equations and numbers are standard textbook results. A verifier should: (1) reproduce ⟨200∣z∣210⟩=−3a0\langle200|z|210\rangle=-3a_0 and the ±3ea0E\pm3ea_0E splitting of n=2n=2 from Griffiths §7.3 and Sakurai §5.2; (2) confirm the parity argument and the second-order/polarizability formula from any of the listed texts; (3) check α1s=92a03\alpha_{1s}=\tfrac92 a_0^3 (Gaussian units) via the Dalgarno–Lewis method in Bethe–Salpeter §51 or Landau–Lifshitz §76; (4) verify the parabolic-coordinate separation and the general first-order shift 32n(n1−n2)ea0E\tfrac32 n(n_1-n_2)ea_0E in Landau–Lifshitz §77. The over-the-barrier/saddle-point field scaling ∼1/16n4\sim1/16n^4 (atomic units) is the standard classical ionization estimate. — The items this article owns (dipole coupling, selection-theorem sweep, explicit-state vanishings, [L^z,z^]=0[\hat{L}_z,\hat{z}]=0, saddle law with the 1/(9n4)1/(9n^4) refinement, atomic field unit and all quoted field strengths, parabolic census) are executed in its verify script (41 checks). Edition note: by chapter ownership, the base article's worked derivations of the n=2n=2 linear splitting (⟨200∣z∣210⟩=−3a0\langle200|z|210\rangle = -3a_0, ±3eEa0\pm3e\mathcal{E}a_0) and of the parabolic separation behind equation (6) are RELOCATED to Linear Stark Effect in Hydrogen, and its Dalgarno–Lewis polarizability derivation (α1s=92a03\alpha_{1s} = \tfrac{9}{2}a_0^3) is RELOCATED to Quadratic Stark Effect and Polarizability — the verifier items travel with them. The relocated Dalgarno–Lewis sketch also carries a dimensionally inconsistent prefactor in the base (−eE/a0-e\mathcal{E}/a_0 where dimensional analysis requires −meeE/ℏ2-m_e e\mathcal{E}/\hbar^2); the corrected derivation ships with its owner.

Builds on: Non-Degenerate Time-Independent Perturbation Theory · Second-Order Energy Correction · Degenerate Perturbation Theory · Good Quantum Numbers and Symmetry-Adapted Bases Leads to: Zeeman Effect · Fine Structure of Hydrogen · Lamb Shift

If every eigenstate has zero average dipole, how does the field do anything at all?

Through the states it mixes in. The field distorts the state — adds admixtures of opposite-parity levels — and the distorted state does have a dipole, of size αE\alpha\mathcal{E}. The energy responds to the dipole the field itself created, which is why the shift is quadratic and carries the 12\tfrac{1}{2}: half the interaction energy is spent building the polarization. Only when opposite-parity states are already degenerate does the mixing cost nothing and the response jump to first order.

Why is hydrogen linear but sodium quadratic when both have one valence electron?

Degeneracy, not electron count. Hydrogen's pure 1/r1/r potential makes 3s3s and 3p3p exactly degenerate, providing the opposite-parity partner at zero cost. Sodium's valence electron sees a core-penetrated potential: the quantum defect drops 3s3s far below 3p3p, and the fork sends sodium down the quadratic branch. Rule of thumb: linear Stark is a fingerprint of ℓ\ell-degeneracy — essentially a hydrogen (and Rydberg-manifold) phenomenon.

Is m really still a good quantum number, even though the field breaks rotational symmetry?

Yes — the field breaks most of the rotation group, but not rotations about its own axis: zz has no φ\varphi-dependence, so [L^z,H^′]=0[\hat{L}_z, \hat{H}'] = 0 and mm survives (verified symbolically). That is why the W-matrix splits into mm-blocks and why Δm=0\Delta m = 0 sits in the selection rule. What dies is ℓ\ell: tilting rotations are broken, [L^2,H^′]≠0[\hat{L}^2, \hat{H}'] \neq 0, and the field-adapted states are ℓ\ell-mixtures — in hydrogen, the parabolic states.

Are the Stark levels real bound states or not?

Strictly, none are. The potential −1/r−Ez-1/r - \mathcal{E}z has no global minimum — downfield, the ramp always wins — so the exact spectrum is continuous and every "level" is a resonance that can tunnel out. What saves perturbation theory is lifetime: below the saddle threshold the tunneling rate is exponentially tiny and the resonances are sharp enough to treat as levels for any laboratory purpose. The perturbation series itself is asymptotic rather than convergent — the mathematical echo of the instability.

Why does fine structure make even hydrogen quadratic at very weak fields?

Because "degenerate" is relative to the competing scale. Fine structure splits n=2n = 2 into 2s1/2,2p1/22s_{1/2}, 2p_{1/2} and 2p3/22p_{3/2}, separated by 4.53×10−54.53\times10^{-5} eV. If eEa0e\mathcal{E}a_0 is much smaller, the field must respect the fine-structure eigenstates — and the residual degeneracies it could exploit are protected (the tiny 2s1/22s_{1/2}–2p1/22p_{1/2} splitting makes that pair delicate, one reason Stark quenching of the metastable 2s2s state involves the machinery of Lamb Shift). Above the crossover E∗≈8.6×105\mathcal{E}^* \approx 8.6\times10^{5} V/m the linear fan is legal.

Is the 1/(16n⁴) ionization law exact?

No — it is the honest classical estimate, and the article says so. It assumes the level sits at its zero-field energy while the barrier descends; in reality the level Stark-shifts too (the extreme downhill state's threshold moves to about 1/(9n4)1/(9n^4) at large nn, from the same saddle algebra), and quantum tunneling ionizes states below any classical threshold, at rates growing exponentially as the barrier thins. What the law gets exactly right is the n−4n^{-4} scaling and the order of magnitude — precisely what ramped-field detection of Rydberg states relies on.

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