University Quantum Mechanics II · Hydrogen in External Fields and Hyperfine Structure · 9.5

The ground-state Stark effect and the polarisability

Here a second-order sum meets a real atom. Parity removes the first order, and the sum that replaces it runs over infinitely many p states and a continuum. You will not evaluate it term by term; you will trap it, between 2.96 and 5.33 in atomic units, tightly enough to trust the quoted 9/2 and to test it against a measurement.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put hydrogen 1s in a uniform field F along z. The electron's energy gains H′ = eFz, and the first-order shift eF⟨1s|z|1s⟩ vanishes because |ψ₁₀₀|² is even and z is odd: a state of definite parity carries no permanent dipole, so the field has nothing to turn. It must first stretch the atom, and that costs a second order. With E⁽²⁾ quoted from Unit 6, the shift is ΔE = −½αₚₒₗF², where αₚₒₗ is a sum over every np₀ state and the whole continuum, each term positive.

The sum cannot be done term by term, but it can be trapped. Shrinking every denominator to E₂ − E₁ and summing the numerators by closure caps αₚₒₗ at (16/3)(4πε₀)a³; the 2p term alone, derived here, is a floor at 2.96 in the same units. The exact (9/2)(4πε₀)a³, quoted from the Dalgarno–Lewis method, sits inside, and at 100 kV/cm it lowers 1s by 5.60 MHz, under two parts in 10⁹ of the binding energy. The model is nonrelativistic: the Dirac equation corrects it at relative order α², and proton recoil by 3mₑ/mₚ.

Where it stops: eFz is unbounded below, so 1s is strictly a resonance that tunnels out on the downfield side, and the series in F is asymptotic. At 100 kV/cm the tunnelling factor is e^{(−3.4 × 10⁴)} and the quadratic law is all there is; it fails only as F nears Fₐₜ/16, about 3 × 10¹⁰ V m⁻¹, where the barrier sinks to the level.

Simple definition: The static polarisability αₚₒₗ of hydrogen 1s is the constant in d = αₚₒₗF, the dipole a weak uniform field induces; because 1s has no permanent dipole, it fixes the whole leading shift, ΔE = −½αₚₒₗF².
Example: With αₚₒₗ = (9/2)(4πε₀)a³ = 7.42 × 10⁻⁴¹ C m² V⁻¹, a field of 100 kV/cm induces d = 7.42 × 10⁻³⁴ C m, the electron cloud's centre pulled 4.6 fm off the proton, and lowers 1s by 3.71 × 10⁻²⁷ J, which is h × 5.60 MHz.

First order vanishes by parityE⁽¹⁾ = eF⟨1s|z|1s⟩ = 0

Derived: |ψ₁₀₀|² is even under r → −r and z is odd, so the integral cancels. No permanent dipole, so the shift starts at F².

e = 1.602 × 10⁻¹⁹ C, F in V m⁻¹, z in m; H′ = eFz for the electron, charge −e, in the potential −Fz

The polarisability as a sum over statesαₚₒₗ = 2e² Σₘ |⟨m|z|1s⟩|² / (Eₘ − E₁), defined by ΔE = −½αₚₒₗF²

Derived from the quoted E⁽²⁾. Every denominator is positive, so every term, continuum included, adds to αₚₒₗ and lowers 1s.

αₚₒₗ in C m² V⁻¹; m runs over np₀ for n ≥ 2 and the εp continuum, the only states z reaches from 1s

Closure caps it from aboveαₚₒₗ ≤ 2e²⟨1s|z²|1s⟩/(E₂ − E₁) = (16/3)(4πε₀)a³

Derived: shrinking each denominator to the smallest one can only enlarge each term, and closure then sums the numerators exactly: 5.33 in units of (4πε₀)a³.

