The Stark Effect

The Stark effect is now a tool wherever a field must be measured or a level moved. Rydberg-atom electrometry inverts this article's scalings: because Stark shifts blow up so fast with $n$, a vapor cell of Rydberg atoms is a calibrated, SI-traceable antenna that reads radio-frequency and microwave amplitudes off Stark-shifted lines, with the $\mathcal{E}_\mathrm{at}/(16n^4)$ ramp as the state-selective detector. In precision timekeeping the enemy version dominates: blackbody radiation at 300 K is a bath of oscillating fields whose time-averaged (quadratic) Stark shift is among the largest systematics an optical clock must subtract at the $10^{-18}$ level. Two more habitats keep it in daily use. In plasmas — from tokamak edges to white-dwarf atmospheres — the fluctuating microfields of nearby ions Stark-broaden emission lines, so a line width becomes a remote thermometer and densitometer, with hydrogen Balmer broadening a standard diagnostic. And in semiconductors the quantum-confined Stark effect drives the electro-absorption modulators that put data onto optical fiber.

Two atoms meet the same field two ways. A generic atom — a sodium ground state, a helium atom, hydrogen $1s$ — is a symmetric cloud with no dipole to offer. The field must first distort it, building an induced dipole $\mu=\alpha\mathcal{E}$ in proportion to the field, and the energy paid is $-\tfrac12\alpha\mathcal{E}^2$, quadratic, the $\tfrac12$ because half the would-be gain goes into distorting the cloud — the same $\tfrac12$ as $\tfrac12 CV^2$ stored while charging a capacitor. Hydrogen's excited shells behave differently: because $2s$ and $2p_0$ sit at one energy with opposite parity, the field mixes them at no cost into two lopsided hybrids, one pointing with the field (energy down), one against (energy up). Those hybrids carry a permanent dipole of fixed size, whose energy $-\boldsymbol\mu\cdot\boldsymbol{\mathcal{E}}$ is linear in $\mathcal{E}$. The first figure shows that fan.

Symmetry referees the affair. The field picks out one axis; rotations about it survive, so $\hat L_z$ commutes with the Hamiltonian and $m$ stays good, while tilting rotations are gone and $\ell$ dies. And because the field couples to position, not spin, the electron's magnetic moment is a bystander — the sharpest contrast with the Zeeman effect. Turn the field high enough and a new regime opens: the well plus the ramp $-e\mathcal{E}z$ tilts into a saddle, the barrier sinks toward the level, and the electron escapes. That threshold falls off ferociously with $n$, and the steepness is what makes it a measurement tool — the punchline of the subject, the field an atom can survive plotted against how highly excited it is.

Everything a static field does descends from one operator. A uniform field derives from the potential $\phi=-\mathcal{E}z$, and an electron of charge $-e$ ($e>0$) carries potential energy $U=(-e)(-\mathcal{E}z)=e\mathcal{E}z$. Because $z=r\cos\theta\propto Y_1^0$, that one added term is parity-odd, $m$-preserving, and reaches only $\ell\to\ell\pm1$.

where: $\mathcal{E}$ — field magnitude along $\hat z$; $z=r\cos\theta$ — electron coordinate along the field; $\boldsymbol\mu=-e\mathbf r$ — electric dipole operator. Physically $\hat H'$ prices displacement: cloud pushed upfield costs energy, downfield saves it, and the transformation content of $z$ decides which states it connects. For a non-degenerate definite-parity state the first-order shift is the average displacement — the permanent dipole — which vanishes because a symmetric cloud is not lopsided before the field arrives. The parity selection theorem is derived in aq-qm-stark-effect; we quote its diagonal corollary.

where: $|n\ell m\rangle$ — a non-degenerate eigenstate of parity $(-1)^\ell$; $\langle z\rangle=0$ because $z$ is odd and the density even. No permanent dipole, no linear shift — the reason the electric splitting waited seventeen years after the magnetic one. When the first-order term dies, the leading response is second order: the field admixes opposite-parity levels, and the distorted state carries a dipole $\alpha\mathcal{E}$ whose energy is quadratic. The sum-over-states $\alpha$ and the exact $\alpha=\tfrac92 a_0^3$ (Gaussian units) belong to aq-qm-quadratic-stark-effect; here we need only its form and sign.

where: $\alpha$ — static scalar polarizability; with $\alpha>0$ the minus sign pushes a ground state down, because polarizing along the field it created is favorable. The generic case: helium, the alkalis, hydrogen $1s$. Hydrogen's excited shells escape the parity veto because opposite-parity states are degenerate there. The legal tool is degenerate perturbation theory: restrict $\hat H'$ to the shell and diagonalize. For $n=2$ the selection rules leave one $2\times2$ block coupling $2s$ to $2p_0$, with $\langle 2s|z|2p_0\rangle=-3a_0$ from aq-qm-linear-stark-effect. Diagonalizing it — the evaluation the spectroscopist owns — gives the observable splitting.

where: the $\pm 3ea_0\mathcal{E}$ eigenstates are the hybrids $(2s\mp2p_0)/\sqrt2$, each with a permanent dipole $\mp 3ea_0$; the $m=\pm1$ states have no partner and ride through unshifted. At $\mathcal{E}=2.5\times10^7$ V/m, $3ea_0\mathcal{E}=3.968\ \mathrm{meV}=32.0\ \mathrm{cm^{-1}}=959.7$ GHz. The general shell obeys the parabolic ladder $\tfrac32 n(n_1-n_2)ea_0\mathcal{E}$. Finally, none of these are true bound states. Add the field's ramp to the Coulomb well along the downfield axis and the potential has no floor, so a barrier — not a wall — holds the electron in. Setting that barrier top equal to the level gives the classical field-ionization threshold.

where: $\mathcal{E}_\mathrm{at}$ — atomic unit of field, the atom's internal scale; $n$ — principal quantum number. The $n^{-4}$ collapse is the operative fact: a ground-state atom holds to $3.21\times10^{10}$ V/m, but an $n=30$ Rydberg atom lets go at $3.97\times10^4$ V/m.

