Potential Step and Reflection

Stand in a lit room at night and look at the window: you see the street and a faint ghost of yourself. Glass transmits most light and reflects a small fraction because the wave speed (equivalently, the refractive index) changes abruptly at the interface, requiring the electromagnetic fields to satisfy boundary conditions. Now replace the light wave with an electron and the pane of glass with a region of higher potential energy. Quantum mechanics makes the same prediction, and it is a scandal by classical standards: a particle with more than enough energy to cross a barrier still has a finite chance of bouncing back. A marble rolling up onto a plateau never returns; an electron meeting the equivalent step sometimes does. The potential step — flat potential here, higher flat potential there, with a sudden jump between — is the simplest scattering problem in quantum mechanics, and your first encounter with states that are not bound but traveling. The machinery is modest: plane waves on each side, two matching conditions at the jump, two lines of algebra. But the bookkeeping teaches something deep — probabilities of reflection and transmission must be ratios of probability flux, not of amplitudes, because the transmitted wave moves at a different speed. Get that factor wrong and probability itself leaks. Below the step the drama sharpens: the particle is turned back with certainty, yet its wavefunction seeps a measurable distance into the wall before dying. That evanescent tail, useless here, becomes the whole point when a second interface is nearby — it is the seed of quantum tunnelling.

Key takeaways

A potential step $V = V_0\,\theta(x)$ is the simplest scattering problem: plane waves on each side of the jump, stitched by continuity of $\psi$ and $\psi'$ at $x = 0$ .
For $E > V_0$ the reflection and transmission coefficients are $R = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^2$ and $T = \frac{4k_1k_2}{(k_1 + k_2)^2}$ , with $R + T = 1$ .
Reflection above the step is purely quantum: it arises from the discontinuity in the wave number at the interface, even when $E>V_0$ . It decreases as $E/V_0$ increases and vanishes only in the classical limit $E/V_0\to\infty$ .
$T$ is a ratio of probability currents, so it carries the flux factor $k_2/k_1$ : $T \neq |C/A|^2$ . Forgetting the factor breaks $R + T = 1$ .
For $E<V_0$ and a single semi-infinite step, reflection is total ( $R=1,\;T=0$ ), yet $\psi$ penetrates the classically forbidden region as an evanescent wave with decay length $1/\kappa$ . This evanescent tail enables tunnelling only when another interface is present (e.g. a finite barrier).
Key idea: whenever the wave number changes discontinuously, continuity of $\psi$ and $\psi'$ requires a reflected wave. Conserved probability current then guarantees $R+T=1$ .

The core insight

Solve the time-independent Schrödinger equation as plane waves in each constant- $V$ region, then enforce continuity of $\psi$ and $\psi'$ at the step; the transmission coefficient is a ratio of probability currents (carrying a $k_2/k_1$ factor), which guarantees $R + T = 1$ .

Where this lives today

Every semiconductor heterojunction—such as GaAs/AlGaAs in a laser diode or Si/SiO $_2$ in a MOS structure—presents carriers with an effective potential step. The resulting quantum reflection influences interface transmission and therefore contributes to contact resistance and device performance. Band-structure engineers speak of "band offsets"; the algebra they run against those offsets is equations (3) and (4), and grading an interface to kill the reflection is matter-wave impedance matching — an anti-reflection coating for electrons. Stack two steps close together and the evanescent tail of one reaches the other: resonant-tunnelling diodes and the entire tunnelling family are step physics composed twice. The cleanest modern displays of the pure step are atomic. Ultracold neutrons with energies below about $100\ \mathrm{neV}$ reflect totally from the effective potential step at a material surface — at any angle of incidence — and so can be stored in bottles; the neutron-lifetime experiments that anchor big-bang nucleosynthesis live on that total reflection. And quantum reflection of cold atoms and Bose–Einstein condensates from surfaces — bouncing off attractive potentials, one of the earliest and clearest genuinely quantum scattering effects — is now a precision tool for probing Casimir–Polder forces.

The physical picture

A plane wave crossing the step is a wave crossing into a new medium. On the left its wavenumber is $k_1 = \sqrt{2mE}/\hbar$ ; on the right, with $V_0$ subtracted from the kinetic budget, it is $k_2 = \sqrt{2m(E - V_0)}/\hbar$ . Different wave number, but the same total energy $E$ : the quantum analogue of an optical refractive-index mismatch. The continuity conditions at the interface therefore require both reflected and transmitted waves. The bigger the mismatch between $k_1$ and $k_2$ , the stronger the reflection. This is why a downward potential step ( $V_0<0$ ) also produces reflection: any discontinuous change in wave number generates partial reflection, irrespective of whether the potential increases or decreases. Ultracold atoms falling toward an attractive surface really do bounce off — quantum reflection from a potential drop, observed in the lab.