⟨z²⟩ = a² with a = 5.29 × 10⁻¹¹ m; E₂ − E₁ = 10.20 eV; (4πε₀)a³ = 1.649 × 10⁻⁴¹ C m² V⁻¹

The 2p term is a floorαₚₒₗ ≥ 2e²|⟨2p₀|z|1s⟩|²/(E₂ − E₁) = (2¹⁹/3¹¹)(4πε₀)a³ = 2.96(4πε₀)a³

Derived here from the n = 1 and n = 2 functions. 2p holds 55% of ⟨z²⟩ and, with the smallest denominator, 66% of αₚₒₗ; the rest is in higher p states and the continuum.

|⟨2p₀|z|1s⟩| = (128√2/243)a = 0.745a; dropping positive terms can only lower the sum

The exact valueαₚₒₗ = (9/2)(4πε₀)a³ = 7.42 × 10⁻⁴¹ C m² V⁻¹

Quoted from the Dalgarno–Lewis method, which trades the sum for one differential equation. It sits 52% above the floor and 16% below the cap.

αₚₒₗ/(4πε₀) = 0.667 × 10⁻³⁰ m³, close to the atom's own volume (4/3)πa³ = 0.621 × 10⁻³⁰ m³

The shift in laboratory unitsΔE/h = −5.60 MHz × (F / 100 kV cm⁻¹)²

Uses the quoted 9/2. Quadratic: 1 MV/cm gives −560 MHz, still only 1.7 × 10⁻⁷ of the 13.6 eV binding, so second order is ample in any laboratory field.

h = 6.626 × 10⁻³⁴ J s; in atomic units ΔE = −(9/4)Eₕ(F/Fₐₜ)², with Fₐₜ = 5.14 × 10¹¹ V m⁻¹

Parity leaves nothing at first order

Put the field along +z. The electron, charge −e, sits in the potential −Fz, so H′ = eFz. Set the electron and proton spins aside, since z acts on neither; then 1s is non-degenerate and its first-order shift is the single diagonal element eF⟨1s|z|1s⟩. Under r → −r, |ψ₁₀₀|² is unchanged while z changes sign, so the integral cancels and E⁽¹⁾ = 0. Physically, a state of definite parity has ⟨r⟩ = 0: there is no permanent dipole for −d⋅F to act on. A second argument shows the whole shift is even in F. Reflecting the atom in the plane z = 0 turns the problem with field +F into the problem with −F and maps 1s onto itself, so E(F) = E(−F) and the series starts at F². Compare the next topic, where 2s and 2p₀ share an energy and an off-diagonal element of z becomes first order. Here the nearest state of opposite parity, 2p₀, lies 10.2 eV above, and the field must pay to mix it in.

The second-order sum, and why every term has one sign

Quote Unit 6 in one line each: E⁽²⁾ = Σₘ |⟨m|H′|n⟩|²/(Eₙ − Eₘ), and for a ground state every denominator is negative, so E⁽²⁾ ≤ 0. With H′ = eFz this is ΔE = −e²F² Σₘ |⟨m|z|1s⟩|²/(Eₘ − E₁), which is −½αₚₒₗF² with αₚₒₗ the sum in the second formula card. Which states enter? z = r cos θ carries l = 1 and m = 0, so from 1s it reaches only the np₀ states, n ≥ 2, and the εp continuum of ionised electrons with l = 1. The continuum lies above E₁ like every bound level, so its terms are positive too and nothing cancels: αₚₒₗ is positive and 1s must go down. αₚₒₗ is also the atom's response coefficient. By the Feynman–Hellmann theorem dE/dF = e⟨z⟩, so the induced dipole d = −e⟨z⟩ equals αₚₒₗF and points along F. Dimensions: e²z²/E is C² m² J⁻¹ = C m² V⁻¹, the unit of d/F. Limit: as F → 0 the shift vanishes as F², with zero slope, as the reflection argument demanded.

The 2p term, derived, and why it is not enough

The first term needs one integral. Quote ψ₁₀₀ = (πa³)^(−1/2) e^(−r/a) and ψ₂₁₀ = (32πa³)^(−1/2) (r/a) e^(−r/2a) cos θ. With z = r cos θ the angular integral is 4π/3 and the radial one is (1/a)∫r⁴e^(−3r/2a) dr = 4!(2a/3)⁵/a, so ⟨2p₀|z|1s⟩ = (4π/3)(768a⁴/243)/(√32 πa³) = (128√2/243)a = 0.745a. Squared, 0.555a². Since E₂ − E₁ = (3/8)e²/(4πε₀a) = 10.20 eV, the term is 2e²(0.555a²)/(E₂ − E₁) = (16/3)(0.555)(4πε₀)a³ = 2.96(4πε₀)a³. Every other term is positive, so this is a floor. The rest fall fast, 3p adding 0.400, 4p 0.132 and 5p 0.060, yet the whole bound series converges to only 3.66 (quoted, from tabulated matrix elements). The remaining 0.84, 19% of αₚₒₗ, lives in the continuum, so no truncation to bound states, however long, reaches the answer. The numerators say the same: 2p carries 55.5% of ⟨z²⟩, the bound states together 71.7% (also tabulated), the continuum 28.3%.

