University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.8
Power-law potentials, and the Langer correction for hydrogen
Here the WKB rule stops being a curiosity and starts producing spectra. One integral gives the ladder of every power-law well, and one honest repair at the origin turns the radial equation into hydrogen's Bohr levels. The skill to take away is telling which part of each answer is exact, which is leading order in ħ, and which is luck.
Build the model
Connect the measurement to the mechanism.
Topic 11.07 turned the connection formulas into a rule, ∫p dx = (n − ½)πħ between two smooth turning points. This topic asks how far that rule reaches. For every power law V ∝ |x|ν one substitution makes the phase integral a fixed number times E(1/2 + 1/ν), so the whole ladder follows at once, Eₙ ∝ (n − ½)q with q = 2ν/(ν + 2), and its spacing is ħ times the classical frequency.
The rule is exact at ν = 2, good to about a percent beyond the lowest level or two for moderate ν, and wrong in the box limit, where the walls bend within the Airy length that the π/4 phase assumes is straight. The radial equation fails more instructively. Its origin is an edge, not a turning point, and near it the centrifugal term varies on the scale of r itself, so WKB with l(l + 1) is never valid there and its energies are wrong, by 9% for hydrogen's 2p.
Rewriting the equation in x = ln r moves the origin to minus infinity and turns l(l + 1) into (l + ½)² exactly. WKB applied in that variable is Langer's rule, and with one table integral it returns hydrogen's −13.6 eV/n² and the oscillator's (2nᵣ + l + 3/2)ħω exactly. The exactness belongs to those two potentials, not to the method: a linear potential misses by a few tenths of a percent, and the WKB wavefunctions are never exact.
- Simple definition
- The Langer correction replaces l(l + 1) by (l + ½)² in the centrifugal term before WKB is applied to the radial equation, so that the WKB solution goes as r(l+1) at the origin and the smooth-turning-point rule (nᵣ + ½)πħ holds.
- Example
- Hydrogen 2p has nᵣ = 0 and l = 1. Keeping l(l + 1) = 2 gives neff = ½ + √2 = 1.914 and −3.71 eV; with (l + ½)² = 2.25 the rule gives n = 2 and −3.40 eV, Unit 2's exact energy.
Derived here from the (n − ½)πħ rule: x = xₜs turns every power law into one integral. ν = 2 returns (n − ½)ħω exactly; ν = 1 is Unit 10's V-shaped well.
V = V₀|x/a|ν with V₀ in J, a in m, m in kg; n = 1, 2, 3, …; Iν = ∫₀¹√(1 − sν) ds: I₁ = 2/3, I₂ = π/4
Derived by differentiating ∮p dx = (2n − 1)πħ with respect to E. A period that grows with E (ν below 2) crowds the levels; one that shrinks (ν above 2) spreads them.
T(E) = ∮dx/v, the classical period in s; ωcl in rad s⁻¹. For V ∝ |x|ν it is ħωcl = qEₙ/(n − ½)
Exact, derived here by the chain rule; the extra ¼ is the Jacobian. WKB applied in x, where the bracket tends to a constant at the origin, is the Langer rule.
u″ = d²u/dr² and φ″ = d²φ/dx²; k² = 2m[E − V(r)]/ħ² in m⁻²; x = ln(r/r₀) is dimensionless and runs over the whole real line, so r = 0 sits at x → −∞ and is no longer an edge
The change adds ħ²/(8mr²), the order in ħ that WKB discards anyway, so the ħ expansion cannot forbid it; the origin demands it, because it makes the WKB tail r(l+1).
r₁, r₂ turning points in m; E, V in J; nᵣ = 0, 1, 2, … counts radial nodes; the naive rule has l(l + 1) for (l + ½)²
Hydrogen uses it directly and the isotropic oscillator after s = r². When b² = ac the roots merge and it gives zero: the circular orbit, where no level can sit.
ar², br and c share one unit; a, b and c positive, b² above ac. Quoted from tables (contour integration); c = 0 gives πb/√a
Derived here from the table integral, identical to Unit 2's. It is order α²μc², so Unit 8's order-α⁴ fine structure lies outside it: Balmer α at 656.47 nm against 656.46 nm measured.
