---
title: The Stark Effect and Field Ionization
module: Atoms in External Fields
moduleNumber: 6
lessonNumber: 3
order: 603
summary: >
  An electric field shifts atomic levels by coupling to the electron's position.
  Parity forbids a first-order shift for a non-degenerate state, so most atoms
  respond only at second order through their polarizability, a quadratic Stark
  shift. Hydrogen is the exception: its accidental degeneracy admits a permanent
  dipole and a linear shift, cleanest in parabolic coordinates. At large fields
  the Coulomb well develops a saddle, and Rydberg states field-ionize at a
  threshold that falls as the fourth power of the principal quantum number.
topics: [Atoms in External Fields]
sources:
  - book: Bransden & Joachain
    ref: "Ch. 9 — Atoms in External Fields; §9.3 The Stark effect, §9.4 Field ionization"
  - book: Foot
    ref: "Ch. 8 — Doppler-free spectroscopy; Stark effect and field ionization of Rydberg states"
  - book: Griffiths & Schroeter
    ref: "Ch. 7 — Time-Independent Perturbation Theory; degenerate perturbation theory"
  - book: Demtröder
    ref: "Ch. 5 — The Hydrogen Atom; §5.2 Atoms in electric fields"
draft: false
---

A static electric field $\vec E = E\hat z$ does to atomic levels what the
magnetic field of the [Zeeman effect](/atomic-physics/atoms-in-external-fields/zeeman-effect)
does, but through a different coupling and with a decisive difference in
symmetry. The perturbation is the electrostatic energy of the electron in the
field,

$$
H' = -\vec d\cdot\vec E = e E\,z,
\qquad \vec d = -e\,\vec r,
$$

with $z = r\cos\theta$ and $e > 0$ the elementary charge. Unlike the magnetic
moment coupling, $H'$ is odd under parity. That single fact splits the theory in
two: for a state of definite parity the first-order shift vanishes and the atom
responds only through its polarizability (the **quadratic** Stark effect), while
hydrogen's degenerate levels mix opposite parities and shift **linearly**. At
the largest fields the perturbative picture fails entirely and the electron
tunnels out.

## Parity and the absence of a first-order shift

A stationary atomic state $\lvert n\,\ell\,m\rangle$ has definite parity
$(-1)^\ell$: under $\vec r \to -\vec r$ the wavefunction picks up that sign. The
operator $z$ is odd. Therefore the integrand of

$$
E^{(1)} = e E\,\langle n\,\ell\,m\lvert z\rvert n\,\ell\,m\rangle
$$

is odd, and the integral over all space vanishes. No atom in a non-degenerate
state of definite parity has a permanent electric dipole moment, and none shows
a linear Stark shift.

> **Theorem (No permanent electric dipole).** For any non-degenerate eigenstate
> of a parity-symmetric Hamiltonian, $\langle \vec d\rangle = 0$, so the
> first-order Stark shift is zero. A linear shift requires either a degeneracy
> that mixes parities (hydrogen) or an externally imposed near-degeneracy.

The magnetic case had no such veto because $\vec\mu \cdot \vec B$ is
parity-even; the electric coupling $\vec d\cdot\vec E$ is parity-odd, and this is
the structural reason the two effects look so different.

## The quadratic Stark effect and polarizability

With the first order gone, the leading shift is second order in the field,

$$
E^{(2)} = e^2 E^2 \sum_{k \neq 0}
\frac{\big\lvert\langle k\lvert z\rvert 0\rangle\big\rvert^2}{E_0 - E_k}
\equiv -\tfrac12\,\alpha\,E^2,
$$

which defines the **static scalar polarizability**

$$
\alpha = 2 e^2 \sum_{k \neq 0}
\frac{\big\lvert\langle k\lvert z\rvert 0\rangle\big\rvert^2}{E_k - E_0} .
$$

The shift is negative for a ground state (every denominator $E_k - E_0 > 0$), so
the field always lowers the ground-state energy. Physically the field induces a
dipole $\langle d_z\rangle = \alpha E$ proportional to the field, and the energy
$-\tfrac12\alpha E^2$ is the work done polarizing the atom, exactly the
$-\tfrac12\alpha E^2$ of a linear dielectric.

> **Definition (Polarizability).** The coefficient $\alpha$ relating the induced
> dipole to the applied field, $\langle d_z\rangle = \alpha E$. It has
> dimensions of volume in Gaussian units and grows steeply with atomic size; for
> hydrogenic states $\alpha \propto n^7$, the fastest-growing of the Rydberg
> scalings.

