WKB tunnelling and the Gamow exponent

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(−κL) with a single κ read off the barrier's peak. 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 since dκ/dx = mV′/(ℏ²κ) it is the same as m|V′| ≪ ℏ²κ³: a limit on the barrier's slope that grows easier to meet the higher V stands above E.

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.

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.

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, R = 9.2 fm) the same line returns 5 × 10¹⁸ s against a measured 4.4 × 10¹⁷ s — about ten times too long, a miss of one decade 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.

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 ∝ V_bias 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.

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: the single exponent exp(−2γ) keeps 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.