Rectangular Barrier and Quantum Tunnelling

Every second inside the Sun, protons that lack anywhere near the energy to touch each other fuse anyway. Every time you save a file to a flash drive, electrons cross an insulating oxide layer that classically should stop them cold. And in 1928 George Gamow explained why uranium leaks alpha particles over billions of years while radium empties itself in days — one mechanism behind all of it: quantum tunnelling, a particle passing through a region it does not have the energy to enter. The rectangular barrier is the cleanest laboratory for this effect. You already have every tool it needs: the Time-Independent Schrödinger Equation in One Dimension taught you to read $\psi''=(2m/\hbar^2)(V-E)\,\psi$ — oscillate where $E>V$ , decay where $E<V$ ; Bound States versus Scattering States put the arriving particle in the continuum; Potential Step and Reflection stitched waves at one wall. Here we stitch at two — and the payoff is the most consequential formula of this chapter: an exact transmission coefficient whose $e^{-2\kappa a}$ heart is so violently sensitive to barrier width that a scanning tunnelling microscope can feel a single atom's height, and so brutally suppressing for macroscopic objects that a tennis ball will never, ever pass through a wall. Mastering this one calculation buys you the working model for the STM, the tunnel diode, flash memory, and — through the Gamow factor it previews — alpha decay and stellar fusion.

Key takeaways

A particle with $E<V_0$ crosses a barrier of width $a$ with probability $T=\left[1+\frac{V_0^2\sinh^2(\kappa a)}{4E(V_0-E)}\right]^{-1}$ , where $\kappa=\sqrt{2m(V_0-E)}/\hbar$ — nonzero, always.
Inside the barrier $\psi$ does not oscillate and does not vanish: it is a mix of decaying and growing exponentials $e^{\pm\kappa x}$ , and its nonzero value at the far wall seeds the transmitted wave.
For an opaque barrier ( $\kappa a\gg 1$ ) the exact result collapses to $T\approx\frac{16E(V_0-E)}{V_0^2}\,e^{-2\kappa a}$ : everything is controlled by one exponential.
That exponential is the STM's superpower: for a typical 4.5 eV work function, the tunnelling current changes by a factor of ≈ 8.8 for every ångström of tip–surface gap.
Above the barrier ( $E>V_0$ ) the same formula with $\sinh\to\sin$ predicts partial reflection — except at resonances $k_2a=n\pi$ , where the barrier turns perfectly transparent.
Key idea: the forbidden region attenuates the wave instead of blocking it, so transmission is exponentially small in the barrier's width and in the square root of its height deficit — small, but never zero.

The core insight

In the classically forbidden region the wavefunction does not oscillate or vanish — it decays exponentially as $e^{-\kappa x}$ , and because this decaying tail is nonzero at the far wall it launches a transmitted wave whose amplitude depends exponentially on the barrier width.

Where this lives today

Every flash memory cell in your phone stores its bit as charge on a floating gate, written and erased by pushing electrons through an oxide barrier via Fowler–Nordheim tunnelling — the exponential law sets both the programming voltage and the ~10-year retention time. Superconducting qubits are engineered Josephson junctions: their Hamiltonian is built from Cooper-pair tunnelling, the junction's $e^{-2\kappa a}$ setting the Josephson energy that designers tune by oxide thickness. And attosecond "attoclock" experiments strip electrons from atoms through a laser-tilted barrier and time the release, sharpening the century-old debate about how long tunnelling takes — a question the stationary picture deliberately does not answer, and which needs Wave Packets and Their Construction to even pose cleanly.

The physical picture

Picture a water wave rolling toward a harbour wall. A classical particle reads the energy books and turns back: crossing would require negative kinetic energy. The quantum wave does something a bookkeeper would never allow — it soaks in. Its amplitude dies off exponentially with depth, with decay length $\kappa^{-1}$ — the evanescent behaviour you met under Potential Step and Reflection. But this barrier, unlike the step, ends. If the wall is thin enough that the amplitude has not fully died by the far face, whatever survives shakes the far side into motion, and a fresh travelling wave walks away with the same energy it arrived with.

