University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.5
The linearised turning point and the Airy functions
WKB is built to fail at the one place a bound particle must visit: the point where it turns round. Zoom in there until the potential is a straight line, solve that problem exactly with the Airy function, and its two tails turn out to be WKB waves already. That is the bridge the connection formulas will cross.
Build the model
Connect the measurement to the mechanism.
WKB fails within a few l of a turning point, because there the local wavelength stops being short compared with the distance over which it changes. Zoom in far enough, though, and any smooth potential looks like its tangent line, V ≈ E + V′(a)(x − a), and a straight-line potential has an exact solution. Measuring distance in units of l = (ħ²/(2mV′(a)))¹⁄³ strips out every constant and leaves Airy's equation ψ″ = zψ, the same for an electron in a field, a neutron above a mirror or an oscillator near its classical edge.
Its decaying solution Ai is finite at the turning point, decays as exp(−2z³⁄²/3) in the forbidden region and oscillates with a π/4 phase on the allowed side; stationary phase on its integral representation derives the oscillating form, and the decaying form is quoted. Both exponents are exactly (1/ħ)∫|p| dx measured from the turning point, so many l out the Airy tails are WKB waves, and Ai becomes a bridge that carries one WKB solution across the point where WKB breaks. The bridge needs ground to land on: a zone with |z| ≫ 1 where the potential is still straight, which exists only if l is small beside the length L = |V′/V″| over which the slope changes.
Since l ∝ ħ²⁄³, that zone opens as ħ → 0, so the matching holds at leading order in ħ. It closes for the lowest levels, at a turning point where V′ = 0, and at a hard wall.
- Simple definition
- Linearising at a turning point a, where V(a) = E, replaces V by its tangent line there; in the variable z = (x − a)/l this turns the Schrödinger equation into Airy's equation ψ″ = zψ, whose solution that decays into the forbidden region is Ai(z).
- Example
- A neutron above a mirror feels V = mgy, exactly linear, so l = (ħ²/(2gm²))¹⁄³ = 5.87 μm and ε = mgl = 0.602 peV. Near a turning height h its wavefunction is proportional to Ai((y − h)/l), which at y = h equals Ai(0) = 0.355, not zero.
The width of the region WKB cannot describe. It scales as ħ²⁄³, so it shrinks as ħ → 0, and it grows without bound where V′(a) → 0.
F: force at the turning point, in N; ħ in J s, m in kg, l in m. Electron in 5.0 × 10⁶ V m⁻¹: 1.97 nm; neutron in gravity: 5.87 μm
Every smooth turning point gives this one parameter-free equation. Its solutions are Ai, which decays into the forbidden side, and Bi, which grows there.
z is dimensionless and positive on the forbidden side when V′(a) is positive; at a left turning point use z = (a − x)/l
Its exponent is (1/ħ)∫|p| dx from the turning point and its prefactor goes as |p|(−1/2): a decaying WKB wave for the tangent line.
Quoted from the asymptotic series, next factor 1 − 5/(72ζ); at z = 2 it gives 0.0360 against the tabulated 0.0349
The π/4 is the phase of a Fresnel integral. The form puts the first zero of Ai at u = 2.320, against the tabulated 2.338.
u = −z, dimensionless. Derived from Ai(z) = (1/π)∫₀^∞ cos(t³/3 + zt) dt at its stationary point t₀ = √u
Why both Airy tails are WKB waves, and why ħ|p′|/p² = 1/(2|z|³⁄²) says WKB fails within about l of the turning point.
p in kg m s⁻¹, l in m; exact for the tangent line, where the WKB amplitude |p|(−1/2) goes as |z|(−1/4)
It exists only if L/l ≥ 0.0023 η(−8/3), which is 6.8 at η = 0.05. An oscillator has L/l = 2¹⁄³(2n + 1)²⁄³, so at that tolerance it needs level n ≥ 6.
