University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.8
Radial expectation values: virial theorem, Kramers' relation
Topic 02.07 handed you normalised radial functions; this one shows how rarely you need to integrate them. One identity from stationarity and one from integrating the radial equation by parts turn ⟨1/r⟩, ⟨r⟩ and ⟨r²⟩ into algebra, and show exactly which inverse moment fine structure needs that the algebra alone cannot reach.
Build the model
Connect the measurement to the mechanism.
A radial expectation value ⟨rˢ⟩ = ∫₀^∞ rˢu² dr is a moment of the radial density, and with topic 02.07's normalised Rₙₗ each one is a factorial integral. For hydrogen two identities make even that unnecessary. The first holds for any potential: in a stationary state the average of a commutator with H vanishes, and choosing r⋅p, the generator of dilations, turns ⟨[H, r⋅p]⟩ = 0 into the virial theorem 2⟨T⟩ = ⟨r⋅∇V⟩.
For the Coulomb potential that means ⟨V⟩ = 2Eₙ, ⟨T⟩ = −Eₙ and ⟨1/r⟩ = Z/(n²a) whatever l is. The second is special to the Coulomb radial equation: multiplied by rˢu′ and by rˢ⁻¹u and integrated by parts, it yields Kramers' three-term recursion among ⟨rˢ⟩, ⟨rˢ⁻¹⟩ and ⟨rˢ⁻²⟩, exact for every s above −(2l + 1). From ⟨r⁰⟩ = 1 it climbs to ⟨r⟩ = (a/2)[3n² − l(l + 1)] and ⟨r²⟩, which at large n become the time averages of a Kepler ellipse.
Downward it stalls: coefficients vanish at s = 0 and s = −1, so ⟨1/r²⟩ must be imported from the Feynman-Hellmann theorem of Unit 6 before s = −1 releases ⟨1/r³⟩, finite only for l ≥ 1. Every result is exact for a spinless, non-relativistic electron and a point nucleus, and the virial theorem marks the edge itself: ⟨T⟩/(μc²) = (Zα)²/(2n²), so relativistic corrections enter at relative order (Zα)², and the inverse moments weight exactly the small-r region where they, and the nucleus's size, act.
- Simple definition
- Radial expectation values ⟨rˢ⟩ are the moments of a state's radial density; for hydrogen the virial theorem fixes ⟨1/r⟩ = 1/(n²a), and Kramers' three-term recursion, derived from the radial equation, generates ⟨r⟩, ⟨r²⟩ and every higher moment by algebra alone.
- Example
- Hydrogen 3p (n = 3, l = 1): Kramers at s = 1 gives ⟨r⟩ = (a/2)(27 − 2) = 12.5a = 662 pm, while the virial theorem gives 1/⟨1/r⟩ = 9a = 477 pm, with a = 52.95 pm. Two different sizes of one state, and neither took an integral.
Derived. For V ∝ rᵏ it reads 2⟨T⟩ = k⟨V⟩: the oscillator (k = 2) splits E equally, and the Coulomb case (k = −1) gives ⟨T⟩ = −E and ⟨V⟩ = 2E.
T and V in J, r in m; any bound stationary state, since ⟨[H, r⋅p]⟩ = 0 and [H, r⋅p] = iħ(r⋅∇V − 2T)
Derived from ⟨V⟩ = 2Eₙ with no integral. For 1s, ∫₀^∞ (4/a³) r e(−2r/a) dr = 1/a confirms it. The Bohr radius n²a/Z survives as a harmonic mean, 1/⟨1/r⟩.
⟨1/r⟩ in m⁻¹; a = 4πε₀ħ²/(μe²) = 52.95 pm for hydrogen; the same for every l of a given n
Derived by integrating the radial equation by parts twice. Each use ties three neighbouring moments, so any two of them fix the third.
Z = 1 (for an ion replace a by a/Z); real s above −(2l + 1); every term has units of mˢ
Derived. It falls as l rises at fixed n, never below n²a, and with l(l + 1) = n²(1 − ε²) it is the Kepler time average n²a(1 + ε²/2) for eccentricity ε.
