Transmission through a rectangular barrier

Derivation · the atom

Transmission through a rectangular barrier

D-011 rectangular-barrier-transmission Home PU-202 Threads waves · matter Depends on D-007 verified

Statement

A particle with energy below a barrier has a non-zero probability of appearing on the far side, and that probability falls exponentially with barrier width.

Why it matters

Tunnelling is the quantum effect with no classical analogue that shows up in real instruments and real matter. The same exponential governs the scanning tunnelling microscope, the Gamow factor for α-decay (PU-304), and solid-state junctions (PU-303). Because the decay rate scales with √m, the effect is dramatic for electrons, marginal for protons, and negligible for anything larger.

Assumptions

The barrier is rectangular and one-dimensional.An idealisation. Real barriers need the WKB treatment of PU-401, which replaces κa with ∫κ(x) dx.

Steady state, single incident energy.A monoenergetic beam, not a wave packet. A packet has an energy spread and its transmission is an average over that spread.

E < V₀.For E > V₀ the algebra is identical with κ → ik′, sinh → sin, giving resonant transmission — the Ramsauer–Townsend effect.

Derivation

ψ_I = e^ikx + r e^−ikx, ψ_II = A e^κx + B e^−κx, ψ_III = t e^ikx

General solution in each region, with k = √(2mE)/ℏ and κ = √(2m(V₀−E))/ℏ. Inside the barrier the solutions are real exponentials, not oscillatory — the wavefunction does not propagate there, it decays. A

ψ and ψ′ continuous at x = 0 and x = a

Four matching conditions for four unknowns. Continuity of ψ is required for the probability density to be well defined; continuity of ψ′ follows from integrating the Schrödinger equation across the step, and holds because the potential jump is finite. A

T ≡ j_trans / j_inc = |t|²

Transmission is defined as a ratio of probability currents (D-007), not of amplitudes. Here the incident and transmitted regions have the same k, so the ratio reduces to |t|² — but that simplification is a coincidence of equal potentials on both sides and must not be assumed generally. B

solve the 4×4 system for t

Routine but tedious linear algebra. Eliminate A and B, then r. The hyperbolic functions appear because the barrier solutions are real exponentials and the combination e^κa ± e^−κa is forced by the two matching points. C

Result

T = [ 1 + V₀² sinh²(κa) / 4E(V₀ − E) ]⁻¹

Thick-barrier limit. For κa ≫ 1, sinh(κa) ≈ ½e^κa, giving T ≈ [16E(V₀−E)/V₀²] e^−2κa.

Reading. The exponential is everything; the prefactor is a detail of order unity. Because κ ∝ √m, tunnelling is dramatic for electrons, marginal for protons, and utterly negligible for anything larger.

Units check. κ has units m⁻¹, so κa is dimensionless and T is a pure number between 0 and 1. ✓

Limiting cases

Thick barrier (κa ≫ 1): T collapses to the exponential form above — the working expression for STM and α-decay.
Thin / weak barrier (κa → 0): sinh²(κa) → 0, so T → 1 — the barrier becomes transparent.
Above the barrier (E > V₀): with κ → ik′ and sinh → sin, T shows resonant maxima at k′a = nπ (Ramsauer–Townsend).

Breaks when

The barrier is not rectangular — use WKB, which reduces to this result for a square profile.
The particle is described by a broad wave packet — transmission must be averaged over the energy distribution, and the sharp resonances above the barrier wash out.
Many-body effects matter, as in real solid-state junctions where image charges lower the effective barrier.

Failure modes

Defining T as an amplitude ratio rather than a current ratio. Wrong whenever the potential differs on the two sides.
Saying the particle "borrows energy" from the uncertainty principle. Nothing is borrowed; the stationary state simply has non-zero amplitude in the classically forbidden region.
Forgetting that T + R = 1 provides a free check on the algebra. Always apply it.

