University Physics IV · Quantum Potentials · 7.9
WKB Tunnelling & the Gamow Exponent
Real barriers curve, and matching plane waves across a curve is hopeless. This lesson trades that algebra for one number — the integral of κ between the turning points — and shows why that exponent beats an exact solution: it turns a 5 MeV spread in alpha energy into a factor of 10²⁴, and it is why an STM feels atoms.
Build the model
Connect the measurement to the mechanism.
Under a barrier the Schrödinger equation has no oscillation left in it: ψ″ = (2m/ℏ²)(V − E)ψ has real exponential solutions, and for constant V the decay constant is κ = √(2m(V − E))/ℏ. WKB takes that answer seriously even where V is not constant. Write ψ = exp(iS/ℏ), expand S in powers of ℏ, keep the first two terms, and what comes out is a wave that decays at the local κ with a 1/√κ amplitude in front.
Everything then collapses into one dimensionless number, the Gamow exponent 2γ = (2/ℏ)∫√(2m(V − E)) dx taken between the classical turning points, with T ≈ exp(−2γ). The cost is precision, and honesty about where the expansion holds: it assumes the barrier changes little over one decay length, which fails exactly at the turning points where κ goes to zero, so the prefactor WKB hands you is worth a factor of a few and nothing more. What it buys is out of all proportion.
Because the answer sits in an exponent, an integral good to a few per cent fixes a rate to within an order of magnitude, and the same one-line calculation reproduces α-decay half-lives from 10⁻⁷ s to 10¹⁷ s, the ångström-scale sensitivity that gives a scanning tunnelling microscope atomic resolution, and the Fowler–Nordheim law for electrons dragged out of a cold metal by a field.
- Simple definition
- The WKB estimate replaces a smooth barrier by one dimensionless number — the Gamow exponent 2γ, twice the integral of the decay constant κ between the classical turning points — and predicts a transmission probability exp(−2γ).
- Example
- For a scanning tunnelling microscope with a 4.00 eV barrier, κ = √(2mφ)/ℏ = 10.2 nm⁻¹, so a rectangular gap gives 2γ = 2κs; pulling the tip back by 0.100 nm adds 2.05 to the exponent and divides the current by 7.8.
Local decay at the local κ, so a curved barrier is handled without ever matching amplitudes across it.
κ in m⁻¹; valid while |dκ/dx| ≪ κ², so the barrier is near-constant over one decay length
One integral replaces the whole matching problem. The prefactor it drops is of order one, while exp(−2γ) spans decades.
x₁ and x₂ are the turning points where V = E; 2γ is a pure number
At 4.00 eV the current falls 7.8-fold per ångström, so a 10 pm corrugation moves it by 19%. That is atomic resolution.
s is the tip-sample gap and φ the barrier height in eV; κ = 10.2 nm⁻¹ at φ = 4.00 eV
The 4/3 replaces the rectangle's 2, so a sloping barrier of the same height and width transmits about 10⁵ times better.
F is the surface field in V m⁻¹ and L the barrier width; B = 6.83 × 10⁹ V m⁻¹ eV⁻³⁄²
At Z′ = 82, E = 8.78 MeV and R = 9.0 fm it returns 2γ = 33.6, so T = 2.6 × 10⁻¹⁵ per assault on the wall.
b = 2Z′ke²/E is the outer turning point and R the contact radius, both in fm
f moves the answer by a factor of a few and 2γ moves it by twenty powers of ten, so get the exponent right first.
f is the assault frequency, about 10²¹ s⁻¹ for an alpha inside a heavy nucleus
Where the exponent comes from
Write ψ = exp(iS(x)/ℏ), which is exact, and the Schrödinger equation becomes (S′)² − iℏS″ = 2m(E − V). Now expand S in powers of ℏ. Leading order gives S′ = ±p(x) with p = √(2m(E − V)) — a plane wave whose wavelength tracks the local momentum. The next order supplies an amplitude going as 1/√p, which is what holds the probability flux constant as the wavelength stretches. Under a barrier E lies below V, p turns imaginary, and the pair becomes ψ ≈ C exp(±∫κ dx)/√κ with κ = √(2m(V − E))/ℏ. The oscillation has become a real exponential whose rate is set locally, so the wave falls off as exp(−∫κ dx) and not as exp(−κ̄L) for some average κ̄. Dropping the rest of the series requires |dκ/dx| ≪ κ², which reads: the barrier must not change appreciably over one decay length 1/κ. That is the whole approximation, and it is a statement about the barrier's shape, not about how tall it is.