Figure

Scattering at a step with E above V₀ — one wave arrives, two leave

Notice the transmitted wave in the figure: longer wavelength, taller amplitude. Both are honest physics. The wavelength stretches because the kinetic energy dropped; the amplitude grows ( $C/A = 2k_1/(k_1+k_2) > 1$ for an upward step) because the wave slows down and piles up, like river water deepening as the current slackens. Yet the transmitted flux — amplitude squared times speed — is smaller than the incident flux. Holding amplitude and flux apart in your head is the key discipline of this problem. Now drop the beam energy below the step. Classically and quantum mechanically the headline is the same — everything comes back — but the wavefunction penetrates into the classically forbidden region with an exponential decay length $1/\kappa$ , before all the flux turns around.

Figure

Below the step — total reflection with a leaky edge

The whole story then fits in one plot. Track the reflected and transmitted fluxes across every energy: total reflection below the step gives way to transmission above it only gradually — there is no sharp classical switch-on — and the two curves sum to one everywhere.

Figure

R and T across the whole energy range — probability always balances

When does the classical intuition return? Two ways. Crank the energy up, $E \gg V_0$ : then $k_1 \approx k_2$ , the mismatch fades, and $R \approx (V_0/4E)^2 \to 0$ . Or smooth the step out over many de Broglie wavelengths: a gradual rise reflects essentially nothing, which is why you have never seen a thrown ball bounce off thin air. Sharp compared to the wavelength is what "sudden" means here — and electrons in nanostructures, with nanometre wavelengths, meet genuinely sudden steps all the time.

The math: two matching conditions, all the physics

The experiment being modelled is a steady beam: a source far to the left fires monoenergetic particles at the step, forever. A steady beam is a stationary state of definite energy, so the whole problem is the time-independent Schrödinger equation with the sharpest possible potential landscape:

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\,\psi = E\,\psi, \qquad V(x) = \begin{cases} 0, & x < 0 \\ V_0, & x > 0 \end{cases} \qquad (1)

where: $V_0$ — step height; $E$ — beam energy, fixed by the source. Note the role reversal from bound-state problems: $E$ is not an unknown to be quantized — the source dials it in, and every positive energy is allowed. The task is therefore to determine the reflected and transmitted amplitudes for that chosen energy. With $V$ constant on each side of the origin, the equation in each region is just the free-particle equation, and its solutions are plane waves. Write the most general physically sensible combination — incident plus reflected on the left, transmitted only on the right:

\psi_{\mathrm{L}} = A\,e^{ik_1x} + B\,e^{-ik_1x}, \qquad \psi_{\mathrm{R}} = C\,e^{ik_2x}, \qquad k_1 = \frac{\sqrt{2mE}}{\hbar},\; k_2 = \frac{\sqrt{2m(E - V_0)}}{\hbar} \qquad (2)

where: $A$ — incident amplitude; $B$ — reflected; $C$ — transmitted. No left-moving wave appears in the transmitted region because we impose the scattering boundary condition of no wave incident from $x=+\infty$ . Why are both $\psi$ and $\psi'$ continuous at $x = 0$ ? Integrate Eq. (1) across an infinitesimal interval containing the step: $-\frac{\hbar^2}{2m}\!\left[\psi'(0^+)-\psi'(0^-)\right] + \int_{-\varepsilon}^{\varepsilon}(V-E)\psi\,dx = 0$ . Since the potential step is finite, the integral vanishes as $\varepsilon \to 0$ , giving $\psi'(0^+) = \psi'(0^-)$ . Together with the continuity of $\psi$ , these are the matching conditions. Only an infinite spike — a delta function — can kink $\psi'$ ; a mere step cannot. These monochromatic plane waves are idealizations and are not square-normalizable; a physical particle is described by a wave packet. But the stationary coefficients $R$ and $T$ are exactly the long-time fractions the packet analysis delivers. And the bookkeeping we build here is one instance of a general law: the current-ratio definitions and the sum-to-one proof for any localized potential belong to Transmission and Reflection Coefficients.

