Quadratic Stark Effect and Polarizability

Place a neutral atom in a uniform static field $\boldsymbol{\mathcal E}$ and the perturbation is the dipole interaction $\hat H'=e\boldsymbol{\mathcal E}\cdot\mathbf r=-\boldsymbol{\mathcal E}\cdot\hat{\mathbf d}$ , with $\hat{\mathbf d}=-e\mathbf r$ the electric dipole operator. For a non-degenerate state of definite parity the first-order shift $\langle n|\hat H'|n\rangle$ vanishes: $\mathbf r$ is odd, its expectation in a parity eigenstate is zero, and the atom has no permanent dipole to align. The leading response is therefore second order — quadratic in the field — and this is the generic case for atomic ground states, in sharp contrast to the degenerate hydrogen levels of Linear Stark Effect in Hydrogen. The second-order shift is conventionally written $\Delta E_n=-\tfrac12\alpha_n\mathcal{E}^2$ , which defines the static dipole polarizability $\alpha_n$ . Physically the field distorts the electron cloud into an induced dipole $\langle\mathbf d\rangle=\alpha\boldsymbol{\mathcal E}$ , and the energy of building that dipole up gradually carries the spring-energy factor of one-half. For a ground state $\alpha>0$ always, because every energy denominator in the defining sum is positive — an atom can only lower its energy by polarizing. Computing $\alpha$ means, in principle, summing over the entire discrete spectrum and the continuum — the central difficulty. Two routes bypass the explicit sum. The Dalgarno–Lewis method recasts the second-order sum as a single inhomogeneous differential equation for the first-order wavefunction correction, and for hydrogen it delivers the exact closed result $\alpha=\tfrac92 a_0^3$ with no truncation. Alternatively, replacing the variable energy denominator by its extremes gives rigorous closure bounds — an upper bound $\tfrac{16}{3}a_0^3$ that brackets the exact value even when the sum is intractable. The quadratic Stark effect is the prototype of linear response in quantum mechanics: the very same machinery — a parity-odd perturbation coupling states at second order — underlies the refractive index of dilute gases, van der Waals attraction, and the dielectric constant of matter. Every number below is machine-verified: the vanishing first-order term, the exact $\tfrac92 a_0^3$ (with the Dalgarno–Lewis residual proven to vanish symbolically), and the closure bound $\tfrac{16}{3}a_0^3$ .

Key takeaways

A non-degenerate atomic state with no permanent dipole has a vanishing first-order Stark shift; the leading effect is the quadratic $\Delta E=-\tfrac12\alpha\mathcal{E}^2$ .
The constant $\alpha$ is the static dipole polarizability, positive for the ground state, built from a sum over virtual dipole transitions to all other states.
The shift is always negative for the ground state: every denominator $E_0-E_m<0$ , so the field-induced polarization lowers the energy.
For hydrogen $\alpha=\tfrac92 a_0^3$ (Gaussian units) exactly, from Dalgarno–Lewis or by closing the perturbation sum — no infinite sum needed.
The closure (sum-rule) method brackets it: $\alpha\le\tfrac{16}{3}a_0^3\approx5.33\,a_0^3$ , only $\sim18.5\%$ above the exact $4.5\,a_0^3$ .
Polarizability bridges perturbation theory to measurables: induced dipole, refractive index, van der Waals $C_6$ , dielectric constant.
Degenerate or parity-mixed states (hydrogen $n\ge2$ ) instead show a linear Stark effect — a different article.
Key idea: a non-degenerate parity eigenstate has no first-order Stark shift; the leading response $-\tfrac12\alpha\mathcal{E}^2$ defines the polarizability $\alpha$ as a sum over virtual dipole transitions, exactly summable for hydrogen to $\tfrac92 a_0^3$ .

The core insight

A non-degenerate parity eigenstate has no first-order Stark shift; the leading response is the second-order shift $-\tfrac12\alpha\mathcal{E}^2$ , which defines the polarizability $\alpha$ as a sum over virtual dipole transitions to all other states — and for hydrogen it can be summed exactly to $\tfrac92 a_0^3$ .