Closure caps the sum from above

The numerators can be summed exactly even though the terms cannot. Completeness, continuum included, gives Σₘ |⟨m|z|1s⟩|² = ⟨1s|z²|1s⟩ − ⟨1s|z|1s⟩² = ⟨z²⟩, and by spherical symmetry ⟨z²⟩ = ⟨r²⟩/3 = a², because ⟨r²⟩ = 3a² in 1s. The denominators cannot be summed, but every one is at least E₂ − E₁. Replace each by that smallest value and every positive term can only grow, so αₚₒₗ ≤ 2e²a²/(E₂ − E₁) = (16/3)(4πε₀)a³ = 5.33(4πε₀)a³. The cap is loose for a reason: 28% of the numerator weight sits in the continuum, at excitation energies of 13.6 eV and up, and closure charges it only 10.2 eV. Tighten it by keeping the first terms exactly and applying closure to what is left at the next gap. Keeping 2p leaves 0.445a² to charge at E₃ − E₁ = 12.09 eV, which adds at most 2.00, so 2.96 ≤ αₚₒₗ ≤ 4.96 in units of (4πε₀)a³. Keeping six p terms narrows the bracket to 3.61 to 4.82, and there it stalls, because no bound term reaches the continuum.

The exact 9/2, checked by size, by α and by experiment

The Dalgarno–Lewis method trades the sum for one inhomogeneous equation for the first-order state, solved for 1s by a polynomial in r times zψ₁₀₀; it returns αₚₒₗ = (9/2)(4πε₀)a³, quoted here. That lies inside every bracket, and it corresponds to a mean excitation energy 2e²a²/αₚₒₗ = 12.09 eV, which happens to equal E₃ − E₁. In SI, αₚₒₗ = 7.42 × 10⁻⁴¹ C m² V⁻¹, and αₚₒₗ/(4πε₀) = 0.667 × 10⁻³⁰ m³, close to the atom's own volume. In atomic units no power of the fine-structure constant appears: the Dirac equation multiplies the result by 1 − (28/27)α² = 1 − 5.5 × 10⁻⁵ (quoted), and proton recoil, which rescales a to a(1 + mₑ/mₚ), by 1.0016. The measurement is optical: shock-tube interferometry at 587 nm (Marlow and Bershader, 1964) gave (4.61 ± 0.07)(4πε₀)a³. For photons of ħω = 2.11 eV, a fifth of the 10.2 eV gap, each 1/(Eₘ − E₁) becomes (Eₘ − E₁)/[(Eₘ − E₁)² − (ħω)²], and the sum gives 4.67 (quoted), within one standard deviation; the static 4.50 lies 1.6 standard deviations below.

No state is bound in a field, and the series is asymptotic

On the downfield side, z → −∞, the potential energy −e²/(4πε₀r) + eFz falls without limit, so every bound state is strictly a resonance that can tunnel out through a barrier. Along that axis the barrier top sits at |z| = √(e/(4πε₀F)), and it sinks to E₁ when F = Fₐₜ/16 = 3.2 × 10¹⁰ V m⁻¹, with Fₐₜ = e/(4πε₀a²) = 5.14 × 10¹¹ V m⁻¹: a classical estimate of where the atom ionises promptly. Below it, the tunnelling rate carries the factor exp(−2Fₐₜ/3F), quoted from a WKB calculation, which at 100 kV/cm is e^{(−3.4 × 10⁴)}: the width is real and unmeasurable. That factor vanishes faster than any power of F, so no order of perturbation theory sees it, and its existence means E(F) is not analytic at F = 0. The series −(9/4)Eₕ(F/Fₐₜ)² − (3555/64)Eₕ(F/Fₐₜ)⁴ − …, the second coefficient quoted, is therefore asymptotic, as in Unit 6. At 100 kV/cm its second term is 9 × 10⁻⁹ of the first, so the quadratic law is all a laboratory sees; by Fₐₜ/16 the ratio is 0.1 and the series is failing.

