University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.9

Tunnelling derived: barrier exponent, splitting and escape

Quantum Mechanics I estimated tunnelling with a slab formula and took ammonia's inversion coupling as a given number. This topic derives both from the connection formulas, adds the escape rate of a trapped state, and keeps the habit that matters most here: saying which part of each answer is leading order in ħ and which part fails first.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Tunnelling is what the connection formulas do when two turning points bound a forbidden stretch. Carry an outgoing wave back through the barrier and its amplitude grows by e^φ, where φ = (1/ħ)∫|p| dx between the turning points; because the WKB amplitude 1/√p already carries constant flux, the transmission is simply T = e^(−2φ). That is Quantum Mechanics I's slab estimate, now derived and placed in the expansion in ħ: the exponent is the leading order, the unit prefactor the next, and it holds only for smooth walls and φ well above 1.

Put two identical wells back to back and the same tail, meeting its mirror image at the midpoint, splits each level by ΔE = (ħω/π)e^(−φ): the coupling Quantum Mechanics I wrote into ammonia by hand, and a function of ħ whose Taylor series vanishes, so no order of perturbation theory about one well can find it. Let the far side open onto a continuum instead and the well empties at Γ = (ω/2π)e^(−2φ), the classical rate of striking the barrier times T. The exponents are robust; the prefactors are not.

Near the top the turning points merge and the parabolic barrier's exact 1/(1 + e^(2φ)) (quoted) takes over, giving ½ where WKB gives 1. Near a well bottom the lowest state is not semiclassical: its Gaussian tail beats the WKB tail by (π/e)¹⁄⁴, so the lowest doublet splits by √(π/e) = 1.075 times the formula as ħ → 0 (quoted in general).

Simple definition: The barrier integral φ = (1/ħ)∫|p| dx, taken across the classically forbidden stretch at the particle's energy, is the pure number that sets, to leading order in ħ, a smooth barrier's transmission e^(−2φ), a double well's splitting (ħω/π)e^(−φ) and a trapped state's escape rate (ω/2π)e^(−2φ).
Example: An electron 1.00 eV below the flat top of a barrier 0.500 nm wide has |p| = √(2m × 1.00 eV), so φ = 2.56 and e^(−2φ) = 6.0 × 10⁻³. A proton, 1836 times heavier, has √1836 = 42.9 times the φ: φ = 110 and e^(−2φ) ≈ 10⁻⁹⁵.

Barrier integralφ = (1/ħ) ∫ₐᵇ √(2m(V(x) − E)) dx

The one number every result here is built from. It grows as √m/ħ: quadruple the mass at fixed V − E and φ doubles, so every e^(−φ) is squared.

a, b: turning points where V = E; m in kg, V and E in J, x in m, ħ in J s; φ is a pure number

Smooth barrier, leading order (derived)T = e^(−2φ)

No ratio of velocities appears: the 1/√p amplitude already carries constant flux. A hard wall loses the prefactor; a rectangle's is 16(E/V₀)(1 − E/V₀).

T and R = 1 − T are pure numbers; needs φ ≳ 2 and each turning point on a smooth, locally linear stretch of V

Inverted-parabola barrier (quoted, exact)T = 1/[1 + e^{(2π(V₀ − E)/(ħω_b))}]

The check on WKB near the top: ½ at E = V₀ where e^(−2φ) gives 1, and within 1.8% of e^(−2φ) once φ = 2.

V = V₀ − ½mω_b²x²; ω_b in s⁻¹, V₀ and E in J; E may lie above V₀; the exponent is 2φ for this shape

Symmetric double-well splitting (derived)ΔE = E₋ − E₊ = (ħω/π) e^(−φ)

Herring's formula with a WKB-normalised one-well state. Half of it is the coupling A that Quantum Mechanics I put into ammonia's two-state Hamiltonian.

ΔE in J; ω in s⁻¹ is the classical frequency in one well at the level's energy; φ runs between the inner turning points

Escape from a metastable well (derived)Γ = (ω/2π) e^(−2φ)

Arrival flux times transmission. A probability leaks out, so the exponent is 2φ; the splitting is set by an amplitude, so it carries φ.