Kramers at s = 1; 1s gives 1.5a = 79.4 pm, 2p gives 5a, and l = n − 1 gives n²a[1 + 1/(2n)]
Derived. With ⟨r⟩ it gives Δr = (a/2)√[n²(n² + 2) − l²(l + 1)²], so a circular state has Δr/⟨r⟩ = 1/√(2n + 1), which shrinks only slowly with n.
⟨r²⟩ in m²; Kramers at s = 2 after s = 1; 1s gives 3a², 2p gives 30a²
Half derived, half quoted. 2p gives 1/(24a³). It is the radial factor of the spin-orbit energy in Unit 8, and it diverges for every s state.
Kramers at s = −1 gives ⟨1/r³⟩ = ⟨1/r²⟩/[l(l + 1)a]; ⟨1/r²⟩ = 1/[(l + ½)n³a²] is quoted from Unit 6
A commutator whose average must vanish
Quantum Mechanics I's Ehrenfest relation, d⟨Q⟩/dt = (i/ħ)⟨[H, Q]⟩, has a sharp corollary: in an eigenstate of H, ⟨ψ|[H, Q]|ψ⟩ = (E − E)⟨ψ|Q|ψ⟩ = 0 for any Q with a finite average. Choose Q = r⋅p, the generator of the dilation r → λr. From [xᵢ, pⱼ] = iħδᵢⱼ, [p², xᵢ] = −2iħpᵢ, so [T, r⋅p] = −2iħT, while [V, pᵢ] = iħ ∂V/∂xᵢ gives [V, r⋅p] = iħ r⋅∇V. Together [H, r⋅p] = iħ(r⋅∇V − 2T), and its vanishing average is the virial theorem 2⟨T⟩ = ⟨r⋅∇V⟩, derived here for any potential and any bound stationary state. Dimensions: r⋅∇V is a length times an energy per length, an energy. For V ∝ rᵏ, r⋅∇V = kV and 2⟨T⟩ = k⟨V⟩: the oscillator (k = 2) has ⟨T⟩ = ⟨V⟩ = E/2, as Quantum Mechanics I found with ladder operators. In a superposition of different energies the commutator's average oscillates, and the theorem holds only on the time average.
The Coulomb case: ⟨V⟩ = 2E and ⟨1/r⟩ for free
For V = −Ze²/(4πε₀r), r⋅∇V = r dV/dr = −V, so the theorem reads 2⟨T⟩ = −⟨V⟩. With ⟨T⟩ + ⟨V⟩ = Eₙ it gives ⟨T⟩ = −Eₙ and ⟨V⟩ = 2Eₙ, and topic 02.05's Eₙ = −Z²e²/(8πε₀an²) then gives ⟨1/r⟩ = Z/(n²a). No wavefunction was used, so l cannot appear: 2s and 2p share ⟨1/r⟩ = 1/(4a). The 1s integral ∫₀^∞ (4/a³) r e(−2r/a) dr = 1/a agrees. Now test ⟨T⟩ = −Eₙ. Against measurement: the model puts hydrogen's 1s ⟨T⟩, and so its binding energy, at 13.598 29 eV, while the measured ionisation energy is 13.598 43 eV, 1.1 × 10⁻⁵ higher. Against α: ⟨T⟩ = ½μ⟨v²⟩ gives vᵣₘₛ/c = Zα/n, so relativistic corrections are of relative order (Zα)²/n², and for 1s the Dirac correction, which deepens the level by (α²/4)|E₁| = 1.8 × 10⁻⁴ eV, less the 1s Lamb shift of 3.4 × 10⁻⁵ eV, closes the 1.5 × 10⁻⁴ eV gap (both quoted, from relativistic quantum mechanics and QED).