Change one variable at a time

Make the relationship visible.

Interactive model

Field F200 kV/cm

p terms kept exactly, k1

Start at k = 0: closure alone caps αₚₒₗ at 5.33 and nothing props it up. Keep 2p and the floor leaps to 2.96 while the cap drops to 4.96. Keep more and both stall, near 3.61 and 4.82 at k = 6, because 19% of αₚₒₗ lies in the continuum, which no bound term reaches.

FLOOR ON αₚₒₗ2.96 (4πε₀)a³

CAP ON αₚₒₗ4.96 (4πε₀)a³

SHIFT ΔE/h, EXACT 9/2-22.39 MHz

CLOUD OFFSET d/e9.26 fm

Live interpretationFLOOR ON αₚₒₗ: 2.96 (4πε₀)a³. CAP ON αₚₒₗ: 4.96 (4πε₀)a³. SHIFT ΔE/h, EXACT 9/2: −22.39 MHz. CLOUD OFFSET d/e: 9.26 fm

Catch the common trap

Explain before calculating.

A student bounds the 1s polarisability by replacing every energy denominator Eₘ − E₁ in the sum over states by E₂ − E₁ = 10.2 eV and summing the numerators by closure, which gives (16/3)(4πε₀)a³ = 5.33(4πε₀)a³. What does that number establish about the exact αₚₒₗ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA beam of ground-state hydrogen atoms passes between parallel plates 2.0 mm apart with 5.0 kV across them. Using the quoted αₚₒₗ = (9/2)(4πε₀)a³, find the field, the induced dipole moment, how far the centre of the electron cloud is pulled off the proton, and the Stark shift of 1s in MHz.

Field: F = V/d = 5.0 × 10³ V ÷ 2.0 × 10⁻³ m = 2.5 × 10⁶ V m⁻¹, or 25 kV/cm.
Polarisability in SI: (4πε₀)a³ = (1.113 × 10⁻¹⁰ F m⁻¹)(5.292 × 10⁻¹¹ m)³ = 1.649 × 10⁻⁴¹ C m² V⁻¹, so αₚₒₗ = 4.5 × 1.649 × 10⁻⁴¹ = 7.42 × 10⁻⁴¹ C m² V⁻¹.
Dipole: d = αₚₒₗF = (7.42 × 10⁻⁴¹)(2.5 × 10⁶) = 1.85 × 10⁻³⁴ C m. Dividing by e gives the offset of the cloud's centre, 1.16 × 10⁻¹⁵ m = 1.16 fm: 2.2 × 10⁻⁵ of a, and about 1.4 times the proton's measured 0.84 fm charge radius.
Shift: ΔE = −½dF = −½(1.85 × 10⁻³⁴ C m)(2.5 × 10⁶ V m⁻¹) = −2.32 × 10⁻²⁸ J, so ΔE/h = −2.32 × 10⁻²⁸ J ÷ 6.626 × 10⁻³⁴ J s = −3.50 × 10⁵ Hz.
Checks: C m × V m⁻¹ = J, an energy; and scaling the 100 kV/cm value, −5.60 MHz × (0.25)² = −0.350 MHz, agrees.

AnswerF = 2.5 × 10⁶ V m⁻¹; d = 1.85 × 10⁻³⁴ C m, a 1.16 fm offset of the electron cloud; ΔE/h = −0.350 MHz.

MediumTighten the bracket on αₚₒₗ by keeping the 2p and 3p terms exactly, given |⟨2p₀|z|1s⟩|² = 0.5549a² and |⟨3p₀|z|1s⟩|² = 0.0890a², and charging all the remaining numerator weight at the next gap, E₄ − E₁. Then bracket the 1s shift at 200 kV/cm and compare with the quoted exact value.