Where this lives today

Polarizability is the microscopic constant behind a surprising amount of macroscopic physics, and the quadratic Stark shift is how it is measured. The dielectric constant and refractive index of a dilute gas follow directly from $\alpha$ through the Clausius–Mossotti relation $P=N\alpha\mathcal{E}_{\text{loc}}$ , making this second-order coefficient a bulk optical property (the EM-track owner develops the macroscopic side). The van der Waals attraction $-C_6/R^6$ between neutral atoms is correlated second-order dipole fluctuation — the same machinery at two atoms — with the static $\alpha$ computed here the zero-frequency limit of the dynamic polarizability $\alpha(\omega)$ that enters the London dispersion formula (owned by the many-body track). Most sharply, the quadratic Stark effect now sets the accuracy of the best clocks on Earth: thermal radiation is an oscillating field, so the blackbody-radiation shift of a clock transition is proportional to the differential polarizability of the two clock states, and evaluating it is essential to optical-clock accuracy at the $10^{-18}$ level. And because $\alpha\sim n^7$ , Rydberg atoms are polarizable enough to serve as microvolt-per-centimetre electrometers and as the interacting qubits of neutral-atom quantum processors, where the field-induced dipole is the engineered interaction — the humble induced dipole of this article, dialed up seven powers of $n$ .

The physical picture

Model the electron cloud as a soft ball of charge tethered to the nucleus by a spring. A field pulls the electrons one way and the nucleus the other, stretching the atom into an induced dipole $\mathbf d=\alpha\boldsymbol{\mathcal E}$ . Because you build that dipole up gradually as the field ramps from zero, the stored energy is $-\tfrac12\alpha\mathcal{E}^2$ — the same factor of one-half as the energy $\tfrac12 kx^2$ of a stretched spring, not the $-\mathbf p\cdot\boldsymbol{\mathcal E}$ of a permanent dipole.

Figure

Induced dipole: a symmetric cloud has no dipole until the field distorts it, and the response energy is quadratic because the dipole itself grows with the field

The sign is forced. For the ground state every excitation energy $E_m-E_0>0$ , so each term in the polarizability sum is positive, $\alpha>0$ , and the energy goes down. There is no way to make a ground-state atom dislike a weak field — polarizing always pays. The size of $\alpha$ is set by how easily the field can virtually promote the electron to nearby opposite-parity states: loosely bound Rydberg atoms, with huge matrix elements and tiny gaps, have polarizabilities scaling as $n^7$ and can be $10^{12}$ times more polarizable than a ground-state atom.

Figure

Second order as a round trip: the field cannot shift a symmetric state directly (first order = 0), but it can via virtual excursions |0⟩ → |m⟩ → |0⟩ to opposite-parity states

Mathematically the two facts are complementary. The vanishing first-order term is a parity statement: the field cannot directly shift a symmetric state. The surviving second-order term is a closure-over-virtual-states statement: it can shift it through round-trip excitations $|0\rangle\to|m\rangle\to|0\rangle$ into the opposite-parity $p$ -states the dipole operator reaches.

From vanishing first order to the polarizability

The perturbation is the dipole coupling, which we align with $\hat z$ :

\hat H' \;=\; e\,\boldsymbol{\mathcal E}\cdot\mathbf r \;=\; -\boldsymbol{\mathcal E}\cdot\hat{\mathbf d},\qquad \hat{\mathbf d}=-e\mathbf r,\qquad \hat H'=e\mathcal{E}z\ \ (\boldsymbol{\mathcal E}=\mathcal{E}\hat z) \qquad (1)

where: $\hat{\mathbf d}=-e\mathbf r$ is the electron's dipole operator and $e>0$ . Read physically: the field couples to the atom only through its dipole, and a symmetric atom has none to offer at zeroth order — which is exactly why the first correction vanishes. That vanishing is a one-line parity argument:

E_n^{(1)}=\langle n|\hat H'|n\rangle = e\mathcal{E}\langle n|z|n\rangle = 0 \qquad (2)

where: $|n\rangle$ is a non-degenerate parity eigenstate, so $|\psi_n|^2$ is even, $z$ is odd, and the integral of an even density times an odd coordinate is zero. No permanent dipole, no linear shift. The leading term is second order, and specializing the standard second-order formula to $\hat H'=e\mathcal{E}z$ isolates the polarizability:

E_n^{(2)} = \sum_{m\neq n}\frac{|\langle m|\hat H'|n\rangle|^2}{E_n^{(0)}-E_m^{(0)}} \;\equiv\; -\tfrac12\,\alpha_n\,\mathcal{E}^2 \qquad (3)

where: the sum runs over all other eigenstates $|m\rangle$ (discrete and continuum) and $E_n^{(0)}-E_m^{(0)}$ is the energy denominator. Consume the second-order machinery — the sign theorem that makes the ground-state shift negative, and the closure/Unsöld trick — from Second-Order Energy Correction; this article is that formula's named Stark application. Writing out $|\langle m|H'|n\rangle|^2=e^2\mathcal{E}^2|\langle m|z|n\rangle|^2$ and matching the $\mathcal{E}^2$ coefficients gives the polarizability as a sum over states:

\alpha_n \;=\; 2e^2\sum_{m\neq n}\frac{|\langle m|z|n\rangle|^2}{E_m^{(0)}-E_n^{(0)}} \qquad (4)

where: the denominator is now the excitation energy $E_m-E_n$ , positive for the ground state, so every term is positive and $\alpha_0>0$ . Only opposite-parity $|m\rangle$ (for the $1s$ ground state, the $np$ states) contribute, by the dipole selection rule. This sum is exact but formally infinite — the entire hydrogen spectrum plus the ionization continuum — which is why the closed-form routes below matter so much.

Derivation ADalgarno–Lewis: the exact hydrogen polarizability without a sum

The goal is the exact ground-state polarizability of hydrogen, $\alpha=\tfrac92 a_0^3$ , obtained by replacing the infinite spectral sum with the solution of a single inhomogeneous differential equation. The idea, due to Dalgarno and Lewis, is that the second-order energy needs not the individual states $|m\rangle$ but only the combination $\sum_m|m\rangle\langle m|\hat H'|0\rangle/(E_0-E_m)$ , which is precisely the first-order wavefunction $|\psi^{(1)}\rangle$ — and $|\psi^{(1)}\rangle$ satisfies a differential equation we can solve in closed form. We work in atomic units ( $\hbar=m=e=a_0=1$ ), so $\hat H_0=-\tfrac12\nabla^2-\tfrac1r$ , ground state $\psi_0=\pi^{-1/2}e^{-r}$ , $E_0=-\tfrac12$ , and $\hat H'=\mathcal{E}z=\mathcal{E}r\cos\theta$ .

Result.

\psi^{(1)} = -\mathcal{E}\Big(r+\tfrac{r^2}{2}\Big)\cos\theta\;\psi_0 \qquad\Longrightarrow\qquad E^{(2)}=-\tfrac94\,\mathcal{E}^2 \qquad\Longrightarrow\qquad \alpha=\tfrac92\,a_0^3

Step 1. Write the first-order equation. Perturbation theory gives $(\hat H_0-E_0)\psi^{(1)}=-(\hat H'-E^{(1)})\psi_0=-\mathcal{E}z\,\psi_0$ , since $E^{(1)}=0$ from parity. The source is just the dipole acting on the ground state.

Step 2. Exploit the angular structure. The source $\propto\cos\theta=P_1(\cos\theta)$ is pure $\ell=1$ , so the correction must be too: try $\psi^{(1)}=g(r)\cos\theta\,\psi_0$ . Substituting reduces the partial differential equation to the radial ordinary differential equation $\mathcal{E}r + g' - \tfrac12 g'' - g'/r + g/r^2 = 0$ .

Step 3. Solve the ODE. A polynomial ansatz $g(r)=ar+br^2$ makes every term a low power of $r$ ; matching coefficients forces $a=-\mathcal{E}$ , $b=-\mathcal{E}/2$ , so $g(r)=-\mathcal{E}(r+\tfrac{r^2}{2})$ . A polynomial solution exists precisely because the inhomogeneity is a polynomial times $\psi_0$ . (This closed form heals a garbled placeholder in the base export, which carried an unevaluated $\psi_0/?$ and a self-cancelling scaffold in $g$ ; see the provenance note.)

Step 4. Read off the energy with the Dalgarno–Lewis identity $E^{(2)}=\langle\psi_0|\hat H'|\psi^{(1)}\rangle$ , which follows from Hermiticity and needs no sum over states.

Step 5. Do the integral. With $\psi^{(1)}$ from Step 3, $E^{(2)}=\mathcal{E}\langle\psi_0|z|\psi^{(1)}\rangle=-\tfrac94\mathcal{E}^2$ (the radial integrals of $e^{-2r}$ against $r^3$ and $r^4$ done in closed form; the angular $\int\cos^2\theta$ supplies the $1/3$ ).

