University Physics IV · Quantum Potentials · 7.8
The Rectangular Barrier & Tunnelling
Two faces, four matching conditions, one exact transmission coefficient. Learn where the sinh² comes from, when the e(−2κL) shorthand is safe, and why a barrier the particle comfortably clears can still turn it back.
Build the model
Connect the measurement to the mechanism.
The rectangular barrier is the smallest problem in which a particle crosses a region it has no energy to enter, and it is solved exactly, which is why every tunnelling estimate is quoted against it. Three regions of constant potential, two abrupt faces. Below the top the equation inside reads ψ″ = κ²ψ with κ = √(2m(V₀−E))/ħ, so the interior solution is not a wave but a pair of real exponentials, rising and falling, and both must be kept because a finite width leaves a far face to reflect from.
Continuity of ψ and ψ′ at each face gives four linear equations; eliminating the two interior amplitudes returns the transmitted flux exactly, T = [1 + V₀²sinh²(κL)/(4E(V₀−E))]⁻¹. Everything this topic teaches lives in that one expression. Once κL passes about 2, sinh² grows as e(2κL)/4 and T collapses to a bounded prefactor times e(−2κL): width and mass sit in an exponent, so doubling L costs four orders of magnitude — which is what spreads α-decay lifetimes over twenty decades and lets an STM resolve single atoms.
Push E above V₀ and κ turns imaginary, sinh becomes sin, and the same formula predicts partial reflection off a barrier the particle clears, broken by energies at which the barrier is perfectly transparent. The cost: unnormalisable plane waves, a square profile no real barrier has, one dimension, and no statement about how long the crossing takes.
- Simple definition
- Barrier tunnelling is the transmission of a particle through a region where its energy lies below the potential, at a rate fixed by how far the wavefunction's real exponential decays across the width.
- Example
- An electron 4.0 eV below the top of a 5.0 eV barrier has κ = 10.2 nm⁻¹. A 0.50 nm barrier passes T = 9.1 × 10⁻⁵ of the incident flux; widen it to 1.00 nm and only 3.2 × 10⁻⁹ gets through — 28 000 times less.
κ fixes how fast T dies with width: 1/2κ = 49 pm at a 4.0 eV deficit.
Both in m⁻¹. For an electron, κ = 5.12 nm⁻¹ × √((V₀−E)/eV).
The only form valid for thin barriers, κL ≲ 1, where the exponential shorthand is 20% out.
L is the barrier width; T is a pure number, and the leading 1 keeps it under 1.
Splits T into a prefactor worth a factor of a few — at most 4 — and an exponent worth decades.
Fractional error below 2e(−2κL): 4% at κL = 2, 0.5% at κL = 3, 27% at κL = 1.
R + T = 1 is the arithmetic check that all four matching equations were solved right.
Ratios of the current j = (ħ/m)⋅Im(ψ*ψ′). Here kₒᵤₜ = kᵢₙ, so T = |t|².
Resonant transmission: a whole number of half-waves fits, and the two face reflections cancel.
κ → ik₂ turns sinh into sin. T = 1 exactly when k₂L = nπ, n = 1, 2, 3…
T runs smoothly through E = V₀, so the two branches need no special case to join them.
β = √(2mV₀L²)/ħ is the barrier's dimensionless strength; nothing diverges here.
Three regions, five amplitudes, four equations
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.
Eliminate the interior and read the sinh²
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.
The thick-barrier limit, and the prefactor it hides
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 it can never shift T by more than a factor of a few, while the exponent shifts it by decades. Know where the shorthand fails. Its fractional error is under 2e(−2κL) — 4% at κL = 2, 0.5% at κL = 3 — but 27% at κL = 1, and for κL below 1 it is worthless. Test κL before you use it, and fall back on the sinh form.
How hard the exponent bites
Take an electron with 4.0 eV of barrier above it: κ = 5.12 × √4.0 = 10.2 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 a few femtometres wide.
Above the barrier: sin replaces sinh, and T can reach 1
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.
What the square barrier is not
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.
Change one variable at a time
Make the relationship visible.
Widen the barrier from 0.10 to 0.80 nm: the tunnelling branch flattens onto the axis while the resonances above the top crowd downwards — one number, β = √(2mV₀L²)/ħ, drives both. The marked point at E = V₀ is only 1/(1 + β²/4), never 1.
BARRIER STRENGTH β5.07
T AT E = V₀/20.0031
T AT THE TOP E = V₀0.135
FIRST RESONANCE E₁2.77 eV
Live interpretationBARRIER STRENGTH β: 5.07. T AT E = V₀/2: 0.0031. T AT THE TOP E = V₀: 0.135. FIRST RESONANCE E₁: 2.77 eV
Catch the common trap
Explain before calculating.
