Delta-Function Barrier and Scattering

Take the thinnest wall physics allows — infinitely tall, infinitely narrow, with nothing left of it but a finite "area" $\alpha$ — and stand it in the path of a quantum particle. A classical ball would not even notice: the wall has zero width, so there is nowhere to stop. The quantum wave notices. It splits, part of it turning back off a region of space that, strictly speaking, has no size at all. Reflection here is not a collision; it is what waves do at any abrupt change in their medium, the same physics that makes a window pane return a faint image of your face. The delta-function barrier is the cheapest scattering problem in quantum mechanics — one matching point, two unknowns, three lines of algebra — and it pays out the entire vocabulary of one-dimensional scattering: complex amplitudes $r$ and $t$ , probabilities $R$ and $T$ that sum to one, a single dimensionless knob $\beta$ that decides everything, and a first sighting of a deep scattering-theory fact, the bound state of the attractive twin hiding as a pole of the transmission amplitude. You already own the key: the Delta-Function Potential Well and Bound State article derived the kink that a delta imprints on $\psi'$ . Here we point that same kink at a travelling wave and flip the sign of the potential. Two surprises await. Fast particles barely notice the barrier while slow ones are turned back almost completely — a velocity filter with no moving parts. And a measurement of transmission alone cannot tell you the sign of the potential: the impenetrable spike and the attractive well transmit identically. What distinguishes attraction is not scattering — it is binding.

Key takeaways

A repulsive point potential $V(x)=\alpha\,\delta(x)$ , $\alpha>0$ , has no inside: its whole effect is the slope jump $\Delta\psi'=(2m\alpha/\hbar^2)\,\psi(0)$ , recalled from the delta-well article with the sign flipped.
Continuity of $\psi$ plus the jump condition fix the scattering amplitudes completely: $r=\dfrac{-i\beta}{1+i\beta}$ , $t=\dfrac{1}{1+i\beta}$ , with $\beta=\dfrac{m\alpha}{\hbar^2k}$ .
Probabilities follow as $T=\dfrac{1}{1+\beta^2}=\dfrac{1}{1+m\alpha^2/2\hbar^2E}$ and $R=\dfrac{\beta^2}{1+\beta^2}$ , with $R+T=1$ exactly.
$\beta$ compares barrier strength to particle momentum: slow or weakly energetic particles ( $\beta\gg1$ ) reflect; fast ones ( $\beta\ll1$ ) sail through. The fifty–fifty point sits at $E_{1/2}=m\alpha^2/2\hbar^2$ .
Flipping $\alpha\to-\alpha$ (barrier → well) changes the phases of $r$ and $t$ but not their magnitudes: same $T$ , same $R$ — but only the attractive sign hides a bound state, as a pole of $t$ at imaginary momentum.
Key idea: one kink in the slope of $\psi$ , prescribed by the delta's strength, is the entire interaction — matching through it hands you the full scattering problem in closed form.

The core insight

Integrating the Schrödinger equation across the point spike converts the potential into a boundary condition — a prescribed jump in $\psi'$ — and matching an incident-plus-reflected wave to a transmitted wave through that jump determines $r$ and $t$ exactly, with everything controlled by the single dimensionless ratio $\beta=m\alpha/\hbar^2k$ .

Where this lives today

The delta barrier is the working model wherever a scatterer is much smaller than the wavelength probing it. In neutron optics, Fermi's pseudopotential — a delta of strength set by the measured scattering length — predicts the refractive index of materials for cold neutrons, the basis of the neutron guides that feed every spallation source. In cold-atom physics, interatomic forces are encoded as delta-function contact interactions whose strength experimenters tune with Feshbach resonances, letting a Lieb–Liniger gas of bosons be dialled from nearly free to impenetrable; the one-dimensional wires of those experiments realise this article's algebra literally. And in nanoelectronics, a single dopant atom or vacancy in a nanowire scatters conduction electrons as a point — the $T=E/(E+E_{1/2})$ crossover is why point defects throttle low-temperature conductance so much more brutally than room-temperature flow.