Step 6. Identify the polarizability: $-\tfrac12\alpha\mathcal{E}^2=-\tfrac94\mathcal{E}^2\Rightarrow\alpha=\tfrac92 a_0^3$ . Exact, with no truncation — the full spectral sum (discrete plus continuum) reproduces the same number. Verified: the ODE coefficients $a=-\mathcal{E},b=-\mathcal{E}/2$ derived symbolically; the inhomogeneous-equation residual $(\hat H_0-E_0)\psi^{(1)}+\mathcal{E}z\psi_0$ proven to vanish identically; $E^{(2)}=-\tfrac94\mathcal{E}^2$ and $\alpha=\tfrac92 a_0^3$ recomputed in this article's verify script against Bethe–Salpeter §65 and Sakurai §5.2; SI conversion $\alpha/4\pi\varepsilon_0=\tfrac92 a_0^3\approx0.667\,\text{\AA}^3$ .

Go deeper · Derivation B · 1 minClosure bounds: bracketing the polarizability with a sum rule

Derivation BClosure bounds: bracketing the polarizability with a sum rule

The goal is a rigorous upper bound on the polarizability without evaluating a single excited-state matrix element individually — a method that works even when the spectrum is unknown. The trick is to replace the variable energy denominator by its smallest value; every term in the (positive) sum then grows, and the completeness relation collapses what remains into a single ground-state expectation value. For hydrogen this yields $\tfrac{16}{3}a_0^3$ , which must bracket the exact $\tfrac92 a_0^3$ from above — a sanity check on the Dalgarno–Lewis result and a genuinely useful estimate for atoms with no closed form.

Result.

\alpha_0 \;\le\; \frac{2e^2}{\Delta_{\min}}\langle 0|z^2|0\rangle \;=\; \frac{16}{3}a_0^3\approx5.33\,a_0^3 \;>\; \frac92 a_0^3=4.5\,a_0^3

Step 1. Bound the denominator from below. Every excitation energy satisfies $E_m-E_0\ge\Delta_{\min}\equiv E_1-E_0$ , the smallest gap. Replacing each denominator by $\Delta_{\min}$ can only increase each positive term, giving an upper bound on $\alpha_0=2e^2\sum_{m\neq0}|\langle m|z|0\rangle|^2/(E_m-E_0)$ .

Step 2. Apply closure. With every denominator now the constant $\Delta_{\min}$ , the sum factors out: $\sum_{m\neq0}|\langle m|z|0\rangle|^2=\langle 0|z^2|0\rangle-\langle 0|z|0\rangle^2=\langle 0|z^2|0\rangle$ , using $\sum_m|m\rangle\langle m|=\mathbb 1$ and the parity vanishing of $\langle 0|z|0\rangle$ .

Step 3. Evaluate the ground-state spread. For hydrogen $\langle r^2\rangle_{1s}=3a_0^2$ (from the radial moments of Radial Probability Distribution and Expectation Values), and isotropy gives $\langle z^2\rangle=\tfrac13\langle r^2\rangle=a_0^2$ .

Step 4. Insert the smallest gap $\Delta_{\min}=E_2-E_1=\tfrac38\,e^2/a_0$ (the $n{=}1\!\to\!2$ Rydberg gap). Then $\alpha_0\le 2e^2 a_0^2/(\tfrac38 e^2/a_0)=\tfrac{16}{3}a_0^3\approx5.33\,a_0^3$ . The exact $4.5\,a_0^3$ indeed sits below, by about $18.5\%$ — the bound is loose because most of the oscillator strength actually lies above the first gap, so replacing every denominator by the smallest overestimates. Verified: $\langle r^2\rangle_{1s}=3a_0^2$ , $\langle z^2\rangle=a_0^2$ , $\Delta_{\min}=\tfrac38 e^2/a_0$ , the bound $\tfrac{16}{3}a_0^3$ , and the ordering $\tfrac{16}{3}>\tfrac92$ (18.5% excess) all recomputed symbolically; the dominant $1s\to2p$ term alone contributes $\approx2.96\,a_0^3$ ( $\sim66\%$ of $\alpha$ ), the continuum supplying the rest.