Derivation AReflection and transmission above the step

The beam has more than enough energy to cross, and we want the exact fractions that reflect and transmit — the numbers detectors on both sides would report. The route is the shortest one in scattering theory: write the plane-wave solution (2) in each region, impose continuity of $\psi$ and $\psi'$ at the join, and solve the resulting two-by-two system for the amplitude ratios. The subtle move comes last: converting amplitudes into measured fractions requires probability currents, and that is where the velocity factor $k_2/k_1$ enters. Watch that step — skipping it is the classic way to break probability conservation.

Step 1. Match values at $x = 0$ : $A + B = C$ . Why: every exponential equals 1 at the origin, so continuity of $\psi$ is just amplitude bookkeeping.

Step 2. Match the derivative at $x=0$ : $\psi_L'(0)=\psi_R'(0)$ , which gives $ik_1(A-B)=ik_2C$ . Why: each plane wave differentiates to $\pm ik$ times itself; cancel the $i$ .

Step 3. Solve the two-by-two system. Substitute $C = A + B$ into the slope equation: $k_1(A - B) = k_2(A + B)$ , so

\frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2}, \qquad \frac{C}{A} = \frac{2k_1}{k_1 + k_2} \qquad (3)

Why: two conditions, two unknown ratios — scattering problems never care about overall normalization.

Step 4. Convert amplitudes to fluxes. The probability current of a plane wave $Ae^{ikx}$ is $j = \frac{\hbar k}{m}|A|^2$ — density times velocity. So

R = \frac{j_{\text{refl}}}{j_{\text{inc}}} = \left|\frac{B}{A}\right|^2 = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^2, \qquad T = \frac{j_{\text{trans}}}{j_{\text{inc}}} = \frac{k_2}{k_1}\left|\frac{C}{A}\right|^2 = \frac{4k_1k_2}{(k_1 + k_2)^2} \qquad (4)

Why: reflected and incident waves share a speed, so the $k$ 's cancel in $R$ — but the transmitted wave moves at $\hbar k_2/m$ , and the flux ratio must carry $k_2/k_1$ .

Step 5. Check conservation. $R + T = \dfrac{(k_1 - k_2)^2 + 4k_1k_2}{(k_1 + k_2)^2} = \dfrac{(k_1 + k_2)^2}{(k_1 + k_2)^2} = 1$ ✓. Why: no probability accumulates at the step in a steady state — whatever arrives, leaves.

Go deeper · Derivation B · 1 minWhy current, not amplitude

Derivation BWhy current, not amplitude

The previous card asserted that transmission is a ratio of fluxes, and this one earns the assertion — it is the conceptual heart of the article. We take the probability current that Probability Current and the Continuity Equation builds from the Schrödinger equation, evaluate it for a plane wave, and read off why a slower wave with a taller amplitude can still deliver fewer particles per second. The moral travels: every scattering problem you will meet, from barriers to three-dimensional cross sections, compares currents.

Step 1. Define the current. $j = \dfrac{\hbar}{m}\,\mathrm{Im}\!\left(\psi^*\dfrac{d\psi}{dx}\right)$ . Why: this is the flux whose conservation law $\partial_t|\psi|^2 + \partial_x j = 0$ follows directly from the Schrödinger equation — it is the only "how much probability flows" there is.

Step 2. Evaluate for a plane wave. $\psi = Ce^{ik_2x}$ gives $\psi^*\psi' = ik_2|C|^2$ , so $j = \dfrac{\hbar k_2}{m}|C|^2$ . Why: amplitude squared alone is a density; multiplying by the speed $\hbar k_2/m$ makes it a rate.

Step 3. Draw the moral. A detector counts arrivals per second — a current, not a density. Comparing $|C|^2$ with $|A|^2$ compares how crowded the waves are; comparing currents compares how many particles per second pass, which is what transmission means. Why: this same lesson — flux ratios, not amplitude ratios — recurs in every scattering problem you will ever solve, from barriers to 3D cross sections.

Go deeper · Derivation C · 1 minTotal reflection below the step

Derivation CTotal reflection below the step

Now drop the beam energy below the step and ask what changes. Classically nothing subtle happens — everything bounces — and the quantum headline agrees, but the mechanism deserves a derivation because the wall acquires a populated skin. The strategy is analytic continuation rather than fresh algebra: the first card's matching never required $k_2$ to be real, so substitute $k_2 \to i\kappa$ , watch the reflected amplitude become a pure phase, and confirm with the current that nothing flows rightward. That phase is the one measurable trace the forbidden region leaves outside.

