
Quantum core · institution-dependent · Interactive lesson
WKB and 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.
UPV-06.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Quantum core · institution-dependent. Your university determines the required depth of these quantum topics, including the use of operators, differential equations and computation. Institution-dependent does not automatically mean optional.
Build the model
Where no exact solution exists, get the exponent right.
The key idea
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.
The complete physical model and assumptions
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.
1Put 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.
2The 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′| / |p|³ ≪ 1, with |p| = ħκ under a barrier. 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 step in the 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 / |p|³ = 0.031, comfortably small, but it grows without bound as p → 0, passing 1 where V − E has fallen to 0.44 eV — the last 0.15 nm of a 1.50 nm barrier, beside the outer turning point. WKB is excellent over ninety per cent of the region and worthless in the rest.
3Turning 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); taking the forbidden side to the right, V′(a) > 0, 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.
4Connection 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.
5A barrier has two turning points and one exponent
Send a wave in from the left, demand a purely outgoing wave on the right, connect at both turning points keeping the branch that decays from a to b, 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, yet because κ vanishes at both, the moving limits add nothing to dγ/dE: the whole energy dependence is κ shrinking, dγ/dE = −(m/ħ²)∫ₐᵇ dx/κ, weighted towards the ends where κ is small. 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.
6The 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 milliseconds against a full diagonalisation and, for a smooth well and n ≥ 2, is usually right to a fraction of a per cent.
Key equations
Read the units, assumptions and limits before substituting.
Semiclassical wave, allowed region
ψ ≈ (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 travelForbidden region
ψ ≈ (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′| / |p|³ ≪ 1, λ̄ = ħ/p
Not “ħ is small”: a step in V breaks WKB at any energy, and p → 0 breaks it always.
dimensionless; λ̄ is the reduced local wavelength, in m, with |λ̄| = 1/κ under a barrierAiry patch at a linear turning point
ψ″ = 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.
V′(a) > 0, forbidden side to the right; 1/α is the patch width in m: 0.23 nm for an electron on a 3.0 V nm⁻¹ rampConnection, 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 outwardGamow exponent and transmission
T ≈ e(−2γ), γ = ∫ₐᵇ κ dxuniformly T = [1 + e(2γ)]⁻¹
The uniform form is exact for a parabolic barrier: 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 dimensionlessInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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.
Live results
- TURNING POINT xₜ
- 0.271 nm
- GAMOW EXPONENT 2γ
- 7.56
- log₁₀ T WKB
- -3.28
- log₁₀ T EXACT
- -3.28
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Electron energy E: 2.0 eV; Barrier half-width w: 0.35 nm; Barrier height V₀: 5.0 eV
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
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 while κ falls everywhere between them, 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.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A 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).
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- 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.
- TWKB = 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 Texact = 3.00 × 2.37 × 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.
Answer. 2γ = 10.65 and TWKB = 2.4 × 10⁻⁵, against the exact 7.1 × 10⁻⁵ — same exponent, prefactor 3.00 missing.
MediumExample 2
A 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- 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 = 7.210 × 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 T by 7.2. That exponential sensitivity, not the prefactor, is what a Fowler–Nordheim plot measures.
Answer. 2γ = 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.
HardExample 3
An 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- 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.
Answer. γ = 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.
Quick check
Select an answer and explain it before checking.
Review & continue
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.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.