Kramers' relation from two integrations by parts
Write the radial equation for Z = 1 as u″ = Wu, with W = l(l + 1)/r² − 2/(ar) + 1/(n²a²): that is Eₙ = −ħ²/(2μa²n²) substituted. Multiply by rˢu′ and integrate from 0 to ∞. The left side is ∫rˢ(u′²/2)′ dr = −(s/2)∫rˢ⁻¹u′² dr; the right side is ∫rˢW(u²/2)′ dr = −½∫(rˢW)′u² dr, a sum of ⟨rˢ⁻³⟩, ⟨rˢ⁻²⟩ and ⟨rˢ⁻¹⟩. Multiply instead by rˢ⁻¹u: the left side becomes −∫rˢ⁻¹u′² dr + ½(s − 1)(s − 2)⟨rˢ⁻³⟩ and the right side ⟨rˢ⁻¹W⟩. Eliminate the unknown ∫rˢ⁻¹u′² dr, shift s by one, and Kramers' relation is left. Each surface term dropped on the way goes as r(s+2l+1) at the nucleus, because u ∝ r(l+1) there, and decays exponentially at infinity, so the relation is exact for every s above −(2l + 1), the same range in which every moment in it converges. Dimensions: all three terms carry mˢ. For a hydrogen-like ion replace a by a/Z.
Climbing the ladder: ⟨1/r⟩, then ⟨r⟩, then ⟨r²⟩
Set s = 0: the last coefficient vanishes and ⟨r⁰⟩/n² = a⟨1/r⟩, so ⟨1/r⟩ = 1/(n²a), the virial result recovered by an independent route. Set s = 1: 2⟨r⟩/n² − 3a + l(l + 1)a²⟨1/r⟩ = 0, so ⟨r⟩ = (a/2)[3n² − l(l + 1)], which is 1.5a = 79.4 pm for 1s, Quantum Mechanics I's integral answer. Set s = 2 and feed in ⟨r⟩: ⟨r²⟩ = (n²a²/2)[5n² + 1 − 3l(l + 1)], 3a² for 1s and 30a² for 2p. Check 2p directly, where u² = r⁴e(−r/a)/(24a⁵): ⟨r⟩ = 5!a/24 = 5a and ⟨r²⟩ = 6!a²/24 = 30a², as the recursion says. Subtracting gives the spread in closed form, Δr = (a/2)√[n²(n² + 2) − l²(l + 1)²], which is 0.866a = 45.9 pm for 1s. Each rung costs one line, and s = 3, 4, … continue without end.
The Kepler orbit hidden in the formulas
A classical Kepler ellipse with energy Eₙ has semi-major axis n²a, and its angular momentum sets the eccentricity ε through L² = ħ²n²(1 − ε²). Its time averages are ⟨1/r⟩ = 1/(n²a), ⟨r⟩ = n²a(1 + ε²/2) and ⟨r²⟩ = n⁴a²(1 + 3ε²/2), quoted from classical mechanics. Replace L²/ħ² by l(l + 1): the quantum ⟨1/r⟩ and ⟨r⟩ match them exactly, and ⟨r²⟩ exceeds the classical value by n²a²/2, a correction of relative order 1/n² that vanishes in the correspondence limit. The circular state l = n − 1 is the sharpest test. Its density peaks exactly at n²a, and Δr/⟨r⟩ = 1/√(2n + 1), which is 0.58 for 1s, 0.26 for n = 7 and still 0.13 for n = 30: quantum circular orbits sharpen only as n(−1/2). An s state is the opposite extreme, ε = 1, a radial orbit, and keeps ⟨r⟩ = 1.5n²a at every n.
Where the ladder stops, and what it hands on
Kramers cannot reach ⟨1/r²⟩. At s = 0 the ⟨rˢ⁻²⟩ coefficient vanishes, and at s = −1 the ⟨rˢ⟩ coefficient does, so s = −1 only ties two unknowns together: a⟨1/r²⟩ = l(l + 1)a²⟨1/r³⟩. One more input is needed, and Unit 6 supplies it: the Feynman-Hellmann theorem applied to the l-dependence of the radial equation gives ⟨1/r²⟩ = 1/[(l + ½)n³a²], quoted here and checked for 1s, where ∫₀^∞ (4/a³)e(−2r/a) dr = 2/a². Then ⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³], or 1/(24a³) for 2p. For l = 0 it is infinite: u² ∝ r² near the nucleus makes ∫u²/r³ dr diverge logarithmically, and s = −1 sits exactly on the edge s = −(2l + 1). This is the moment spin-orbit coupling needs in Unit 8. Since a ∝ 1/α, it scales as α³ and the energy it produces is of order α⁴mc², α² times the binding. Inverse moments weight small r, where the non-relativistic, point-nucleus model is least safe.