η in rad. Lower edge: Ai's asymptotic error 5/(48|z|³⁄²). Upper edge: tangent-line phase error |z|⁵⁄² l/(10L)
Zoom in until the potential is straight
Near a point where V(a) = E, expand V(x) ≈ E + V′(a)(x − a) and drop the curvature. The Schrödinger equation becomes ħ²ψ″ = 2mV′(a)(x − a)ψ. Put x − a = lz and choose l³ = ħ²/(2mV′(a)) so the constants cancel: ψ″(z) = zψ, derived with no further assumption. Dimensions check: (J s)² ÷ (kg × J m⁻¹) = J s² m kg⁻¹ = m³. The energy scale is ε = V′(a)l. An electron against 5.0 × 10⁶ V m⁻¹ has l = 1.97 nm and ε = 9.84 meV; a neutron in gravity, V′ = mg, has l = 5.87 μm and ε = 0.602 peV. That last case is exactly linear and has been measured: the neutron-mirror experiments at the ILL, from 2002 on, measured the height of the lowest state as 12.2 ± 1.8 (systematic) ± 0.7 (statistical) μm, against the predicted 2.338 l = 13.7 μm, with 2.338 the first zero of Ai quoted from a table. Limits check the form: a hard wall, V′ → ∞, squeezes l to zero; a flat top, V′ → 0, sends it to infinity, where linearising fails.
In momentum space the equation is first order
A straight-line potential is special because position acts on a momentum-space wavefunction as a derivative, so Airy's equation, second order in z, is first order in the conjugate variable. Write ψ(z) = (1/(2π))∫Y(t) e(izt) dt, where t is the momentum in units of ħ/l. Then ψ″ contributes −t²Y, and zψ, after one integration by parts, contributes iY′(t). Airy's equation demands Y′ = it²Y, so Y = e(it³/3); the odd part of the integrand cancels, leaving Ai(z) = (1/π)∫₀^∞ cos(t³/3 + zt) dt, derived, with the overall constant fixed by convention. The second solution, Bi, has quoted forms too: Bi(z) ≈ e(+ζ)/(√π z¹⁄⁴) on the forbidden side, growing and with twice Ai's prefactor, and Bi(−u) ≈ π(−1/2)u(−1/4) cos(ζ + π/4) on the allowed side, a quarter period out of step with Ai. The turning-point values Ai(0) = 0.3550, Ai′(0) = −0.2588 and Bi(0) = 0.6149 are quoted from tables, and the Wronskian of the pair is 1/π. When the forbidden region runs to infinity, any admixture of Bi diverges, so the state is pure Ai: at z = 0 it is neither zero nor singular, and it falls to a tenth of Ai(0) by z = 2.
Stationary phase finds the oscillating tail
Set z = −u with u positive. The phase φ(t) = t³/3 − ut is stationary where t² = u, at t₀ = √u, with φ(t₀) = −(2/3)u³⁄² = −ζ and φ″(t₀) = 2√u. Away from t₀ the cosine oscillates ever faster and cancels itself; near it the integrand is cos(√u s² − ζ), with s = t − t₀. Expand the cosine of the difference and use the Fresnel integrals ∫cos(as²) ds = ∫sin(as²) ds = √(π/(2a)) over the whole line: the result is √(π/a) cos(ζ − π/4) with a = √u, so Ai(−u) ≈ π(−1/2)u(−1/4) sin(ζ + π/4), derived. The π/4 comes from the equal cosine and sine parts of the Fresnel integral, the trace of the phase's curvature at t₀. The step needs the cubic term to be negligible across the stationary region, whose width is (2√u)(−1/2), and that requires u ≫ 1. For positive z, φ′ = t² + z never vanishes, nothing real is stationary and the integral is exponentially small; the form e(−ζ)/(2√π z¹⁄⁴) comes from a saddle at t = i√z and is quoted here.
Airy's tails are WKB waves
For the tangent line the classical momentum is p = √(2mV′(a)(a − x)) = (ħ/l)|z|¹⁄², so (1/ħ)∫ₓᵃ p dx′ = (2/3)|z|³⁄² = ζ, exactly the Airy phase, and the WKB amplitude p(−1/2) ∝ |z|(−1/4) is exactly the Airy prefactor. The forbidden side works the same way with |p|. Both asymptotic forms are therefore WKB waves, and Ai supplies the two numbers WKB cannot: the relative amplitude, π(−1/2) against 1/(2√π), a factor of 2, and the phase π/4. The next lesson turns those into connection formulas. Why WKB fails here is now quantitative: ħ|p′|/p² = 1/(2|z|³⁄²), which is 0.5 at |z| = 1. The true phase error of the asymptotic form, 5/(72ζ) = 5/(48|z|³⁄²) from the quoted next term, is 4.8 times smaller: 0.10 rad at |z| = 1, 0.037 at 2 and 0.020 at 3. And where the WKB density, proportional to 1/p like a classical particle's dwell time, diverges at the turning point, Ai rounds it into a finite peak: |Ai|² reaches its tabulated maximum, 0.287, at z = −1.02, about one l inside.