The exponent is an integral, not a product
Because κ ∝ √(V − E), a barrier that sags away from its peak has a much smaller κ over most of its width, and that difference is compounded exponentially. Take the triangular barrier a field pulls at a metal surface, V(x) = φ − eFx. It ends where V = 0, at L = φ/eF, and along it κ(x) = κ₀√(1 − x/L) with κ₀ = √(2mφ)/ℏ. The integral is elementary: γ = ∫₀ᴸ κ₀√(1 − x/L) dx = (2/3)κ₀L, so 2γ = (4/3)κ₀L, exactly two-thirds of the rectangle's 2κ₀L. Put numbers on it. Tungsten at φ = 4.50 eV in F = 3.00 V nm⁻¹ gives L = 1.50 nm and κ₀ = 10.87 nm⁻¹, so 2γ = 21.7 and T = 3.6 × 10⁻¹⁰. A rectangle of the same height and the same width would give 2γ = 32.6 and T = 6.9 × 10⁻¹⁵. The slope of the barrier is worth 5 × 10⁴ in the emitted current, and it lives entirely inside the exponent.
Turning points set the limits and break the method
The limits of the Gamow integral are the points where V = E, and those are precisely where WKB stops working. There κ goes to zero, the 1/√κ amplitude diverges, the local wavelength becomes infinite, and |dκ/dx| ≪ κ² fails no matter how gently V varies. The repair is to linearise V about each turning point, solve that exactly with Airy functions, and match the asymptotic forms onto the WKB waves on either side. Those connection formulae are what supply the prefactor. For a thick barrier — anything with 2γ above about 5 — the prefactor is of order unity: the exact rectangular result is 16E(V − E)/V² times exp(−2κL), and that front factor never exceeds 4. So quote WKB honestly: the exponent is trustworthy, the prefactor is not. A 1% error in an exponent of 90 is 0.9, a factor of 2.5 in the rate; a factor-of-4 error in the prefactor is nothing beside it.
Alpha decay: Gamow's 1928 calculation
Model the alpha as already formed inside the nucleus, carrying the energy E fixed by the Q-value, rattling against the wall at an assault frequency f = v/2R of order 10²¹ s⁻¹, and escaping with probability exp(−2γ) each time it arrives. Then λ = fT. Outside the contact radius R the potential is pure Coulomb, V(r) = 2Z′ke²/r, and the outer turning point sits at b = 2Z′ke²/E. The integral closes: 2γ = 2(√(2mE)/ℏ)b[arccos√x − √(x(1 − x))] with x = R/b. For ²¹²Po (Z′ = 82, E = 8.78 MeV, R = 9.0 fm) it gives 2γ = 33.6 and t½ = 0.23 μs against a measured 0.30 μs. For ²³²Th (Z′ = 88, E = 4.01 MeV) the same line returns 5 × 10¹⁸ s against a measured 4.4 × 10¹⁷ s — ten times too long, and still inside one power of ten across a range of twenty-four. When b ≫ R the bracket tends to π/2 and 2γ becomes proportional to Z′/√E, which is exactly why Geiger and Nuttall found log t½ linear in 1/√E.