k = 1/(4πε₀); μ reduced mass in kg; α = 1/137.036; 13.598 eV/n². For l ≥ 1, keeping l(l + 1) gives neff = nᵣ + ½ + √(l(l + 1))
One substitution quantises every power law
Take V = V₀|x/a|ν and topic 11.07's rule, ∫p dx = (n − ½)πħ between the turning points ±xₜ, where xₜ = a(E/V₀)(1/ν). Put x = xₜs and the integrand becomes √(2mE)√(1 − sν), so ∫p dx = 2xₜ√(2mE) Iν with Iν = ∫₀¹√(1 − sν) ds, a pure number. Energy now enters only through xₜ√E, which goes as E(1/2 + 1/ν); setting it equal to (n − ½)πħ gives Eₙ ∝ (n − ½)q with q = 2ν/(ν + 2), derived here to leading order in ħ. Iν is a Beta function quoted from tables, but you need only I₁ = 2/3, I₂ = π/4 and I → 1 as ν → ∞. Dimensions: ħ/(a√(mV₀)) is J s over kg m² s⁻¹, a pure number. At ν = 2, with V₀/a² = ½mω², the formula gives (n − ½)ħω exactly. At ν = 1, Unit 10's V = F|x|, it gives [3π(n − ½)/4]²⁄³ in units of (ħ²F²/2m)¹⁄³: 1.115 for n = 1 against the exact 1.0188 (9.5% high), and 2.320 for n = 2 against 2.3381 (0.8% low), which is the bouncer's first level.
Level spacing is set by the classical period
Differentiate the rule. The closed-orbit integral ∮p dx = (2n − 1)πħ has E-derivative ∮dx/v = T(E), the classical period, so neighbouring levels sit ΔE ≈ 2πħ/T = ħωcl apart: the correspondence principle, derived. For a power law ∮p dx ∝ E(1/2 + 1/ν), so T ∝ E(1/ν − 1/2). At ν = 2 the period ignores amplitude and the ladder is even. Below ν = 2 the period lengthens with energy and levels crowd: WKB gives the V-shaped well a spacing ratio (E₄ − E₃)/(E₂ − E₁) of 0.68. Above ν = 2 they spread: 1.46 for the quartic, 2.10 at ν = 10. Mass enters only through ħ²/m, so Eₙ ∝ m(−q/2), a scaling that is exact, not merely WKB. Data test it: the measured 2S–1S gaps of charmonium, 589 MeV, and bottomonium, 563 MeV, differ by under 5% although the b quark is about three times heavier than the c (quoted). That needs ν ≈ 0.09, a nearly logarithmic potential; a Coulomb law would predict about 1800 MeV for bottomonium and a linear one about 400 MeV.
The box limit: right exponent, wrong offset
As ν grows, q → 2 and Iν → 1, so the formula tends to (n − ½)²π²ħ²/(8ma²), while the infinite well of width 2a has n²π²ħ²/(8ma²). The exponent is right, the offset wrong: the ground state comes out a quarter of the truth. The fault is in the connection formula. Its π/4 assumes V is a straight line across the Airy length ℓ = (ħ²/2mF)¹⁄³ around the turning point, with F = V′(xₜ), while a power-law wall bends over a distance xₜ/(ν − 1). So (ν − 1)ℓ/xₜ must be small. With the exact levels it is 0.47 for the quartic at n = 5, but 3.8 at ν = 10, n = 1 and still 1.1 at n = 5. Numerical diagonalisation agrees: at ν = 10 WKB is 43% low for n = 1, 10% low for n = 2 and 1% low for n = 5. For any finite ν the rule improves as n grows, but ν → ∞ at fixed n never recovers the box, whose hard walls supply π/2 each, as topic 11.07 showed. The oscillator's ratio is 0.79 at n = 1, yet it is exact: a property of that potential, not a licence.
At the origin the radial equation has no turning point
The radial equation for u = rR is one-dimensional on r ≥ 0, with Veff = V + ħ²l(l + 1)/(2mr²) and u(0) = 0. Applying topic 11.07's rule to it naively fails two ways. For l = 0 there is no inner turning point, so r = 0 must act as a hard wall and the rule becomes (nᵣ + ¾)πħ; for hydrogen that puts 1s at −24.2 eV, because near a 1/r singularity the local wavelength goes as √r and its slope dλ/dr diverges. For l ≥ 1 the centrifugal barrier does make an inner turning point, but not a WKB one. Near the origin κ ≈ √(l(l + 1))/r, so the validity measure |d(1/κ)/dr| = 1/√(l(l + 1)) never shrinks however close to the origin you go: 0.71 at l = 1. The WKB tail κ(−1/2) exp(∫κ dr) grows as r to the power ½ + √(l(l + 1)), 1.914 at l = 1, where Unit 2 found u ∝ r(l+1). The energies inherit the error: 2p lands at −3.71 eV and 2s at −4.44 eV against −3.40 eV for both, so even the n² degeneracy is lost.
