University Physics V · Finite Potential Wells and Quantum Tunneling · 6.7

WKB & the Gamow Exponent

A general V(x) has no closed-form solution. WKB does the next best thing: it makes the phase the unknown, hands you a wavefunction wherever the potential is smooth, and buys you the exponent of the tunnelling probability — at the price of collapsing exactly where the classical particle stops.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Write ψ(x) = exp(iS(x)/ħ). That is not yet an approximation, only a change of unknown, and it turns the Schrödinger equation into (S′)² − iħS″ = p² with p = √(2m(E−V)). Expand S in powers of ħ. At leading order S₀′ = ±p, so the phase is the classical action; at the next order S₁ = (i/2) ln p, so the amplitude is p^(−1/2) and |ψ|² ∝ 1/v — the particle is found where it moves slowly.

Under a barrier p is imaginary and the same two branches become growing and decaying exponentials in κ = √(2m(V−E))/ħ. Everything rests on the term thrown away, ħ|S₀″| ≪ (S₀′)², which is the statement that the local de Broglie wavelength changes by little over its own length. That fails at every classical turning point, where p → 0, so the allowed and forbidden solutions cannot be joined by the series that produced them. The repair is local: linearise V about the turning point, solve the resulting Airy equation exactly, and match its asymptotics onto both WKB branches.

What comes back is a π/4 of phase per soft turning point, the quantisation rule ∫p dx = (n + ½)πħ for a well, and for a barrier the Gamow factor T ≈ exp(−2γ) with γ = ∫κ dx across the forbidden interval. The cost is stated plainly: the exponent is trustworthy, the prefactor is not, and nothing here tells you how long tunnelling takes.

Simple definition: The WKB approximation solves the one-dimensional Schrödinger equation by putting the unknown in the exponent, ψ ≈ p(x)^(−1/2) exp(±(i/ħ)∫p dx), and is valid wherever the local de Broglie wavelength changes by little over its own length.
Example: An electron 3.0 eV below a barrier top has κ = √(2m × 3.0 eV)/ħ = 8.9 nm⁻¹, so 0.50 nm of forbidden region gives 2γ = 2κL = 8.9 and T ≈ e^(−8.9) = 1.4 × 10⁻⁴: four orders of magnitude fixed by the exponent alone.

Semiclassical wave, allowed regionψ ≈ (C/√p) exp(±(i/ħ)∫p dx′), p(x) = √(2m(E − V))

|ψ|² ∝ 1/p is the classical dwell-time density, and it is what keeps the current constant.

p in kg m s⁻¹; C set by normalisation; the sign picks the direction of travel

Forbidden regionψ ≈ (C/√κ) exp(±∫κ dx′), κ(x) = √(2m(V − E))/ħ

Both signs solve the equation; which survives is a boundary condition, never the series.

κ in m⁻¹; for an electron κ = 5.12 nm⁻¹ × √((V − E)/eV)

Validity condition|dλ̄/dx| = ħ m |V′| / p³ ≪ 1, λ̄ = ħ/p

Not “ħ is small”: a steep V breaks WKB at any energy, and p → 0 breaks it always.

dimensionless; λ̄ is the reduced local wavelength, in m

Airy patch at a linear turning pointψ″ = α³(x − a)ψ, α = (2m|V′(a)|/ħ²)¹⁄³, ψ = Ai(α(x − a))

The one place an exact solution is cheap, and the only source of the π/4.

1/α is the patch width in m: 0.23 nm for an electron on a 3.0 V nm⁻¹ ramp

Connection, forbidden side to the right(1/√κ) exp(−∫ₐˣ κ dx′) ⟷ (2/√k) cos(∫ₓᵃ k dx′ − π/4)

Used the other way, an error of one part in e^(2γ) feeds the growing branch and wins.

k = p/ħ in m⁻¹, a the turning point; read it from the decaying side outward

Gamow exponent and transmissionT ≈ e^(−2γ), γ = ∫ₐᵇ κ dxuniformly T = [1 + e^(2γ)]⁻¹

The uniform form is exact for a parabolic top: it gives T = ½ at E = V₀, where e^(−2γ) gives 1.

