The potential step and probability current

Multiply the time-dependent Schrödinger equation by ψ*, subtract its complex conjugate multiplied by ψ, and the potential term cancels because V is real. What survives is ∂|ψ|²/∂t = −∂j/∂x, with j = (ħ/m) Im(ψ* ∂ψ/∂x). That is a conservation law: probability is not created or destroyed, it flows, and j is the flow. For one plane wave Ae^(ikx) it evaluates to j = (ħk/m)|A|², exactly density × speed, since ħk/m is the group velocity. A wavefunction that is real, or real up to a constant phase, gives j = 0 — nothing is going anywhere. Every statement about reflection and transmission below is a statement about j, because in this problem j, not the density, is what stays the same on both sides of the step.

Take V = 0 for x < 0 and V = V₀ > 0 for x > 0, and send particles in from the left with E > V₀. On the left ψ = e^(ik₁x) + r e^(−ik₁x) with k₁ = √(2mE)/ħ; on the right ψ = t e^(ik₂x) with k₂ = √(2m(E − V₀))/ħ. There is no e^(−ik₂x) term on the right, and that is a physical choice, not algebra: nothing is sending waves back from x = +∞. Because V is finite everywhere, both ψ and ψ′ are continuous at x = 0. Continuity of ψ gives 1 + r = t; continuity of ψ′ gives k₁(1 − r) = k₂t. Two equations, two unknowns: r = (k₁ − k₂)/(k₁ + k₂) and t = 2k₁/(k₁ + k₂). Notice t > 1 whenever k₂ < k₁ — the transmitted amplitude exceeds the incident one, and nothing has gone wrong.

The incident current is (ħk₁/m)⋅1, the reflected current is (ħk₁/m)|r|², and the transmitted current is (ħk₂/m)|t|². Divide each by the incident current: R = |r|² and T = (k₂/k₁)|t|². That wavenumber factor is not decoration — it is the transmitted particles' lower speed. Take E = 1.25 eV onto a step of V₀ = 1.00 eV. Then k₂/k₁ = √(0.25/1.25) = 0.4472, r = 0.3820 and t = 1.3820, so |t|² = 1.910: the probability density beyond the step is nearly double the incident density. Yet T = 0.4472 × 1.910 = 0.854, and R = 0.146. Density piles up precisely because the flow has slowed; the current getting through is smaller than the current arriving.

For a stationary state ∂|ψ|²/∂t = 0, so ∂j/∂x = 0 and the current is the same number everywhere, including across the step. On the left ψ is a sum of two counter-propagating waves, and their interference contributes nothing to the current: for ψ = Ae^(ikx) + Be^(−ikx) the current is exactly (ħk/m)(|A|² − |B|²), the cross terms cancelling identically. So the left current is (ħk₁/m)(1 − |r|²), the right current is (ħk₂/m)|t|², and equating them is R + T = 1. Algebraically (k₁ − k₂)² + 4k₁k₂ = (k₁ + k₂)², so it holds for every E and V₀. Use it as a check: if your R and T do not sum to one, you either dropped the k₂/k₁ factor or squared a complex amplitude without taking its modulus.

For E < V₀ the right-hand equation has real exponential solutions, and boundedness as x → ∞ kills the growing one, leaving ψ = t e^(−κx) with κ = √(2m(V₀ − E))/ħ. Now r = (k₁ − iκ)/(k₁ + iκ) is a complex number over its own conjugate, so |r| = 1 exactly: R = 1, T = 0, everything comes back. That is not the same as nothing entering. The density just inside is |t|² = 4E/V₀ times the incident density and falls by 1/e over 1/2κ. For an electron at E = 0.60 eV against V₀ = 1.00 eV, κ = 3.24 nm⁻¹, so the density at the edge is 2.40 times the incident value and has fallen to 14% of that by 0.30 nm in. This evanescent wave carries no current — ψ there is real up to a constant phase — yet it is exactly what lets a thin barrier be tunnelled. Total reflection also costs a phase: r = e^{(−2i arctan(κ/k₁))}, here −78.5°.

Classically nothing reflects when E > V₀, and nothing reflects off a drop at all. Quantum mechanically the only thing that matters is that k changes: R = ((k₁ − k₂)/(k₁ + k₂))² is nonzero whenever k₂ ≠ k₁, in either direction. An electron of E = 1.00 eV running off a cliff where the potential falls by 3.00 eV has k₂/k₁ = √(4.00/1.00) = 2, so R = (1/3)² = 0.111 — one in nine bounces back off a step it is speeding down. That is quantum reflection, and it is the same effect that makes an abrupt change of refractive index reflect light. One last debt: these plane waves are not normalisable, since |ψ|² is constant over an infinite line. The honest object is a wave packet with a narrow spread in k, and R and T are then the probabilities of finding it moving left or right long after the collision — provided R(E) barely varies across that spread.