Figure

The three regions — oscillation, exponential decay, and a smaller wave reborn on the far side

Three features of this picture carry all the physics. First, the transmitted wave has the same wavelength as the incident one — no energy is lost tunnelling; only amplitude is paid as toll. Second, inside the wall the curvature relation $\psi''=+\kappa^2\psi$ bends the wavefunction away from the axis — the signature of a forbidden region, opposite to the oscillatory bending outside. Third, the toll is exponential. That fact deserves its own graph.

Figure

Transmission vs energy for a 10 eV, 0.5 nm barrier — the exponential climb below V₀ and the resonance spikes above

Below $V_0$ the transmission climbs by roughly half a decade per electron-volt — that near-straight line on a log scale is the law $T\sim e^{-2\kappa a}$ at work. Above $V_0$ , where a classical particle would sail through with certainty, the wave still reflects off the two potential edges — except at special energies where the front- and back-wall reflections cancel and $T=1$ exactly. Both surprises are one statement: abrupt changes in potential scatter waves, and interference bookkeeping decides how much gets through.

Setting up the three regions

The potential is a rectangular bump on an otherwise free line. A particle of mass $m$ and energy $0<E<V_0$ arrives from the left:

V(x)=\begin{cases}0, & x<0\\ V_0, & 0<x<a\\ 0, & x>a\end{cases} \qquad (1)

where: $V_0>0$ — barrier height; $a$ — barrier width. Real barriers (an oxide layer, a vacuum gap) are never perfectly rectangular, but any short-range bump behaves this way, and the rectangle is the one we can solve exactly. In each region the potential is constant, so the Time-Independent Schrödinger Equation in One Dimension hands us the solutions immediately. Outside, $E>V$ and the solutions oscillate; inside, $E<V_0$ and they are real exponentials:

\psi_{\mathrm{I}}=e^{ikx}+r\,e^{-ikx},\qquad \psi_{\mathrm{II}}=C\,e^{\kappa x}+D\,e^{-\kappa x},\qquad \psi_{\mathrm{III}}=t\,e^{ikx} \qquad (2)

where: $k=\sqrt{2mE}/\hbar$ — the outside wavenumber, identical in I and III because the potential returns to zero; $\kappa=\sqrt{2m(V_0-E)}/\hbar$ — the inside decay rate, whose inverse is the penetration depth; $r,t$ — reflection and transmission amplitudes (incident amplitude normalised to 1); $C,D$ — interior coefficients. Region III has no left-mover: nothing comes in from $+\infty$ . And we must keep both interior exponentials — the barrier is finite, so the growing $e^{+\kappa x}$ stays bounded and is genuinely needed at the far wall. The wavefunction and its derivative must be continuous at $x=0$ and $x=a$ (a finite jump in $V$ cannot kink $\psi$ — that takes the delta spike of Delta-Function Potential Well and Bound State). Four matching equations, four unknowns $r,C,D,t$ : fully determined.

Derivation AExact transmission through the rectangular barrier

We want the probability that the particle emerges on the far side: the transmission coefficient $T=|t|^2$ , where $t$ is the amplitude of the outgoing wave in region III. The strategy is direct — impose continuity of $\psi$ and $\psi'$ at both walls, eliminate the interior coefficients $C$ and $D$ first (they are nuisance variables), and express $t$ in terms of $k$ , $\kappa$ and $a$ alone. The one thing to watch is bookkeeping discipline: work at the far wall first, so each interior coefficient is expressed through $t$ , then feed the results into the near-wall equations. Since region III shares the wavenumber $k$ of region I, the currents on the two sides carry the same velocity factor, and $T=|t|^2$ needs no correction — the general current-ratio doctrine is the business of the Transmission and Reflection Coefficients article.