a, b are the roots of V(x) = E at that energy; γ is dimensionless

Put the unknown in the exponent

Nothing is approximated by writing ψ(x) = exp(iS(x)/ħ); S is simply a new unknown, complex in general. Substituting into ψ″ = −(p²/ħ²)ψ, with p(x) = √(2m(E − V)), gives the exact nonlinear equation (S′)² − iħS″ = p². Now expand S = S₀ + ħS₁ + ħ²S₂ + … and collect powers of ħ. At order ħ⁰, (S₀′)² = p², so S₀ = ±∫p dx′ — the classical action, which is where the word semiclassical comes from. At order ħ¹, 2S₀′S₁′ = iS₀″, so S₁ = (i/2) ln p. Exponentiating both terms gives ψ ≈ p^(−1/2) exp(±(i/ħ)∫p dx′). The amplitude is not decoration: |ψ|² ∝ 1/p ∝ 1/v is the classical dwell-time density, and it is exactly the factor that holds the probability current j = (ħ/m) Im(ψ*ψ′) constant along x, as a stationary state in one dimension demands.

The term you dropped is the validity condition

Truncating after S₁ is legitimate only if ħ|S₀″| ≪ (S₀′)², that is ħ|p′| ≪ p². Write λ̄ = ħ/p for the reduced local wavelength and the same statement reads |dλ̄/dx| ≪ 1: the wavelength must change by little over one wavelength. Expanded, it is ħ m |V′| / p³ ≪ 1. Read what that does and does not say. It is not “ħ is small” — ħ is a constant. It is a joint claim about the gradient of V and the momentum at that energy, so a steep potential breaks WKB at any energy while a smooth one supports it even for a light particle. Take the field-emission barrier of the medium example: at the metal surface ħ m eF / p³ = 0.031, comfortably small, but it grows without bound as p → 0, passing 1 when the barrier top is still 0.44 eV away — the last 0.15 nm of a 1.50 nm barrier. WKB is excellent over ninety per cent of the region and worthless in the rest.

Turning points: linearise, then solve exactly

At x = a where V(a) = E the momentum vanishes, the amplitude p^(−1/2) diverges, and the criterion fails outright. The cure is local. Over a small neighbourhood V(x) ≈ E + V′(a)(x − a), so the Schrödinger equation becomes ψ″ = α³(x − a)ψ with α = (2m|V′(a)|/ħ²)¹⁄³: the Airy equation, whose bounded solution is Ai(α(x − a)). Its scale 1/α sets the patch width — for an electron on a 3.0 V nm⁻¹ ramp, (ħ²/2m eF)¹⁄³ = 0.23 nm. Away from a, Ai has known asymptotics: a decaying exponential on the forbidden side, and on the allowed side an oscillation carrying an extra −π/4 of phase. Matching those asymptotics onto the two WKB branches in the overlap window — far enough from a for Airy's asymptotics, near enough that the linearisation still holds — is the entire content of the connection formulas. Where no such window exists, there is nothing to connect.

Connection formulas point one way

Under a barrier the two solutions are exp(+∫κ dx) and exp(−∫κ dx), and they differ by the vast factor e^(2γ). Start on the allowed side and propagate inward and an error of one part in e^(2γ) in your coefficients seeds the growing branch, which then swamps the decaying one you wanted; impose decay where decay is required and carry the solution outward instead, and no such contamination arises. That is why the safe statement is (1/√κ) exp(−∫ₐˣ κ dx′) → (2/√k) cos(∫ₓᵃ k dx′ − π/4), read from the forbidden side out. The same asymmetry appears numerically as the transfer matrix's e^(2κL) condition number. It is also the reason WKB delivers a reliable barrier exponent and an unreliable prefactor: the exponent is just the integral of κ, while the prefactor is precisely the coefficient bookkeeping that the growing branch corrupts.