An overlap zone needs a smooth potential
Ai is exact only for the tangent line. Restore ½V″(a)(x − a)² and |p| becomes (ħ/l)|z|¹⁄²(1 ± |z| l/(2L))¹⁄², with L = |V′(a)/V″(a)|; expanding to first order and integrating from the turning point shifts the phase by |z|⁵⁄² l/(10L) and the amplitude by a fraction |z| l/(8L), derived; beyond |z| ≈ 1.2 the phase shift is the larger. Matching needs |z| large enough for Ai's asymptotic error 5/(48|z|³⁄²) and small enough for the tangent-line error. Their sum is least at |z| = (5L/8l)¹⁄⁴, where it is (1/6)(5L/8l)(−3/8). For oscillator level n, L is the turning point a and L/l = 2¹⁄³(2n + 1)²⁄³: 1.26 at n = 0, where the whole well is 2.5 l wide and even the best mismatch is 0.18 rad; 19.5 at n = 30, with 0.065 rad; 68.5 at n = 200, with 0.041 rad. Above a mirror V″ = 0 and the zone is unlimited. Because L/l ∝ ħ(−2/3), the zone always opens as ħ → 0, and no power of α enters. This is a pointwise mismatch between leading-order forms, not the error in the energies the connection formulas give.
Change one variable at a time
Make the relationship visible.
Start at n = 0: the well is only 2.5 l wide and the dashed hump lies on Ai nowhere. Raise n and the shaded overlap opens, reaching |z| = 4.1 at n = 200, while the dot, the best matching point, creeps out only as (L/l)¹⁄⁴.
SMOOTHNESS L/l19.5
WKB HOLDS FOR |z| ≥1.63
TANGENT LINE HOLDS FOR |z| ≤2.49
PHASE ERROR AT BEST MATCH0.065 rad
Live interpretationSMOOTHNESS L/l: 19.5. WKB HOLDS FOR |z| ≥: 1.63. TANGENT LINE HOLDS FOR |z| ≤: 2.49. PHASE ERROR AT BEST MATCH: 0.065 rad
Catch the common trap
Explain before calculating.
Near the right turning point of an oscillator's level n = 100, L/l = 43. Taking 5/(48|z|³⁄²) as the error of Ai's asymptotic form and |z|⁵⁄² l/(10L) as the tangent line's, where can Ai be matched to the WKB wave with each phase error at most 0.05 rad?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron climbs against a uniform electric field of 5.0 × 10⁶ V m⁻¹, so near its turning point V′(a) = eE. Find the Airy length l and the energy scale ε = V′(a)l, confirm that l is a length, and find how far from the turning point Ai's asymptotic form, the WKB wave, is good to 0.05 rad.
- Slope: V′(a) = eE = (1.602 × 10⁻¹⁹ C)(5.0 × 10⁶ V m⁻¹) = 8.01 × 10⁻¹³ N.
- l³ = ħ²/(2mV′(a)) = (1.0546 × 10⁻³⁴ J s)² ÷ (2 × 9.109 × 10⁻³¹ kg × 8.01 × 10⁻¹³ N) = 1.112 × 10⁻⁶⁸ ÷ 1.459 × 10⁻⁴² = 7.62 × 10⁻²⁷ m³, so l = 1.97 × 10⁻⁹ m.
- Dimensions: J² s² ÷ (kg J m⁻¹) = J s² m kg⁻¹, and with J = kg m² s⁻² this is m³, whose cube root is a length.
- Energy scale: ε = V′(a)l = 8.01 × 10⁻¹³ N × 1.97 × 10⁻⁹ m = 1.58 × 10⁻²¹ J = 9.84 meV.
