University Physics IV · Quantum Potentials · 7.7
The Potential Step & Probability Current
Reflection and transmission are bookkeeping on a current, not on an amplitude. Here you build j from the wavefunction, match a plane wave across a sudden change in potential, and find out why a step turns back particles that carry more than enough energy to cross it.
Build the model
Connect the measurement to the mechanism.
A potential step is the simplest place a quantum particle meets a boundary, and it is where the gap between a probability amplitude and a probability flow first bites. Solve ψ″ = 2m(V − E)ψ/ħ² on each side and you get plane waves with two different wavenumbers, k₁ = √(2mE)/ħ before the step and k₂ = √(2m(E − V₀))/ħ beyond it; continuity of ψ and ψ′ at the edge then fixes the reflected and transmitted amplitudes in one line, r = (k₁ − k₂)/(k₁ + k₂) and t = 2k₁/(k₁ + k₂). What you may not do is call |t|² the transmitted fraction.
The conserved quantity is the probability current j = (ħ/m) Im(ψ* ∂ψ/∂x), which for a plane wave is density times speed, so the transmitted flux carries an extra factor k₂/k₁: T = (k₂/k₁)|t|², and only then does R + T = 1 come out exactly. The model charges you two idealisations for that clarity. A genuinely abrupt step is fiction — real interfaces smooth over a few atomic spacings — and a plane wave is not normalisable, so R and T are ratios of steady currents standing in for the probabilities that a real, momentum-spread wave packet ends up moving backwards or forwards long after it has struck the edge.
- Simple definition
- The reflection and transmission coefficients of a potential step are the fractions of the incident probability current that flow back from it and onward past it, R = |r|² and T = (k₂/k₁)|t|², and they always sum to one.
- Example
- Electrons of 2.00 eV crossing a 1.00 eV step have k₂/k₁ = 0.7071, r = 0.1716 and t = 1.1716, so R = 0.0294 and T = 0.7071 × 1.3726 = 0.9706 — and 0.0294 + 0.9706 = 1.
It is what ∂|ψ|²/∂t + ∂j/∂x = 0 conserves, so it is what R and T must be built from.
ψ in m(−1/2) and m the particle mass in kg, so j is a probability per second
Two regions with equal |A|² but different k carry different current. That single fact is the whole lesson.
ħk/m is the group speed in m s⁻¹; |A|² is a density in m⁻¹
One energy, two wavelengths: the particle moves slower past an up-step, so its wave is stretched.
For an electron k = 5.123 √(E/eV) nm⁻¹; k₂ turns imaginary once E < V₀
Two continuity conditions, two unknowns — no integral and no normalisation anywhere.
Dimensionless amplitude ratios, with the incident amplitude set to 1
R + T = 1 identically, which is the arithmetic check on every step problem you will solve.
Both dimensionless; the factor k₂/k₁ is the speed ratio and is never optional
The wave leaks in but carries no current, and the reflection costs a phase of −2 arctan(κ/k₁).
κ in m⁻¹; the density falls by 1/e over 1/2κ, about 0.15 nm when V₀ − E = 0.40 eV
Where the current comes from
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 is the only thing that is conserved.
Solve each side, then match at the edge
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.
T is a ratio of currents, so it carries k₂/k₁
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.
R + T = 1 is conservation, not a coincidence
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.
Below the step: all of it returns, some of it enters
For E < V₀ the right-hand equation has real exponential solutions, and normalisability 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°.
Steps reflect downhill too, and plane waves are a fiction
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.
Change one variable at a time
Make the relationship visible.
Hold V0 at 1.00 eV and slide E from 0.60 to 1.20 eV. The through arrow appears only once E passes V0, yet the density just past the step is above 1 on both sides of that crossing: density is not flux, and the missing factor is k2/k1.
REFLECTED FLUX R0.177
TRANSMITTED FLUX T0.823
DENSITY JUMP |t|²2.02
SPEED RATIO k₂/k₁0.408
Live interpretationREFLECTED FLUX R: 0.177. TRANSMITTED FLUX T: 0.823. DENSITY JUMP |t|²: 2.02. SPEED RATIO k₂/k₁: 0.408
Catch the common trap
Explain before calculating.
