University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.8

The Feynman-Hellmann theorem: hydrogen's ⟨1/r²⟩ and ⟨1/r³⟩

Most of the averages Unit 8 needs would each cost a page of Laguerre integrals. This lesson replaces them with one line of calculus: prove that an eigenvalue's slope is the average of the Hamiltonian's slope, then turn hydrogen's l into a dial and read ⟨1/r²⟩, and from it ⟨1/r³⟩, off the Bohr formula.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An eigenvalue that depends on a parameter already knows an expectation value, and the Feynman-Hellmann theorem reads it off: for a normalised eigenstate of a Hermitian H(λ), dEₙ/dλ = ⟨n|∂H/∂λ|n⟩, exactly and at every λ, because the change in the state cancels against its fixed norm. The corrected state, the hard object of perturbation theory, never appears. So every closed-form spectrum becomes a table of averages: the oscillator's ω hands over ⟨x²⟩, the Coulomb strength e² hands over ⟨1/r⟩, and hydrogen's l, which the radial equation accepts as any real number above −½, hands over ⟨1/r²⟩ = 1/[(l + ½)n³a²], provided the derivative is taken along a family of fixed radial node count nᵣ rather than fixed n.

Kramers' relation at s = −1 turns that into ⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³], and the two moments are the radial content of Unit 8's relativistic and spin-orbit corrections, both of order α⁴mc². The theorem stops where its hypotheses do. An approximate state gives a slope wrong at first order in its error; a degenerate level must first be resolved into good states; a parameter hiding in a boundary, like a well's width, must be moved into the operator by rescaling; and for l = 0 the ⟨1/r³⟩ it leads to is infinite, the s-state problem Unit 8 resolves with the Darwin term.

Simple definition: The Feynman-Hellmann theorem says that for a normalised eigenstate |n(λ)⟩ of a Hermitian Hamiltonian H(λ), the slope of the eigenvalue equals the expectation value of the slope of the Hamiltonian: dEₙ/dλ = ⟨n(λ)|∂H/∂λ|n(λ)⟩.
Example: In hydrogen's radial equation ∂H/∂l = ħ²(2l + 1)/(2μr²) and E = −13.598 eV/(nᵣ + l + 1)², so for 2p (nᵣ = 0, l = 1) the slope 2 × 13.598 eV/2³ = 3.400 eV gives ⟨1/r²⟩ = 1/(12a²) = 2.97 × 10¹⁹ m⁻².

Feynman-Hellmann theoremdEₙ/dλ = ⟨n(λ)|∂H/∂λ|n(λ)⟩

Derived, and exact at every λ: the state's derivative cancels against d⟨n|n⟩/dλ = 0. For H₀ + λH′ at λ = 0 it is Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩.

Eₙ in J; λ any real parameter, in its own unit; |n(λ)⟩ a normalised eigenstate of Hermitian H(λ), a good state if degenerate

The l-derivative, for any central potential⟨1/r²⟩ = [2μ/((2l + 1)ħ²)] (∂E/∂l) at fixed nᵣ

Derived. Any spectrum known as a smooth function of l at fixed node count hands over ⟨1/r²⟩; the isotropic oscillator, with ∂E/∂l = ħω, gives 1/[(l + ½)b²], b = √(ħ/mω).

⟨1/r²⟩ in m⁻², E in J, μ in kg, ħ in J s; nᵣ counts radial nodes; from ∂Hₗ/∂l = ħ²(2l + 1)/(2μr²)

Coulomb levels, smooth in l at fixed nᵣE(nᵣ, l) = −ħ²/[2μa²(nᵣ + l + 1)²]

Quoted from Unit 2's truncation, whose recursion never needs l to be an integer. At fixed nᵣ, ∂E/∂l = ħ²/(μa²n³) with n = nᵣ + l + 1: 3.400 eV for 2p.

