University Physics V · Finite Potential Wells and Quantum Tunneling · 6.5
The Rectangular Barrier & Its Opaque Limit
Almost everything students believe about tunnelling comes from one exponential. This is where that exponential is earned: four matching conditions, one collapse of the algebra, an exact T(E) good at every energy, and only then the opaque limit — plus the sine branch above the barrier that classical mechanics gets flatly wrong.
Build the model
Connect the measurement to the mechanism.
A rectangular barrier is the one tunnelling problem that solves exactly, and it is worth solving that way because every approximation used afterwards is a limit of this one answer. The model is three regions of constant potential, a single energy E, and a boundary condition put in by hand: a unit-amplitude wave incident from the left, a reflected amplitude r, an evanescent pair inside, and one outgoing amplitude t on the right, with nothing returning from infinity. Continuity of ψ and ψ′ at both faces gives four linear equations, and because k² + κ² = 2mV₀/ħ² is fixed by the barrier alone, the algebra collapses to T = [1 + V₀² sinh²(κL)/(4E(V₀ − E))]⁻¹.
What it costs is worth naming. The state is not normalisable, so T is a ratio of probability currents, not the probability of any single event; the sharp corners are a fiction no material supplies; and the calculation is time-independent, so it fixes how much flux gets through and says nothing whatever about how long that took. Push it to κL ≫ 1 and it becomes the exponential everyone quotes, 16E(V₀ − E)/V₀² · e(−2κL), whose prefactor is part of the limit rather than decoration.
Continue it past E = V₀ and sinh turns into sin: a particle that can clear the barrier is still reflected, except at qL = nπ, where the barrier is perfectly transparent.
- Simple definition
- The transmission coefficient of a rectangular barrier is the ratio of transmitted to incident probability current for a beam of fixed energy E, obtained by matching ψ and ψ′ at both faces of a constant potential V₀ of width L.
- Example
- An electron at E = 2.0 eV meeting V₀ = 5.0 eV across L = 0.30 nm has κ = 8.87 nm⁻¹ and κL = 2.66, so T = [1 + (25/24)sinh²2.66]⁻¹ = 1.85 × 10⁻² — one part in 54 of the incident current.
Exact at every energy below the barrier — no approximation anywhere in it.
κ = √(2m(V₀ − E))/ħ in m⁻¹; L in m; E and V₀ in any one energy unit
V₀ − E = 3.0 eV gives κ = 8.87 nm⁻¹, so ψ decays by e every 1.13 Å.
for an electron κ/nm⁻¹ = 5.123 √((V₀ − E)/eV), since ħ²/2mₑ = 0.0381 eV nm²
Relative error ≈ (16E(V₀ − E)/V₀² − 2) e(−2κL): 2 × 10⁻⁵ at κL = 5.73, but 0.90% at κL = 2.66.
the prefactor is dimensionless and never exceeds 4, reached at E = V₀/2
T = 1 exactly at qL = nπ; the minima sit at qL = (n + ½)π.
q = √(2m(E − V₀))/ħ is the interior wavenumber, not the asymptotic k = √(2mE)/ħ
arg t carries the phase shift; the poles of t are the next topic's bound states.
k = √(2mE)/ħ in both leads, which is why T = |t|² and R = 1 − T
At κ = 11.5 nm⁻¹ that is 0.10 nm — the exponential the STM lives on.
κ in m⁻¹; one decade of T costs ΔL = ln10 / 2κ
Three regions, four conditions, one deliberate choice
Write V(x) = V₀ on 0 < x < L and zero outside, then solve Ĥ|ψ⟩ = E|ψ⟩ region by region. For 0 < E < V₀ the solutions are ψI = e(ikx) + r e(−ikx) with k = √(2mE)/ħ, ψII = C e(κx) + D e(−κx) with κ = √(2m(V₀ − E))/ħ, and ψIII = t e(ikx). Only three of those four amplitudes are forced by the differential equation; the fourth is a choice. Dropping the incoming wave on the right is the scattering boundary condition, and it picks one of the two degenerate continuum states at this energy — the other is the mirror solution incident from the right. Because V is bounded, both ψ and ψ′ are continuous at x = 0 and x = L, giving four equations for r, C, D and t once the incident amplitude is normalised to 1. None of these states lies in L²; they are δ-normalised, so every quantity you extract from them is a ratio.