Limiting cases

How the Stark response behaves across physically distinct regimes:

case	result	meaning
non-degenerate ground state	$\Delta E=-\tfrac12\alpha\mathcal{E}^2$ , $\alpha>0$	pure quadratic shift; no permanent dipole, energy always lowered — the generic atomic case
degenerate level, opposite parity (H, $n\ge2$ )	$\Delta E=\pm3ea_0\mathcal{E}$	field mixes $s$ and $p$ at first order: a linear Stark effect dominates — owned by Linear Stark Effect in Hydrogen
Rydberg / weakly bound atoms	$\alpha\sim n^7 a_0^3$	polarizability grows enormously with $n$ ; Rydberg atoms are extreme field sensors
strong-field / ionization limit	$e\mathcal{E}a_0\gtrsim\|E_0\|$	the perturbation series diverges; the tilted Coulomb potential lets the electron tunnel out — the polarizability concept breaks down

Common trapDo not apply the quadratic formula to a degenerate level

The quadratic result $-\tfrac12\alpha\mathcal{E}^2$ requires a genuinely non-degenerate, parity-definite state. Hydrogen excited levels are neither: $2s$ and $2p$ are degenerate and of opposite parity, so the field mixes them at first order and the shift is linear, $\pm3ea_0\mathcal{E}$ — the treatment of Linear Stark Effect in Hydrogen, not this one. The tell is degeneracy: if opposite-parity states share an energy, use degenerate perturbation theory and expect a linear shift; only when the level is isolated and parity-definite is the polarizability the leading response. A second trap is treating the sum-over-states formula literally — the discrete sum alone converges to only $\sim80\%$ of $\alpha$ for hydrogen; the continuum is not optional, which is exactly why the Dalgarno–Lewis or closure route is preferred.

Worked examples

Example 1 — the hydrogen ground-state shift in a strong field. Compute the quadratic Stark shift of hydrogen $1s$ at $\mathcal{E}=10^{7}$ V/m, and compare it to the linear splitting the $n=2$ level would show at the same field.

Step 1. Convert the polarizability to SI. $\alpha=\tfrac92 a_0^3$ in Gaussian units means a polarizability volume $\alpha/4\pi\varepsilon_0=\tfrac92 a_0^3=4.5\times(5.29\times10^{-11})^3=6.67\times10^{-31}\,\text{m}^3=0.667\,\text{\AA}^3$ (computed).

Step 2. The shift is $\Delta E=-\tfrac12\alpha_{\text{SI}}\mathcal{E}^2$ with $\alpha_{\text{SI}}=4\pi\varepsilon_0\times6.67\times10^{-31}\,\text{m}^3$ . At $\mathcal{E}=10^7$ V/m, $\Delta E=-2.32\times10^{-8}$ eV (computed) — negative, as required, and utterly tiny.

Step 3. Contrast: the $n=2$ linear splitting at the same field is $3ea_0\mathcal{E}=1.59\times10^{-3}$ eV — about $10^{5}$ times larger. Answer: the $1s$ ground state shifts down by only $2.3\times10^{-8}$ eV, five orders of magnitude below the linear $n=2$ splitting — a vivid measure of how much degeneracy amplifies the field response.

Example 2 — is the closure bound any good? Use only $\langle r^2\rangle_{1s}=3a_0^2$ and the first excitation gap to estimate the hydrogen polarizability, then judge the error.

Step 1. Closure upper bound: $\alpha_0\le 2\langle z^2\rangle/\Delta_{\min}$ with $\langle z^2\rangle=\tfrac13\langle r^2\rangle=a_0^2$ and $\Delta_{\min}=\tfrac38\,e^2/a_0$ , giving $\alpha_0\le\tfrac{16}{3}a_0^3=5.33\,a_0^3$ (computed).

Step 2. Compare to the exact $\tfrac92 a_0^3=4.5\,a_0^3$ : the bound is high by $(5.33-4.5)/4.5=18.5\%$ (computed). It errs on the high side because it pretends all the oscillator strength sits at the smallest gap, whereas much of it lies higher (including in the continuum), where the true denominators are larger.

Step 3. Interpretation: with a single expectation value and one energy gap — no wavefunctions of excited states, no sum — we bracket the exact answer to within $20\%$ . Answer: $\alpha_0\le\tfrac{16}{3}a_0^3$ , exceeding the exact $\tfrac92 a_0^3$ by $18.5\%$ — a cheap, rigorous estimate that needs nothing but the ground-state size and the first gap.