For the hydrogen ground state the sum can be evaluated exactly by the
Dalgarno-Lewis method, which replaces the infinite sum by the solution of an
inhomogeneous differential equation. The result is[^dl]

$$
\alpha_{1s} = \frac{9}{2}\,a_0^3
\quad(\text{Gaussian}),
\qquad
E^{(2)} = -\frac{9}{4}\,a_0^3\,E^2 .
$$

A crude bound makes the scale plausible without the full machinery: keeping only
the dominant $\ell = 1$ intermediate states and bounding every denominator below
by the first excitation energy $E_2 - E_1 = \tfrac34\,(13.6\ \text{eV})$, the
closure relation $\sum_k \lvert\langle k\lvert z\rvert 0\rangle\rvert^2 = \langle z^2\rangle = a_0^2$
gives $\alpha \lesssim 8a_0^3/3 \approx 2.7\,a_0^3$, the right order and a lower
bound to the exact $4.5\,a_0^3$.

$$
% caption: The quadratic Stark shift $-\tfrac12\alpha E^2$ is a downward parabola
% in the field; its curvature is the polarizability $\alpha$, and the induced
% dipole $\alpha E$ grows linearly with the field.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (-3.2,0) -- (3.2,0) node[right] {E};
\draw[->, black] (0,-2.7) -- (0,0.8) node[above] {energy};
% downward parabola: y = -0.32 x^2
\draw[acc, very thick] plot[domain=-2.85:2.85, samples=60] (\x, {-0.32*\x*\x});
\node[acc, anchor=west] at (1.4,-1.35) {downward parabola};
\end{tikzpicture}
$$

## The linear Stark effect in hydrogen

Hydrogen escapes the parity veto because its $n$-level is degenerate across
$\ell$: within a given $n$, states of opposite parity share the same unperturbed
energy, and degenerate perturbation theory mixes them. Take $n = 2$. The four
states $\{\lvert 200\rangle, \lvert 211\rangle, \lvert 210\rangle, \lvert 21,\!-\!1\rangle\}$
are degenerate, and $H' = eEz$ connects only states with $\Delta\ell = \pm1$ and
$\Delta m = 0$. The single nonzero matrix element is

$$
\langle 200\lvert z\rvert 210\rangle = -3a_0 ,
$$

evaluated from the $2s$ and $2p_0$ radial functions.

> **Worked example.** Evaluate $\langle 200\lvert z\rvert 210\rangle$ with
> $\psi_{200} = N(2 - r/a_0)e^{-r/2a_0}$,
> $\psi_{210} = N(r/a_0)e^{-r/2a_0}\cos\theta$, and $N^2 = 1/(32\pi a_0^3)$.
> Writing $z = r\cos\theta$, the two $\cos\theta$ factors combine and the angular
> part is
> $$
> \int_0^{2\pi}\!\!\d\phi \int_0^\pi \cos^2\theta\,\sin\theta\,\d\theta
> = 2\pi\cdot\tfrac23 = \frac{4\pi}{3}.
> $$
> The radial part, with $u = r/a_0$, is
> $$
> \int_0^\infty \big(2 - u\big)\,u^4\,e^{-u}\,a_0^4\,\d u
> = a_0^4\big(2\cdot 4! - 5!\big) = a_0^4(48 - 120) = -72\,a_0^4 .
> $$
> Assembling, $\langle 200\lvert z\rvert 210\rangle
> = N^2\cdot\tfrac{4\pi}{3}\cdot(-72\,a_0^4)
> = \tfrac{1}{32\pi a_0^3}(-96\pi a_0^4) = -3a_0$, the promised result. The sign
> is a phase convention; the magnitude $3a_0$ is what sets the splitting.

In the $m = 0$ subspace
$\{\lvert 200\rangle, \lvert 210\rangle\}$ the perturbation is

$$
H' = eE\begin{pmatrix} 0 & -3a_0 \\ -3a_0 & 0 \end{pmatrix},
$$

with eigenvalues $\mp 3 e a_0 E$ and eigenvectors
$\tfrac{1}{\sqrt2}\big(\lvert 200\rangle \mp \lvert 210\rangle\big)$. The two
$m = \pm1$ states are untouched at first order. The $n = 2$ level splits into
three:

$$
E^{(1)} = +3 e a_0 E,\quad 0,\quad -3 e a_0 E .
$$

The shift is **linear** in the field. The mixed eigenstates are $sp$ hybrids
with charge displaced along the field, carrying a **permanent** dipole
$\langle d_z\rangle = \mp\,3 e a_0$; the field merely orients an intrinsic dipole
that the degeneracy made available.