The STM turns the exponent into a ruler
A scanning tunnelling microscope puts a vacuum gap s between a sharp tip and a surface, so the barrier height is roughly the work function φ and the current runs as I ∝ Vbias exp(−2κs) with κ = 0.512√(φ/eV) Å⁻¹. At φ = 4.00 eV that is 2κ = 2.05 Å⁻¹: retracting one ångström divides the current by 7.8, and a corrugation of 10 pm still moves it by 19%. Nothing else in the instrument has that gain, which is why a feedback loop holds I fixed and records the z that does it. But be careful what the resulting image is. Constant current is a contour of constant tunnelling probability, and by the Tersoff–Hamann result that is a contour of constant local density of states at the Fermi level — not a map of where the nuclei sit. Adsorbates that deplete states at the Fermi level can image as depressions. One-dimensional WKB gives the gap dependence and says nothing about which states carry the current.
Field emission, and where one dimension runs out
Apply a field to a cold metal surface and an electron at the Fermi level sees the triangle φ − eFx of width φ/eF. The exponent is Bφ³⁄²/F, so the emitted current density is J = aF²/φ times exp(−Bφ³⁄²/F), and a Fowler–Nordheim plot of ln(J/F²) against 1/F is a straight line whose slope returns φ³⁄². Reality then adds the image charge, −e²/16πε₀x, which rounds the top of the triangle and lowers it by the Schottky amount; the correction multiplies the exponent by a factor v(y) of about 0.7 to 0.9, worth some 10³ in current. Those are the honest limits of the whole method: WKB carries no phase information, so it cannot produce resonant transmission above a barrier; it knows nothing of band structure, effective mass or many-body screening; and in three dimensions an alpha with l ≠ 0 meets an extra centrifugal term l(l + 1)ℏ²/2μr² that raises the barrier and lengthens the life.
Change one variable at a time
Make the relationship visible.
Hold Z′ at 82 and R at 9.0 fm and drag the alpha energy from 9.0 down to 4.0 MeV: b slides from 26 fm out to 59 fm, 2γ climbs from 32 to 84, and log₁₀(t½/s) walks from −7 to +15 — a factor of 2.2 in energy paying for twenty-two orders of magnitude in lifetime.
OUTER TURNING POINT b26.8 fm
GAMOW EXPONENT 2γ33.5
log₁₀ T PER ASSAULT-14.6
log₁₀ (t½ / s)-6.7
Live interpretationOUTER TURNING POINT b: 26.8 fm. GAMOW EXPONENT 2γ: 33.5. log₁₀ T PER ASSAULT: −14.6. log₁₀ (t½ / s): −6.7
Catch the common trap
Explain before calculating.
An STM tip sits 0.500 nm above a surface whose barrier height is 4.00 eV, giving κ = √(2mφ)/ℏ = 10.2 nm⁻¹. The tip is withdrawn to 0.600 nm at fixed bias. By what factor does the tunnelling current fall?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn STM tip images a gold surface whose effective barrier height is 5.00 eV. Find the decay constant κ, then find the factor by which the tunnelling current changes as the tip, held at fixed height, passes over a single atomic step 0.20 nm high. Take the electron rest energy as 0.511 MeV and ℏc = 197.3 eV nm.
- κ = √(2mφ)/ℏ is easiest written as √(2mc²φ)/ℏc, so every quantity is an energy: √(2 × 0.511 × 10⁶ eV × 5.00 eV) = √(5.11 × 10⁶ eV²) = 2261 eV.
- Divide by ℏc: κ = 2261 eV ÷ 197.3 eV nm = 11.46 nm⁻¹. The shortcut κ = 0.512 √(φ/eV) Å⁻¹ gives the same 1.15 Å⁻¹.
- The probability is the squared amplitude and the amplitude decays as exp(−κs), so T ∝ exp(−2κs). Taking a ratio kills every prefactor and leaves only the change of gap: I₂/I₁ = exp(−2κΔs).
- Riding over the step at fixed height shrinks the gap by Δs = 0.20 nm, so the exponent changes by 2κΔs = 2 × 11.46 × 0.20 = 4.58, and I₂/I₁ = exp(+4.58) = 98.
Answerκ = 11.5 nm⁻¹, and one atomic step multiplies the current by about 98 — two orders of magnitude, which is why an STM is run in constant-current mode with a feedback loop rather than at fixed height.
