University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.6

The linear Stark effect in n = 2: a splitting of ±3eaF

This is where hydrogen's accidental degeneracy stops being a curiosity and starts moving levels. The calculation is short: one integral, one 2 × 2 matrix, a splitting of 6eaF. The real lesson is knowing when to believe it, so we size it against the fine structure and the Lamb shift the field has to beat.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An electric field moves a non-degenerate level only at second order, because a state of definite parity has no dipole to turn towards the field; that is why the 1s shift grows as F². Hydrogen's n = 2 is the exception, and the reason is the Coulomb degeneracy of 2s with 2p. The perturbation eFz connects only states of opposite parity and equal m, so inside n = 2 its single matrix element is ⟨2s|z|2p₀⟩ = −3a, computed here from Unit 2's functions. Degenerate perturbation theory then diagonalises a 2 × 2 block.

The field selects (2s + 2p₀)/√2 and (2s − 2p₀)/√2, whose charge centroids sit 3a to either side of the proton, and moves them by −3eaF and +3eaF, while 2p₊₁ and 2p₋₁ stay where they were. The 7.6 D dipole belongs to those mixtures, not to 2s or 2p₀; degeneracy is what lets an arbitrarily weak field assemble it. So the model holds only while the Stark coupling dwarfs every splitting inside n = 2 and is dwarfed by the gap to other shells. The 45 μeV fine structure is overcome near 3 kV/cm.

Below a floor near 0.5 kV/cm, set by the 4.4 μeV Lamb shift, the 2S₁/₂ shift is quadratic, yet its small 2P admixture still cuts the metastable lifetime from 0.12 s to microseconds at 10 V/cm. At the top, second-order mixing with other shells reaches 1% of the linear shift only near 1.8 MV/cm.

Simple definition: The linear Stark effect in hydrogen n = 2 is the first-order shift, −3eaF for (2s + 2p₀)/√2 and +3eaF for (2s − 2p₀)/√2, that a field F produces by mixing the degenerate 2s and 2p₀ states, while 2p₊₁ and 2p₋₁ stay unshifted.
Example: At F = 10 kV/cm, eaF = (5.29 × 10⁻¹¹ m)(1.0 × 10⁶ V m⁻¹)e = 52.9 μeV, so the two m = 0 states move by ∓159 μeV (∓38.4 GHz), 3.5 times the 45.4 μeV fine-structure interval, and end up 318 μeV apart.

The one matrix element that survives⟨2s|z|2p₀⟩ = −3a

Derived here: an angular factor 4π/3 times a radial integral of −72a⁴, over 32πa³. Equal m and opposite parity forbid every other element inside n = 2.

a = 5.29 × 10⁻¹¹ m, the Bohr radius; the sign follows the phases of ψ₂₀₀ and ψ₂₁₀ and cancels from every energy

First-order energies of the n = 2 levelE = E₂ ∓ 3eaF for (2s ± 2p₀)/√2, and E = E₂ for 2p₊₁ and 2p₋₁

Derived by diagonalising the 2 × 2 block with zeros on the diagonal and −3eaF off it. 3eaF/h = 3.84 GHz for every kV/cm.

e = 1.602 × 10⁻¹⁹ C, F in V m⁻¹, energies in J; E₂ = −3.40 eV, and eaF in eV equals a × F in V

The dipole the degeneracy permits|d| = 3ea = 2.54 × 10⁻²⁹ C m

Derived: −d⋅F reproduces ∓3eaF. Neither 2s nor 2p₀ has it, since each is a parity eigenstate; it exists only in their equal mixtures.

C m; 7.6 D with 1 D = 3.34 × 10⁻³⁰ C m, four times water's measured 1.85 D; ⟨z⟩ = ∓3a in the two good states

Spin shares the coupling between two partnersW₁ = √3 eaF to 2P₁/₂ and W₂ = √6 eaF to 2P₃/₂, with W₁² + W₂² = (3eaF)²

Derived: z acts only on the mₗ = 0 part of each 2P state. It sets both crossover fields and the quench rate once fine structure is on.

couplings from 2S₁/₂ at equal mⱼ = ±½, in J or μeV; √3 and √6 carry the Clebsch–Gordan weights ⅓ and ⅔ of mₗ = 0

Below the Lamb shift: a quadratic 2S₁/₂ shiftΔE(2S₁/₂) ≈ 3(eaF)²/L − 6(eaF)²/(Δ − L)

Derived at second order: 2P₁/₂ below pushes 2S₁/₂ up, 2P₃/₂ above pushes it down. At 100 V/cm the net is +36.5 MHz, not ±384 MHz.