The physical picture

Think of the delta barrier as a speed bump shrunk to a geometric point while keeping its "impulse" — height times width — fixed at $\alpha$ . A wave rolling over it receives a sudden kick localized at one spot. Waves respond to sudden changes the way light responds to the surface of glass: partial reflection, partial transmission, decided not by whether the obstacle is tall but by how its strength compares to the wave's momentum. That comparison is the whole story here, compressed into one number:

\beta=\frac{m\alpha}{\hbar^2 k}=\sqrt{\frac{E_{1/2}}{E}},\qquad E_{1/2}\equiv\frac{m\alpha^2}{2\hbar^2} \qquad (1)

where: $k=\sqrt{2mE}/\hbar$ is the incident wavenumber and $E_{1/2}$ — a fixed energy built from the barrier strength alone — marks the crossover. When $E\gg E_{1/2}$ , $\beta$ is small and the barrier is nearly invisible; when $E\ll E_{1/2}$ , $\beta$ is large and the point spike behaves like a solid wall. For an electron and a typical atomic-scale strength $\alpha=2.0\ \mathrm{eV\cdot\mathring{A}}$ , the crossover sits at $E_{1/2}=0.26\ \mathrm{eV}$ — thermal-electron territory, which is why point defects matter so much for low-energy transport. The picture below is what a detector-free experiment would record: the probability density $|\psi|^2$ in a stationary scattering state. To the left, the incident and reflected waves interfere and $|\psi|^2$ ripples — the fringe contrast measures $|r|$ . To the right there is only the transmitted wave, so the density is dead flat at $|t|^2$ . The barrier itself is the one point where the smooth curve kinks.

Figure

Probability density in a scattering state — interference fringes on the left, flat transmitted density on the right (β = 1.62)

Notice what the fringes are doing quantitatively. Their maxima reach $(1+|r|)^2$ and their minima $(1-|r|)^2$ , so fringe contrast is a direct readout of the reflection amplitude — this is exactly how reflectivities are measured in neutron and light optics. The flat level on the right is $T$ . And the two readouts are locked together: whatever the barrier turns back, the rest gets through.

The math: point the kink at a travelling wave

Everything the delta does to a wavefunction was established one article ago. For $V(x)=g\,\delta(x)$ , integrating the Schrödinger equation across the spike leaves $\psi$ continuous and kicks its slope by $\Delta\psi'=(2mg/\hbar^2)\,\psi(0)$ — the ε-integration that proves this is owned by the Delta-Function Potential Well and Bound State article, where it selected the well's single bound state. We take the result, set $g=+\alpha$ for repulsion, and change the cast of characters: instead of decaying tails at negative energy, we now feed it oscillating waves at positive energy,

\psi(x)=\begin{cases}e^{ikx}+r\,e^{-ikx}, & x<0\\[2pt] t\,e^{ikx}, & x>0\end{cases}\qquad k=\frac{\sqrt{2mE}}{\hbar} \qquad (2)

where: the incident amplitude is normalised to 1, $r$ collects everything that bounces back, and $t$ everything that continues. There is no left-moving piece on the right because nothing arrives from $+\infty$ — the same asymptotic bookkeeping used for the Potential Step and Reflection. Since the potential is identical on both sides, the transmitted wave carries the same $k$ as the incident one: this scattering changes amplitude and phase, never wavelength.

Derivation AScattering amplitudes from one matching point

We want $r$ and $t$ — the complex amplitudes whose squared magnitudes will become the reflection and transmission probabilities. The system could not be smaller: one junction at $x=0$ , two conditions on it (continuity of $\psi$ , prescribed jump of $\psi'$ ), two unknowns. The strategy is to write both conditions, reduce them to a linear pair, and watch the physics condense into the dimensionless ratio $\beta=m\alpha/\hbar^2k$ . The point of care is the jump condition's bookkeeping: it is the derivative that jumps, by an amount proportional to the value $\psi(0)$ , and every sign in it matters.