MediumA tungsten field emitter has work function φ = 4.50 eV and sits in a surface field F = 3.00 V nm⁻¹, so an electron at the Fermi level meets the triangular barrier V(x) = φ − eFx. Find the barrier width, the Gamow exponent and T; compare with a rectangle of the same height and width; then repeat at F = 5.00 V nm⁻¹.
- The barrier ends where V = 0, so L = φ/eF = 4.50 eV ÷ 3.00 eV nm⁻¹ = 1.50 nm, and along it κ(x) = √(2m(φ − eFx))/ℏ = κ₀√(1 − x/L).
- κ₀ = √(2 × 0.511 × 10⁶ × 4.50) eV ÷ 197.3 eV nm = 2145 ÷ 197.3 = 10.87 nm⁻¹.
- γ = ∫₀ᴸ κ₀√(1 − x/L) dx = (2/3)κ₀L, so 2γ = (4/3) × 10.87 × 1.50 = 21.7 and T = exp(−21.7) = 3.6 × 10⁻¹⁰.
- A rectangle of the same height φ and the same width L would give 2γ = 2κ₀L = 32.6 and T = 6.9 × 10⁻¹⁵. Sloping the barrier is worth 5 × 10⁴ in the current, and the whole effect sits in the exponent.
- At F = 5.00 V nm⁻¹ the width falls to L = 0.900 nm, so 2γ = (4/3) × 10.87 × 0.900 = 13.0 and T = 2.2 × 10⁻⁶: a 67% rise in field multiplies the emission by 6 × 10³.
AnswerL = 1.50 nm, 2γ = 21.7 and T = 3.6 × 10⁻¹⁰; the same-height rectangle gives 6.9 × 10⁻¹⁵, and F = 5.00 V nm⁻¹ gives T = 2.2 × 10⁻⁶.
Hard²¹²Po α-decays to ²⁰⁸Pb with an alpha kinetic energy of 8.78 MeV. Take the daughter charge as Z′ = 82 and the contact radius as R = 9.0 fm, with 2Z′ke² = 236 MeV fm, ℏc = 197.3 MeV fm and the alpha rest energy 3727 MeV. Estimate the transmission per assault, the decay constant and the half-life, and compare with the measured 0.30 μs.
- Outer turning point: b = 2Z′ke²/E = 236 ÷ 8.78 = 26.9 fm, so x = R/b = 9.0 ÷ 26.9 = 0.335 and the forbidden stretch is 17.9 fm wide.
- Wavenumber scale: √(2mE)/ℏ = √(2 × 3727 × 8.78) MeV ÷ 197.3 MeV fm = 255.8 ÷ 197.3 = 1.297 fm⁻¹.
- Closed form: 2γ = 2 × 1.297 × 26.9 × [arccos√0.335 − √(0.335 × 0.665)] = 69.8 × [0.9536 − 0.4720] = 69.8 × 0.4816 = 33.6.
- T = exp(−33.6) = 2.6 × 10⁻¹⁵, so the alpha escapes on roughly one wall collision in 4 × 10¹⁴.
- Assault frequency: v = c√(2E/mc²) = c√(2 × 8.78 ÷ 3727) = 0.0686c = 2.06 × 10⁷ m s⁻¹, so f = v/2R = 2.06 × 10⁷ ÷ (1.8 × 10⁻¹⁴ m) = 1.14 × 10²¹ s⁻¹.
- λ = fT = 1.14 × 10²¹ × 2.6 × 10⁻¹⁵ = 3.0 × 10⁶ s⁻¹, and t½ = ln2/λ = 0.693 ÷ (3.0 × 10⁶) = 2.3 × 10⁻⁷ s.
AnswerT ≈ 2.6 × 10⁻¹⁵, λ ≈ 3.0 × 10⁶ s⁻¹ and t½ ≈ 0.23 μs against the measured 0.30 μs — 23% low, from a model whose only real input is one integral.