L = 4.375 μeV (1057.8 MHz, measured) and Δ = 45.36 μeV (10.969 GHz); valid while √3 eaF ≪ L, so F well below 0.48 kV/cm

Field quenching of metastable 2sΓ(F) = [3(eaF)²/L² + 6(eaF)²/(Δ − L)²] Γ₂ₚ

Derived from the first-order 2P admixture: at 10 V/cm the 2s lifetime falls from 0.122 s (quoted, two-photon decay) to 3.6 μs.

Γ₂ₚ = 1/(1.596 ns) = 6.27 × 10⁸ s⁻¹, quoted ahead of Unit 13; Γ(F) = 2.81 × 10³ s⁻¹ × (F in V/cm)²

Symmetry leaves one matrix element in n = 2

Put the field along +z. The electron, charge −e, sits in the potential −Fz, so H′ = eFz. Without spin, n = 2 holds four degenerate states, 2s, 2p₋₁, 2p₀ and 2p₊₁, and degenerate perturbation theory needs the 4 × 4 matrix of H′ among them. Two symmetries empty almost all of it. z commutes with L_z, so ⟨n l m|z|n l′ m′⟩ = 0 unless m′ = m. z is odd under parity while ψₙₗₘ has parity (−1)ˡ, so an element vanishes unless l + l′ is odd, which kills every diagonal entry, ⟨2s|z|2s⟩ included. Only ⟨2s|z|2p₀⟩ and its conjugate survive. That is the difference from 1s in the previous topic: there the lone diagonal element was the whole first-order answer and parity made it zero; here the Coulomb l-degeneracy puts a state of opposite parity at the same energy, and an off-diagonal element becomes first order. One more symmetry: reflection in any plane containing F leaves eFz unchanged but turns m into −m, so a Stark level depends on |m| only, unlike a Zeeman level.

Computing ⟨2s|z|2p₀⟩ from the n = 2 functions

Quote Unit 2's real, normalised functions: ψ₂₀₀ = (32πa³)^(−1/2) (2 − r/a) e^(−r/2a) and ψ₂₁₀ = (32πa³)^(−1/2) (r/a) e^(−r/2a) cos θ. With z = r cos θ and ρ = r/a the integral separates. The angular part is ∫cos²θ dΩ = 4π/3. The radial part is ∫(2 − r/a)(r/a) r e^(−r/a) r² dr = a⁴ ∫(2 − ρ)ρ⁴ e^(−ρ) dρ = a⁴(2 × 4! − 5!) = −72a⁴. Assembled, ⟨2s|z|2p₀⟩ = (4π/3)(−72a⁴)/(32πa³) = −3a. Three checks. Dimensions: a length, as a matrix element of z must be. Size: 2s and 2p have mean radii 6a and 5a, so a displacement of 3a is a large but sensible fraction of the orbit. Sign: 2 − ρ changes sign at the 2s node, ρ = 2, and the outer lobe dominates, −72.54a⁴ against +0.54a⁴ from inside the node. Flipping the phase of either function flips the sign, but every energy below depends only on the eigenvalues ±3a of this block, never on the sign.

Diagonalising: two states move and two stay

In the pair (2s, 2p₀) the block has zeros on the diagonal and −3eaF off it, so its eigenvalues are ∓3eaF with eigenvectors (2s ± 2p₀)/√2, while 2p₊₁ and 2p₋₁ have no partner and stay at E₂ = −3.40 eV to first order. The lower state (2s + 2p₀)/√2 has ⟨z⟩ = ⟨2s|z|2p₀⟩ = −3a: its charge sits on the −z side, where the force −eF pushes the electron, its dipole d = −e⟨r⟩ = +3ea ẑ points along F, and −d⋅F = −3eaF reproduces the eigenvalue. The size is 3ea = 2.54 × 10⁻²⁹ C m = 7.6 D, four times the measured 1.85 D of a water molecule. In frequency, 3eaF/h = 3.84 GHz for every kV/cm. Spin only doubles each level, giving 2 + 2 + 4 states. A quoted cross-check: solving the Coulomb problem in parabolic coordinates gives E = Eₙ + (3/2)n(n₁ − n₂)eaF for any n, and n = 2 allows n₁ − n₂ = −1, 0 and +1, which is ∓3eaF and 0 again. Limit: F → 0 returns the degenerate level.