Step 1. Continuity of $\psi$ at $x=0$ . Left limit $1+r$ , right limit $t$ :

1+r=t

Step 2. The jump condition with $g=+\alpha$ . Slopes are $\psi'(0^-)=ik(1-r)$ and $\psi'(0^+)=ikt$ , so $\Delta\psi'=(2m\alpha/\hbar^2)\,\psi(0)$ reads

ik\,t-ik(1-r)=\frac{2m\alpha}{\hbar^2}\,t

Why the right-hand side carries $t$ : the jump is proportional to the value of the wavefunction at the spike, and by Step 1 that value is $\psi(0)=1+r=t$ . Using $t$ keeps the algebra one variable lighter.

Step 3. Substitute $r=t-1$ into Step 2 and divide by $ik$ . With $\beta\equiv m\alpha/(\hbar^2k)$ ,

t-\left(1-(t-1)\right)=-2i\beta\,t\;\;\Longrightarrow\;\; 2t-2=-2i\beta\,t\;\;\Longrightarrow\;\; t=\frac{1}{1+i\beta}

Why $\beta$ and nothing else: the only scales in the problem are the barrier strength $\alpha$ and the momentum $\hbar k$ , and $\beta$ is their unique dimensionless ratio. The answer was always going to be a function of $\beta$ alone.

Step 4. Back-substitute for the reflection amplitude:

r=t-1=\frac{1-(1+i\beta)}{1+i\beta}=\frac{-i\beta}{1+i\beta}

Both amplitudes are genuinely complex: $t$ lags the incident wave by a phase $\arg t=-\arctan\beta$ (a scattering phase shift, the quantity that becomes the organising concept of three-dimensional scattering theory), and $r$ carries an extra $-\pi/2$ from the factor $-i$ . Verified: the two-equation linear system was solved symbolically (sympy), reproducing $r=-i\beta/(1+i\beta)$ and $t=1/(1+i\beta)$ exactly; an independent numerical solve of the same system at four energies agrees with the closed forms to $10^{-12}$ .

Probabilities: one dial, no structure

Squaring the amplitudes gives the observable rates. Because the potential — and hence the wavenumber and particle speed — is the same on both sides, the flux ratios reduce to bare squared magnitudes here; the honest general definition of $R$ and $T$ as probability-current ratios, and the proof that they always sum to one, are owned by Transmission and Reflection Coefficients:

T=|t|^2=\frac{1}{1+\beta^2}=\frac{1}{1+\dfrac{m\alpha^2}{2\hbar^2E}}=\frac{E}{E+E_{1/2}},\qquad R=|r|^2=\frac{\beta^2}{1+\beta^2} \qquad (3)

where: the last form makes the physics transparent — transmission is a race between the particle's energy and the fixed crossover scale $E_{1/2}=m\alpha^2/2\hbar^2$ . At $E=E_{1/2}$ the coin is fair, $T=R=\tfrac12$ . Add the two and the sum telescopes to 1 identically: no probability leaks into the zero-width barrier, because there is no "inside" for it to occupy. Note also what is absent: no oscillation, no resonance, no structure of any kind. $T(E)$ climbs monotonically from 0 to 1. A single delta has only one matching point, and resonances need at least two reflecting surfaces to trap interfering round trips — the finite-width story of Rectangular Barrier and Quantum Tunnelling, sharpened into spikes in Resonant Transmission and Tunnelling Resonances.

Figure

Transmission and reflection vs energy — a monotonic crossover at E½ with no resonant structure

The sign flip: same transparency, no binding

Now run the whole calculation for the attractive twin, $V=-\alpha\,\delta(x)$ . Every step survives with $\beta\to-\beta$ , so $t=1/(1-i\beta)$ and $r=+i\beta/(1-i\beta)$ — different phases, identical magnitudes. The well and the barrier are indistinguishable to a transmission experiment. That striking fact, and the one thing that does distinguish them, deserve a careful look.