Step 1. Continuity at the near wall $x=0$ , for $\psi$ and $\psi'$ :

1+r=C+D, \qquad ik(1-r)=\kappa(C-D)

Step 2. Continuity at the far wall $x=a$ :

C e^{\kappa a}+D e^{-\kappa a}=t\,e^{ika}, \qquad \kappa\left(C e^{\kappa a}-D e^{-\kappa a}\right)=ik\,t\,e^{ika}

Step 3. Solve the far-wall pair for the interior coefficients:

C=\frac{t}{2}\Big(1+\frac{ik}{\kappa}\Big)e^{(ik-\kappa)a},\qquad D=\frac{t}{2}\Big(1-\frac{ik}{\kappa}\Big)e^{(ik+\kappa)a}

Why solve at $x=a$ first: the far-wall equations involve only $C,D,t$ — two equations, and we want two of the three unknowns in terms of the third. The near-wall equations then become two equations for the remaining pair $r$ and $t$ .

Step 4. Add and subtract the near-wall pair to isolate the incident amplitude, substitute Step 3, and collect the exponentials into hyperbolic functions:

\frac{1}{t}=e^{ika}\left[\cosh(\kappa a)+\frac{i}{2}\Big(\frac{\kappa}{k}-\frac{k}{\kappa}\Big)\sinh(\kappa a)\right]

Why hyperbolic functions appear: $C$ carries $e^{-\kappa a}$ and $D$ carries $e^{+\kappa a}$ , so every combination of the two produces $(e^{\kappa a}\pm e^{-\kappa a})/2$ — that is, $\cosh$ and $\sinh$ .

Step 5. Take the modulus squared. The pure phase $e^{ika}$ drops out, and $\cosh^2=1+\sinh^2$ merges the two terms:

\frac{1}{T}=\frac{1}{|t|^2}=1+\left[1+\frac{1}{4}\Big(\frac{\kappa}{k}-\frac{k}{\kappa}\Big)^2\right]\sinh^2(\kappa a) =1+\frac{(k^2+\kappa^2)^2}{4k^2\kappa^2}\,\sinh^2(\kappa a)

Step 6. Translate back to energies. From the definitions, $k^2+\kappa^2=2mV_0/\hbar^2$ and $k^2\kappa^2=(2m/\hbar^2)^2E(V_0-E)$ , so the prefactor is $V_0^2/[4E(V_0-E)]$ and

T=\left[1+\frac{V_0^2\,\sinh^2(\kappa a)}{4E(V_0-E)}\right]^{-1}

Why the answer had to look like this: at $a\to 0$ the barrier disappears and indeed $T\to 1$ ; for $\kappa a$ large the $\sinh^2$ term dominates and $T$ is crushed exponentially; and every quantity enters through the two dimensionless combinations $\kappa a$ and $E/V_0$ — nothing else could appear. *Verified: the four matching equations were solved symbolically (sympy) and the result matches this closed form identically; an independent numerical solve of the 4×4 linear system agrees to 10 significant figures at five test energies below and above the barrier, and $R+T=1$ holds in every case.*

Reading the formula: the exponential law

The exact $T$ is compact, but its working regime deserves its own statement. In almost every application — an STM gap, an oxide layer, a nuclear Coulomb wall — the barrier is opaque: $\kappa a\gg 1$ , transmission far below unity. There the formula simplifies into the version worth memorising.

Go deeper · Derivation B · 1 minThe opaque barrier and the sixteen prefactor

Derivation BThe opaque barrier and the sixteen prefactor

We extract the limit of the exact transmission when $\kappa a\gg 1$ — the regime of alpha decay ( $T\sim10^{-39}$ per knock) and of STM gaps ( $T\sim10^{-5}$ ). The goal is to see that a single exponential controls everything, to get the honest prefactor in front of it, and to check the opposite extreme — the barrier top — where the exponential language stops applying. The one subtlety is knowing when the 1 in $1/T$ may be dropped: precisely when transmission is small.

Step 1. For $\kappa a\gg1$ , keep only the growing half of the hyperbolic sine: $\sinh(\kappa a)=\tfrac12\!\left(e^{\kappa a}-e^{-\kappa a}\right)\approx\tfrac12 e^{\kappa a}$ , so $\sinh^2(\kappa a)\approx\tfrac14 e^{2\kappa a}$ .

