University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.6
Deriving the connection formulas and their π/4 phase
Every WKB result that crosses a smooth turning point, from Bohr–Sommerfeld levels to tunnelling rates, borrows one phase and one factor of 2 from this lesson. We derive both by matching WKB to the exact solution on a straight ramp, then meet the two ways they are misused: read backwards, and used where the ramp is not straight.
Build the model
Connect the measurement to the mechanism.
WKB fails at a turning point, but only inside a zone a few Airy lengths ℓ = (ħ²/(2m|V′(a)|))¹⁄³ wide, and if V stays close to linear over a somewhat wider zone, the Schrödinger equation there is Airy's and is solved exactly. The method is an overlap. A few ℓ out on either side, both descriptions hold at once, and on a straight ramp they are literally the same functions: the WKB phase measured from the turning point is exactly ζ = (2/3)|z|³⁄², the variable of the Airy asymptotics. Matching is then bookkeeping.
A bound state must decay into the barrier, which leaves Ai alone, and Ai's two asymptotic forms say that (C/√|p|) e(−∫|p| dx/ħ) continues into the well as (2C/√p) cos(∫p dx/ħ − π/4). The 2 is an incident and a reflected wave of amplitude C each; the π/4 says a smooth turning point reflects with a phase lag of π/2, half a hard wall's π, because the wave runs into the forbidden region before it turns. Bi supplies the partner formula for the growing solution. Three limits bound the result.
It is leading order in ħ, and the next phase correction, 5/(72ζ), is quoted from Ai's asymptotic series. It needs an isolated turning point, so it fails where two turning points close up, near a well bottom or a barrier top. And each formula is safe in one direction only, because a growing exponential swamps any decaying one that was meant to be there.
- Simple definition
- A connection formula is the rule, derived by matching WKB to the Airy solution at a smooth turning point, that says which standing wave in the allowed region a given exponential in the forbidden region continues into.
- Example
- A tail (C/√|p|)e(−Φ) becomes (2C/√p)cos(θ − π/4) inside. At 3.0ℓ into the well, θ = (2/3)(3.0)³⁄² = 3.46 rad, so ψ = −1.790C/√p; the exact Airy solution there gives −1.767C/√p, 1.3% away.
Derived, exact for linear V. The WKB solutions become e(±ζ), cos ζ and sin ζ over |z|¹⁄⁴: the shapes of Ai's and Bi's large-|z| forms, so matching reduces to reading off coefficients.
z = (x − a)/ℓ, dimensionless; ℓ = (ħ²/(2m|V′(a)|))¹⁄³ in m; p in kg m s⁻¹; under the barrier |p| gives the same ζ
Derived from Ai, to leading order in ħ. The 2 is incident plus reflected wave; the π/4 is the phase every smooth turning point contributes from here on. Safe from the barrier into the well only.
well at x ≤ a; |C|² in kg s⁻¹, so |ψ|² is in m⁻¹; for a well on the right, swap the limits: both integrals start at a
Derived from Bi, whose two asymptotic prefactors are equal, so there is no factor 2. Safe from the well into the barrier only: read backwards, it cannot see the decaying admixture, hidden under the growth, that sets the well's cosine.
same limits and units as the decaying formula; D plays the role of C
Derived by splitting 2cos(θ − π/4) into e(iπ/4)e(−iθ) + e(−iπ/4)e(iθ). |r| = 1, total reflection; a hard wall would give r = −1, twice the phase lag.
r is reflected over incident amplitude, both referred to x = a, for time dependence e(−iEt/ħ); dimensionless
Derived by connecting cos(θ − π/4 − ε) piece by piece. Equal at Φ = ½ ln(1/(2 tan ε)): only 3.1 e-folds for ε = 10⁻³, which is why the decaying formula runs one way.
ε is the error in the well-side phase, in rad; Φ = (1/ħ)∫ₐˣ|p| dx′ is dimensionless
The assumption, not a result. For the oscillator's nth state 3ℓ/xₜ = 2.38(2n + 1)⁻²⁄³: 2.4 at n = 0, where no overlap exists, and 0.31 at n = 10.