J; a = 4πε₀ħ²/(μe²) = 52.95 pm; ħ²/(2μa²) = 13.598 eV; nᵣ = 0, 1, 2, …; l any real number above −½

Mean inverse square radius⟨1/r²⟩ = 1/[(l + ½)n³a²]

Derived by equating the two slopes. It falls with l at fixed n because low-l states reach the nucleus, and it equals the classical Kepler average of 1/r² with L = (l + ½)ħ.

m⁻²; 1s gives 2/a² = 7.13 × 10²⁰ m⁻², 2p gives 1/(12a²); for a hydrogen-like ion replace a by a/Z

Mean inverse radius, from the Coulomb strength⟨1/r⟩ = 1/(n²a)

Derived as a cross-check: it repeats topic 02.08's virial result by an independent route, and |nlm⟩ is already a good basis because 1/r is rotationally invariant.

m⁻¹; dial q = e²/(4πε₀) in J m, with ∂H/∂q = −1/r and Eₙ = −μq²/(2ħ²n²), so ∂Eₙ/∂q = 2Eₙ/q

Mean inverse cube radius, l ≥ 1 only⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³]

Derived from ⟨1/r²⟩. 2p gives 1/(24a³) = 2.81 × 10²⁹ m⁻³. It scales as α³, which makes spin-orbit coupling of order α⁴mc²; for l = 0 the integral diverges.

m⁻³; from Kramers' relation at s = −1, quoted from topic 02.08: a⟨1/r²⟩ = l(l + 1)a²⟨1/r³⟩

The proof: the state's derivative cancels

Let H(λ) be Hermitian, with H(λ)|n(λ)⟩ = Eₙ(λ)|n(λ)⟩ and ⟨n(λ)|n(λ)⟩ = 1 for every λ. Then Eₙ = ⟨n|H|n⟩, and the product rule gives three terms, ⟨∂n|H|n⟩ + ⟨n|∂H/∂λ|n⟩ + ⟨n|H|∂n⟩, with |∂n⟩ = d|n⟩/dλ. H acting to the right gives Eₙ|n⟩, and acting to the left gives Eₙ⟨n| because H is Hermitian and Eₙ real, so the outer terms add to Eₙ d⟨n|n⟩/dλ = 0. What survives is dEₙ/dλ = ⟨n|∂H/∂λ|n⟩, derived and exact at every λ. For H = H₀ + λH′ it reads dEₙ/dλ = ⟨n(λ)|H′|n(λ)⟩: at λ = 0 that is topic 06.02's Eₙ⁽¹⁾, and one more derivative at λ = 0 gives 2Re⟨n⁽⁰⁾|H′|n⁽¹⁾⟩ = 2Eₙ⁽²⁾, topic 06.04's second order. Test it on Quantum Mechanics I's oscillator: ∂H/∂ω = mωx² and Eₙ = (n + ½)ħω, so ⟨x²⟩ = (n + ½)ħ/(mω), the ladder-operator result, in units of J s/(kg s⁻¹) = m².

Three hypotheses, and what fails without each

Each line of the proof leaned on something. First, |n⟩ must be an exact eigenstate: with an approximate state, ⟨∂H/∂λ⟩ misses dEₙ/dλ at first order in the state's error, although the energy itself would be wrong only at second order. Second, in a degenerate level the branches Eₖ(λ) have different slopes, and the theorem speaks only for topic 06.05's good states, which diagonalise ∂H/∂λ inside the level. If ∂H/∂λ couples a two-fold pair by W and has no diagonal elements, the branches have slopes ±|W|, while either unperturbed state returns 0, their average. Third, λ must sit in the operator, not in its domain. The infinite well's H = p²/2m contains no L, yet E = n²π²ħ²/(2mL²) depends on L through the walls. Rescale x = Lξ onto a fixed interval: H = −(ħ²/2mL²)d²/dξ², so ∂H/∂L = −2H/L and dE/dL = −2E/L, the theorem restored. The force on a wall is then −dE/dL = 2E/L: 1.20 × 10⁻¹⁰ N for an electron's ground state in a 1.00 nm well.

Why l may be varied continuously, at fixed nᵣ

In the separated problem the radial Hamiltonian Hₗ = pᵣ²/2μ + ħ²l(l + 1)/(2μr²) − e²/(4πε₀r) contains l only as a number in the centrifugal term. Unit 2's series solution never needed that number to be an integer: the solution starting as r^(l+1) exists for any real l above −½, the recursion runs unchanged, and termination after nᵣ steps gives E(nᵣ, l) = −ħ²/[2μa²(nᵣ + l + 1)²], a smooth function of l. Integer l comes from the angular equation, where Unit 1's Legendre series terminates only for integer l. Two facts make the theorem safe here. Each level of a fixed-l radial problem is non-degenerate, since a one-dimensional bound state with u(0) = 0 is unique, so hydrogen's l-degeneracy never enters. And the eigenstate that deforms continuously with l keeps its node count nᵣ, so along it n = nᵣ + l + 1 grows with l. Holding n fixed would need nᵣ to change by a fraction, which no eigenstate of Hₗ can do, so ∂Eₙ/∂l = 0 at fixed n is not a Feynman-Hellmann derivative at all.