Langer's replacement, from r = r₀eˣ
Change variable to x = ln(r/r₀), r₀ any fixed length, and write u = e(x/2)φ. The chain rule turns u″ + [k² − l(l + 1)/r²]u = 0, with k² = 2m(E − V)/ħ², into φ″ + [r²k² − (l + ½)²]φ = 0. Nothing has been approximated; the ¼ is the Jacobian of the change of variable. In x the problem lives on the whole real line. The origin sits at x → −∞, where the bracket tends to the constant −(l + ½)², and WKB is exact for a constant, so both turning points are smooth and the ordinary (nᵣ + ½)πħ applies. Converting the phase back, ∫√(r²k² − (l + ½)²) dx = ∫√(k² − (l + ½)²/r²) dr: the Langer rule, derived here in outline. The tail is φ ∝ r(l+½) as x → −∞, so u = r¹⁄²φ ∝ r(l+1), Unit 2's behaviour. In r the replacement adds ħ²/(8mr²) to the potential, a term of order ħ², which is precisely the order WKB throws away. The ħ expansion alone therefore cannot choose between l(l + 1) and (l + ½)²; the origin chooses.
Hydrogen and the oscillator come out exact, little else
With the Langer term the hydrogen phase integral is ∫√(2mEr² + 2mke²r − ħ²(l + ½)²) dr/r, the table form with a = 2m|E|, b = mke² and c = ħ²(l + ½)². It equals π[mke²/√(2m|E|) − ħ(l + ½)], and setting that to (nᵣ + ½)πħ gives E = −m(ke²)²/(2ħ²n²) with n = nᵣ + l + 1: Unit 2's energies, exactly, with m the reduced mass. Checks: mke²/ħ is a momentum, so the dimensions close; E = −α²μc²/(2n²) is order α², 13.598 eV/n², and Balmer α lands at 656.47 nm against 656.46 nm measured, the rest being Unit 8's order-α⁴ fine structure. With s = r² the isotropic oscillator gives ħω(2nᵣ + l + 3/2), also exact. There the luck ends. For V = Fr numerical diagonalisation puts the Langer rule 0.49% high for 1s and 0.12% for 1p. Even for hydrogen the WKB wavefunction is not exact: it diverges at the 1s turning points, 0.13 a₀ and 1.87 a₀, where the true 1s is smooth.
Change one variable at a time
Make the relationship visible.
Slide ν from 1 to 10: the levels crowd, space evenly at ν = 2, then spread like a box's. On the right the solid Langer curve is the same for every l in these units; at l = 0 the dashed curve loses its barrier and its level falls to −24 eV. Raise l and the naive error falls from 9% to 1%.
WKB EXPONENT q = 2ν/(ν + 2)1.33
SPACING (E₄ − E₃)/(E₂ − E₁)1.46
HYDROGEN Eₙ, LANGER = EXACT-3.400 eV
NAIVE l(l + 1) ERROR9.2 %
Live interpretationWKB EXPONENT q = 2ν/(ν + 2): 1.33. SPACING (E₄ − E₃)/(E₂ − E₁): 1.46. HYDROGEN Eₙ, LANGER = EXACT: −3.400 eV. NAIVE l(l + 1) ERROR: 9.2 %
Catch the common trap
Explain before calculating.
A hydrogen 3p state has nᵣ = 1 and l = 1. You quantise the radial phase integral as (nᵣ + ½)πħ and read off E = −13.6 eV/neff². Which statement is right?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyNumerical diagonalisation for a particle in the quartic well V = V₀(x/a)⁴ gives E₂ = 12.00 meV, E₄ = 36.78 meV, E₅ = 51.36 meV and E₆ = 67.07 meV. Predict E₅ from E₂ alone with the WKB power law, compare, and estimate the level spacing near n = 5 from the classical frequency.
- For ν = 4 the exponent is q = 2ν/(ν + 2) = 8/6 = 4/3, so Eₙ ∝ (n − ½)⁴⁄³, and V₀, a, m and I₄ all cancel in a ratio.
- E₅/E₂ = (4.5/1.5)⁴⁄³ = 3⁴⁄³ = 4.327, so E₅ ≈ 12.00 meV × 4.327 = 51.92 meV.
- Against the numerical 51.36 meV the prediction is 1.1% high. WKB's own E₅ is only 0.17% low but its E₂ is 1.26% low, so scaling from the true E₂ carries that error up the ladder: the lowest levels are the least semiclassical.