Worked number — why the STM works

An electron with V₀ − E = 4.0 eV gives κ = √(2 × 9.11×10⁻³¹ × 6.41×10⁻¹⁹)/1.055×10⁻³⁴ ≈ 1.02 × 10¹⁰ m⁻¹. Over a 0.5 nm gap, 2κa ≈ 10.2, so T ~ 10⁻⁵ with the prefactor. Increase the gap by just 0.1 nm and 2κa rises by 2.0, cutting the current by a factor of about 7.4. That extreme sensitivity to distance is precisely what gives the scanning tunnelling microscope atomic resolution — the instrument is a direct readout of this exponential.

Run the check — plot T(E) across V₀ →

Discussion

Classically a particle with energy E below a barrier of height V₀ simply turns around: the region inside the barrier is forbidden because it would demand a negative kinetic energy. Quantum mechanically the stationary state does not vanish at the barrier face — it continues as a decaying real exponential e^−κx and re-emerges on the far side with a small but non-zero amplitude. Nothing propagates through the barrier in the ordinary sense; the wavefunction is evanescent inside it, carrying no oscillation and no net current there, yet the boundary matching forces a finite transmitted wave beyond it. The transmitted flux is real, steady, and measurable.

The single most important feature of the result is the exponential factor e^−2κa in the thick-barrier limit. Because κ = √(2m(V₀−E))/ħ, the transmission is exponentially sensitive to three things: the barrier width a, the height deficit V₀−E, and the particle mass m. A change in width of a single atomic spacing can alter T by orders of magnitude, and the √m dependence is why tunnelling dominates the behaviour of electrons, is marginal for protons, and is unmeasurably small for anything heavier still. This is not a small perturbation on classical behaviour; it is an all-or-nothing regime governed entirely by the exponent.

That extreme sensitivity is the working principle of several technologies. In the scanning tunnelling microscope (STM) a bias drives electrons across the vacuum gap between a sharp tip and a surface; because the current tracks e^−2κa, a height change of 0.1 nm changes it by nearly an order of magnitude, giving atomic vertical resolution. In alpha decay the emitted particle tunnels out through the Coulomb barrier confining it to the nucleus; the enormous range of decay half-lives — from microseconds to billions of years — is almost entirely the exponential responding to small changes in barrier height, the content of the Gamow factor (PU-304). The tunnel (Esaki) diode exploits interband tunnelling in a heavily doped junction to produce a negative-differential-resistance region used in fast oscillators, and the Josephson junction rests on the coherent tunnelling of Cooper pairs.

Common misconceptions. The particle does not "borrow" energy from the uncertainty principle to climb the barrier and pay it back afterward — the stationary state has a single definite energy E throughout, and the barrier region simply hosts a non-zero, non-propagating amplitude. Nor is the particle ever measured with negative kinetic energy: a position measurement that localises it inside the barrier necessarily injects, through the collapse, enough momentum spread to make its energy consistent with being there. Two further traps are worth naming at this level. First, T is a ratio of probability currents, not of amplitudes; the two coincide here only because the potential is equal on both sides, and the identification fails the moment the exit region differs from the entry region. Second, transmission is not exponentially small for all energies — for E > V₀ the same algebra with κ → ik′ turns sinh into sin, and T oscillates back up to exactly 1 at the resonances k′a = nπ — the Ramsauer–Townsend effect.

Worked examples

Example 1 — an electron through a nanometre-scale barrier. An electron of energy E = 1.0 eV is incident on a rectangular barrier of height V₀ = 5.0 eV and width a = 0.40 nm. Find the transmission probability. Use m_e = 9.11×10⁻³¹ kg, ħ = 1.055×10⁻³⁴ J·s.

κ = √(2m_e(V₀−E)) / ħ

Set up the decay constant symbolically before inserting numbers. The relevant energy is the deficit V₀−E = 4.0 eV.

V₀−E = 4.0 eV × 1.602×10⁻¹⁹ J/eV = 6.41×10⁻¹⁹ J

Convert the energy deficit to joules so SI constants can be used.