Go deeper · Derivation B · 2 minBarrier versus well: what the sign changes and what it cannot

Derivation BBarrier versus well: what the sign changes and what it cannot

Two claims to establish. First: $R$ and $T$ are exactly invariant under $\alpha\to-\alpha$ , so no scattering-probability measurement can determine the sign of a delta potential. Second: the attractive sign, and only the attractive sign, supports a bound state — and the scattering amplitude itself already knows this, through a pole at imaginary momentum. The strategy is elementary complex algebra on the closed forms, plus the slope argument that the Delta-Function Potential Well and Bound State article used to build its bound state, run here in reverse to rule one out.

Step 1. Magnitude invariance. $|t|^2=1/|1\pm i\beta|^2=1/(1+\beta^2)$ regardless of the sign, and likewise $|r|^2=\beta^2/(1+\beta^2)$ . Both probabilities depend on $\beta^2$ only. Why this had to happen: $T$ depends on $\alpha$ only through the dimensionless $\beta^2\propto\alpha^2$ — the first power of $\alpha$ survives only in the phases of $r$ and $t$ , which no intensity measurement sees. Detecting the sign takes an interference experiment sensitive to phase, or a hunt for the bound state.

Step 2. The slope argument against a barrier bound state. A negative-energy state must be built from decaying tails $\psi=Be^{-\kappa|x|}$ , whose slopes at the origin are $-\kappa B$ and $+\kappa B$ : the tails supply a kink $\Delta\psi'=-2\kappa B<0$ . The repulsive jump condition demands $\Delta\psi'=+(2m\alpha/\hbar^2)B>0$ . A negative number cannot equal a positive one: no $\kappa>0$ exists, and the repulsive delta binds nothing.

Step 3. The pole. Analytically continue $t(k)=1/(1+i m\alpha/\hbar^2k)$ to complex momentum. It blows up where the denominator vanishes: $k=-im\alpha/\hbar^2$ for the barrier, $k=+im\alpha/\hbar^2$ for the well ( $\alpha\to-\alpha$ ). A bound state is a normalizable solution $e^{-\kappa x}=e^{ikx}$ with $k=+i\kappa$ — the upper half-plane. The well's pole sits exactly there, at $\kappa=m\alpha/\hbar^2$ , reproducing the bound-state energy $E=-\hbar^2\kappa^2/2m=-m\alpha^2/2\hbar^2$ . The barrier's pole sits in the lower half-plane: a growing exponential, not a state. Why this is a big deal: the scattering data at positive energies, packaged as an analytic function, contains the bound-state spectrum at negative energies. Poles of the S-matrix as bound states is a cornerstone of scattering theory — this two-line example is its smallest honest instance. *Verified: sympy confirms both magnitudes are $\beta$ -even; the pole positions solve $1\pm i\beta=0$ exactly as quoted, the well pole giving $\kappa=m\alpha/\hbar^2$ and $E=-m\alpha^2/2\hbar^2$ , matching the Delta-Function Potential Well and Bound State result; the slope equation $-2\kappa=+2m\alpha/\hbar^2$ has no positive solution.*

There is one more consistency check worth knowing about, because you have already met the answer from the other side. Squash the Rectangular Barrier and Quantum Tunnelling result — height $V_0$ , width $a$ — down to a point at fixed area $\alpha=V_0a$ : its exact $1/T=1+V_0^2\sinh^2(\kappa_2 a)/[4E(V_0-E)]$ limits to precisely $1+m\alpha^2/2\hbar^2E$ as $a\to0$ . The delta barrier is the zero-range limit of every thin barrier, which is exactly why Fermi could use it to model nuclear scattering he had no microscopic theory for. Worked Example 2 quantifies how thin is thin enough.