Fine structure decides when ±3eaF is the answer

The spinless result assumes n = 2 is exactly degenerate, and Unit 8 showed it is not: 2P₃/₂ lies Δ = 45.36 μeV (10.969 GHz, measured) above 2P₁/₂. The Stark coupling must dwarf Δ before ±3eaF describes the atom, and 3eaF = Δ at 2.86 kV/cm. In powers of α: eaF = (F/Fₐₜ)α²mc² with Fₐₜ = e/(4πε₀a²) = 5.14 × 10¹¹ V m⁻¹, while Dirac's Δ = α⁴mc²/32, so the crossover is F = α²Fₐₜ/96 = 2.85 kV/cm, 0.2% below the value from the measured interval. With spin, 2S₁/₂ couples to two partners of the same mⱼ, and the Clebsch–Gordan weights give √3 eaF to 2P₁/₂ and √6 eaF to 2P₃/₂, which the figure diagonalises exactly. At high field the outer levels tend to ±3eaF plus a common offset above 2P₁/₂, half the Lamb shift plus Δ/3, about 17 μeV, so their separation exceeds 6eaF by only 1.2% at 10 kV/cm. At the far end, mixing with other n, quoted from the parabolic solution as −84F² hartree with F in atomic units, reaches 1% of the linear shift only near 1.8 MV/cm.

Below the Lamb shift the 2s shift turns quadratic

QED lifts the last degeneracy: 2S₁/₂ lies L = 1057.8 MHz = 4.375 μeV above 2P₁/₂, a measured splitting of order α⁵mc² ln(1/α), quoted. Near zero field 2S₁/₂ is a two-level problem with 2P₁/₂: levels L apart coupled by W₁ = √3 eaF, with energies ±√(L²/4 + W₁²) about their midpoint. For W₁ ≫ L that is linear in F; for W₁ ≪ L it expands to L/2 + W₁²/L, a quadratic shift of 2S₁/₂ upward, away from the level below. W₁ = L at 0.48 kV/cm. At 100 V/cm, second order gives +W₁²/L − W₂²/(Δ − L) = +0.192 − 0.041 = +0.151 μeV, or +36.5 MHz, and the exact 3 × 3 answer is +35.0 MHz; the spinless formula would have promised ±384 MHz. The shift is quadratic like 1s's but upward, a negative polarisability, because the dominant neighbour lies below. In real hydrogen the lowest level's dipole is 0.07ea at 10 V/cm, 1.8ea at 0.5 kV/cm and 2.9ea at 5 kV/cm: the full ±3ea needs the Lamb shift and the fine structure both beaten.

The admixture that quenches metastable 2s

In zero field 2S₁/₂ has no electric-dipole route to 1s and lives 0.122 s by two-photon decay (quoted). In a field, first-order perturbation theory mixes in 2P₁/₂ with amplitude η₁ = W₁/L and 2P₃/₂ with η₂ = W₂/(Δ − L), and the dressed state inherits their decay: Γ(F) = (η₁² + η₂²)Γ₂ₚ, with the 1.596 ns 2P lifetime quoted ahead of Unit 13. Numerically Γ(F) = 2.81 × 10³ s⁻¹ × (F in V/cm)², 98% of it through 2P₁/₂. At 10 V/cm that is 2.8 × 10⁵ s⁻¹: the lifetime falls from 0.122 s to 3.6 μs, a factor of 3.4 × 10⁴, while the level itself has moved by only 0.37 MHz. Quenching matches the natural decay at just 0.054 V/cm, which is why metastable-hydrogen experiments shield stray fields. Checks: η is an energy over an energy; Γ ∝ F², as a rate built from a first-order amplitude must be; and ħΓ₂ₚ = 0.41 μeV is small against L, so the neglected width changes Γ(F) by 0.2%. The spinless coupling 3eaF would have made the rate three times too large.