V′ in N, V″ in N m⁻¹, ℓ in m; δ ≤ 0.1 needs |x − a| ≥ 2.9ℓ, and V′ may barely change over that distance
Why WKB cannot be carried through the turning point
Near a turning point x = a WKB offers two solutions on each side: (1/√p)e(±iθ) in the well and (1/√|p|)e(±Φ) under the barrier, with θ and Φ the phase and barrier integrals measured from a. A physical state fixes the combination on one side, and the task is to find what it becomes on the other. WKB alone cannot say. Both amplitudes diverge at a, and the validity parameter from the WKB lesson, δ = ½(ℓ/|x − a|)³⁄², exceeds 0.1 everywhere within 2.9ℓ of it. So solve that zone exactly instead. Over a few ℓ, V ≈ E + V′(a)(x − a), and the Schrödinger equation becomes ψ″ = zψ with z = (x − a)/ℓ, whose solutions Ai and Bi the last lesson built. The plan is to find a band on each side far enough out for WKB and close enough in for the straight line, where both hold, and match there. That band exists only if V′ barely changes over a few ℓ, the one assumption everything below rests on.
On a straight ramp, WKB and Airy are the same functions
Put the linear V into p = √(2m(E − V)): p = √(2mV′(a)(a − x)) = (ħ/ℓ)√|z|, since ħ²/ℓ² = 2mV′(a)ℓ. The phase from the turning point is then θ = ∫√s ds from 0 to |z|, which is (2/3)|z|³⁄² exactly, and under the barrier Φ has the same form; call either ζ. So on the ramp the WKB solutions are e(±ζ)/z¹⁄⁴ for z ≥ 0 and cos ζ, sin ζ over |z|¹⁄⁴ for z ≤ 0. These are exactly the shapes of the large-|z| Airy forms quoted from the last lesson, which follow from Ai's integral representation by steepest descent: Ai → e(−ζ)/(2√π z¹⁄⁴) and Bi → e(+ζ)/(√π z¹⁄⁴) for z ≫ 1, while Ai → cos(ζ − π/4)/(√π|z|¹⁄⁴) and Bi → −sin(ζ − π/4)/(√π|z|¹⁄⁴) for z ≪ −1. Here they are inputs, not results. The overlap is real but not perfect: at z = 3, where δ = 0.096, the e(−ζ) form exceeds the exact Ai by 1.8%, and at z = 2 by 3.1%.
Matching the decaying solution gives the 2 and the π/4
A bound state must decay under the barrier, ψ = (C/√|p|)e(−Φ). In the Airy zone the exact solution is A Ai + B Bi, and since Bi grows as e(+ζ) while the tail has no growing part, B = 0. On the barrier side, (C/√|p|)e(−ζ) = C√(ℓ/ħ) z⁻¹⁄⁴ e(−ζ) must equal A e(−ζ)/(2√π z¹⁄⁴), so A = 2√π C√(ℓ/ħ). Now read the same A Ai on the well side: A cos(ζ − π/4)/(√π|z|¹⁄⁴) = 2C√(ℓ/ħ)|z|⁻¹⁄⁴ cos(ζ − π/4) = (2C/√p) cos(θ − π/4). That is the connection formula, derived to leading order in ħ, and neither number was put in by hand: the 2 is the ratio of Ai's prefactors, 1/√π against 1/(2√π), and the π/4 is Ai's phase. Test it where it is hardest, at the nodes. cos(θ − π/4) vanishes at θ = 3π/4 and 7π/4, which is |z| = 2.320 and 4.082; Ai vanishes at 2.338 and 4.088. The first node is only 0.8% off although it lies where δ = 0.14.
The growing solution, the mirror, and the reflection phase
Matching B Bi the same way gives the partner: (D/√|p|)e(+Φ) under the barrier continues as −(D/√p) sin(θ − π/4) in the well, with no factor 2 because Bi's two prefactors are equal. For a well on the right of its turning point, reflect x → 2a − x: nothing changes provided both integrals run from the turning point and are positive, so the π/4 never flips sign. Now split the standing wave into travelling waves. With time dependence e(−iEt/ħ), e(−iθ) moves toward the turning point and e(+iθ) away from it, and 2cos(θ − π/4) = e(iπ/4)e(−iθ) + e(−iπ/4)e(iθ). The reflection amplitude referred to x = a is r = −i. |r| = 1, as it must be, since a lone decaying exponential carries no current. A hard wall at a forces sin θ and r = −1: a smooth turning point costs half the phase, because the wave runs into the barrier before turning round. Two such points put π/4 + π/4 into a well, the ½ in (n + ½)ħω of the next lesson. That ½ was first measured as an isotope shift in boron monoxide's 0–0 band, which would vanish without it (Mulliken, 1924).