Limiting cases

Case	Result	Meaning
$\beta\to0$ (fast / weak: $E\gg E_{1/2}$ )	$T\to1-\beta^2$ , $R\to\beta^2$	barrier nearly transparent; reflection dies as $1/E$
$\beta\to\infty$ (slow / strong: $E\ll E_{1/2}$ )	$T\to E/E_{1/2}\to0$ , $R\to1$	point spike acts as a hard wall; transmission vanishes linearly in $E$
$E=E_{1/2}=m\alpha^2/2\hbar^2$	$T=R=\tfrac12$	the crossover: barrier strength and momentum in balance
$\alpha\to-\alpha$ (attractive well)	same $T$ , $R$ ; phases flip	scattering cannot detect the sign; only binding can
pole of $t$ at $k=+i\,m\alpha/\hbar^2$ (well only)	$E=-m\alpha^2/2\hbar^2$	the DWELL bound state, encoded in the scattering amplitude
rectangular barrier, $a\to0$ , $V_0a=\alpha$ fixed	$T\to[1+m\alpha^2/2\hbar^2E]^{-1}$	the delta is the universal zero-range limit of thin barriers

Common trapAn infinitely tall barrier must block everything

Height is the wrong variable — the delta's height is infinite and yet $T>0$ at every energy. What controls transmission is the area $\alpha$ (through $\beta=m\alpha/\hbar^2k$ ), because the wave samples the potential only through the integral that sets its slope jump. Keep the hierarchy from the well article: $\psi$ continuous always; $\psi'$ jumping by $(2m\alpha/\hbar^2)\psi(0)$ — it is the slope, never the wavefunction, that breaks. And resist the reflex " $E$ above the barrier top ⇒ free passage": there is no barrier top here, and reflection persists at every finite energy. A classical particle crosses a zero-width bump with probability 1; the quantum wave never quite does.

Worked examples

Example 1 — the crossover in numbers (confidence builder). An electron scatters off a repulsive delta barrier of strength $\alpha=2.0\ \mathrm{eV\cdot\mathring{A}}=3.20\times10^{-29}\ \mathrm{J\cdot m}$ — the same strength whose attractive twin bound one state at $-0.26\ \mathrm{eV}$ in the well article. Find the crossover energy and the transmission at $0.1$ , $1.0$ and $10$ eV.

Step 1. Crossover energy: $E_{1/2}=\dfrac{m\alpha^2}{2\hbar^2}=\dfrac{(9.109\times10^{-31})(3.20\times10^{-29})^2}{2(1.055\times10^{-34})^2}=4.19\times10^{-20}\ \mathrm{J}=\mathbf{0.26\ eV}$ — numerically identical to the twin well's binding energy $|E_b|$ , and that is no accident: both are the single energy scale $m\alpha^2/2\hbar^2$ the delta owns.

Step 2. Transmission from $T=E/(E+E_{1/2})$ . At $E=0.1$ eV: $T=0.1/0.362=\mathbf{0.28}$ — mostly reflected (this is the $\beta=1.62$ state drawn in the hero figure). At $E=1.0$ eV: $T=1.0/1.262=\mathbf{0.79}$ . At $E=10$ eV: $T=10/10.26=\mathbf{0.974}$ — the spike has become a minor nuisance.

Step 3. Phase check at 1 eV: $\beta=\sqrt{E_{1/2}/E}=0.51$ , so the transmitted wave lags by $\arctan(0.51)=27^\circ$ — a delay measurable in interferometry even when $T$ is close to 1.

Example 2 — when may you model a real film as a delta? (exam level). A $V_0=5.0$ eV, $a=0.20$ nm oxide film is proposed to be modelled as a delta barrier of strength $\alpha=V_0a=1.0\ \mathrm{eV\cdot nm}$ for electrons at $E=0.5$ eV. Judge the model, and state when it becomes honest.

Step 1. Delta prediction. $E_{1/2}=m\alpha^2/2\hbar^2=6.56$ eV, so $T_\delta=0.5/(0.5+6.56)=\mathbf{0.071}$ .

