The rectangular barrier and tunnelling

Put the faces at x = 0 and x = L, with V = V₀ between them and V = 0 outside. To the left E > V, so ψ = A e^(ikx) + B e^(−ikx) with k = √(2mE)/ħ — an incident beam and a reflected one. Inside, for E < V₀, the equation is ψ″ = +κ²ψ, whose solutions are real exponentials C e^(κx) + D e^(−κx): nothing oscillates in a classically forbidden region. Both terms are needed. The rising one is dropped only for a barrier of infinite width; here the far face is a finite distance away and reflects. To the right, ψ = F e^(ikx) alone, because nothing beyond the barrier sends a wave back — that single choice is what makes this a scattering problem rather than a bound-state one. Only ratios matter, so set A = 1, and continuity of ψ and ψ′ at both faces gives four linear equations for B, C, D and F. Note that ψ″ does jump at each face, since V does; ψ and ψ′ do not, because V is finite.

The four equations are linear, so eliminate the interior amplitudes C and D and solve for F. Taking the modulus squared gives T = |F|²/|A|² = [1 + V₀² sinh²(κL)/(4E(V₀−E))]⁻¹. Three things are worth noticing straight away. It is written as 1/(1 + something non-negative), so 0 < T ≤ 1 automatically and no algebra slip can hand you T > 1. R = 1 − T comes out of the same solution, which is the standard check on the work. And only two dimensionless numbers appear: divide through by V₀² and T = [1 + sinh²(κL)/(4u(1 − u))]⁻¹ with u = E/V₀. Height, width and mass never act separately — they enter through the single strength β = √(2mV₀L²)/ħ, with κL = β√(1 − u). Two barriers with the same β have the same transmission curve.

Since sinh(κL) = (e^(κL) − e^(−κL))/2, once κL is past about 2 the second term is under 2% of the first, so sinh²(κL) → e^(2κL)/4 and T → [16E(V₀−E)/V₀²] e^(−2κL). The 2 in the exponent is not decoration: the amplitude falls as e^(−κL) across the width, and probability is amplitude squared. Write the prefactor as 16u(1 − u): it peaks at 4 when E = V₀/2 and vanishes at both ends, so away from those ends it shifts T by only a factor of a few, while the exponent shifts it by decades. Know where the shorthand fails. Its fractional error is at most 2e^(−2κL) + e^(−4κL) — 4% at κL = 2, 0.5% at κL = 3 — but 29% at κL = 1, and for κL below 1 it is worthless. Test κL before you use it, and fall back on the sinh form.

Take an electron with 4.0 eV of barrier above it: κ = 5.123 × √4.0 = 10.25 nm⁻¹, so 2κ = 20.5 nm⁻¹. Every extra 0.10 nm of width multiplies T by e^(−2.05) = 0.13 — a factor of 7.8 lost per ångström. That one number is the scanning tunnelling microscope: an atom-sized bump bringing the surface 0.1 nm closer to the tip carries about eight times the current, so a feedback loop holding the current fixed traces out single atoms. Mass rides in the same exponent, through κ ∝ √m. At the same 4.0 eV deficit a proton's κ is √1836 = 42.9 times an electron's, so across 0.50 nm the electron's T = 9 × 10⁻⁵ becomes about 10⁻¹⁹⁰. Heavy particles do not tunnel through nanometre barriers at all; alpha decay works only because the nuclear barrier is some tens of femtometres wide.

Let E > V₀. Then κ = √(2m(V₀−E))/ħ is imaginary; write κ = ik₂ with k₂ = √(2m(E−V₀))/ħ real and use sinh(iθ) = i sin θ. The same expression continues into T = [1 + V₀² sin²(k₂L)/(4E(E−V₀))]⁻¹, and two classical expectations break at once. A particle with E > V₀ is now reflected part of the time: at V₀ = 3.0 eV, L = 1.0 nm and E = 3.85 eV the barrier turns back 41% of the incident flux. And T hits exactly 1 whenever sin(k₂L) = 0, that is k₂L = nπ, or L = n⋅λ₂/2 — a whole number of half-wavelengths inside. The wave reflected from the far face then comes back exactly out of phase with the one reflected from the near face and cancels it: the mechanism of an anti-reflection coating, and of the Ramsauer–Townsend minimum in electron–noble-gas scattering.

Four costs. Plane waves are not normalisable, so this is a steady-state statement about a beam, not about a wave packet; a real packet carries a spread of energies and smears the sharp resonances into shallow ripples. Real barriers are not square: a metal surface is rounded by the image potential and an alpha particle sees a Coulomb tail, and for those you integrate κ(x) between the turning points instead — that is the WKB exponent. The model is one-dimensional and single-particle, with no transverse momentum, no other electrons and no phonons; inside a semiconductor heterostructure m must also be swapped for an effective mass. And the calculation returns amplitudes and fluxes but never a time: how long the particle spends in the barrier is not a question this model answers, and the rival definitions of tunnelling time disagree.