Change one variable at a time

Make the relationship visible.

Interactive model

Field F as log₁₀(F in kV/cm)0.00

Fine structure Δ, off or on45.36 μeV

Lamb shift L, off or on4.375 μeV

Set both switches to 0 and the curves sit on +3, 0 and −3 at every field: the spinless ±3eaF. Fine structure alone leaves ±1.73 = ±√3 at low field, the 2S₁/₂–2P₁/₂ pair; the Lamb shift pulls even that towards zero below about 0.5 kV/cm, roughly halving it by 0.14 kV/cm, and ±3 returns only above about 5 kV/cm.

FIELD F1000 V/cm

STARK SHIFT 3eaF15.88 μeV

3eaF / Δ0.35

√3 eaF / L2.09

Live interpretationFIELD F: 1000 V/cm. STARK SHIFT 3eaF: 15.88 μeV. 3eaF / Δ: 0.35. √3 eaF / L: 2.09

Catch the common trap

Explain before calculating.

Misconception check

It is tempting to read ±3eaF as 2s moving one way and 2p₀ the other, or as proof that excited hydrogen carries a permanent electric dipole. Both readings are wrong. 2s and 2p₀ are parity eigenstates, so ⟨2s|z|2s⟩ = ⟨2p₀|z|2p₀⟩ = 0: neither has a dipole, and neither is a stationary state in the field. The states that shift are the equal mixtures (2s ± 2p₀)/√2, whose charge sits 3a off-centre, and the field can choose them at first order only because 2s and 2p₀ share an energy. Break that tie and the linear shift goes with it. Real hydrogen's Lamb shift puts 2S₁/₂ 4.375 μeV above 2P₁/₂, so below about 0.5 kV/cm the 2S₁/₂ shift is quadratic: +36.5 MHz at 100 V/cm, not ±384 MHz. A permanent moment would shift linearly at any field, however weak; this one is induced, and degeneracy is what makes the inducing cheap.

Hydrogen in n = 2 sits in a field F = 20 kV/cm along z. There 3eaF = 318 μeV, seven times the fine-structure interval, so spin can be ignored. Which describes the first-order Stark pattern of the level?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySuppose microwave spectroscopy of n = 2 hydrogen in a uniform field finds the two outermost Stark levels 92.1 GHz apart. Treating the atom without spin, find F, and decide whether neglecting the 45.4 μeV fine structure was reasonable.

The outermost levels are (2s + 2p₀)/√2 at E₂ − 3eaF and (2s − 2p₀)/√2 at E₂ + 3eaF, so they are 6eaF apart.
Convert: 6eaF = h × 92.1 GHz = (4.1357 × 10⁻¹⁵ eV s)(9.21 × 10¹⁰ s⁻¹) = 3.809 × 10⁻⁴ eV, so eaF = 6.348 × 10⁻⁵ eV = 63.5 μeV.
eaF in eV equals a × F in volts, so F = 6.348 × 10⁻⁵ V ÷ 5.292 × 10⁻¹¹ m = 1.200 × 10⁶ V m⁻¹ = 12.0 kV/cm. Dimensions: C × m × V m⁻¹ = J, an energy.
Regime: 3eaF = 190 μeV is 4.2 times the 45.4 μeV fine structure, so the spinless pattern is a fair start. Fine structure moves both outer levels by nearly the same amount, centring the pair 17 μeV above 2P₁/₂, so it cancels from their separation; the exact 3 × 3 diagonalisation puts them 0.9% further apart than 6eaF at this field.

AnswerF = 12.0 kV/cm (1.20 × 10⁶ V m⁻¹) in the spinless model; including fine structure lowers it by 0.9%, to 11.9 kV/cm.