- The asymptotic form's phase error is 5/(48|z|³⁄²), quoted from its next term. Setting it to 0.05 gives |z| = (5/2.4)²⁄³ = 1.63, that is 1.63 × 1.97 nm = 3.2 nm from the turning point. The cruder test ħ|p′|/p² = 1/(2|z|³⁄²) is still 0.24 there, 4.8 times the true error.
Answerl = 1.97 nm and ε = 9.84 meV; the WKB form is good to 0.05 rad everywhere more than about 3.2 nm from the turning point, on either side.
MediumUse the oscillating form Ai(−u) ≈ π(−1/2)u(−1/4) sin(ζ + π/4), with ζ = (2/3)u³⁄², to predict the first two zeros of Ai, and compare them with the tabulated 2.3381 and 4.0879. What would a phase of 0 or π/2 in place of π/4 predict for the first zero? Then correct the first zero with the next term in the phase, −5/(72ζ), which is quoted.
- A zero needs ζ + π/4 = kπ, so ζₖ = (k − ¼)π and uₖ = [(3/2)(k − ¼)π]²⁄³.
- k = 1: (3/2)(3π/4) = 3.5343, so u₁ = 3.5343²⁄³ = 2.3203, 0.76% below 2.3381. k = 2: (3/2)(7π/4) = 8.2467, so u₂ = 4.0818, 0.15% below 4.0879. The error shrinks as u grows, as an asymptotic form's should.
- With phase 0, sin ζ first vanishes at ζ = π: u = (3π/2)²⁄³ = 2.811, 20% high. With π/2, cos ζ first vanishes at ζ = π/2: u = (3π/4)²⁄³ = 1.771, 24% low. The tabulated zero pins the constant close to π/4.
- Correction: at the first zero ζ = 3π/4 = 2.356, so the phase falls short by 5/(72 × 2.356) = 0.0295 rad. Since dζ/du = √u = 1.523, the zero moves out by 0.0295/1.523 = 0.0193, to u₁ = 2.3396, 0.06% from the table.
Answeru₁ = 2.320 and u₂ = 4.082, 0.76% and 0.15% below the tabulated zeros; with the correction u₁ = 2.340. A phase of 0 or π/2 would miss the first zero by 20% or 24%.
HardAn electron in a harmonic trap with ħω = 1.00 meV occupies level n = 30. At its right turning point find a, l and L = |V′(a)/V″(a)|. Then find the zone where Ai can be matched to the WKB wave with each phase error at most 0.05 rad, the best matching point and its mismatch, and the lowest level for which such a zone exists at all.
- Oscillator length: b = ħ/√(mħω) = 1.0546 × 10⁻³⁴ J s ÷ √(9.109 × 10⁻³¹ kg × 1.602 × 10⁻²² J) = 8.73 nm. The turning point is a = √(2n + 1) b = √61 × 8.73 nm = 68.2 nm.
- With ω = 1.519 × 10¹² s⁻¹, V′(a) = mω²a = 1.43 × 10⁻¹³ N, so l = (ħ²/(2mV′(a)))¹⁄³ = 3.49 nm = 0.400 b. Since V″ = mω², L = a = 68.2 nm and L/l = 19.5, agreeing with 2¹⁄³(61)²⁄³ = 19.5.
- Ai's asymptotic form is good to 0.05 rad for |z| ≥ (5/(48 × 0.05))²⁄³ = 1.63, and the tangent line for |z| ≤ (10 × 0.05 × 19.5)²⁄⁵ = 2.49. The zone runs from 1.63 l to 2.49 l, that is 5.7 nm to 8.7 nm inside the turning point.
- The summed error 5/(48|z|³⁄²) + |z|⁵⁄² l/(10L) is least at |z| = (5L/8l)¹⁄⁴ = 1.87, 6.5 nm from a, where it equals 1/(6 × 1.87³⁄²) = 0.065 rad.
- A zone at 0.05 rad needs (10 × 0.05 L/l)²⁄⁵ ≥ 1.63, that is L/l ≥ 6.80, so (2n + 1)²⁄³ ≥ 5.40 and 2n + 1 ≥ 12.5. Level 5 gives L/l = 6.23 and fails; level 6 gives 6.97 and passes.
Answera = 68.2 nm, l = 3.49 nm, L/l = 19.5. Matching works 5.7 to 8.7 nm inside the turning point, best at 6.5 nm with a 0.065 rad mismatch; a zone first exists at n = 6.