Change one variable at a time
Make the relationship visible.
Hold l = 0 and raise nᵣ: nodes appear, yet ⟨r⟩ stays at 1.5 n²a. Then set nᵣ = 0 and raise l: the solid line slides toward the dashed n²a, the bar narrows, and the single peak sits exactly on n²a, the circular orbit.
PRINCIPAL n = nᵣ + l + 13
⟨r⟩ = (a/2)[3n² − l(l + 1)]12.5 a
1/⟨1/r⟩ = n²a (VIRIAL)9 a
SPREAD Δr FROM ⟨r²⟩4.87 a
Live interpretationPRINCIPAL n = nᵣ + l + 1: 3. ⟨r⟩ = (a/2)[3n² − l(l + 1)]: 12.5 a. 1/⟨1/r⟩ = n²a (VIRIAL): 9 a. SPREAD Δr FROM ⟨r²⟩: 4.87 a
Catch the common trap
Explain before calculating.
Starting from ⟨r⁰⟩ = 1 and using Kramers' relation alone, with no other input, which radial moments of hydrogen's 3p state (n = 3, l = 1) can you obtain?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA He⁺ ion (Z = 2, nuclear mass 7294.30 mₑ) is in its 2p state. Using only the virial theorem and the Bohr energy, find ⟨T⟩ and ⟨V⟩ in eV, the harmonic-mean radius 1/⟨1/r⟩ in pm, and the rms relative speed as a fraction of c. Take 13.6057 eV for the Rydberg energy, a₀ = 52.9177 pm and mₑc² = 510 999 eV.
- Reduced mass: μ/mₑ = 1/(1 + 1/7294.30) = 0.999863. The Bohr energy, derived in topic 02.05, is E₂ = −Z²(μ/mₑ)(13.6057 eV)/n² = −4 × 0.999863 × 13.6057 eV/4 = −13.6038 eV.
- Coulomb potential: r⋅∇V = −V, so 2⟨T⟩ = −⟨V⟩. With ⟨T⟩ + ⟨V⟩ = E₂ this gives ⟨T⟩ = −E₂ = 13.604 eV and ⟨V⟩ = 2E₂ = −27.208 eV, which sum back to E₂.
- ⟨V⟩ = −(Ze²/4πε₀)⟨1/r⟩ and E₂ = −Z²e²/(8πε₀an²) give ⟨1/r⟩ = Z/(n²a), with a = a₀(mₑ/μ) = 52.9177 pm × 1.000137 = 52.925 pm. So 1/⟨1/r⟩ = n²a/Z = 4 × 52.925 pm/2 = 105.85 pm, and 2s has the same value, since no l entered.
- ⟨T⟩ = ½μ⟨v²⟩ gives vᵣₘₛ/c = √(2⟨T⟩/μc²) = √(2 × 13.6038 eV/510 929 eV) = 7.297 × 10⁻³, which is Zα/n = 2α/2 = α, or 2.19 × 10⁶ m s⁻¹.
- Checks: every step equates energies or pure numbers, and (v/c)² = 5.3 × 10⁻⁵ is the relative size, of order (Zα)²/n², of the relativistic corrections this model leaves out.
Answer⟨T⟩ = 13.604 eV, ⟨V⟩ = −27.208 eV, 1/⟨1/r⟩ = 105.85 pm and vᵣₘₛ = 7.297 × 10⁻³ c = 2.19 × 10⁶ m s⁻¹, exactly hydrogen 1s's value, since vᵣₘₛ/c = Zα/n and Z/n = 1 for both.
MediumFor hydrogen's 3s and 3d states, use Kramers' relation to find ⟨r⟩, ⟨r²⟩ and the spread Δr, in units of a and in pm with a = 52.947 pm. Confirm the 3d values by a direct ratio of integrals, and say which radius the two states share and why.
- Both have n = 3, so s = 0, or the virial theorem, gives ⟨1/r⟩ = 1/(9a): both have harmonic-mean radius 9a = 476.5 pm, because ⟨1/r⟩ is fixed by E₃ alone.