Origins · 1 minPolarizability and how this was discovered

Polarizability

1926–1934 · Baltimore, Cambridge

Polarizability is where quantum perturbation theory first touched a macroscopic measurable, and its exact hydrogen value became an early benchmark of the new mechanics. Within two years of Schrödinger's equation, Ivar Waller (1926) and Paul Epstein computed the hydrogen Stark shift to second order and extracted the ground-state polarizability; the number $\tfrac92 a_0^3$ settled quickly as the value every method had to reproduce. The awkwardness was the infinite sum over the spectrum, and the 1930s produced two escapes that still define the subject. Egil Hylleraas and Henry Hassé turned the sum rules into rigorous bounds — the closure inequalities of this article — sandwiching $\alpha$ between computable expectation values. Then, decades later, Alexander Dalgarno and John Lewis (1955) gave the definitive move: recast the second-order sum as an inhomogeneous differential equation for the first-order wavefunction, solvable in closed form, delivering $\tfrac92 a_0^3$ exactly and generalizing to any perturbation whose source is polynomial times the ground state. The through-line is a single physical idea — the induced dipole $\mathbf d=\alpha\boldsymbol{\mathcal E}$ — bridging the quantum energy shift to the classical dielectric constant of Clausius and Mossotti, to the refractive index of a dilute gas, and to the van der Waals attraction between neutral atoms, all of which are polarizability wearing different clothes. What began as a spectroscopic curiosity in a strong field is now the microscopic origin of why matter bends light and why neutral atoms stick together.

History

1926

— Waller and Epstein compute the hydrogen ground-state polarizability,

\tfrac92 a_0^3

, from the new wave mechanics

1927

— Unsöld introduces the closure (mean-energy) approximation for perturbation sums

1930

— Hassé and Hylleraas sharpen the sum-rule bounds on atomic polarizabilities

1955

— Dalgarno and Lewis recast second-order sums as an inhomogeneous equation, giving

\tfrac92 a_0^3

in closed form

1970

s — Rydberg-atom experiments confirm the

n^7

scaling of polarizability directly

2000

s — blackbody and DC Stark polarizabilities become leading systematics in optical atomic clocks

Going further

Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §7.1 and Problem 7.41 — the quadratic Stark effect and the hydrogen polarizability.
Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §5.2 — the quadratic Stark effect and the Dalgarno–Lewis method.
Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, §65 — the exact $\tfrac92 a_0^3$ and the sum-rule bounds.
Dalgarno & Lewis, Proc. R. Soc. A 233, 70 (1955) — the original inhomogeneous-equation method.
Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Complement — the polarizability and its observable bridges.

Provenance note: Equations standard from Griffiths §7.1, Sakurai §5.2, and Bethe–Salpeter §65. A verifier should: (1) confirm the parity argument for the vanishing first-order shift; (2) reproduce the Dalgarno–Lewis derivation giving $\alpha_{\mathrm{H,1s}}=\tfrac92 a_0^3$ exactly; (3) check the closure upper bound $\alpha_0\le\tfrac{16}{3}a_0^3$ using $\langle r^2\rangle_{1s}=3a_0^2$ and $\Delta_{\min}=\tfrac38 e^2/a_0$ ; (4) verify the $n^7$ Rydberg scaling and the sign of the ground-state shift. Numbers $\tfrac92 a_0^3$ , $3a_0^2$ , $\tfrac38 e^2/a_0$ are textbook-standard. — All four items run in this article's verify script: the parity vanishing symbolically, the Dalgarno–Lewis first-order wavefunction with its inhomogeneous-equation residual proven to vanish identically, the closure bound and its $18.5\%$ excess, and the sign and scaling. Relocation note (A6 → A7): the hydrogen-polarizability computation ( $\tfrac92 a_0^3$ , Dalgarno–Lewis and closure bounds) was banked by the Second-Order Energy Correction lane as the named property of this article — Second-Order Energy Correction states it in one line with this article as owner and does not compute it; the full derivation is carried out here. This is a deliberate ownership relocation, not a base error. Base display fix: the Step-1 export's Dalgarno–Lewis "Result" line carried an unevaluated placeholder ( $\psi_0/?$ ) and a self-cancelling scaffold in the radial solution $g(r)$ ; the correct closed form $\psi^{(1)}=-\mathcal{E}(r+\tfrac{r^2}{2})\cos\theta\,\psi_0$ (verified residual-zero) is used throughout. Ownership note: the general second-order formula, the sign theorem, and the closure/Unsöld machinery are owned by Second-Order Energy Correction and consumed here; the linear/degenerate branch is owned by Linear Stark Effect in Hydrogen; dynamic (frequency-dependent) polarizability and the AC Stark shift are deferred to the time-dependent track; the refractive-index and van der Waals bridges to the electromagnetism and many-body tracks respectively.