κ = √(2 × 9.11×10⁻³¹ × 6.41×10⁻¹⁹) / 1.055×10⁻³⁴ = 1.02×10¹⁰ m⁻¹

Numerator √(1.168×10⁻⁴⁸) = 1.081×10⁻²⁴ kg·m/s; divide by ħ.

κa = 1.02×10¹⁰ × 0.40×10⁻⁹ = 4.10

Dimensionless exponent. Since κa > 1 the barrier is moderately thick but the full formula is still used.

sinh(4.10) = ½(e^4.10 − e^−4.10) = 30.1, sinh²(κa) = 905

Evaluate the hyperbolic term that carries the barrier dependence.

T = [ 1 + V₀² sinh²(κa) / (4E(V₀−E)) ]⁻¹ = [ 1 + (25)(905) / (4·1.0·4.0) ]⁻¹

The energy ratio uses eV² in both numerator and denominator, so the units cancel: V₀²/4E(V₀−E) = 25/16 = 1.5625.

T = [ 1 + 1.5625 × 905 ]⁻¹ = (1 + 1414)⁻¹ = 7.1×10⁻⁴

Cross-check against the thick-barrier form: 16E(V₀−E)/V₀² = 2.56, times e^−2×4.10 = 2.76×10⁻⁴, gives 7.1×10⁻⁴. ✓

T ≈ 7.1×10⁻⁴

Answer. About 1 electron in 1400 is transmitted — a pure dimensionless probability. Roughly seven electrons in ten thousand cross a 0.40 nm, 4 eV barrier; widen it to 0.50 nm and T falls by a further factor of ≈7.7.

Example 2 — an alpha particle in a nuclear-scale barrier. Model the confinement of an alpha particle (m_α = 6.64×10⁻²⁷ kg) as a rectangular barrier of height V₀ = 10.0 MeV, with the particle at E = 5.0 MeV and a barrier width a = 20 fm. Estimate T.

V₀−E = 5.0 MeV = 5.0×10⁶ × 1.602×10⁻¹⁹ J = 8.01×10⁻¹³ J

Convert the energy deficit to joules. Note the deficit is a million times larger than in Example 1, but so is the mass.

κ = √(2 × 6.64×10⁻²⁷ × 8.01×10⁻¹³) / 1.055×10⁻³⁴ = 9.78×10¹⁴ m⁻¹

Numerator √(1.064×10⁻³⁸) = 1.032×10⁻¹⁹ kg·m/s.

κa = 9.78×10¹⁴ × 20×10⁻¹⁵ = 19.6 (≫ 1)

The exponent is large, so the thick-barrier limit applies and the full sinh formula reduces to the exponential form.

T ≈ [16E(V₀−E)/V₀²] e^−2κa = [16(5)(5)/(10)²] e^−39.1 = 4.0 × e^−39.1

The prefactor 16E(V₀−E)/V₀² is an order-unity number; the exponent 2κa = 39.1 dominates completely.

e^−39.1 = 1.0×10⁻¹⁷ ⇒ T ≈ 4 × 1.0×10⁻¹⁷ = 4×10⁻¹⁷

Using e^−39.1 = 10^{−39.1/2.303} = 10^−16.99.

T ≈ 4×10⁻¹⁷ per encounter

Answer. A vanishingly small dimensionless probability per collision with the wall. Multiplied by the ∼10²¹ wall-assaults per second an alpha makes inside a nucleus, this still yields an escape rate — and the steepness of e^−2κa in the deficit and width is exactly why measured alpha half-lives span more than twenty orders of magnitude. (A rectangular model overestimates T; the true Coulomb barrier requires the WKB integral of PU-401.)