- s = 1: ⟨r⟩ = (a/2)[27 − l(l + 1)]. For 3s (l = 0) that is 13.5a = 714.8 pm; for 3d (l = 2) it is (a/2)(27 − 6) = 10.5a = 555.9 pm.
- s = 2: ⟨r²⟩ = (9a²/2)[46 − 3l(l + 1)], so 207a² for 3s and (9/2)(28)a² = 126a² for 3d. Then Δr = √(⟨r²⟩ − ⟨r⟩²): for 3s √(207 − 182.25) a = 4.975a = 263.4 pm, for 3d √(126 − 110.25) a = 3.969a = 210.1 pm.
- Direct check for 3d, which is nodeless: u ∝ r³e(−r/3a), so u² ∝ r⁶e(−br) with b = 2/(3a), and the normalisation cancels in a ratio. ⟨r⟩ = 7!b⁻⁸/(6!b⁻⁷) = 7/b = 10.5a and ⟨r²⟩ = 8!b⁻⁹/(6!b⁻⁷) = 56/b² = 126a². Both agree.
- Checks: both means exceed 9a, as Jensen's inequality requires, and the Kepler average 9a(1 + ε²/2) with ε² = 1 − 6/9 = 1/3 is exactly 10.5a for 3d.
Answer3s: ⟨r⟩ = 13.5a = 715 pm, ⟨r²⟩ = 207a², Δr = 4.97a = 263 pm. 3d: ⟨r⟩ = 10.5a = 556 pm, ⟨r²⟩ = 126a², Δr = 3.97a = 210 pm. Both share 1/⟨1/r⟩ = 9a = 477 pm.
HardHydrogen's 2p level splits into j = 3/2 and j = 1/2 levels, measured 10.969 GHz apart. Treat the proton as infinitely heavy, so a = a₀ = 52.918 pm, and take the spin-orbit operator [e²/(8πε₀mₑ²c²)](1/r³) L⋅S as quoted from Unit 8. Use Kramers' relation to get ⟨1/r³⟩ for 2p, predict the splitting in GHz, and explain why the same route fails for 2s.
- Kramers at s = −1 is allowed for l = 1, since −1 lies above −(2l + 1) = −3. The (s + 1) term drops out, leaving a⟨1/r²⟩ − ¼[3² − 1]a²⟨1/r³⟩ = 0, so ⟨1/r³⟩ = ⟨1/r²⟩/(2a).
- ⟨1/r²⟩ = 1/[(l + ½)n³a²] = 1/(12a²) is quoted from the Feynman-Hellmann theorem of Unit 6. Directly, with u² = r⁴e(−r/a)/(24a⁵): ∫₀^∞ r²e(−r/a) dr/(24a⁵) = 2a³/(24a⁵) = 1/(12a²). So ⟨1/r³⟩ = 1/(24a³) = 2.812 × 10²⁹ m⁻³.
- L⋅S = (ħ²/2)[j(j + 1) − l(l + 1) − ¾], with ¾ from the electron's spin ½, is +ħ²/2 for j = 3/2 and −ħ² for j = 1/2, a difference of 3ħ²/2. So ΔE = (3ħ²/2)[e²/(8πε₀mₑ²c²)]/(24a³).
- With e²/4πε₀ = αħc and a = ħ/(mₑcα) this is ΔE = α⁴mₑc²/32 = (7.2974 × 10⁻³)⁴ × 510 999 eV/32 = 4.528 × 10⁻⁵ eV, so ΔE/h = 10.949 GHz. Order: α⁴mc², which is α²/4 = 1.3 × 10⁻⁵ of the 3.40 eV binding of n = 2.
- The prediction is 0.18% below the measured 10.969 GHz: the electron's anomalous magnetic moment (QED, quoted) raises it by 0.23% and nuclear recoil (also quoted) lowers it by 0.05%. For 2s, s = −1 = −(2l + 1) lies outside Kramers' range and ⟨1/r³⟩ diverges, but L⋅S = 0, so there is no spin-orbit splitting to find.
Answer⟨1/r³⟩ = 1/(24a³) = 2.81 × 10²⁹ m⁻³ and ΔE = α⁴mₑc²/32 = 4.53 × 10⁻⁵ eV, a splitting of 10.95 GHz against the measured 10.969 GHz.