The algebra collapses because k² + κ² is fixed
Eliminate C and D and the transmission amplitude is t = e(−ikL)/[cosh κL + (i/2)(κ/k − k/κ) sinh κL]. Take the modulus squared: |1/t|² = cosh²κL + ¼(κ/k − k/κ)² sinh²κL = 1 + sinh²κL · [1 + (κ² − k²)²/(4k²κ²)] = 1 + sinh²κL · (k² + κ²)²/(4k²κ²). Now the barrier does the work. k² + κ² = 2mV₀/ħ² is independent of E, so (k² + κ²)²/(4k²κ²) = V₀²/(4E(V₀ − E)) and T = |t|² = [1 + V₀² sinh²(κL)/(4E(V₀ − E))]⁻¹. Two details deserve saying out loud. T is defined as jₜᵣₐₙₛ/jinc with j = (ħ/m)Im(ψ*ψ′), and only because the two leads share the same k does that flux ratio reduce to |t|². R = |r|² follows the same way, and R + T = 1 is the stationary current being divergence-free, not an extra assumption.
The opaque limit, and exactly how wrong it is
For κL ≫ 1, sinh κL → e(κL)/2 and sinh²κL → e(2κL)/4, so T → 16E(V₀ − E)/V₀² · e(−2κL). That prefactor is dimensionless, peaks at 4 when E = V₀/2, and vanishes at both ends — everything with real range in it sits in the exponent. Numbers: with ħ²/2mₑ = 0.0381 eV nm², so κ/nm⁻¹ = 5.123√(ΔE/eV), an electron at E = 5.0 eV under V₀ = 10.0 eV with L = 0.50 nm has κ = 11.46 nm⁻¹ and κL = 5.73, and the exact T = 4.2353 × 10⁻⁵ against 4.2354 × 10⁻⁵ from the limit. The limit always overestimates, by a relative (16E(V₀ − E)/V₀² − 2)e(−2κL): 2 × 10⁻⁵ here, 0.90% at κL = 2.66, and useless below κL ≈ 2. The same exponent gives d ln T/dL = −2κ, so one decade of transmission costs ΔL = ln10/2κ = 0.10 nm at this κ.
Cross E = V₀ and sinh becomes sin
Nothing in the algebra breaks at E = V₀; κ simply becomes imaginary. Put κ = iq with q = √(2m(E − V₀))/ħ: sinh²(iqL) = −sin²(qL), the factor V₀ − E flips sign with it, and T = [1 + V₀² sin²(qL)/(4E(E − V₀))]⁻¹. Two readings follow. First, a particle with more than enough energy is still reflected — an electron at E = 6.0 eV over V₀ = 5.0 eV with L = 0.30 nm has qL = 1.537 rad and T = 0.490, so 51% of the flux comes back off a barrier it could classically walk across. Second, T returns to exactly 1 whenever sin qL = 0, that is qL = nπ: an integer number of interior half-wavelengths fits between the faces and the two reflected amplitudes cancel. The wavenumber in that condition is the interior q, never the asymptotic k. For the same barrier the first resonance sits at E = 9.18 eV; widen L to 0.80 nm and three of them, at 5.59, 7.35 and 10.29 eV, crowd into the same window.