A barrier has two turning points and one exponent

Send a wave in from the left, demand a purely outgoing wave on the right, run the decaying solution inward from each turning point, and the transmitted amplitude emerges suppressed by e^(−γ) with γ = ∫ₐᵇ κ dx between the roots of V(x) = E. Squaring gives T ≈ e^(−2γ). Two cautions follow. First, a and b move with E, and most of T's energy dependence is those turning points sliding together, not κ shrinking. Second, the prefactor. For a 4.00 eV rectangular barrier 0.600 nm wide at E = 1.00 eV, WKB returns 2γ = 10.65 and T = 2.4 × 10⁻⁵, while the exact opaque result is 7.1 × 10⁻⁵ — a factor 16E(V₀ − E)/V₀² = 3.00 that WKB cannot see, because vertical walls are not linear turning points and no Airy patch applies to them. Kemble's uniform formula T = [1 + e^(2γ)]⁻¹ repairs the near-top behaviour and is exact for a parabolic barrier.

The same patch at both ends quantises a well

For a well bounded by two linear turning points a and b, demanding decay outside both and consistency between them gives ∫ₐᵇ p dx = (n + ½)πħ with n = 0, 1, 2, … — Bohr–Sommerfeld, with the half Bohr's original rule lacked. The ½ is phase bookkeeping: each soft turning point costs π/4. Change the boundary condition and the count changes. A hard wall forces a node, costing π/2, so the infinite square well obeys ∫p dx = nπħ and returns Eₙ = n²π²ħ²/2mL² exactly; for the harmonic oscillator the soft rule is exact too, giving (n + ½)ħω. In Python this is a quadrature inside a root find: locate a and b with brentq on V(x) − E, form φ(E) = ∫ₐᵇ p dx/ħ − (n + ½)π with scipy.integrate.quad, then brentq on φ. It costs microseconds against a full diagonalisation and, for a smooth well and n ≥ 2, is usually right to a fraction of a per cent.

Change one variable at a time

Make the relationship visible.

Interactive model

Electron energy E2.0 eV

Barrier half-width w0.35 nm

Barrier height V05.0 eV

Raise E until the two dots meet. The WKB estimate climbs to T = 1 while the exact value stops at ½ — that gap is the connection formula failing once the two Airy patches overlap.

TURNING POINT xₜ0.271 nm

GAMOW EXPONENT 2γ7.56

log10 T WKB-3.28

log10 T EXACT-3.28

Live interpretationTURNING POINT xₜ: 0.271 nm. GAMOW EXPONENT 2γ: 7.56. log10 T WKB: −3.28. log10 T EXACT: −3.28

Catch the common trap

Explain before calculating.

Misconception check

Two errors travel together here. The first is reading WKB as an expansion in a small ħ, so its failure at a turning point looks like a technicality a finer grid would cure. ħ is a constant; the real condition is |dλ̄/dx| ≪ 1, a claim about the gradient of V at that energy, and it can fail for a tall wide barrier that turns over sharply. The second is treating the forbidden interval as a fixed barrier width read off the drawing. The limits in γ = ∫ₐᵇ κ dx are the roots of V(x) = E: raise E and they slide inward, which is most of why T rises, and multiplying the drawn width by κ at the peak overestimates γ badly. And the number that emerges is a one-figure claim: for a 4.00 eV, 0.600 nm rectangular barrier at E = 1.00 eV, e^(−2γ) returns 2.4 × 10⁻⁵ where the exact answer is 7.1 × 10⁻⁵. The exponent is right; the missing factor of three is the prefactor WKB never had.

An electron meets a smooth barrier whose peak is V₀ = 5.00 eV, carrying E = 4.90 eV, so its two classical turning points sit only 0.12 nm apart. What is the correct verdict on the estimate T ≈ e^(−2γ) at this energy?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA rectangular barrier of height V₀ = 4.00 eV and width L = 0.600 nm faces an electron of energy E = 1.00 eV. Find the Gamow exponent and the WKB transmission, then compare with the exact opaque-limit result T = [16E(V₀ − E)/V₀²] e^(−2κL).