Electrons of energy E = 1.25 eV meet an upward potential step of height V₀ = 1.00 eV. Matching ψ and ψ′ at the edge gives a transmitted amplitude t = 1.382, and beyond the step the wavenumber ratio is k₂/k₁ = 0.4472. What fraction of the incident probability current crosses the step?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyElectrons of kinetic energy E = 2.00 eV travel along a flat region and meet an upward potential step of height V₀ = 1.00 eV. Find the reflection and transmission coefficients, and check them.
- Both regions are classically allowed, so both wavenumbers are real and only their ratio is needed: k₂/k₁ = √((E − V₀)/E) = √(1.00/2.00) = 0.7071.
- Matching ψ and ψ′ at x = 0 gives r = (k₁ − k₂)/(k₁ + k₂). Dividing top and bottom by k₁: r = (1 − 0.7071)/(1 + 0.7071) = 0.2929/1.7071 = 0.1716.
- The reflected wave shares the incident wave's speed, so R needs no correction factor: R = |r|² = 0.1716² = 0.0294.
- T is a flux ratio and does need one. t = 2/(1 + 0.7071) = 1.1716, so |t|² = 1.3726, and T = (k₂/k₁)|t|² = 0.7071 × 1.3726 = 0.9706.
- Check the ledger: R + T = 0.0294 + 0.9706 = 1.0000.
AnswerR = 0.0294 and T = 0.9706 — about 2.9% of the electrons bounce off a step they carry twice the energy to cross, which no classical particle would do.
MediumThe same electrons are slowed to E = 0.60 eV before meeting the same V₀ = 1.00 eV step. Find R and T, the probability density just inside the step relative to the incident density, the distance over which that density falls by 1/e, and the phase of the reflected wave. Use k = 5.123 √(E/eV) nm⁻¹ for an electron.
- Now E < V₀, so k₂ is imaginary; write k₂ = iκ with κ = √(2m(V₀ − E))/ħ = 5.123√0.40 = 3.240 nm⁻¹, while k₁ = 5.123√0.60 = 3.968 nm⁻¹.
- Then r = (k₁ − iκ)/(k₁ + iκ) is a complex number divided by its own conjugate, so |r| = 1 exactly. R = 1 and T = 0: every electron is reflected.
- Reflection is not the whole story. t = 2k₁/(k₁ + iκ), so |t|² = 4k₁²/(k₁² + κ²) = 4E/V₀ = 2.40 — the density just inside the step is 2.40 times the incident density.
- It dies fast. |ψ|² ∝ e(−2κx) falls by 1/e over 1/(2κ) = 1/6.480 = 0.154 nm, and at x = 0.30 nm it is e(−1.944) = 0.143 of its edge value.
- The reflected wave is phase-shifted, not simply turned round: r = e(−2i arctan(κ/k₁)), and arctan(3.240/3.968) = arctan(0.8165) = 39.2°, so the phase is −78.5°.
AnswerR = 1 and T = 0, yet |ψ|² at the edge is 2.40 times the incident density and falls by 1/e in 0.154 nm. The reflected wave returns with a phase of −78.5°.
HardFor an upward step of height V₀ = 1.000 eV, find the incident energy at which exactly half the incident probability current is transmitted. Then state the probability density just beyond the step, and explain how a density near 3 can accompany a transmitted flux of only 0.5.
- Write u = k₂/k₁. Then T = 4k₁k₂/(k₁ + k₂)² = 4u/(1 + u)². Setting T = 1/2 gives (1 + u)² = 8u, that is u² − 6u + 1 = 0.
- The roots are u = 3 ± 2√2. Only u = 3 − 2√2 = 0.17157 lies below 1, as it must for an upward step where k₂ < k₁.
- u² = (E − V₀)/E = 0.029437, so V₀/E = 1 − 0.029437 = 0.970563 and E = 1.000/0.970563 = 1.0303 eV — barely 3% above the step, yet half the current still crosses.
- Amplitudes follow: r = (1 − u)/(1 + u) = 0.82843/1.17157 = 0.70711, so R = 0.5000; and t = 2/(1 + u) = 1.70711, so |t|² = 2.9142.
- Flux check: T = u|t|² = 0.17157 × 2.9142 = 0.5000, and R + T = 1.0000 exactly.
AnswerE = 1.0303 eV. The density beyond the step is 2.914 times the incident density while only half the current crosses, because the transmitted electrons crawl at 0.1716 of the incident speed: 2.9142 × 0.17157 = 0.5000.