$$
% caption: The hydrogen $n=2$ level fans into three components linear in the
% field: $+3ea_0E$, the unshifted $m=\pm1$ pair, and $-3ea_0E$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,-2.2) -- (0,2.2) node[above] {energy};
\draw[->, black] (0,0) -- (4.0,0) node[right] {E};
% degenerate at E=0
\fill[black] (0,0) circle (2pt);
% three fanning lines
\draw[acc, very thick] (0,0) -- (3.4,1.7);
\draw[acc, very thick] (0,0) -- (3.4,-1.7);
\draw[black, very thick] (0,0) -- (3.4,0);
\node[acc, anchor=west] at (3.45,1.7) {$+3ea_0E$};
\node[acc, anchor=west] at (3.45,-1.7) {shifted down};
\node[black, anchor=west] at (3.45,-0.4) {central pair};
\end{tikzpicture}
$$

### Parabolic coordinates and the general level

The exact separation of the hydrogen Stark problem uses **parabolic
coordinates** $\xi = r + z$, $\eta = r - z$, $\phi$, in which the Schrödinger
equation with a uniform field separates just as spherical coordinates separate
the field-free atom. The states are labelled by two parabolic quantum numbers
$n_1, n_2 \geq 0$ and the azimuthal $m$, with

$$
n = n_1 + n_2 + \lvert m\rvert + 1 .
$$

First-order perturbation theory in this basis gives the linear shift in closed
form for every level,[^bj-stark]

$$
E^{(1)} = \frac{3}{2}\,n\,(n_1 - n_2)\,e a_0 E .
$$

For $n = 2$ the allowed triples $(n_1, n_2, m)$ are $(1,0,0)$, $(0,1,0)$, and
$(0,0,\pm1)$, giving $n_1 - n_2 = +1, -1, 0, 0$ and shifts
$+3ea_0E,\ -3ea_0E,\ 0,\ 0$, matching the degenerate-perturbation result. The
permanent dipole of a general parabolic state is
$\langle d_z\rangle = -\partial E^{(1)}/\partial E = -\tfrac32 n(n_1 - n_2)e a_0$,
which grows as $n^2$: high-$n$ hydrogen carries an enormous field-alignable
dipole.

$$
% caption: A parabolic Stark state displaces the electron cloud along the field,
% giving a body-fixed dipole set by $n_1-n_2$; the two lobes sit up- and down-field.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% field
\draw[->, black!70, very thick] (-3.0,-1.8) -- (-3.0,1.8) node[above] {E};
% nucleus
\fill[black] (0,0) circle (2pt);
\node[anchor=north] at (0,-0.1) {nucleus};
% displaced cloud (two ellipses, larger up)
\draw[black, very thick] (0,0.9) ellipse (1.1 and 0.7);
\draw[thick, black] (0,-0.8) ellipse (0.7 and 0.45);
% dipole arrow
\draw[->, acc, very thick] (2.0,-0.9) -- (2.0,0.9) node[right] {dipole};
\node[black, anchor=north] at (0,-1.9) {larger lobe points along E: charge displaced};
\end{tikzpicture}
$$

The linear-in-$n$ shift and the parity mixing are two views of the same fact:
the accidental $\ell$-degeneracy of the pure Coulomb problem, itself a
consequence of the conserved Runge-Lenz vector, is what lets hydrogen carry a
dipole that no other atom in its ground configuration can.

## Field ionization of Rydberg states

Add the Stark potential to the Coulomb well and the total potential energy of
the electron along the field axis, in atomic units (energies in hartree,
lengths in $a_0$, field in units of $E_a = 5.142\times10^{11}\ \text{V m}^{-1}$),
is

$$
U(z) = -\frac{1}{\lvert z\rvert} + F z .
$$

On the down-field side the two terms conspire to pull the electron away: writing
$z = -s$ with $s > 0$, $U = -1/s - F s$ falls without bound. But it must first
climb over a barrier. Setting $\frac{\d U}{\d s} = 1/s^2 - F = 0$ locates
the saddle at $s_0 = F^{-1/2}$, where the barrier top sits at

$$
U_{\text{saddle}} = -2\sqrt{F} .
$$

A state of binding energy $E_n = -1/(2n^2)$ is classically trapped only while it
lies below the barrier top. It escapes over the barrier once
$E_n \geq U_{\text{saddle}}$, i.e. $-1/(2n^2) \geq -2\sqrt F$, which gives the
**classical field-ionization threshold**