⟨1/r²⟩ from the slope, and three checks

The operator's slope is ∂Hₗ/∂l = ħ²(2l + 1)/(2μr²); the eigenvalue's slope at fixed nᵣ is ∂E/∂l = ħ²/[μa²(nᵣ + l + 1)³] = ħ²/(μa²n³). Equate them: ⟨1/r²⟩ = 2/[(2l + 1)n³a²] = 1/[(l + ½)n³a²], derived. Units: m⁻², the inverse of a². Direct integral: for 1s, u² = (4/a³)r²e^(−2r/a), and ∫(4/a³)e^(−2r/a) dr = 2/a², which the formula gives at l = 0, n = 1. Correspondence: a classical Kepler orbit of energy Eₙ and angular momentum L has the time average ⟨1/r²⟩ = ħ/(Ln³a²), because dt/r² = μ dθ/L (quoted from classical mechanics). The quantum result is exactly this with L = (l + ½)ħ, and for the circular state l = n − 1 at large n it tends to 1/(n²a)², the inverse square of the orbit's radius. Order in α: 1/a² ∝ α², so the mean centrifugal energy ħ²l(l + 1)⟨1/r²⟩/(2μ) is of order α²mc², the binding scale, as a share of ⟨T⟩ must be.

The e² dial, and what two moments give together

The theorem accepts any parameter the closed form contains. Take q = e²/(4πε₀) as the dial, holding μ, ħ, nᵣ and l fixed: ∂H/∂q = −1/r and Eₙ = −μq²/(2ħ²n²), so ∂Eₙ/∂q = 2Eₙ/q and ⟨1/r⟩ = μq/(ħ²n²) = 1/(n²a). That is topic 02.08's virial result, reached by an independent route, and for an ion the same dial with q → Zq gives Z/(n²a). The full level is n²-fold degenerate, but −1/r is rotationally invariant and hence diagonal in |nlm⟩, which is already a good basis. Now combine the two moments. With V = −q/r, ⟨V⟩² = q²/(n⁴a²) and ⟨V²⟩ = q²⟨1/r²⟩, so ⟨V²⟩/⟨V⟩² = n/(l + ½). For 1s the ratio is 2: the potential energy fluctuates by ΔV = 27.2 eV about its mean of −27.2 eV, while circular states at large n approach a sharp orbit with ΔV → 0. ⟨V²⟩ is the input Unit 8's relativistic kinetic correction needs, and divided by mc² it is of order α⁴mc².

Kramers at s = −1: ⟨1/r³⟩, and the s-state edge

Kramers' relation, quoted from topic 02.08, ties three neighbouring moments: (s + 1)⟨rˢ⟩/n² − (2s + 1)a⟨rˢ⁻¹⟩ + (s/4)[(2l + 1)² − s²]a²⟨rˢ⁻²⟩ = 0 for s above −(2l + 1). At s = −1 the first term vanishes, leaving a⟨1/r²⟩ = l(l + 1)a²⟨1/r³⟩, one equation in two unknowns that the Feynman-Hellmann ⟨1/r²⟩ closes: ⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³], derived for l ≥ 1. For 2p, u² = r⁴e^(−r/a)/(24a⁵) and ∫r e^(−r/a) dr/(24a⁵) = 1/(24a³), as the formula says. For l = 0, s = −1 sits on the edge −(2l + 1), and u ∝ r near the nucleus makes ∫u²/r³ dr diverge logarithmically. Since 1/a³ ∝ α³, Unit 8's spin-orbit energy e²⟨1/r³⟩⟨S⋅L⟩/(8πε₀m²c²), quoted, is of order α⁴mc². For 2p with an infinitely heavy nucleus, ⟨S⋅L⟩ differs by 3ħ²/2 between j = 3/2 and j = ½, a splitting α⁴mc²/32 = 4.528 × 10⁻⁵ eV, or 10.949 GHz, against the measured 10.969 GHz. The 0.18% gap is the electron's anomalous magnetic moment, +0.23% from QED, less a reduced-mass correction, both quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

Radial nodes nᵣ0

Orbital number l, varied continuously1.0

Start at nᵣ = 0, l = 1, the 2p state: the slope is 3.400 eV, and the faint flat line meets the dashed nᵣ = 1 curve at l = 0, the degenerate 2s. Slide l between integers: the radial equation keeps a smooth eigenvalue, which is all the derivative needs. Raise nᵣ and every slope falls as 1/n³.