E = V₀ is a removable singularity, not a wall
At E = V₀ both κ and V₀ − E vanish and the exact formula reads 0/0. That is a fact about the expression, not about the physics. Expand it: sinh²(κL) → κ²L² while V₀ − E = ħ²κ²/2m, the κ² cancels, and V₀² sinh²(κL)/(4E(V₀ − E)) → mV₀L²/(2ħ²), so T(E = V₀) = [1 + mV₀L²/(2ħ²)]⁻¹. For V₀ = 5.0 eV and L = 0.30 nm that group is V₀L²/(4 × 0.0381 eV nm²) = 2.953, giving T = 0.253: a particle whose energy exactly matches the barrier still gets a quarter of its flux through and loses three quarters. In Python, evaluate this branch from the limit rather than the ratio — within about 10⁻³ eV of V₀ the numerator and the denominator each lose most of their significant digits, and sinh(κ*L)**2/(V0 − E) returns noise long before it returns a nan you would notice.
|t|² is not a clock
T answers how much, never how long. A stationary scattering state is a steady beam at one energy — it has no beginning, so it cannot time anything. What t carries beyond its modulus is a phase, and differentiating that gives the Wigner phase time τ = ħ d(arg t)/dE. For an opaque barrier the phase time saturates, stopping its growth with L, which is the Hartman effect; it is not superluminal signalling, because a delay read off a stationary phase means anything only for a packet narrow in energy and barely reshaped in transit. A real tunnelling-time question needs a wave packet, a stated definition — phase time, dwell time and Larmor time disagree with each other — and a check that the packet's energy spread does not straddle a resonance. None of that is inside T(E), and estimating L/v from an interior velocity is meaningless: below the barrier there is no real wavenumber to build a velocity from.
Change one variable at a time
Make the relationship visible.
Hold V₀ at 5 eV and drag L from 0.30 to 0.80 nm: below the barrier the solid curve sinks nearly four decades, while above it the lone resonance at 9.18 eV becomes three, at 5.59, 7.35 and 10.29 eV. Then watch the dashed curve dive to the floor as E → V₀, exactly where its limit stops being true.
INTERIOR κ OR q8.57 nm⁻¹
κL OR qL2.57
log₁₀ T EXACT-1.643
log₁₀ T OPAQUE-1.638
Live interpretationINTERIOR κ OR q: 8.57 nm⁻¹. κL OR qL: 2.57. log₁₀ T EXACT: −1.643. log₁₀ T OPAQUE: −1.638
Catch the common trap
Explain before calculating.
An electron beam with E > V₀ crosses a rectangular barrier of width L. For which condition is the transmission exactly 1?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron of energy E = 2.0 eV meets a rectangular barrier of height V₀ = 5.0 eV and width L = 0.30 nm. Find κ and κL, evaluate the exact transmission, and compare it with the opaque-limit estimate. Use ħ²/2mₑ = 0.0381 eV nm².
- Decay constant: κ = √(2m(V₀ − E))/ħ = √((V₀ − E)/(ħ²/2mₑ)) = √(3.0/0.0381) nm⁻¹ = 8.87 nm⁻¹, so κL = 8.87 × 0.30 = 2.662.
- Exact formula: T = [1 + V₀² sinh²(κL)/(4E(V₀ − E))]⁻¹. Its prefactor is V₀²/(4E(V₀ − E)) = 25/(4 × 2.0 × 3.0) = 25/24 = 1.0417.
- sinh(2.662) = 7.128, so sinh²(κL) = 50.81 and 1/T = 1 + 1.0417 × 50.81 = 53.93, giving T = 1.854 × 10⁻².
- Opaque estimate: 16E(V₀ − E)/V₀² = 16 × 2.0 × 3.0/25 = 3.84, and e(−2κL) = e(−5.324) = 4.872 × 10⁻³, so T ≈ 3.84 × 4.872 × 10⁻³ = 1.871 × 10⁻².
- The estimate is high by 0.90%, matching (16E(V₀ − E)/V₀² − 2)e(−2κL) = 1.84 × 4.872 × 10⁻³ = 0.0090. At κL = 2.66 this barrier is not yet opaque.