- Other exponents do far worse: the oscillator's q = 1 gives 12.00 × 3 = 36.0 meV, 30% low, and the box's n² gives 12.00 × 25/4 = 75.0 meV, 46% high.
- Spacing: ħωcl = qEₙ/(n − ½) = (4/3)(51.92 meV)/4.5 = 15.4 meV, between the numerical E₅ − E₄ = 14.58 meV and E₆ − E₅ = 15.71 meV, as a derivative at n = 5 should be.
AnswerE₅ ≈ 51.9 meV from the WKB exponent 4/3, 1.1% above the numerical 51.36 meV; the local spacing ħωcl ≈ 15.4 meV.
MediumPredict the vacuum wavelength of hydrogen's 3d → 2p line, the strongest component of Balmer α, from radial WKB: first keeping l(l + 1), then with the Langer term. Use E = −13.5983 eV/neff², which includes the reduced mass, compare with the measured 656.46 nm, and say what sets the remaining difference.
- The table integral turns (nᵣ + ½)πħ into neff = nᵣ + ½ + √(l(l + 1)) with l(l + 1) kept, or neff = nᵣ + l + 1 with (l + ½)². Both states here have nᵣ = 0.
- Naive: 3d has neff = ½ + √6 = 2.9495, so E = −13.5983/8.6995 = −1.5631 eV; 2p has neff = ½ + √2 = 1.9142, so E = −3.7111 eV. ΔE = 2.1480 eV and λ = hc/ΔE = 1239.842 eV nm/2.1480 eV = 577.2 nm.
- Langer: neff = 3 and 2, so ΔE = 13.5983 eV × (1/4 − 1/9) = 13.5983 eV × 5/36 = 1.88865 eV and λ = 1239.842/1.88865 = 656.47 nm.
- Measured: 656.46 nm in vacuum (656.28 nm in air). The Langer value agrees to about 1 part in 10⁵; the naive one is 12% short, a yellow line where the atom emits red.
- The residual is fine structure, Unit 8's subject. The energies used are order α²μc² and non-relativistic by construction; the order-α⁴ corrections shift lines by a fraction of order α² = 5.3 × 10⁻⁵, the size of the gap.
AnswerNaive WKB 577 nm; Langer WKB 656.47 nm; measured 656.46 nm. The residual, about 1 part in 10⁵, is fine structure, order α² relative.
HardFor V = ½mω²r²: (a) use s = r² and the table integral to show that the Langer rule gives E = ħω(2nᵣ + l + 3/2); (b) find the naive energies of the 1p, 1d and 2s levels, the 2s from the hard-wall rule; (c) for a nucleon in a shell-model well with ħω = 14.0 MeV, give the naive 1p shift and 2s–1d splitting, and say which measured numbers the Langer spectrum accounts for.
- Write Λ² for the centrifugal coefficient. With s = r², dr = ds/(2√s), so ∫√(2mE − m²ω²r² − ħ²Λ²/r²) dr = ½∫√(−m²ω²s² + 2mEs − ħ²Λ²) ds/s.
- Table form with a = m²ω², b = mE, c = ħ²Λ²: the integral is ½π(E/ω − ħΛ). Setting it to (nᵣ + ½)πħ gives E = ħω(2nᵣ + 1 + Λ). With Λ = l + ½, E = ħω(2nᵣ + l + 3/2), the exact (N + 3/2)ħω that Cartesian separation gives, with N = 2nᵣ + l.
- Naive, Λ = √(l(l + 1)): 1p is (1 + √2)ħω = 2.414ħω and 1d is (1 + √6)ħω = 3.449ħω. For 2s (l = 0) the wall rule sets ∫₀ p dr = πE/(2ω) equal to (1 + ¾)πħ, giving 3.5ħω, exact, because an l = 0 state is an odd one-dimensional state that already vanishes at r = 0.
- At ħω = 14.0 MeV: naive 1p is 33.80 MeV, 1.20 MeV below the exact 35.0 MeV, and the N = 2 shell splits, 1d at 48.29 MeV against 2s at 49.0 MeV, a gap of 0.71 MeV.
- Langer keeps each shell degenerate, holding (N + 1)(N + 2) nucleons of one kind with spin: 2, 6 and 12, closing at 2, 8 and 20, the measured magic numbers (quoted). The next measured one is 28, not 40: spin–orbit coupling, which no central potential contains.
AnswerE = ħω(2nᵣ + l + 3/2) exactly; naive 1p = 2.414ħω (1.20 MeV low at 14.0 MeV), 1d = 3.449ħω against 2s = 3.5ħω, a 0.71 MeV split; the Langer shells close at 2, 8 and 20.