Step 1. Replace the transmitted wave. For $E < V_0$ , define $\kappa = \sqrt{2m(V_0 - E)}/\hbar$ ; the right-region solution that stays finite is $\psi_{\mathrm{R}} = Ce^{-\kappa x}$ . Why: with $E < V$ the equation reads $\psi'' = +\kappa^2\psi$ — solutions curve away from the axis, and the growing exponential is unphysical at $x \to +\infty$ .

Step 2. Reuse the algebra. Every step of Derivation 1 goes through with the substitution $k_2 \to i\kappa$ :

\frac{B}{A} = \frac{k_1 - i\kappa}{k_1 + i\kappa} \qquad (5)

Why: analytic continuation saves re-deriving — the matching conditions never asked $k_2$ to be real.

Step 3. Take the modulus. Numerator and denominator are complex conjugates, so $|B/A|^2 = \dfrac{k_1^2 + \kappa^2}{k_1^2 + \kappa^2} = 1$ : total reflection, $R = 1$ . Why: the reflected wave carries all the incident flux — but with a phase shift, $B/A = e^{-2i\arctan(\kappa/k_1)}$ , which delays the bounced packet.

Step 4. Confirm nothing flows right. For the real exponential, $\psi^*\psi' = -\kappa|C|^2e^{-2\kappa x}$ is real, so $j_{\text{trans}} = \frac{\hbar}{m}\mathrm{Im}(\cdots) = 0$ and $T = 0$ exactly. Why: although the wavefunction is nonzero inside the classically forbidden region, its probability current vanishes. The state stores probability density but transports no probability. The particle can be found a depth $\sim 1/\kappa$ inside, but in steady state every one that enters comes back out. *The identity $|(k_1 - i\kappa)/(k_1 + i\kappa)| = 1$ is worth confirming directly: numerator and denominator are complex conjugates, so their moduli are equal.*

Limiting cases

Case	Result	Meaning
$E \gg V_0$	$R \approx \left(\dfrac{V_0}{4E}\right)^2 \to 0$	wavenumbers nearly match; the step is barely felt — the classical limit
$E \to V_0^+$	$k_2 \to 0$ , $R \to 1$	The transmitted wave has $k_2\to0$ , so its group velocity and transmitted current vanish continuously
$E < V_0$	$R = 1$ , $T = 0$ , tail $e^{-\kappa x}$	total reflection with penetration depth $1/\kappa$ — classically forbidden, quantum-mechanically populated
$V_0 < 0$ (cliff down)	$R = \left(\dfrac{k_1 - k_2}{k_1 + k_2}\right)^2 > 0$	even a potential drop reflects — discontinuity, not direction, causes reflection
step smoothed over $\gg \lambda$	$R \to 0$	adiabatic passage: gradual changes do not reflect, which is why classical balls never bounce off hills they can climb

Common trap“T is just |C/A|² — amplitude squared is probability”

The most instructive wrong answer in scattering theory. Amplitude squared is a probability density; transmission compares rates. The transmitted wave travels at $\hbar k_2/m$ , slower than the incident wave's $\hbar k_1/m$ (for an upward step), so its taller amplitude does not mean more particles per second. Take $E$ just above $V_0$ : the bare ratio gives $|C/A|^2 \to 4$ — "400% transmission" — while the honest flux ratio $T = (k_2/k_1)|C/A|^2 \to 0$ . Only the current bookkeeping satisfies $R + T = 1$ . And its mirror-image trap for $E < V_0$ : a nonzero $\psi$ inside the wall does not mean nonzero transmission — the evanescent tail is a real exponential and carries identically zero current. Presence is not passage.

Worked examples

1 · A fast electron meets a low step (confidence builder)

An electron with $E = 5.0\ \mathrm{eV}$ approaches a step of height $V_0 = 3.0\ \mathrm{eV}$ . Classically it always crosses. Quantum mechanically, what fraction reflects?

Step 1. Only the ratio of wavenumbers matters: $\dfrac{k_2}{k_1} = \sqrt{\dfrac{E - V_0}{E}} = \sqrt{\dfrac{2.0}{5.0}} = 0.632$ .

Step 2. Reflection from (4): $R = \left(\dfrac{1 - 0.632}{1 + 0.632}\right)^2 = \left(\dfrac{0.368}{1.632}\right)^2 = (0.225)^2 = \boxed{0.051}$ .