Γ in s⁻¹ and lifetime τ = 1/Γ; ω/2π is the rate at which the trapped particle arrives at the barrier

Lowest doublet as ħ → 0 (quoted in general)ΔE₀ → √(π/e) (ħω/π) e^(−φ) = 1.075 (ħω/π) e^(−φ)

A harmonic bottom's Gaussian tail is (π/e)¹⁄⁴ times the WKB tail; the splitting and the escape rate both go as its square.

ΔE₀ in J, symbols as above; level n carries √(2π)(n + ½)^(n+½) e^(−(n+½))/n!: 1.075, 1.028, 1.017, tending to 1

Carry the transmitted wave back through the barrier

Quote topic 11.06: where the forbidden side lies left of a turning point, (2/√p)cos(θ − π/4) joins the exponential decaying into the barrier and (1/√p)sin(θ − π/4) joins minus the one growing into it, with θ = (1/ħ)∫p dx from the turning point. Beyond the barrier put only a transmitted wave, (F/√p)e^(iθ) = (F e^(iπ/4)/√p)[cos(θ − π/4) + i sin(θ − π/4)]. Followed back through the barrier, the sine part becomes an exponential that is e^φ times larger at the far turning point a, where φ = (1/ħ)∫ₐᵇ|p| dx; the cosine part has shrunk by e^(−φ) and is dropped. At a the growing piece is the solution that decays towards b, so it joins a cosine on the incident side: incident and reflected waves, each of magnitude |F|e^φ/√p. The flux of (C/√p)e^(±iθ) is |C|²/m at every x, because |ψ|² ∝ 1/v, so T = |F|²/(|F|e^φ)² = e^(−2φ) (derived), with no ratio of velocities as at a step, and R = 1 − T ≈ 1.

Leading order in ħ, and what a hard wall costs

|p| dx is an action, so φ = S/ħ is a pure number growing as 1/ħ and as √m: ln T = −2S/ħ + c + O(ħ). The exponent is the leading order of topic 11.04's expansion; c = 0, the unit prefactor, is the next order, right only where each turning point is a smooth, locally linear crossing. Tunnelling microscopes test the exponent: for a work function W = 4.5 eV, κ = √(2mW)/ħ = 10.9 nm⁻¹, and e^(−2κd) falls 8.8-fold per 0.1 nm, the decade per ångström they measure. A rectangle tests the prefactor. Quantum Mechanics I's exact T = [1 + V₀² sinh²(κL)/(4E(V₀ − E))]⁻¹ tends, for κL ≫ 1, to 16(E/V₀)(1 − E/V₀)e^(−2κL), while WKB, blind to vertical walls, returns the slab estimate e^(−2κL): the same exponent, with a lost prefactor that is 3 at E = V₀/4 and never above 4. That is the right priority. A barrier integral φ = 20 known only to ±1 leaves T uncertain by a factor e² = 7.4 either way, and a deuteron, twice a proton's mass, has √2 times its φ, so T_d = Tₚ^(√2).

At the top the turning points merge

The Airy matching needs each turning point on a stretch of V that is linear over a few Airy lengths ℓ_F = (ħ²/2mF)¹⁄³, F = |V′| there, with the two turning points much further apart. Near its maximum every smooth barrier is an inverted parabola, V = V₀ − ½mω_b²x²; its integrand is a semicircle, so φ = π(V₀ − E)/(ħω_b) (derived). The turning points ±xₜ then lie xₜ/ℓ_F = (8φ²/π²)¹⁄³ Airy lengths from the top, only 1.5 at φ = 2, and the linear zones collide as φ falls toward 1. For this shape the transmission is known exactly, T = 1/(1 + e^(2φ)) (quoted, from parabolic-cylinder functions), valid above the top too, where φ is negative. WKB overshoots it by the factor 1 + e^(−2φ): 13.5% at φ = 1, 1.8% at φ = 2, and at the top it gives 1 where the truth is ½. Above the top the exact result keeps a reflection 1/(1 + e^(2|φ|)) ≈ e^(−2|φ|) that WKB, with no turning point to reflect from, misses at every order in ħ.