Each formula runs in one direction only
Exponentials make the formulas lopsided. Suppose the phase in the well is π/4 + ε instead of π/4. Then 2C cos(θ − π/4 − ε) = 2C cos ε cos(θ − π/4) + 2C sin ε sin(θ − π/4), and connecting each piece gives, under the barrier, (C/√|p|)[cos ε e(−Φ) − 2 sin ε e(+Φ)]. The growing piece overtakes the decaying one once 2 tan ε e(2Φ) exceeds 1: at Φ = 1.96 for ε = 0.01 rad, and even ε = 10⁻⁶ only postpones it to Φ = 6.6. So a standing wave in the well cannot be used to infer decay; the decaying formula is trustworthy from the barrier into the well only. The growing formula is the mirror case. A decaying admixture C e(−Φ) beneath D e(+Φ), with C as large as D/2, is smaller by a further e(−2Φ), below any measurement or rounding error, yet in the well it becomes a cosine of amplitude 2C, as large as the sine. So the growing formula is safe from the well into the barrier and never back. A numerical integration says the same: march a bound state into a forbidden region and it diverges, so start it on the far side and integrate back toward the well.
Where the formulas stop, and what they are worth
The matching needed V linear over the band, 3ℓ|V″(a)| ≪ |V′(a)|, and behind that an isolated turning point. In the oscillator 3ℓ/xₜ = 2.38(2n + 1)⁻²⁄³: 0.31 at n = 10 but 2.4 at n = 0, where the two turning points are 2.5ℓ apart and each sits inside the other's Airy zone. At a barrier top they merge, and connecting through a barrier whose integral Θ is below about 1 fails: the parabolic-barrier result, quoted from parabolic cylinder functions, gives T = 1/(1 + e(2Θ)), ½ at the top where e(−2Θ) says 1. Order: θ is order 1/ħ, the π/4 and 1/√p order ħ⁰, and the first neglected term, quoted from Ai's asymptotic series, shifts the phase by 5/(72ζ) = (5/24)δ, order ħ. At the first node that is 0.029 rad; the exact node phase is 3π/4 + 0.027. Dimensions: every exponent and cosine argument is an action over ħ. Limit: steepening a straight ramp only shrinks ℓ, and the phase stays π/4; the hard wall's π/2 returns only when V stops being linear within an Airy length.
Change one variable at a time
Make the relationship visible.
Move φ to 50°: the well side only slides, but D = −0.087 and the solid curve dives through zero at z ≈ 1.1. Near 135° it is the reverse: 10° either way barely moves the barrier side. At 90° the WKB node sits on the turning point, a hard wall's phase. Then shrink the Airy zone.
FIRST WKB NODE z₁-2.32
DECAYING AMPLITUDE C0.500
GROWING AMPLITUDE D0.000
δ AT THE ZONE EDGE0.272
Live interpretationFIRST WKB NODE z₁: −2.32. DECAYING AMPLITUDE C: 0.500. GROWING AMPLITUDE D: 0.000. δ AT THE ZONE EDGE: 0.272
Catch the common trap
Explain before calculating.
A wave in the allowed region travels toward a smooth turning point x = a, beyond which V rises linearly through E. Split the connection formula's standing wave (2C/√p) cos(θ − π/4), θ = (1/ħ)∫ₓᵃ p dx′, into incident and reflected waves. With both referred to x = a, what is the reflection amplitude r?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron's bound wavefunction decays beyond a turning point where its potential energy rises through E with slope V′ = 0.10 eV nm⁻¹. Find the Airy length ℓ, use the connection formula to locate the first two nodes on the allowed side, and compare with the exact Airy zeros at 2.338ℓ and 4.088ℓ.
- V′ = 0.10 × 1.602 × 10⁻¹⁹ J ÷ 10⁻⁹ m = 1.602 × 10⁻¹¹ N. With ħ²/(2mₑ) = 6.104 × 10⁻³⁹ J m², ℓ³ = 6.104 × 10⁻³⁹ ÷ 1.602 × 10⁻¹¹ = 3.810 × 10⁻²⁸ m³, so ℓ = 7.25 × 10⁻¹⁰ m = 0.725 nm.
- The decaying tail connects to (2C/√p) cos(θ − π/4), which vanishes where θ − π/4 = π/2 and 3π/2, that is at θ = 3π/4 and 7π/4.
- On the ramp θ = (2/3)|z|³⁄², so |z| = (3θ/2)²⁄³: (9π/8)²⁄³ = 2.320 and (21π/8)²⁄³ = 4.082.
- Distances: 2.320 × 0.725 nm = 1.68 nm and 4.082 × 0.725 nm = 2.96 nm inside the turning point. The exact zeros give 1.70 nm and 2.96 nm, so the errors are 0.013 nm (0.8%) and 0.004 nm (0.15%).
- Check the conditions: the local wavelength at the first node is 2πℓ/√2.320 = 2.99 nm and δ = ½(2.320)⁻³⁄² = 0.14 there, so the node is barely half a wavelength in and the formula still holds to 1%. A hard-wall phase would have put a node at the turning point itself.