Answerκ = 8.87 nm⁻¹, κL = 2.662, exact T = 1.854 × 10⁻² — about 1 in 54. The opaque limit gives 1.871 × 10⁻², 0.90% too high.
MediumKeep the same barrier, V₀ = 5.0 eV and L = 0.30 nm, but send the electron in at E = 6.0 eV, above the top. (a) Find the transmission. (b) Find the lowest energy above V₀ at which the barrier is perfectly transparent.
- Above the barrier κ = iq, and the interior wavenumber is q = √(2m(E − V₀))/ħ = √(1.0/0.0381) nm⁻¹ = 5.123 nm⁻¹, so qL = 5.123 × 0.30 = 1.537 rad — just short of π/2.
- The sinh in the exact result becomes a sine: T = [1 + V₀² sin²(qL)/(4E(E − V₀))]⁻¹, with V₀²/(4E(E − V₀)) = 25/(4 × 6.0 × 1.0) = 1.0417.
- sin(1.537) = 0.99943, so sin²(qL) = 0.99885 and 1/T = 1 + 1.0417 × 0.99885 = 2.0405, giving T = 0.490. Just over half the flux is reflected by a barrier the electron could classically walk across.
- (b) T = 1 needs sin qL = 0, so the first resonance is at qL = π, that is q = π/0.30 nm = 10.472 nm⁻¹.
- Convert back: E − V₀ = (ħ²/2mₑ)q² = 0.0381 × 10.472² = 4.178 eV, so E = 5.0 + 4.178 = 9.18 eV. The condition fixes the interior q, never the asymptotic k, which is 15.5 nm⁻¹ at that energy.
Answer(a) T = 0.490 at E = 6.0 eV, so 51% is reflected. (b) The first perfect-transmission resonance is at E = 9.18 eV, where exactly one interior half-wavelength spans the barrier.
HardAn electron at E = 1.0 eV faces a barrier of height V₀ = 4.0 eV. (a) What width gives T = 1.0 × 10⁻⁶, and how good is the opaque form there? (b) By what factor does T change if that width grows by 0.010 nm? (c) Take the same width and the same barrier, but a proton, with mₚ/mₑ = 1836.
- κ = √(3.0/0.0381) nm⁻¹ = 8.874 nm⁻¹, and the prefactor is 16E(V₀ − E)/V₀² = 16 × 1.0 × 3.0/16 = 3.00.
- (a) Solve 3.00 e(−2κL) = 1.0 × 10⁻⁶: 2κL = ln(3.0 × 10⁶) = 14.914, so L = 14.914/(2 × 8.874) = 0.8404 nm.
- Check it against the exact formula at that width: κL = 7.4571, sinh²(κL) = 7.5000 × 10⁵ and V₀²/(4E(V₀ − E)) = 16/12 = 1.3333, so 1/T = 1 + 1.0000000 × 10⁶ and T = 9.999997 × 10⁻⁷. The opaque form is high by 3 × 10⁻⁷ relative, which is exactly (3.00 − 2)e(−14.914).
- (b) d ln T/dL = −2κ, so ΔL = 0.010 nm multiplies T by e(−2 × 8.874 × 0.010) = e(−0.1775) = 0.837: one tenth of an ångström costs 16% of the current, and a whole ångström costs a factor of 5.9.
- (c) Mass enters only through κ ∝ √m: κₚ = 8.874 × √1836 = 8.874 × 42.85 = 380.2 nm⁻¹, so 2κₚ L = 2 × 380.2 × 0.8404 = 639.1 and log₁₀T = log₁₀3.00 − 639.1/ln10 = 0.48 − 277.6 = −277.1.
Answer(a) L = 0.8404 nm, where the opaque form is accurate to 3 parts in 10⁷. (b) T falls to 0.837 of its value — a 16% drop for 0.1 Å. (c) The proton gets T ≈ 10⁻²⁷⁷, 271 decades below the electron on the same barrier.