The forbidden interval is the whole barrier and κ is constant, so γ = κL with κ = √(2m(V₀ − E))/ħ = 5.123 nm⁻¹ × √((V₀ − E)/eV) = 5.123 × √3.00 = 8.87 nm⁻¹.
γ = κL = 8.87 nm⁻¹ × 0.600 nm = 5.32, so the Gamow exponent is 2γ = 10.65.
T_WKB = e^(−10.65) = 2.4 × 10⁻⁵.
The exact prefactor is 16E(V₀ − E)/V₀² = 16 × 1.00 × 3.00 / 16.0 = 3.00, giving T_exact = 3.00 × 2.4 × 10⁻⁵ = 7.1 × 10⁻⁵.
WKB got the exponent exactly right and the size wrong by a factor of three: a vertical wall is not a linear turning point, so no Airy patch supplies its coefficient.

Answer2γ = 10.65 and T_WKB = 2.4 × 10⁻⁵, against the exact 7.1 × 10⁻⁵ — same exponent, prefactor 3.00 missing.

MediumA metal of work function Φ = 4.50 eV sits in a field F = 3.00 V nm⁻¹, so an electron at the Fermi level sees the triangular barrier V(x) = Φ − eFx measured from its own energy. Find 2γ and the transmission, then find what a 10% rise in F does to it.

Turning point where V = E = 0: xₜ = Φ/(eF) = 4.50 eV ÷ 3.00 eV nm⁻¹ = 1.50 nm.
γ = (1/ħ)∫₀^{xₜ} √(2m(Φ − eFx)) dx = (2/3)√(2m) Φ³⁄² / (ħ eF), the Fowler–Nordheim exponent.
Numbers: √(2m) = 1.350 × 10⁻¹⁵ kg¹⁄²; Φ = 7.209 × 10⁻¹⁹ J so Φ³⁄² = 6.122 × 10⁻²⁸; ħeF = (1.055 × 10⁻³⁴)(4.807 × 10⁻¹⁰) = 5.069 × 10⁻⁴⁴. Then γ = (2/3)(1.350 × 10⁻¹⁵)(6.122 × 10⁻²⁸)/(5.069 × 10⁻⁴⁴) = 10.87.
2γ = 21.74, so T ≈ e^(−21.74) = 3.6 × 10⁻¹⁰.
γ ∝ 1/F, so F = 3.30 V nm⁻¹ gives γ = 10.87/1.10 = 9.88, 2γ = 19.8 and T = 2.6 × 10⁻⁹ — a 10% field change multiplies the current by 7.2. That exponential sensitivity, not the prefactor, is what a Fowler–Nordheim plot measures.

Answer2γ = 21.7 and T ≈ 3.6 × 10⁻¹⁰; at F = 3.30 V nm⁻¹, T = 2.6 × 10⁻⁹, larger by a factor of 7.2.

HardAn inverted-oscillator barrier V(x) = V₀ − ½mω²x² has ħω = 0.500 eV. Take an electron with V₀ − E = 1.20 eV. Find γ in closed form, evaluate T, then test the formula at the top of the barrier and 0.300 eV above it.

Turning points at x = ±a with a = √(2(V₀ − E)/(mω²)), so V − E = (V₀ − E)(1 − x²/a²) between them.
γ = (1/ħ)∫₋ₐ⁺ᵃ √(2m(V₀ − E)) √(1 − x²/a²) dx = (1/ħ)√(2m(V₀ − E)) × (πa/2) = π(V₀ − E)/(ħω): every mass and length cancels.
γ = π × 1.20/0.500 = 7.54, so 2γ = 15.08 and T ≈ e^(−15.08) = 2.8 × 10⁻⁷.
The exact transmission for this barrier is T = [1 + e^(2γ)]⁻¹. Since e^(15.08) = 3.5 × 10⁶ ≫ 1 the two agree to four figures: for an opaque parabolic barrier the Gamow exponent is not an approximation at all.
At E = V₀, γ = 0: the exact result is T = ½ while e^(−2γ) = 1. At 0.300 eV above the top, γ = −1.885 and T = [1 + e^(−3.77)]⁻¹ = 0.977, so 2.3% reflects off a barrier the electron clears — while e^(−2γ) = 43 is not a probability at all.

Answerγ = π(V₀ − E)/(ħω) = 7.54 and T = 2.8 × 10⁻⁷; at the top T = ½, not 1, and 0.300 eV above it T = 0.977.