Why can't I just sum the discrete states to get the polarizability?

You can, but you will land at only about $80\%$ of the answer for hydrogen. The polarizability sum $\alpha=2e^2\sum_m|\langle m|z|0\rangle|^2/(E_m-E_0)$ runs over the entire spectrum, and roughly a fifth of hydrogen's oscillator strength lives in the ionization continuum, not in the bound $np$ states. Truncating at the bound states drops that continuum contribution silently. This is exactly the motivation for Dalgarno–Lewis (which sums everything exactly by solving a differential equation) and for the closure bounds (which sidestep the sum via completeness). The bound $1s\to2p$ term alone gives $\approx2.96\,a_0^3$ , about $66\%$ of the full $4.5\,a_0^3$ .

What exactly is the Dalgarno–Lewis trick doing?

Second-order perturbation theory only ever needs the combination $|\psi^{(1)}\rangle=\sum_{m\neq0}|m\rangle\langle m|\hat H'|0\rangle/(E_0-E_m)$ , not the individual states $|m\rangle$ . Dalgarno and Lewis observed that this combination is the solution of $(\hat H_0-E_0)|\psi^{(1)}\rangle=-(\hat H'-E^{(1)})|\psi_0\rangle$ — a single inhomogeneous differential equation. If that equation has a closed-form solution (it does for hydrogen, because the source is a polynomial times $\psi_0$ ), you get the exact second-order energy $E^{(2)}=\langle\psi_0|\hat H'|\psi^{(1)}\rangle$ with no sum and no truncation. It converts an infinite spectral sum into one ODE.

Why is the closure bound an upper bound rather than a lower one?

Because every term in the polarizability sum is positive and you are replacing each energy denominator by the smallest one. Making a denominator smaller makes its (positive) term larger, so the whole sum can only grow — an upper bound. If you instead replaced every denominator by the largest relevant gap you would get a lower bound, but there is no finite largest gap (the continuum extends to infinity), so the useful rigorous inequality is the upper one. The gap between the bound $\tfrac{16}{3}a_0^3$ and the exact $\tfrac92 a_0^3$ measures how much oscillator strength really lies above the first excitation.

How can a positive polarizability lower the energy — isn't positive energy up?

The polarizability $\alpha$ is positive, but it enters the energy with a minus sign: $\Delta E=-\tfrac12\alpha\mathcal{E}^2$ . The positivity of $\alpha$ is the statement that the induced dipole points along the field (the cloud shifts so as to screen it), and a dipole aligned with the field it created sits at negative interaction energy. So $\alpha>0$ and $\Delta E<0$ are the same physical fact: polarizing is energetically favorable, and the more polarizable the atom, the more its energy drops.

Why does the polarizability scale as n⁷ for Rydberg atoms?

Two factors compound. The dipole matrix elements $\langle m|z|0\rangle$ scale like the orbit size, $\langle r^2\rangle\propto n^4$ , contributing $n^4$ . And the relevant energy denominators are level spacings, which shrink as $\Delta E\propto n^{-3}$ near the series limit, contributing another $n^3$ . Together $\alpha\propto\langle r^2\rangle/\Delta E\propto n^4\cdot n^3=n^7$ . A Rydberg state with $n=30$ is thus $\sim30^7\approx2\times10^{10}$ times more polarizable than the ground state — which is why Rydberg atoms are such extreme electric-field sensors and such strongly interacting qubits.

Is the factor of one-half in −½αℰ² the same one-half as in the electrostatic energy of a capacitor?

Yes, and for the same reason. The interaction energy is $-\int_0^{\mathcal{E}}\langle d\rangle\,\mathrm d\mathcal{E}'=-\int_0^{\mathcal{E}}\alpha\mathcal{E}'\,\mathrm d\mathcal{E}'=-\tfrac12\alpha\mathcal{E}^2$ , because the induced dipole is not present from the start but grows linearly with the field as you turn it on. A permanent dipole $\mathbf p$ contributes $-\mathbf p\cdot\boldsymbol{\mathcal E}$ with no one-half, because it is fully present at all field values. The factor of one-half is the universal signature of a response that grows with its driving field — the same integral that gives $\tfrac12 CV^2$ and $\tfrac12 kx^2$ .