An electron of energy E = 1.0 eV meets a rectangular barrier of height V₀ = 5.0 eV and width 0.50 nm, giving a transmission coefficient of 9.1 × 10⁻⁵. The width is doubled to 1.00 nm, everything else unchanged. What is the new transmission coefficient?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron of energy 2.0 eV meets a rectangular barrier of height 6.0 eV and width 0.40 nm. Find the decay constant κ inside the barrier, check that the thick-barrier formula is safe to use, and estimate the transmission coefficient.
- The barrier stands V₀ − E = 6.0 − 2.0 = 4.0 eV above the electron's energy, and κ = √(2m(V₀−E))/ħ. For an electron that is κ = 5.12 nm⁻¹ × √((V₀−E)/eV) = 5.12 × 2.0 = 10.2 nm⁻¹.
- Test the limit before trusting it: κL = 10.2 × 0.40 = 4.10, well past 2, so the thick-barrier form is good to far better than 1%.
- Prefactor: 16E(V₀−E)/V₀² = 16 × 2.0 × 4.0 / 6.0² = 128/36 = 3.56 — a pure number, below its ceiling of 4.
- Exponent: 2κL = 8.20, and e(−8.20) = 2.76 × 10⁻⁴. So T = 3.56 × 2.76 × 10⁻⁴ = 9.8 × 10⁻⁴.
- Check against the exact form: sinh(4.10) = 30.1, so T = [1 + 6.0² × 30.1²/(4 × 2.0 × 4.0)]⁻¹ = 9.79 × 10⁻⁴ — 0.04% below the estimate.
Answerκ = 10.2 nm⁻¹ and T ≈ 9.8 × 10⁻⁴. About one electron in a thousand crosses; the other 999 are reflected.
MediumA scanning tunnelling microscope tip sits 0.50 nm above a metal surface, across an apparent barrier standing 4.0 eV above the electron energy. The tip is withdrawn to 0.60 nm at fixed bias. By what factor does the tunnelling current fall, and what change of gap would halve it?
- The current is proportional to T, and with E and V₀ unchanged the prefactor 16E(V₀−E)/V₀² is identical at both gaps, so it cancels in the ratio and only e(−2κL) survives. That is why the STM's exponential law holds even though nobody knows the prefactor accurately.
- κ = 5.12 × √4.0 = 10.25 nm⁻¹, so 2κ = 20.5 nm⁻¹.
- I(0.60)/I(0.50) = e(−2κ × 0.10 nm) = e(−2.049) = 0.129, so the current falls to 12.9% of its former value — a factor of 7.8 for one ångström of retraction.
- For a halving, solve e(−2κ ΔL) = ½: ΔL = ln 2/(2κ) = 0.693/20.5 nm⁻¹ = 0.0338 nm = 34 pm.
AnswerThe current falls by a factor of 7.8, to 12.9%. A gap change of only 34 pm halves it — the sensitivity that lets an STM image single atoms.
HardAn electron beam is fired at a 1.0 nm wide barrier of height V₀ = 3.0 eV, at energies above the top. Find the two lowest energies at which the barrier is perfectly transparent, and the worst transmission between them.
- Above the top κ turns imaginary. With k₂ = √(2m(E−V₀))/ħ, the transmission is T = [1 + V₀² sin²(k₂L)/(4E(E−V₀))]⁻¹.
- T = 1 needs sin(k₂L) = 0, so k₂L = nπ: the barrier must hold a whole number of half-wavelengths, and the reflections from its two faces then cancel.
- For an electron k₂ = 5.12 nm⁻¹ × √((E−V₀)/eV), so with L = 1.0 nm the condition √((E−V₀)/eV) = nπ/5.12 gives E − V₀ = 0.376n² eV. For n = 1, E = 3.38 eV; for n = 2, E = 4.50 eV.
- The worst point between them has sin²(k₂L) = 1, i.e. k₂L = 3π/2, so E − V₀ = 0.376 × 1.5² = 0.846 eV and E = 3.85 eV.
- There T = [1 + 3.0²/(4 × 3.85 × 0.846)]⁻¹ = [1 + 9.00/13.02]⁻¹ = 0.591.
AnswerPerfect transmission at E = 3.38 eV and E = 4.50 eV. At E = 3.85 eV, T = 0.59: the barrier reflects 41% of a beam that classically clears it every time.