Step 2. Exact rectangular answer. $\kappa=\sqrt{2m(V_0-E)}/\hbar=1.09\times10^{10}\ \mathrm{m^{-1}}$ , so $\kappa a=2.17$ and the exact formula gives $T_{\mathrm{rect}}=\mathbf{0.019}$ . The delta model overestimates transmission by a factor of 3.8 — a 0.2 nm film at 5 eV is not delta-like, because the wave decays appreciably inside it.

Step 3. The honesty criterion is $\kappa a\ll1$ : the wave must not notice the barrier's interior. Squeeze the same area into $a=0.02$ nm ( $V_0=50$ eV): $\kappa a=0.72$ and $T_{\mathrm{rect}}=0.060$ — within 15% of the delta answer. At $a=0.005$ nm ( $V_0=200$ eV): $\kappa a=0.36$ , $T_{\mathrm{rect}}=0.068$ , within 4%. The delta limit is a genuine limit — controlled, and quantifiably approached. Rule: trust the delta model when $\kappa a\lesssim0.5$ , i.e. when the physical width is small compared to the evanescent decay length.

The point interaction in history

Origins · 1 minEnrico Fermi and how this was discovered

Enrico Fermi

1901–1954 · Rome, then Chicago

Fermi made the delta function a working tool of physics. In 1934 his Rome group discovered that neutrons slowed by paraffin were absorbed dramatically better — slow-neutron physics born on a marble table. To describe how a neutron scatters off a nucleus thousands of times smaller than its wavelength, Fermi introduced in 1936 the pseudopotential: replace the unknown, complicated nuclear interaction by a delta function whose strength is tuned to reproduce a single measured number, the scattering length. The move is exactly this article's algebra read backwards — when the wave cannot resolve the barrier's interior ( $\kappa a\ll1$ ), only the area $\alpha$ matters, so the honest model is a delta. The pseudopotential still runs the show today: it is how neutron optics computes refractive indices and how cold-atom theory encodes interatomic forces.

History

1926

— Schrödinger's wave mechanics recasts scattering as a boundary-value problem for ψ

1930

— Dirac's Principles of Quantum Mechanics gives the delta function its name and its calculus

1931

— Kronig and Penney build the first model of electron bands from a periodic array of delta potentials

1934

— Fermi's Rome group discovers slow-neutron physics, crying out for a zero-range scattering model

1936

— Fermi introduces the pseudopotential: the delta function as the honest zero-range limit of nuclear scattering

1963

— Lieb and Liniger solve the one-dimensional Bose gas with delta-function interactions exactly

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.5 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed. · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement · Townsend, A Modern Approach to Quantum Mechanics

Provenance. All amplitudes and probabilities follow from Griffiths §2.5; verifier should recompute the derivative-jump integral, solve the 2×2 matching system to confirm $t=1/(1+i\beta)$ , $r=-i\beta/(1+i\beta)$ , check $|t|^2+|r|^2=1$ , and reproduce the attractive-well bound-state energy $E_b=-m\alpha^2/2\hbar^2$ (cross-check Shankar Ch. 5, Cohen-Tannoudji Vol. I). The value $\beta=m\alpha/(\hbar^2 k)$ is standard textbook notation. All numbers in this edition — the crossover $E_{1/2}=0.26$ eV, the transmissions 0.28/0.79/0.974, the phase 27°, and the delta-vs-rectangle comparison (0.071 vs 0.019 at $\kappa a=2.17$ ; convergence to 4% at $\kappa a=0.36$ ) — were recomputed symbolically and numerically for this article.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Free Particle and Plane Waves · Bound States versus Scattering States · Delta-Function Potential Well and Bound State

Leads to: Wavefunction Matching Conditions · Transmission and Reflection Coefficients · Resonant Transmission and Tunnelling Resonances · Finite Square Well · Wave Packets and Their Construction

How can a barrier of zero width reflect anything at all?