n = nᵣ + l + 12.0

ENERGY E-3.400 eV

SLOPE ∂E/∂l AT FIXED nᵣ3.400 eV

⟨1/r²⟩ IN UNITS OF 1/a²0.0833

Live interpretationn = nᵣ + l + 1: 2.0. ENERGY E: −3.400 eV. SLOPE ∂E/∂l AT FIXED nᵣ: 3.400 eV. ⟨1/r²⟩ IN UNITS OF 1/a²: 0.0833

Catch the common trap

Explain before calculating.

Misconception check

The tempting move is to differentiate the famous Eₙ = −13.6 eV/n² with respect to l, find no l in it, and conclude either that ⟨1/r²⟩ = 0 or that the theorem fails for hydrogen. Neither follows. ∂Hₗ/∂l = ħ²(2l + 1)/(2μr²) is a positive operator for every l above −½, so its average cannot vanish: a zero means the derivative was taken along the wrong path. The theorem differentiates one eigenvalue of one Hamiltonian as the Hamiltonian changes, and the eigenstate of Hₗ that deforms continuously with l keeps its number of radial nodes, so n = nᵣ + l + 1 rises with l and the energy rises with it. The l-degeneracy is a coincidence between different radial families at integer l, 2s being the nᵣ = 1 family at l = 0 and 2p the nᵣ = 0 family at l = 1, not a flat direction of either. Hold nᵣ fixed, not n: for 2p the slope is 3.400 eV and ⟨1/r²⟩ = 1/(12a²).

For hydrogen's 3p state (n = 3, l = 1), differentiating at fixed nᵣ gives ⟨1/r²⟩ = 2/(81a²), and Kramers' relation at s = −1 reads a⟨1/r²⟩ = l(l + 1)a²⟨1/r³⟩. What is ⟨1/r³⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen's 3d state has n = 3 and l = 2. Find ⟨1/r²⟩ and ⟨1/r³⟩ in terms of a and in SI units, with a = 52.95 pm, and confirm both with the factorial integral ∫₀^∞ rᵏe^(−r/b) dr = k! bᵏ⁺¹, given that u ∝ r³e^(−r/3a).

Radial family first: nᵣ = n − l − 1 = 0, a nodeless state. The Feynman-Hellmann result at fixed nᵣ gives ⟨1/r²⟩ = 1/[(l + ½)n³a²] = 1/(2.5 × 27a²) = 1/(67.5a²).
Kramers at s = −1: ⟨1/r³⟩ = ⟨1/r²⟩/[l(l + 1)a] = 1/(6 × 67.5a³) = 1/(405a³). It is finite because l = 2 is above zero.
Check: u² ∝ r⁶e^(−2r/3a), so b = 3a/2. Then ⟨1/r²⟩ = 4! b⁵/(6! b⁷) = 1/(30b²) = 4/(270a²) = 1/(67.5a²), and ⟨1/r³⟩ = 3! b⁴/(6! b⁷) = 1/(120b³) = 8/(3240a³) = 1/(405a³). Both agree.
SI: a² = 2.804 × 10⁻²¹ m² and a³ = 1.485 × 10⁻³¹ m³, so ⟨1/r²⟩ = 5.28 × 10¹⁸ m⁻² and ⟨1/r³⟩ = 1.66 × 10²⁸ m⁻³.
Sense check: ⟨1/r²⟩^(−1/2) = √67.5 a = 8.22a = 435 pm, inside the circular-orbit radius n²a = 9a = 477 pm, because 1/r² weights the inner flank of the density.

Answer⟨1/r²⟩ = 1/(67.5a²) = 5.28 × 10¹⁸ m⁻² and ⟨1/r³⟩ = 1/(405a³) = 1.66 × 10²⁸ m⁻³, both confirmed by direct integration.

MediumHydrogen's kinetic energy divides as ⟨T⟩ = ⟨pᵣ²⟩/2μ + ⟨ħ²l(l + 1)/(2μr²)⟩, a radial part plus a centrifugal part W. Using the Feynman-Hellmann ⟨1/r²⟩ and ⟨T⟩ = −Eₙ from the virial theorem (quoted from topic 02.08), find W for 3s, 3p and 3d in eV and its share of ⟨T⟩, then the share for a circular state l = n − 1 and its large-n limit.