Two wells: Herring's formula gives the splitting

In a symmetric V let ψ₀ be the level localised in the right well, normalised there, and ψ± = [ψ₀(x) ± ψ₀(−x)]/√2. Take ψ₊ times ψ₀'s Schrödinger equation minus ψ₀ times ψ₊'s, integrate over positive x and use ψ₊′(0) = 0: E₊ = E₀ − (ħ²/m)ψ₀(0)ψ₀′(0). The odd state, with ψ₋(0) = 0, sits as far above, so ΔE = (2ħ²/m)ψ₀(0)ψ₀′(0), Herring's formula (derived, up to the overlap of ψ₀(x) with ψ₀(−x)). WKB supplies ψ₀. In the well ψ₀ = (C/√p)cos(θ − π/4); cos² averages ½, so normalisation reads (C²/2m)∫dx/v = C²T_cl/4m = 1, giving C² = 2mω/π with ω = 2π/T_cl at the level's energy. Into the barrier the connection formula continues it as (C/2√|p|)e^{(−(1/ħ)∫ₓᵇ|p| dx′)}, b the right well's inner turning point, so at the midpoint ψ₀′ = (|p|/ħ)ψ₀ and ψ₀(0)² = (C²/4|p(0)|)e^(−φ). Then ΔE = (ħ/2m)C²e^(−φ) = (ħω/π)e^(−φ) (derived), φ taken across the whole barrier, −b to b.

A coupling no perturbation series can see

Half the splitting is the matrix element Quantum Mechanics I wrote into ammonia by hand. In the basis of the two localised states, ⟨L|H|R⟩ = −A with A = ΔE/2 = (ħω/2π)e^(−φ), and a particle started in one well is found in the other with probability sin²(ΔE t/2ħ), with certainty first at t = πħ/ΔE. That two-state model is itself leading order: it keeps one doublet, fair while ΔE is far below the ħω to the next. Now write φ = S/ħ. Every derivative of e^(−S/ħ) with respect to ħ vanishes as ħ → 0⁺, so its Taylor series is identically zero: expanding about one well, in powers of ħ or of an anharmonic coupling, gives ψ₊ and ψ₋ the same energy at every order. The splitting is non-analytic, beyond all orders. It is also violently sensitive to whatever scales S: halve ħ, or quadruple the mass, and e^(−φ) is squared, taking 4.5 × 10⁻⁵ at φ = 10 to 2.1 × 10⁻⁹ at φ = 20. Ammonia's measured 0.793 cm⁻¹ (23.8 GHz) is the textbook case, and the medium example reads its φ ≈ 6 off the formula.

Escape, and the prefactor near a well bottom

Open the far side onto a continuum. The well's standing wave (C/√p)cos(θ − π/4) is two travelling waves of amplitude C/2, so it brings flux C²/4m = ω/2π to the barrier per unit probability in the well: the classical arrival rate. A fraction e^(−2φ) passes, so Γ = (ω/2π)e^(−2φ) and τ = 1/Γ (derived). Both prefactors rest on the WKB tail, which is wrong for the lowest state: at E = ħω/2 a harmonic bottom's turning points are one oscillator length ℓ = √(ħ/mω) from the minimum, inside the Airy zone. Far out, the WKB exponent is x²/2ℓ² − ¼ − ½ln(2x/ℓ), and the exact Gaussian tail comes out (π/e)¹⁄⁴ = 1.037 times the WKB tail (derived); ΔE and Γ go as its square. So the lowest doublet's true splitting is √(π/e) = 1.075 times the formula as ħ → 0 (quoted in general; the hard example derives it for the double oscillator). Level n carries √(2π)(n + ½)^(n+½)e^(−(n+½))/n!, 1.028 for n = 1 and tending to 1 by Stirling's formula: the prefactor fails at the bottom while the exponent stands.

Change one variable at a time

Make the relationship visible.

Interactive model

Well offset a/ℓ3.0

Doublet n0

Slide a from 2 to 4: the barrier rises only from 2 to 8 ħω, yet the n = 0 splitting falls five decades, because φ grows as a². Then switch to n = 1: the band shrinks and the splitting is 17 times larger at a = 3, but at a = 2 its φ = 0.35 is far too small to trust.

BARRIER INTEGRAL φ6.72

log₁₀ (ΔE/ħω), WKB-3.42

CROSSING TIME πħ/ΔE1305 well periods

BARRIER HEIGHT V₀4.50 ħω

Live interpretationBARRIER INTEGRAL φ: 6.72. log₁₀ (ΔE/ħω), WKB: −3.42. CROSSING TIME πħ/ΔE: 1305 well periods. BARRIER HEIGHT V₀: 4.50 ħω

Catch the common trap

Explain before calculating.