MediumA hydrogen atom in 2S₁/₂ sits in a stray field F = 50 V/cm. Using the spin-resolved couplings √3 eaF to 2P₁/₂, which lies L = 4.375 μeV below, and √6 eaF to 2P₃/₂, which lies 40.99 μeV above, find the shift of 2S₁/₂ in MHz and how much the field changes the 2S₁/₂–2P₁/₂ interval, and compare with what the spinless ±3eaF would predict.

eaF = (5.292 × 10⁻¹¹ m)(5.00 × 10³ V m⁻¹)e = 0.2646 μeV, so W₁ = √3 eaF = 0.458 μeV and W₂ = √6 eaF = 0.648 μeV. W₁/L = 0.105, small enough for second order.
The level below pushes 2S₁/₂ up by W₁²/L = 0.2100/4.375 = 0.04801 μeV; the level above pushes it down by W₂²/(Δ − L) = 0.4200/40.99 = 0.01025 μeV. Net: +0.03776 μeV, which with 1 μeV = 241.8 MHz is +9.13 MHz. Exact diagonalisation of the 3 × 3 block gives +9.03 MHz; the 1.1% gap is the next order, of relative size (W₁/L)² = 0.011.
z cannot link 2P₁/₂ to 2P₃/₂, which has the same parity, so the only partner of 2P₁/₂ is 2S₁/₂ and it is pushed down by the same W₁²/L = 0.04801 μeV. The 2S₁/₂–2P₁/₂ interval therefore widens by 0.04801 + 0.03776 = 0.08577 μeV = 20.7 MHz, 2.0% of the 1057.8 MHz Lamb shift; the exact block gives 20.5 MHz.
Checks: (eaF)²/L is an energy, and both shifts grow as F², so each is a quarter of its value at 100 V/cm, where the 2S₁/₂ shift is +36.5 MHz. The spinless formula would claim ±3eaF = ±0.794 μeV = ±192 MHz, twenty times larger and of either sign.

AnswerΔE(2S₁/₂) ≈ +0.03776 μeV = +9.13 MHz at second order (+9.03 MHz exactly), and the 2S₁/₂–2P₁/₂ interval widens by 20.7 MHz, against ±192 MHz from the spinless ±3eaF.

HardMetastable H(2S₁/₂) atoms at 8.0 × 10³ m s⁻¹ cross a 2.0 cm gap between plates. Find the quench rate as a function of F, the field that quenches 90% of the atoms in the gap, and the field at which quenching merely equals the 8.23 s⁻¹ two-photon decay. Take Γ₂ₚ = 1/(1.596 ns), L = 4.375 μeV and Δ − L = 40.99 μeV.

First order mixes in 2P₁/₂ with amplitude η₁ = √3 eaF/L and 2P₃/₂ with η₂ = √6 eaF/(Δ − L). Both 2P levels decay to 1s at Γ₂ₚ = 6.266 × 10⁸ s⁻¹, so Γ(F) = (η₁² + η₂²)Γ₂ₚ.
At 1 V/cm, eaF = 5.292 × 10⁻³ μeV and (eaF)² = 2.800 × 10⁻⁵ μeV², so η₁² + η₂² = 2.800 × 10⁻⁵ × (3/19.14 + 6/1680) = 4.489 × 10⁻⁶. Then Γ(F) = 2.81 × 10³ s⁻¹ × (F in V/cm)², 2.2% of it through 2P₃/₂.
The transit time is 0.020 m ÷ 8.0 × 10³ m s⁻¹ = 2.5 μs. Survival e^(−Γt) = 0.10 needs Γ = ln 10 ÷ 2.5 μs = 9.21 × 10⁵ s⁻¹, so F² = 9.21 × 10⁵ ÷ 2813 = 327 and F = 18.1 V/cm.
Validity: at 18.1 V/cm, W₁/L = 0.038, and ħΓ₂ₚ/2 = 0.21 μeV is small against L, a 0.2% correction, so the first-order admixture holds. Dimensions: η is an energy ratio, and Γ(F) takes s⁻¹ from Γ₂ₚ.
Natural decay is matched when 2813 F² = 8.23, so F² = 2.93 × 10⁻³ and F = 0.054 V/cm. A stray field of 0.3 V/cm already quenches at 253 s⁻¹, thirty times the natural rate.

AnswerΓ(F) = 2.81 × 10³ s⁻¹ × (F in V/cm)²; F ≈ 18 V/cm quenches 90% in the gap (quench lifetime 1.1 μs); quenching equals the natural decay at only 0.054 V/cm.