Example 1 — what Stark's grating spectrograph could see in 1913 (a real instrument number). In an ordinary discharge tube free charges screen any field, so Stark sent canal rays — fast hydrogen atoms — through the gap behind a perforated cathode, where nothing screened his tens of kilovolts per centimeter. Take his working field $\mathcal{E}=2.5\times10^7$ V/m and ask whether the split is resolvable on a grating.

Step 1. Each shifted $n=2$ component sits at $3ea_0\mathcal{E}=3.968\ \mathrm{meV}=32.0\ \mathrm{cm^{-1}}=959.7$ GHz, so a Balmer line spreads over tens of $\mathrm{cm^{-1}}$, symmetric about center.

Step 2. Convert to a wavelength split at $\mathrm{H}\beta$ (486.1 nm, the $n=4\to n=2$ line Stark studied). With $\tilde\nu=1/\lambda=20\,572\ \mathrm{cm^{-1}}$ and $\Delta\tilde\nu=32.0\ \mathrm{cm^{-1}}$, the component moves by $\Delta\lambda=\lambda\,\Delta\tilde\nu/\tilde\nu=\mathbf{0.756\ nm}$.

Step 3. The resolving power required is $R=\tilde\nu/\Delta\tilde\nu=20\,572/32.0=\mathbf{643}$, while a modest grating reaches $R\sim10^4$. The $32\ \mathrm{cm^{-1}}$ / $0.76$ nm splitting at $\mathrm{H}\beta$ needs only $R\approx640$ — which is why Stark, with a good grating and an un-screened field, caught in 1913 what Voigt's theory had called invisible.

Example 2 — selective field ionization: telling $n=30$ from $n=31$. A Rydberg experiment reads out the principal quantum number by slowly ramping an electric field and recording when the atom ionizes. How well separated are adjacent levels?

Step 1. Thresholds from equation (5): $\mathcal{E}_c(30)=\mathcal{E}_\mathrm{at}/(16\cdot30^4)=\mathbf{3.97\times10^4\ V/m}$ and $\mathcal{E}_c(31)=\mathcal{E}_\mathrm{at}/(16\cdot31^4)=\mathbf{3.48\times10^4\ V/m}$ — both a few tens of volts across a millimeter.

Step 2. Their ratio is $\mathcal{E}_c(30)/\mathcal{E}_c(31)=(31/30)^4=\mathbf{1.14}$: the $n=30$ atom needs a $14\%$ higher field than $n=31$. So $n=31$ ionizes first, $n=30$ a little later, and a time-gated detector separates them cleanly.

Step 3. The gap scales as $\approx 4/n$, so higher manifolds are harder to resolve. A $14\%$ threshold difference between $n=30$ and $n=31$ lets a ramped field sort Rydberg populations state by state — the standard read-out of Rydberg spectroscopy.

The Stark effect carries one man's name across two legacies — a decisive experiment that helped launch quantum theory, and a later campaign against it.

Texts. Foot, Atomic Physics, §8.3 · Demtröder, Laser Spectroscopy, on Stark spectroscopy of Rydberg atoms · Bransden and Joachain, Physics of Atoms and Molecules, Ch. 6 · Gallagher, Rydberg Atoms, for selective field ionization · Griffiths and Schroeter, Introduction to Quantum Mechanics, §7.3.

Provenance. The perturbation machinery and parity selection theorem are recalled from aq-qm-stark-effect; the $-3a_0$ matrix element, the $\pm 3ea_0\mathcal{E}$ splitting, and the parabolic ladder from aq-qm-linear-stark-effect; the polarizability sum and exact $\alpha=\tfrac92 a_0^3$ from aq-qm-quadratic-stark-effect. This article's verify script reproduces every evaluated number: the $2\times2$ eigenvalues $\pm 3ea_0\mathcal{E}\to 3.968\ \mathrm{meV}=32.0\ \mathrm{cm^{-1}}=959.7$ GHz at $2.5\times10^7$ V/m; the $\mathrm{H}\beta$ split $0.756$ nm at $R\approx643$; $\alpha=7.42\times10^{-41}\ \mathrm{C\cdot m^2/V}$ and the $2.315\ \mu\mathrm{eV}$ shift at $10^8$ V/m; the saddle law with $\mathcal{E}_\mathrm{at}=5.142\times10^{11}$ V/m, $\mathcal{E}_c(1)=3.21\times10^{10}$, $\mathcal{E}_c(30)=3.97\times10^4$ V/m. A base slip is corrected in the text: the extreme-state shift is $\tfrac32 n(n_1-n_2)ea_0\mathcal{E}$ (so $n=2$ gives $\pm 3ea_0\mathcal{E}$), not the twice-too-large "$\pm3n\,ea_0\mathcal{E}$" the dry base tabulated.

Prerequisites. Solution of the Schrödinger Equation for Hydrogen · Hydrogenic Wavefunctions and Radial Distributions · Quantum Numbers and Orbital Angular Momentum · Fine Structure and Spin-Orbit Coupling. Leads to. The Zeeman Effect · The Lamb Shift · Multi-electron Atoms and the Central-Field Approximation · Rydberg Atoms.