Misconception check

The tempting picture of tunnelling between two wells is repeated escape: the particle strikes the barrier ω/2π times a second, gets through with probability e^(−2φ) each time, and so crosses in about (2π/ω)e^(2φ). That is right when the far side is a continuum, because each transmitted piece of amplitude travels away and never returns, so probabilities add. Between two identical wells the transmitted amplitude lands in a state of exactly the same energy and adds in phase with what earlier bounces left there. Amplitude of order e^(−φ) accumulates coherently, the transfer is a resonance, and it completes in πħ/ΔE = (π²/ω)e^φ. For φ = 7 the incoherent picture is about 700 times too slow. Detune one well by an energy ε much larger than ΔE, with an applied bias for instance, and the resonance is lost: an isolated particle then reaches the other well only with probability of order (ΔE/ε)², and never completely. Only when something takes up the energy mismatch, a continuum on the far side or an environment such as phonons or collisions, does the crossing become incoherent hopping, at a rate that scales as the square of the coupling, e^(−2φ).

A symmetric double well's lowest doublet has barrier integral φ = 7.0, and the classical frequency in each well is ω = 5.0 × 10¹³ s⁻¹. A particle is prepared in the left well. Using the WKB splitting formula, when is it first certain to be found in the right well?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyNear its top, a smooth barrier met by an electron is V(x) = V₀ − ½mω_b²x² with ħω_b = 0.100 eV. Find φ in terms of V₀ − E, then compare the WKB transmission e^(−2φ) with the exact parabolic result 1/(1 + e^(2φ)) (quoted) for electrons 0.100 eV and 0.020 eV below the top, and give the barrier's width in the first case.

Turning points: V = E at x = ±xₜ with xₜ² = 2(V₀ − E)/(mω_b²). The integrand √(2m(V − E)) = mω_b√(xₜ² − x²) is a semicircle of radius xₜ, so φ = (mω_b/ħ)(πxₜ²/2) = π(V₀ − E)/(ħω_b). Dimensions: energy over energy, a pure number.
At 0.100 eV below the top, φ = π = 3.14. WKB: e^(−2π) = 1.87 × 10⁻³. Exact: 1/(1 + 535.5) = 1.86 × 10⁻³. They differ by the factor 1 + e^(−2φ) = 1.002.
At 0.020 eV below, φ = 0.2π = 0.628. WKB: e^(−1.257) = 0.285. Exact: 1/(1 + 3.51) = 0.222. WKB is 28% high: the turning points are too close for separate Airy matchings.
Width at 0.100 eV below: ω_b = 0.100 eV/ħ = 1.52 × 10¹⁴ s⁻¹, so xₜ = √(2 × 1.60 × 10⁻²⁰ J/(9.11 × 10⁻³¹ kg × (1.52 × 10¹⁴ s⁻¹)²)) = 1.23 nm, and the barrier is 2.47 nm wide.
Limit: at E = V₀ the exact result is ½ while e^(−2φ) = 1, so the WKB estimate should not be trusted within about ħω_b/π of the top, where φ falls below 1.

Answerφ = π(V₀ − E)/(ħω_b). At 0.100 eV below: T = 1.87 × 10⁻³ (WKB) against 1.86 × 10⁻³ (exact), barrier 2.47 nm wide. At 0.020 eV below: 0.285 against 0.222.

MediumAmmonia's ground inversion doublet is split by a measured 0.793 cm⁻¹, and its umbrella quantum is about ħω = 950 cm⁻¹ (energies in wavenumbers, E = hcν̃). Infer the barrier integral φ₀ of the ground doublet using the lowest-doublet factor 1.075, find how long the nitrogen takes to tunnel through the plane of hydrogens, and use the first excited doublet's measured splitting of about 36 cm⁻¹, with factor 1.028, to find how much φ falls per quantum.