Step 3. Transmission: $T = \dfrac{4(0.632)}{(1.632)^2} = \dfrac{2.528}{2.663} = 0.949$ , and $R + T = 0.051 + 0.949 = 1$ ✓.

Step 4. Read the result: one electron in twenty turns back from a wall it could easily climb. In a beam of $10^{12}$ electrons per second, that is $5\times10^{10}$ reflections per second — easily measurable, and measured routinely at semiconductor junctions.

2 · Below the step: how deep does the ghost reach? (exam level)

An electron with $E = 1.0\ \mathrm{eV}$ hits a step of height $V_0 = 5.0\ \mathrm{eV}$ . Find the penetration depth, the relative probability density $0.30\ \mathrm{nm}$ inside the wall, and the phase of the reflected wave.

Step 1. Decay constant: $\kappa = \dfrac{\sqrt{2m(V_0 - E)}}{\hbar} = \dfrac{\sqrt{2(9.11\times10^{-31})(4.0\times1.602\times10^{-19})}}{1.055\times10^{-34}} = 1.02\times10^{10}\ \mathrm{m^{-1}}$ .

Step 2. Penetration depth: $1/\kappa = \boxed{0.098\ \mathrm{nm}}$ — about half an atomic diameter. The electron's presence fades within one atom of the interface.

Step 3. Density $0.30$ nm inside, relative to the interface: $e^{-2\kappa x} = e^{-2(1.02\times10^{10})(3.0\times10^{-10})} = e^{-6.1} = \boxed{0.0022}$ — suppressed five-hundred-fold. If a second interface sat within a tenth of a nanometre, though, the tail would still be strong there — and the particle would tunnel, the story Rectangular Barrier and Quantum Tunnelling takes up.

Step 4. Reflected phase from (5): the reflection amplitude is $\dfrac{B}{A} = \dfrac{k_1 - i\kappa}{k_1 + i\kappa}$ , so $\arg\!\left(\dfrac{B}{A}\right) = -2\arctan\!\left(\dfrac{\kappa}{k_1}\right)$ . With $\kappa/k_1 \simeq 2.0$ , $\arg(B/A) \approx -2.21\ \mathrm{rad} \approx \boxed{-127^\circ}$ .

Step 5. Interpretation: although $R = 1$ , the reflected wave acquires a non-trivial phase shift. For a wave packet this phase produces a small reflection delay (the Wigner delay time). Everything measurable about the forbidden region is encoded in that phase.

Origins · 1 minMax Born and how this was discovered

Max Born

1882–1970 · Göttingen, then Edinburgh

Born read Schrödinger's wave mechanics and asked the question the theory could not duck: what is $\psi$ ? His answer came in a 1926 paper on collision problems — scattering, exactly the physics of this article. Treating the scattered wave's amplitude, Born proposed that $|\psi|^2$ measures the probability of finding the particle, adding in a famous footnote that probability goes with the square, not the amplitude itself. Reflection and transmission coefficients — fractions of a beam counted at detectors — were the birthplace of the interpretation now stamped on all of quantum mechanics. Einstein's letter objecting that God "does not play dice" was addressed to Born; the two argued warmly about it for three decades. Born's statistical interpretation earned the Nobel Prize in 1954, embarrassingly late.

History

1926

— Schrödinger's wave equation makes piecewise-constant potentials solvable by matching plane waves at the joins.

1926

— Born, analyzing collision problems, reads

|\psi|^2

as probability — reflection and transmission become measurable fractions of a beam.

1928

— Gamow, and independently Gurney & Condon, stretch step-and-tail physics into tunnelling, explaining alpha decay.

1970

— Esaki and Tsu design semiconductor superlattices: engineered stacks of potential steps at heterojunctions.