$$
\;F_{\text{ion}} = \frac{1}{16\,n^4}\ \ (\text{atomic units})\;
$$

$$
% caption: The Coulomb well tilted by the field forms a saddle on the down-field
% side with barrier top at $U=-2\sqrt{F}$; a level above it ionizes over the
% barrier, one just below tunnels through.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (-3.4,0) -- (3.4,0) node[right] {z};
\draw[->, black] (0,-2.6) -- (0,1.4) node[above] {U};
% tilted coulomb: piecewise smooth. Draw right side (uphill from -1/z + Fz)
\draw[black, very thick] plot[domain=0.35:3.0, samples=60] (\x, {-1/\x + 0.55*\x});
% left side (down-field): -1/|z| + F z, z<0  => -1/(-z) ... use s
\draw[acc, very thick] plot[domain=0.35:3.0, samples=60] (-\x, {-1/\x - 0.55*\x});
% saddle marker on left
\fill[acc] (-1.35,-1.48) circle (1.6pt);
\node[acc, anchor=south] at (-1.5,-1.4) {saddle};
% a bound level
\draw[black, thick, dashed] (-2.6,-0.9) -- (1.6,-0.9);
\node[black, anchor=west] at (1.65,-0.9) {level above barrier: ionizes};
\node[black, anchor=north] at (0,-2.75) {saddle sets the barrier top};
\end{tikzpicture}
$$

The $n^{-4}$ scaling is the practical heart of Rydberg-atom detection.
Converting to laboratory units, $F_{\text{ion}} = 3.21\times10^{10}\,n^{-4}\ \text{V m}^{-1}$,
so an $n = 30$ state ionizes near $4\times10^{4}\ \text{V m}^{-1}$, about
$400\ \text{V cm}^{-1}$, a field trivially produced between two plates. Because
the threshold is sharp and $n$-selective, ramping the field and recording the
voltage at which electrons appear reads out the principal quantum number of a
Rydberg population directly.

> **Worked example.** Compare the ionizing field of the ground state and an
> $n = 50$ Rydberg state. The ground state needs $F \approx 1/16 \approx 0.06$
> atomic units, or $3\times10^{10}\ \text{V m}^{-1}$, comparable to the internal
> field an electron feels and far beyond ordinary laboratory reach. The $n = 50$
> state needs $F = 1/(16\cdot 50^4) = 3.2\times10^{-9}$ atomic units, or about
> $16\ \text{V cm}^{-1}$. The ratio is $50^4 = 6.25\times10^6$: raising the
> principal quantum number from $1$ to $50$ drops the ionizing field by nearly
> seven orders of magnitude.

The classical threshold slightly overestimates the field because a state just
below the barrier still tunnels through it at a finite rate; the tunnelling rate
rises so steeply near threshold, however, that the classical estimate is
accurate to a few percent for the reddest Stark state and remains the working
formula. Field ionization, the quadratic polarizability, and the linear
hydrogenic shift together make the atom a calibrated probe of static electric
fields, closing the treatment of atoms in external fields.

$$
% caption: The ionizing field falls as $n^{-4}$; laboratory fields ionize states
% above $n\approx 15$, placing all Rydberg states within easy experimental reach.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axes (log-log schematic)
\draw[->, black] (0,0) -- (5.4,0) node[right] {log n};
\draw[->, black] (0,0) -- (0,3.6) node[above] {log F};
% straight line slope -4 (decreasing): from high left to low right
\draw[acc, very thick] (0.4,3.3) -- (5.0,0.4);
\node[acc, anchor=south west] at (3.3,2.35) {slope $=$ minus four};
% lab reach band
\draw[dashed, black] (0,1.1) -- (5.2,1.1);
\node[black, anchor=west] at (0.2,0.75) {laboratory reach};
\end{tikzpicture}
$$

[^dl]: The exact hydrogen ground-state polarizability $\alpha_{1s} = \tfrac92 a_0^3$
    follows from the Dalgarno-Lewis / Sternheimer method, which solves an
    inhomogeneous equation for the first-order wavefunction rather than summing
    over states. See **Bransden & Joachain**, _Physics of Atoms and Molecules_,
    §9.3, and the closure-bound argument for the lower bound.
[^bj-stark]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, §9.3 —
    separation of the hydrogen Stark problem in parabolic coordinates, the
    quantum numbers $(n_1, n_2, m)$, and the first-order linear shift
    $E^{(1)} = \tfrac32 n(n_1 - n_2)e a_0 E$; §9.4 for the saddle-point
    field-ionization threshold $F_{\text{ion}} = 1/16n^4$. See also **Foot**,
    _Atomic Physics_, Ch. 8, for field ionization of Rydberg states.