Invert ΔE = 1.075(ħω/π)e^(−φ₀): e^(−φ₀) = π × 0.793/(1.075 × 950) = 2.44 × 10⁻³, so φ₀ = 6.02. Without the factor, φ₀ = 5.94: the prefactor correction moves φ by ln 1.075 = 0.072, small beside φ itself.
Tunnelling time: the two-state result P = sin²(ΔE t/2ħ) completes at t = πħ/ΔE = 1/(2cΔν̃) = 1/(2 × 2.998 × 10¹⁰ cm s⁻¹ × 0.793 cm⁻¹) = 2.10 × 10⁻¹¹ s, since ΔE/h = 23.8 GHz.
Compare with one umbrella period, 1/(cν̃) = 1/(2.998 × 10¹⁰ cm s⁻¹ × 950 cm⁻¹) = 3.51 × 10⁻¹⁴ s: the nitrogen vibrates about 600 times per crossing, as a small e^(−φ) demands.
First excited doublet, keeping ω = 950 cm⁻¹: e^(−φ₁) = π × 36/(1.028 × 950) = 0.116, so φ₁ = 2.16. One quantum of energy lowered the barrier integral by 3.86, and the splitting rose 45-fold.
Validity: φ₀ ≈ 6 is comfortably large; φ₁ ≈ 2.2 puts the excited doublet, about 1.5ħω ≈ 1400 cm⁻¹ above the minimum, near the top of the fitted barrier of about 2000 cm⁻¹, where the formula is at its limit and ω itself is falling.

Answerφ₀ ≈ 6.0; the nitrogen tunnels across in about 21 ps, some 600 umbrella periods; φ₁ ≈ 2.2, so one quantum lowers φ by about 3.9 and raises the splitting 45-fold.

HardTwo oscillator wells of frequency ω are joined at a cusp, V(x) = ½mω²(|x| − a)², with a = 3ℓ and ℓ = √(ħ/mω). For the lowest doublet find ΔE from the WKB formula, then from Herring's formula using the exact right-well Gaussian for x ≥ 0, compare both with the numerically diagonalised 4.197 × 10⁻⁴ ħω, and show that their ratio tends to √(π/e) as a/ℓ → ∞.

The level sits at E = ½ħω up to exponentially small shifts, so the inner turning points are at |x| = a − ℓ. With s = a − |x|, |p| = mω√(s² − ℓ²), and φ = (2mω/ħ)∫ from ℓ to a of √(s² − ℓ²) ds = A√(A² − 1) − ln(A + √(A² − 1)), where A = a/ℓ.
At A = 3: φ = 8.4853 − 1.7627 = 6.7225. The classical frequency is ω at every energy, so ΔE_WKB = (ħω/π)e^(−6.7225) = 3.831 × 10⁻⁴ ħω.
Herring: for x ≥ 0, V is exactly the right oscillator, so ψ₀ = (mω/πħ)¹⁄⁴e^{(−(x − a)²/2ℓ²)} and ψ₀′(0) = (a/ℓ²)ψ₀(0). Then ΔE = (2ħ²/m)(a/ℓ²)(mω/πħ)¹⁄²e^(−A²) = (2/√π)A e^(−A²) ħω = 3.385 × 1.234 × 10⁻⁴ ħω = 4.178 × 10⁻⁴ ħω.
Against the numerical 4.197 × 10⁻⁴ ħω, Herring's value is 0.5% low and WKB's 8.7% low; Herring over WKB is 1.091.
Large A: A√(A² − 1) → A² − ½ − 1/(8A²) and ln(A + √(A² − 1)) → ln 2A − 1/(4A²), so e^(−φ) → 2A√e e^(−A²) and ΔE_WKB → (2√e/π)A e^(−A²) ħω. The ratio tends to (2/√π)/(2√e/π) = √(π/e) = 1.075; at A = 3 the leftover e^(1/(8A²)) = 1.014 gives 1.090.
Checks: both results are ħω times pure numbers; the barrier ½ħωA² = 4.5ħω gives φ = 6.7 ≫ 1; and both vanish faster than any power of ħ as A = a√(mω/ħ) grows, sharing the exponent A² and differing only in the prefactor.

AnswerΔE_WKB = 3.83 × 10⁻⁴ ħω and ΔE_Herring = 4.18 × 10⁻⁴ ħω, against 4.20 × 10⁻⁴ ħω numerically: WKB is 8.7% low, and the ratio of the two tends to √(π/e) = 1.075.