The Stark Effect and Field Ionization

A static electric field $E = E \overset{z}{^}$ does to atomic levels what the magnetic field of the 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^{'} = - d \cdot E = e E z, d = - e r,

with $z = r cos θ$ 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 $∣ n ℓ m ⟩$ has definite parity $(- 1)^{ℓ}$ : under $r \to - r$ the wavefunction picks up that sign. The operator $z$ is odd. Therefore the integrand of

E^{(1)} = e E ⟨ n ℓ m ∣ z ∣ n ℓ m ⟩

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.

The magnetic case had no such veto because $μ \cdot B$ is parity-even; the electric coupling $d \cdot E$ 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} k \neq = 0 \sum \frac{⟨ k ∣ z ∣ 0 ⟩ ^{2}}{E _{0} - E _{k}} \equiv - \frac{1}{2} α E^{2},

which defines the static scalar polarizability

α = 2 e^{2} k \neq = 0 \sum \frac{⟨ k ∣ z ∣ 0 ⟩ ^{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 $⟨ d_{z} ⟩ = α E$ proportional to the field, and the energy $- \frac{1}{2} α E^{2}$ is the work done polarizing the atom, exactly the $- \frac{1}{2} α E^{2}$ of a linear dielectric.

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 is¹

α_{1 s} = \frac{9}{2} a_{0}^{3} (Gaussian), 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 $ℓ = 1$ intermediate states and bounding every denominator below by the first excitation energy $E_{2} - E_{1} = \frac{3}{4} (13.6 eV)$ , the closure relation $\sum_{k} ∣⟨ k ∣ z ∣ 0 ⟩ ∣^{2} = ⟨ z^{2} ⟩ = a_{0}^{2}$ gives $α ≲ 8 a_{0}^{3} /3 \approx 2.7 a_{0}^{3}$ , the right order and a lower bound to the exact $4.5 a_{0}^{3}$ .

The quadratic Stark shift

- \frac{1}{2} α E^{2}

is a downward parabola in the field; its curvature is the polarizability

α

, and the induced dipole

α E

grows linearly with the field.

The linear Stark effect in hydrogen

Hydrogen escapes the parity veto because its $n$ -level is degenerate across $ℓ$ : 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 ${∣ 200 ⟩, ∣ 211 ⟩, ∣ 210 ⟩, ∣ 21, - 1 ⟩}$ are degenerate, and $H^{'} = e E z$ connects only states with $Δ ℓ = \pm 1$ and $Δ m = 0$ . The single nonzero matrix element is

⟨ 200 ∣ z ∣ 210 ⟩ = - 3 a_{0},

evaluated from the $2 s$ and $2 p_{0}$ radial functions.

In the $m = 0$ subspace ${∣ 200 ⟩, ∣ 210 ⟩}$ the perturbation is

H^{'} = e E (0 - 3 a_{0} - 3 a_{0} 0),

with eigenvalues $\mp 3 e a_{0} E$ and eigenvectors $\frac{1}{2} (∣ 200 ⟩ \mp ∣ 210 ⟩)$ . The two $m = \pm 1$ states are untouched at first order. The $n = 2$ level splits into three:

E^{(1)} = + 3 e a_{0} E, 0, - 3 e a_{0} E .

The shift is linear in the field. The mixed eigenstates are $s p$ hybrids with charge displaced along the field, carrying a permanent dipole $⟨ d_{z} ⟩ = \mp 3 e a_{0}$ ; the field merely orients an intrinsic dipole that the degeneracy made available.

The hydrogen

n = 2

level fans into three components linear in the field:

+ 3 e a_{0} E

, the unshifted

m = \pm 1

pair, and

- 3 e a_{0} E

Parabolic coordinates and the general level

The exact separation of the hydrogen Stark problem uses parabolic coordinates $ξ = r + z$ , $η = r - z$ , $ϕ$ , 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} + ∣ m ∣ + 1.

First-order perturbation theory in this basis gives the linear shift in closed form for every level,²

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 $(0, 0, \pm 1)$ , giving $n_{1} - n_{2} = + 1, - 1, 0, 0$ and shifts $+ 3 e a_{0} E, - 3 e a_{0} E, 0, 0$ , matching the degenerate-perturbation result. The permanent dipole of a general parabolic state is $⟨ d_{z} ⟩ = - \partial E^{(1)} / \partial E = - \frac{3}{2} n (n_{1} - n_{2}) e a_{0}$ , which grows as $n^{2}$ : high- $n$ hydrogen carries an enormous field-alignable dipole.

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.

The linear-in- $n$ shift and the parity mixing are two views of the same fact: the accidental $ℓ$ -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 $E_{a} = 5.142 \times 1 0^{11} V m^{- 1}$ ), is

U (z) = - \frac{1}{∣ z ∣} + 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_{saddle} = - 2 F .

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

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

The Coulomb well tilted by the field forms a saddle on the down-field side with barrier top at

U = - 2 F

; a level above it ionizes over the barrier, one just below tunnels through.

The $n^{- 4}$ scaling is the practical heart of Rydberg-atom detection. Converting to laboratory units, $F_{ion} = 3.21 \times 1 0^{10} n^{- 4} V m^{- 1}$ , so an $n = 30$ state ionizes near $4 \times 1 0^{4} V m^{- 1}$ , about $400 V cm^{- 1}$ , 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.

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.

The ionizing field falls as

n^{- 4}

; laboratory fields ionize states above

n \approx 15

, placing all Rydberg states within easy experimental reach.

The exact hydrogen ground-state polarizability $α_{1 s} = \frac{9}{2} a_{0}^{3}$ 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. ↩
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 $E^{(1)} = \frac{3}{2} n (n_{1} - n_{2}) e a_{0} E$ ; §9.4 for the saddle-point field-ionization threshold $F_{ion} = 1/16 n^{4}$ . See also Foot, Atomic Physics, Ch. 8, for field ionization of Rydberg states. ↩