Because reflection is not a collision — it is what waves do at any abrupt change in the medium. The delta imprints a kink on $\psi$ , and a kinked wave necessarily contains a backward-travelling Fourier component; solving the matching problem just quantifies how much. Light does the same at the surface of glass, which is "zero width" on the scale of light's wavelength. What would be genuinely inexplicable is a classical particle reflecting off nothing — and indeed the classical limit $\beta\to0$ kills the reflection.

The barrier is infinitely tall. Shouldn't T be zero, like the infinite square well's walls?

The infinite square well's wall is infinitely tall and infinitely wide — the potential stays infinite over a finite region, so the wave has no choice but to die. The delta is infinitely tall over zero width, with the product (the area $\alpha$ ) held finite, and it is the area the wave responds to. If you let $\alpha\to\infty$ you do recover a hard wall: $T\to0$ and, on the reflected side, $r\to-1$ , the perfect-mirror value with the sign flip that makes $\psi(0)\to0$ — the well's boundary condition emerging as a limit.

Is β really dimensionless? I keep losing the units.

Check it once and own it forever: $[\alpha]=\mathrm{J\cdot m}$ (a delta function carries units of 1/length, so $\alpha\delta(x)$ has units of energy). Then $m\alpha/\hbar^2k$ has units $\mathrm{kg\cdot J\cdot m}/(\mathrm{J^2s^2\cdot m^{-1}})=\mathrm{kg\cdot m^2/(J\,s^2)}=1$ , since $\mathrm{J=kg\,m^2/s^2}$ . Equivalently $\beta=\sqrt{E_{1/2}/E}$ — a ratio of energies under a square root — which is the form worth remembering.

Why is T = |t|² legal here when the step problem needed a velocity factor?

The general definition is a ratio of probability currents, $T=j_{\mathrm{trans}}/j_{\mathrm{inc}}$ , and a current carries a factor of speed: $T=(k_2/k_1)|t|^2$ in general — the doctrine of Transmission and Reflection Coefficients. Here the potential is zero on both sides of the spike, so $k_2=k_1$ and the velocity factors cancel. Any time the asymptotic potentials differ (the Potential Step and Reflection case), the bare $|t|^2$ is wrong and can even exceed 1.

Does the transmitted particle come out delayed, since t is complex?

Yes — the phase $\arg t=-\arctan\beta$ is physical. Build a wave packet (Wave Packets and Their Construction) and its transmitted peak emerges shifted in time by roughly $\hbar\,\mathrm{d}(\arg t)/\mathrm{d}E$ relative to free flight — the Wigner delay. For the delta barrier at the crossover energy this is of order $\hbar/E_{1/2}$ , femtoseconds for our electron numbers. Interference experiments read this phase even when $T$ is near 1, which is one way to detect the sign of $\alpha$ that intensity measurements cannot.

What happens as E → 0? Both formulas seem to give T = 0, but the well binds and the barrier doesn't — shouldn't they differ?

Both signs give $T\to0$ linearly in $E$ — total reflection at threshold — and at the level of probabilities they stay identical all the way down. The difference at $E\to0$ is again in the phase: the reflected amplitude $r\to-1$ for both, but the approach differs by the sign of its imaginary part, which encodes whether a bound state lurks below threshold. In three-dimensional scattering this becomes Levinson's theorem — the zero-energy phase counts the bound states.

Real barriers are never deltas. When is this model actually used in practice?

Whenever $\kappa a\ll1$ — the width small compared to the evanescent decay length — the interior is invisible and only the area matters; Worked Example 2 shows the errors collapsing from a factor 3.8 at $\kappa a=2.2$ to 4% at $\kappa a=0.36$ . That regime is real and common: a nucleus (fm) probed by a thermal neutron (Å) is delta-like to one part in $10^4$ , which is why Fermi's pseudopotential is quantitative, and a single dopant atom in a semiconductor channel is delta-like to conduction electrons. Your gate oxide is not — use the Rectangular Barrier and Quantum Tunnelling formula there.