Answerℓ = 0.725 nm; nodes 1.68 nm and 2.96 nm inside the turning point, against exact 1.70 nm and 2.96 nm (errors 0.8% and 0.15%).
MediumAn electron meets a ramp of slope 1.0 eV nm⁻¹, so ℓ = 0.336 nm. An approximate calculation gives the well-side wave as (2C/√p) cos(θ − π/4 − ε) with ε = 1.0 × 10⁻³ rad. Connect it into the barrier, find the depth Φ where the spurious growing branch equals the decaying one, convert it to a distance beyond the turning point, and find their ratio at Φ = 5.
- Split the cosine: 2C cos(θ − π/4 − ε) = 2C cos ε cos(θ − π/4) + 2C sin ε sin(θ − π/4).
- Connect each piece. The cosine comes from (C cos ε/√|p|)e(−Φ). The sine is −(D/√p) sin(θ − π/4) with D = −2C sin ε, which connects to (D/√|p|)e(+Φ). So under the barrier ψ = (C/√|p|)[cos ε e(−Φ) − 2 sin ε e(+Φ)].
- The magnitudes are equal when 2 tan ε e(2Φ) = 1: Φ = ½ ln(1/(2 × 1.0 × 10⁻³)) = ½ ln 500 = 3.11.
- On the ramp Φ = (2/3)z³⁄², so z = (3Φ/2)²⁄³ = (4.66)²⁄³ = 2.79, which is 2.79 × 0.336 nm = 0.94 nm beyond the turning point.
- At Φ = 5 the ratio is 2 × 1.0 × 10⁻³ × e¹⁰ = 44, at z = 7.5²⁄³ = 3.83, or 1.29 nm in. The exact functions at that point, sin ε Bi(z) ÷ cos ε Ai(z), give 45.3: the 3% gap is the asymptotic forms' own error there.
- So a phase error of one milliradian owns the barrier region within a nanometre. The decaying formula must be run from the barrier into the well; to compute the tail, integrate in from the far side.
Answerψ = (C/√|p|)[cos ε e(−Φ) − 2 sin ε e(+Φ)]; the branches are equal at Φ = 3.1, 0.94 nm past the turning point, and the growing one is 44 times larger at Φ = 5 (1.29 nm in).
HardAn oscillator with length unit b = √(ħ/(mω)) is in the state n = 2: turning points ±√5 b, outermost node at b/√2 = 0.7071b. Predict that node from the right turning point in two ways: (a) replace V by its tangent line there, so the node is 2.338ℓ inside; (b) keep the true phase θ(x), the integral of p/ħ from x to xₜ, and impose θ = 3π/4. Repeat for n = 10, whose outermost node is 3.4362b, and say which assumption costs more.
- xₜ = √5 b = 2.2361b. There V′ = mω²xₜ, so ℓ³ = ħ²/(2m²ω²xₜ) = b⁴/(2xₜ) and ℓ = (2√5)⁻¹⁄³ b = 0.6070b; the overlap measure 3ℓ/xₜ is 0.81, far from small.
- (a) Tangent line: node at 2.2361b − 2.338 × 0.6070b = 2.2361b − 1.4192b = 0.8169b. Its depth below the turning point, 1.419b, is 7.2% short of the true 1.529b.
- (b) With u = x/b, p = (ħ/b)√(5 − u²) and θ(u) = ½[5 arccos(u/√5) − u√(5 − u²)]. Solving θ = 3π/4 = 2.356 numerically gives u = 0.7149: a depth of 1.521b, only 0.5% short.
- n = 10: xₜ = √21 b = 4.5826b and ℓ = (2√21)⁻¹⁄³ b = 0.47784b, so 3ℓ/xₜ = 0.31. Tangent line: 4.5826b − 2.3381 × 0.47784b = 4.5826b − 1.1172b = 3.4654b, depth 1.117b against the exact 1.146b, 2.5% short. True phase: 3.4448b, depth 0.75% short.
- Interpretation: the tangent line is the costly step, and its error falls from 7.2% to 2.5% as 3ℓ/xₜ falls from 0.81 to 0.31. The π/4 survives even at n = 2, because it is fixed within an Airy length of the turning point while ∫p dx carries the curvature. The residual 0.5 to 0.8% is the next-order phase: on a pure ramp θ = 3π/4 misses Ai's first zero by the same 0.76%.
Answern = 2: tangent line 0.817b (depth 7.2% short), true phase with π/4 0.715b (0.5% short), exact 0.707b. n = 10: 3.465b (2.5% short) and 3.445b (0.75% short), exact 3.436b. Linearising V costs far more than the π/4 does.