Potential barriers & quantum tunnelling

Where U = U₀ exceeds E, the time-independent Schrödinger equation −(ħ²/2m)ψ″ + U₀ψ = Eψ rearranges to ψ″ = κ²ψ, with κ = √(2m(U₀ − E))/ħ. Its general solution is C e^(−κx) + D e^(κx): no oscillation and no wavelength, only a length 1/κ. For an electron 2.0 eV below the top, κ = √(2 × 9.109 × 10⁻³¹ kg × 3.204 × 10⁻¹⁹ J)/(1.055 × 10⁻³⁴ J s) = 7.24 × 10⁹ m⁻¹, so |ψ|² falls by a factor e every 1/(2κ) = 69 pm. In a finite well the growing term was discarded because the region ran to infinity. A barrier has a far face, so D stays, and it is not optional: a single real exponential has zero probability current j = (ħ/m) Im(ψ*ψ′), so a decaying tail alone could carry nothing through. The current inside comes from the interference of the two terms, j = (2ħκ/m) Im(C*D), and it must equal the transmitted current at every x.

Send in A e^(ikx) from the left, with k = √(2mE)/ħ. When κL ≫ 1 the two faces act almost independently, so build T from three factors. At the entrance, matching ψ and ψ′ onto a decaying exponential, as at a single step, leaves an amplitude 2k/(k + iκ) per unit A just inside. Across the barrier that amplitude shrinks by e^(−κL). At the exit, matching the decaying and growing pieces onto an outgoing wave F e^(ikx) multiplies it by 2κ/(κ − ik). Transmission is a ratio of probability currents, (ħk/m)|F|² over (ħk/m)|A|², and with the same k on both sides it is |F/A|² = [4k²/(k² + κ²)][4κ²/(k² + κ²)] e^(−2κL) = 16(E/U₀)(1 − E/U₀) e^(−2κL). For E = 2.0 eV, U₀ = 5.0 eV and L = 0.50 nm: κ = 8.87 nm⁻¹, e^(−8.87) = 1.40 × 10⁻⁴, the prefactor is 3.84, and T ≈ 5.38 × 10⁻⁴, matching the exact solution of all four boundary conditions to three figures.

Written out, the estimate is ln T ≈ −(2L/ħ)√(2m(U₀ − E)). The logarithm of T is linear in L but only square-root in m and in U₀ − E, so the logarithmic sensitivities differ by a factor of two: ∂(ln T)/∂(ln L) = −2κL, while ∂(ln T)/∂(ln m) = −κL. Take the barrier above, where 2κL = 8.87. Doubling the width doubles the exponent to 17.7 and cuts T by a further e^(8.87) ≈ 7100; doubling U₀ − E raises the exponent only to 8.87 × √2 = 12.5, a factor of about 40. The root still bites when the exponent is large: send a proton at the same barrier and κ grows by √1836 = 42.9, giving 2κL = 380 and T ≈ 10⁻¹⁶⁵. That is the correspondence check. With ħ in the denominator, anything that makes the system more classical, whether more mass or a wider or higher barrier, drives the exponent up without limit, and the classical verdict T = 0 for E below U₀ is recovered as the limit.

An α particle inside a nucleus meets the Coulomb barrier U(r) = 2Ze²/(4πε₀r) beyond the contact radius R, with Z the daughter's charge number. For Z = 82 and R = 9.0 fm the top is 2 × 82 × 1.440 MeV fm ÷ 9.0 fm = 26.2 MeV, yet emitted α energies lie between about 4 and 9 MeV, so a 6.0 MeV α is forbidden all the way out to b = 39.4 fm. No single κ describes that. Across any short stretch dx the amplitude falls by e^(−κ(x)dx), so over the whole forbidden region it falls by exp(−∫κ dx) and the probability by the square, valid while U changes little over 1/κ. For the Coulomb shape the integral is the Gamow exponent 2G, and for R ≪ b it tends to πb√(2mE)/ħ, proportional to Z/√E because b ∝ Z/E. For Z = 82, 2G is 38.4 at 8 MeV and 83.7 at 4 MeV: a factor of two in energy moves the half-life by about 10²⁰. The nucleus is three-dimensional, but with no angular momentum u = rψ obeys a one-dimensional equation in r, so the same machinery applies.

A scanning tunnelling microscope holds a sharp metal tip a few tenths of a nanometre above a conducting surface. Electrons near the Fermi level on either side face a barrier whose height above them is roughly the mean work function φ, about 4 to 5 eV, and whose width is the gap d. A bias of a few tens of millivolts, small beside φ, opens a narrow window of energies in which filled states on one side face empty states on the other, so the current is proportional to the bias times the transmission of that window: I ∝ V e^(−2κd), with κ = √(2mₑφ)/ħ ≈ 11 nm⁻¹. The exponent does two jobs. Vertically, a feedback loop adjusts the tip height to hold I fixed while scanning, and the record of that height is the topograph. Laterally, a tip tens of nanometres across still ends in a single atom nearest the surface, and the exponent hands almost all the current to it. Two cautions: the image charge rounds and lowers the barrier at small gaps, and a bias comparable with φ tilts it into a trapezoid that needs ∫κ dx.

The estimate describes a thick, one-dimensional, static barrier, and each of those words marks a boundary. Thick: at E = U₀/2 the exact rectangular result, T = [1 + U₀² sinh²(κL)/(4E(U₀ − E))]⁻¹, agrees with the two-face formula to 4% at κL = 2 but differs by 29% at κL = 1. For an electron of 1.0 eV meeting a barrier 1.5 eV high and 0.10 nm wide, κL = 0.36: the exact T is 0.87, the bare e^(−2κL) gives 0.48, and the two-face formula returns 1.72, an impossible transmission. Near the top κ → 0 and the bare estimate tends to 1, while the exact T tends to [1 + mL²U₀/(2ħ²)]⁻¹, which is 0.38 for an electron at 1.0 eV and a 0.50 nm barrier: reflection survives even at the top. One-dimensional and static: real barriers are rounded, tips and nuclei are three-dimensional, a barrier that changes during the crossing can exchange energy, and an α particle must first form inside the nucleus, a probability no barrier calculation contains.