Gaps in units of e²/(4πε₀a) = 27.21 eV: E₂ − E₁ = 3/8 (10.20 eV), E₃ − E₁ = 4/9 (12.09 eV) and E₄ − E₁ = 15/32 (12.76 eV). In those units each term 2e²|z|²/(Eₘ − E₁) becomes 2(|z|²/a²) divided by the gap, in units of (4πε₀)a³.
Kept terms: 2 × 0.5549/(3/8) = 2.960 and 2 × 0.0890/(4/9) = 0.400, so the floor is 3.360.
Remainder: closure gives a total numerator of a², so 1 − 0.5549 − 0.0890 = 0.3561 of it is left, in 4p, higher p states and the continuum, all at gaps of at least 15/32. Charging it there adds at most 2 × 0.3561/(15/32) = 1.519, so the cap is 4.879.
Shift per unit of αₚₒₗ: ½(1.649 × 10⁻⁴¹ C m² V⁻¹)(2.0 × 10⁷ V m⁻¹)² = 3.298 × 10⁻²⁷ J, which is 4.977 MHz after dividing by h. So ΔE/h lies between −3.360 × 4.977 = −16.7 MHz and −4.879 × 4.977 = −24.3 MHz.
The quoted 9/2 gives −4.5 × 4.977 = −22.4 MHz, inside the bracket and three quarters of the way from floor to cap. Against 2p alone with closure alone, 2.96 to 5.33, the width has shrunk from 2.37 to 1.52, and the continuum is what keeps it from shrinking much further.

Answer3.36 ≤ αₚₒₗ/[(4πε₀)a³] ≤ 4.88, so −24.3 MHz ≤ ΔE/h ≤ −16.7 MHz at 200 kV/cm; the exact 9/2 gives −22.4 MHz.

HardThe quadratic law ΔE = −(9/4)Eₕ(F/Fₐₜ)² is the first term of a series whose next term is −(3555/64)Eₕ(F/Fₐₜ)⁴ (quoted). Find the field at which the classical barrier on the downfield side sinks to the unperturbed 1s energy, the ratio of the F⁴ term to the F² term there and at 100 kV/cm, and the size of the tunnelling factor exp(−2Fₐₜ/3F) at 100 kV/cm. Say where the quadratic Stark law can be trusted.

Along the downfield axis, z = −s with s positive, the electron's potential energy is V(s) = −e²/(4πε₀s) − eFs. dV/ds = 0 at s* = √(e/(4πε₀F)), where V* = −2√(e³F/(4πε₀)).
Set V* = E₁ = −e²/(8πε₀a) and square: 4e³F/(4πε₀) = e⁴/[4(4πε₀)²a²], so F = (1/16)e/(4πε₀a²) = Fₐₜ/16. With Fₐₜ = 5.142 × 10¹¹ V m⁻¹ that is 3.21 × 10¹⁰ V m⁻¹, or 321 MV/cm, and the barrier then sits at s* = 4a.
Series ratio: [(3555/64)/(9/4)](F/Fₐₜ)² = 24.69(F/Fₐₜ)². At Fₐₜ/16 it is 24.69/256 = 0.096: the quadratic shift, −(9/4)(1/256)Eₕ = −0.239 eV, is already being corrected by 10%. At 100 kV/cm, F/Fₐₜ = 1.945 × 10⁻⁵ and the ratio is 9.3 × 10⁻⁹.
Tunnelling at 100 kV/cm: 2Fₐₜ/3F = 2(5.142 × 10¹¹)/(3 × 10⁷) = 3.43 × 10⁴, so the factor is e^{(−34 300)}, about 10^{(−14 900)}. The level has a width in principle and none any experiment could see.
Reading it: the F² term is the whole shift to 1 part in 10⁸ at 100 kV/cm, and to 1 part in 10⁶ even at 1 MV/cm, and the level is sharp. The series still cannot converge, because a width of that exponential form is invisible at every order in F; it is asymptotic, and it breaks down as F approaches Fₐₜ/16, where classically the electron spills over the barrier.

AnswerF = Fₐₜ/16 ≈ 3.2 × 10¹⁰ V m⁻¹ (321 MV/cm), barrier at 4a; the F⁴/F² ratio is 0.096 there and 9.3 × 10⁻⁹ at 100 kV/cm, where the tunnelling factor is e^{(−3.4 × 10⁴)}. Trust the quadratic law at any laboratory field; it fails only near Fₐₜ/16.