W = ħ²l(l + 1)⟨1/r²⟩/(2μ) = [ħ²/(2μa²)] l(l + 1)/[(l + ½)n³] = R l(l + 1)/[(l + ½)n³], with R = ħ²/(2μa²) = 13.598 eV. R carries the unit; the rest is a pure number.
n³ = 27. 3s: l(l + 1) = 0, so W = 0. 3p: W = 2/(1.5 × 27) × 13.598 eV = 0.672 eV. 3d: W = 6/(2.5 × 27) × 13.598 eV = 1.209 eV.
Virial theorem, quoted: ⟨T⟩ = −E₃ = 13.598 eV/9 = 1.511 eV for every l. The shares W/⟨T⟩ = l(l + 1)/[(l + ½)n] are 0, 0.444 and 0.800, and the radial parts are what remain: 1.511, 0.839 and 0.302 eV.
Circular state l = n − 1: W/⟨T⟩ = (n − 1)n/[(n − ½)n] = 1 − 1/(2n − 1). That is 0.800 at n = 3, 0.947 at n = 10 and 0.990 at n = 50, tending to 1, the classical circular orbit with no radial motion.
The missing 1/(2n − 1) is radial zero-point motion, which no bound state can shed: the kinetic counterpart of the radial spread Δr/⟨r⟩ = 1/√(2n + 1) that topic 02.08 found for circular states.

AnswerW = 0, 0.672 and 1.209 eV for 3s, 3p and 3d, that is 0, 44.4% and 80.0% of ⟨T⟩ = 1.511 eV; a circular state's share is (n − 1)/(n − ½), tending to 1.

HardIn the shell model a nucleon in oxygen-16 moves in an isotropic oscillator with ħω = 14.0 MeV (a fitted value), whose levels E = ħω(2nᵣ + l + 3/2) are quoted from Quantum Mechanics I. Using the Feynman-Hellmann theorem with l and then with ω as the dial, find ⟨1/r²⟩ and ⟨r²⟩ for a 0p nucleon (nᵣ = 0, l = 1), check ⟨1/r²⟩ with a Gaussian integral, and evaluate both and the mean centrifugal energy with mc² = 939 MeV and ħc = 197.3 MeV fm.

l as the dial: the radial Hamiltonian holds l only in ħ²l(l + 1)/(2mr²), and at fixed nᵣ, ∂E/∂l = ħω. So ħ²(2l + 1)⟨1/r²⟩/(2m) = ħω, giving ⟨1/r²⟩ = 2mω/[(2l + 1)ħ] = 1/[(l + ½)b²] with b = √(ħ/mω), the oscillator's twin of hydrogen's 1/[(l + ½)n³a²].
ω as the dial: ∂H/∂ω = mωr² and ∂E/∂ω = ħ(2nᵣ + l + 3/2), so ⟨r²⟩ = (2nᵣ + l + 3/2)b². For 0p, ⟨1/r²⟩ = 2/(3b²) and ⟨r²⟩ = 5b²/2.
Check: the 0p radial function is u ∝ r²e^{(−r²/2b²)}, so ⟨1/r²⟩ = ∫r²e^(−r²/b²) dr/∫r⁴e^(−r²/b²) dr = (√π b³/4)/(3√π b⁵/8) = 2/(3b²). The product ⟨r²⟩⟨1/r²⟩ = 5/3 exceeds 1, as the Cauchy-Schwarz inequality requires of any state spread in r.
Numbers: b² = (ħc)²/(mc² ħω) = (197.3 MeV fm)²/(939 MeV × 14.0 MeV) = 2.961 fm², so b = 1.721 fm, ⟨1/r²⟩ = 0.2251 fm⁻² and ⟨r²⟩ = 7.403 fm², a root-mean-square radius of 2.72 fm.
Centrifugal energy: ħ²l(l + 1)⟨1/r²⟩/(2m) = ħω l(l + 1)/(2l + 1) = (2/3)(14.0 MeV) = 9.33 MeV. With the oscillator's virial split ⟨T⟩ = E/2 = 17.5 MeV (quoted), that is 8/15, or 53%, of the kinetic energy.

Answer⟨1/r²⟩ = 2/(3b²) = 0.225 fm⁻² and ⟨r²⟩ = 5b²/2 = 7.40 fm² (2.72 fm root-mean-square), with a mean centrifugal energy of 9.33 MeV, two-thirds of ħω.