2001

— Shimizu observes quantum reflection of ultracold neon atoms from a solid surface — atoms bouncing off an attractive potential, exactly as step physics predicts.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §2.5 · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Complement H_I · Shankar, Principles of Quantum Mechanics, §5.4 · Sakurai & Napolitano, Modern Quantum Mechanics, §2.5. The amplitude ratios $B/A = (k_1 - k_2)/(k_1 + k_2)$ and $C/A = 2k_1/(k_1 + k_2)$ follow from the two matching conditions, and $R + T = 1$ holds algebraically. For $E < V_0$ : $|B/A| = 1$ and $T = 0$ with $\kappa = \sqrt{2m(V_0 - E)}/\hbar$ . Key numerical benchmarks: $k_2/k_1 = 0.632$ , $R = 0.051$ , $T = 0.949$ for $E = 5$ eV, $V_0 = 3$ eV; $\kappa = 1.02\times10^{10}\ \mathrm{m^{-1}}$ , $1/\kappa = 0.098$ nm, $e^{-2\kappa x} = 0.0022$ at $x = 0.3$ nm, and reflected phase $-2\arctan 2 \approx -127^\circ$ for $E = 1$ eV, $V_0 = 5$ eV.

Merge note. Merged with the v3 pilot article (v3-potential-step-reflection) at Step-2; AllQFT depth grafted onto the v3 narrative spine.

Prerequisites. Time-independent Schrödinger equation in 1D · free-particle plane waves · bound vs. scattering states · wave packets (for what "a particle arrives" rigorously means). Leads to. The rectangular barrier and quantum tunnelling · the finite square well's scattering states · wave-packet dynamics and group velocity · transfer matrices and band structure.

The transmitted amplitude can exceed the incident one — how is that not more than 100% probability?

Because amplitude squared is a density, not a rate. On an upward step the transmitted wave slows and piles up: $C/A = 2k_1/(k_1+k_2)$ approaches 2 near threshold, so the density beyond the step nears four times the incident density. But a detector counts arrivals per second — density times speed — and the flux $(\hbar k_2/m)|C|^2$ stays below the incident flux. River water deepens where the current slackens.

Where does the reflected wave physically come from? Nothing hits anything at the step.

Reflection needs no impact and no backward force. A wave reflects wherever its medium changes on a scale comparable to its wavelength — for matter waves, wherever the local wavenumber $k = \sqrt{2m(E-V)}/\hbar$ changes abruptly. The step is pure mismatch, the same thing glass does to light with zero contact mechanics. Smooth the rise over many de Broglie wavelengths and the reflection genuinely disappears — why macroscopic objects never bounce off hills they can climb.

If T is exactly zero below the step, is the evanescent tail measurable at all, even in principle?

Yes — through the phase, and through proximity. The reflected amplitude $e^{-2i\arctan(\kappa/k_1)}$ delays a reflected packet as if it had ventured a distance $\sim 1/\kappa$ inside and back, and time-resolved reflection reads that delay. Bring a second interface within a few $1/\kappa$ and the tail stops being a ghost: flux crosses — tunnelling. Optics runs the same experiment as frustrated total internal reflection: press a second prism close and light jumps the forbidden gap.

Why is the derivative of ψ continuous at the step when it jumps at a delta potential?

Integrate the Schrödinger equation across the discontinuity: the change in $\psi'$ equals $\tfrac{2m}{\hbar^2}\int (V-E)\psi\,dx$ over the crossing interval. A step keeps $V$ finite, so the integral vanishes as the interval shrinks — no kink. A delta puts infinite $V$ inside and leaves a finite residue, hence the jump. The hierarchy: $\psi$ continuous always; $\psi'$ continuous unless the potential is infinitely singular.

Does the step ever transmit more than a classical particle would?

Above the step, never: the classical answer is $T = 1$ , and quantum mechanics only subtracts, $T = 4k_1k_2/(k_1+k_2)^2 < 1$ . The step's surprises are all on the reflection side — including reflecting from a drop, where a classical particle would simply speed up and go on. Beating expectation needs interference between two boundaries: the resonance physics of barriers and wells, where $T$ climbs back to exactly 1 at special energies but never beyond.

R depends only on the ratio E over V₀ — no ħ, no mass. Where did the quantum mechanics go?

Into the idealization. For a perfectly sharp step the only energies around are $E$ and $V_0$ , so the answer can only depend on their ratio — $\hbar$ has nothing to combine with. It hides in the word "sharp": real potentials rise over some length $d$ , and reflection requires $k_1 d \lesssim 1$ , a comparison that contains $\hbar$ through the de Broglie wavelength. That is why nanometre junctions feel the step formula and thrown balls do not.

Potential Step and Reflection

Where this lives today

The physical picture

The math: two matching conditions, all the physics

Limiting cases

Worked examples

1 · A fast electron meets a low step (confidence builder)

2 · Below the step: how deep does the ghost reach? (exam level)

Going further

Before this

Where this leads