The rectangular barrier and its opaque limit

Write V(x) = V₀ on 0 < x < L and zero outside, then solve Ĥ|ψ⟩ = E|ψ⟩ in the position basis, ψ(x) = ⟨x|ψ⟩, region by region. For 0 < E < V₀ the solutions are ψ_I = e^(ikx) + r e^(−ikx) with k = √(2mE)/ħ, ψ_II = C e^(κx) + D e^(−κx) with κ = √(2m(V₀ − E))/ħ, and ψ_III = t e^(ikx). Outside the barrier the equation allows four travelling waves, e^(±ikx) on each side, and only three appear here; leaving out 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.

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/j_inc 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.

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 = 4.2354 × 10⁻⁵ against 4.2355 × 10⁻⁵ from the limit. The limit misses by a relative (16E(V₀ − E)/V₀² − 2)e^(−2κL), high when the prefactor exceeds 2 and low when it falls below 2: 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.

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.

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 point from the limit rather than the ratio: at E = V₀ exactly, sinh²(κL)/(V₀ − E) is 0/0, which NumPy returns as nan and plain floats raise as ZeroDivisionError, so an energy grid that lands on V₀ leaves a hole in the curve. Anywhere else the ratio is safe, because the same V₀ − E sets both κ and the denominator.

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 + kL)/dE, the kL undoing the free-flight phase e^(−ikL) that this convention builds into t. 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.