Problems

Compute the barrier decay constant κ and the penetration depth 1/κ for an electron facing an energy deficit V₀−E = 3.0 eV.
Solution

V₀−E = 3.0 × 1.602×10⁻¹⁹ = 4.81×10⁻¹⁹ J. Then 2m_e(V₀−E) = 2 × 9.11×10⁻³¹ × 4.81×10⁻¹⁹ = 8.76×10⁻⁴⁹. So κ = √(8.76×10⁻⁴⁹)/1.055×10⁻³⁴ = 9.36×10⁻²⁵/1.055×10⁻³⁴ = 8.9×10⁹ m⁻¹. The penetration depth is 1/κ = 0.11 nm — roughly one atomic spacing, which is why nanometre gaps already suppress the current heavily.
An electron of energy E = 3.0 eV meets a barrier V₀ = 8.0 eV of width a = 0.30 nm. Find T from the full formula.
Solution

Deficit V₀−E = 5.0 eV = 8.01×10⁻¹⁹ J. κ = √(2·9.11×10⁻³¹·8.01×10⁻¹⁹)/1.055×10⁻³⁴ = 1.145×10¹⁰ m⁻¹. Then κa = 1.145×10¹⁰ × 0.30×10⁻⁹ = 3.44, so sinh(3.44) = 15.5 and sinh² = 240. The energy factor is V₀²/4E(V₀−E) = 64/(4·3·5) = 64/60 = 1.067. Hence T = [1 + 1.067×240]⁻¹ = (1 + 256)⁻¹ = 3.9×10⁻³.
For the barrier of Problem 2, the width is increased from 0.30 nm to 0.50 nm. By what factor does the tunnelling current change? Comment on what this shows.
Solution

In the thick-barrier regime T ∝ e^−2κa, so the ratio is T(0.50)/T(0.30) = e^−2κΔa with Δa = 0.20 nm. Here 2κΔa = 2 × 1.145×10¹⁰ × 0.20×10⁻⁹ = 4.58, giving a factor e^−4.58 = 1.0×10⁻². The current drops by about a factor of 100 for a change of only two-fifths of a nanometre. This exponential lever is exactly what gives the STM its vertical sensitivity: the tunnelling current is a direct, steep readout of tip–surface separation.
Above the barrier (E > V₀) the transmission becomes T = [1 + V₀²sin²(k′a)/(4E(E−V₀))]⁻¹ with k′ = √(2m(E−V₀))/ħ. Show that T = 1 at resonances, and find the electron energy above V₀ for the first resonance of a barrier of width a = 0.50 nm.
Solution

T = 1 requires sin(k′a) = 0, i.e. k′a = nπ for integer n ≥ 1 — the barrier width is a whole number of half wavelengths, so the internally reflected waves interfere destructively and the barrier becomes perfectly transparent (Ramsauer–Townsend). First resonance n = 1: k′ = π/a = π/(0.50×10⁻⁹) = 6.28×10⁹ m⁻¹. Then E−V₀ = ħ²k′²/2m = (1.055×10⁻³⁴)²(6.28×10⁹)²/(2·9.11×10⁻³¹) = 2.41×10⁻¹⁹ J = 1.5 eV above V₀.
A proton and an electron each face the same barrier: V₀−E = 4.0 eV, width a = 0.20 nm. Compare their transmission exponents 2κa and hence their transmission probabilities. (m_p/m_e = 1836.)
Solution

For the electron, κ_e = √(2·9.11×10⁻³¹·6.41×10⁻¹⁹)/1.055×10⁻³⁴ = 1.024×10¹⁰ m⁻¹, so 2κ_ea = 2×1.024×10¹⁰×0.20×10⁻⁹ = 4.10 and T_e ∼ e^−4.10 ≈ 1.7×10⁻² (a few percent with the prefactor). Since κ ∝ √m, the proton exponent scales by √1836 = 42.8: 2κ_pa = 42.8 × 4.10 = 176, so T_p ∼ e⁻¹⁷⁶ ≈ 10⁻⁷⁶. Identical barrier, but the proton's transmission is smaller by some 74 orders of magnitude — the √m in the exponent is why tunnelling is an electron phenomenon and effectively forbidden for heavier particles.