Step 2. The second term in $1/T$ is then enormous compared to 1, so

\frac{1}{T}\approx \frac{V_0^2}{16\,E(V_0-E)}\,e^{2\kappa a} \qquad\Longrightarrow\qquad T\approx \frac{16\,E(V_0-E)}{V_0^2}\,e^{-2\kappa a}

Why the exponent is $2\kappa a$ and not $\kappa a$ : the amplitude decays as $e^{-\kappa a}$ across the wall, and probability is amplitude squared.

Step 3. Read the prefactor. $16E(V_0-E)/V_0^2$ peaks at $E=V_0/2$ , where it equals exactly 4, and it never strays far from order one anywhere in the middle of the barrier. All the violent dependence — on width, on mass, on energy deficit — lives in $e^{-2\kappa a}$ with $\kappa=\sqrt{2m(V_0-E)}/\hbar$ . Why this matters physically: because $\kappa\propto\sqrt{m}$ , tunnelling discriminates savagely by mass — electrons tunnel readily, protons grudgingly, tennis balls never. And because $a$ sits in the exponent, transmission is a precision ruler for distance.

Step 4. Check the opposite limit so we know the formula's boundary of validity. As $E\to V_0$ the decay rate $\kappa\to0$ and $\sinh(\kappa a)\to\kappa a$ , giving the exact barrier-top value

T(E=V_0)=\left[1+\frac{mV_0a^2}{2\hbar^2}\right]^{-1}

which is finite and smooth — for our worked electron barrier below (10 eV, 0.5 nm) it evaluates to $T=0.058$ . There is no drama at the top of the barrier, just a crossover from $\sinh$ to $\sin$ behaviour. Verified: sympy confirms the ratio of exact to approximate $T$ tends to 1 as $\kappa a\to\infty$ , and the barrier-top limit follows from the exact formula by l'Hôpital; numerically, at $\kappa a=5.7$ (the worked example) the opaque formula is already accurate to 0.3%.

The prefactor-times-exponential structure is the seed of the Gamow factor: for a barrier of varying height the exponent generalises to $2\int\kappa(x)\,\mathrm{d}x$ across the forbidden region — the workhorse of alpha-decay theory, derived properly in Bohr–Sommerfeld and Semiclassical Quantization in 1D and again in The WKB Approximation at full strength. Here you are seeing its exactly solvable ancestor.

Above the barrier: reflection where none should be

Push the energy above $V_0$ and the interior region becomes classically allowed: $\kappa^2=2m(V_0-E)/\hbar^2$ turns negative, so $\kappa=ik_2$ with $k_2=\sqrt{2m(E-V_0)}/\hbar$ real. Nothing in the algebra cared whether $\kappa$ was real, so the exact answer analytically continues: $\sinh(ik_2a)=i\sin(k_2a)$ flips $\sinh^2\to-\sin^2$ , and $(V_0-E)\to-(E-V_0)$ flips the denominator, leaving

T=\left[1+\frac{V_0^2\,\sin^2(k_2 a)}{4E(E-V_0)}\right]^{-1} \qquad (3)

where: $k_2$ — the wavenumber inside the barrier region, shorter than $k$ because the particle slows down on top of the bump. The two sign flips cancel, so $T\le 1$ stays manifestly sensible. A classical particle with $E>V_0$ crosses with certainty; the wave instead reflects off both edges of the barrier, and the two reflected wavelets interfere. Mostly they add and $T<1$ . But when

k_2\,a=n\pi \qquad\Longleftrightarrow\qquad E_n=V_0+\frac{n^2\pi^2\hbar^2}{2ma^2} \qquad (4)

where: $n=1,2,3,\dots$ — an integer number of half-wavelengths fitting exactly across the barrier — the two edge reflections cancel and the barrier becomes perfectly transparent, $T=1$ . Look at the offsets $n^2\pi^2\hbar^2/2ma^2$ : they are precisely the Infinite Square Well energies of a box the width of the barrier — the same standing-wave geometry that quantises a bound particle picks out the transparent energies of a flying one. For the 10 eV, 0.5 nm barrier of the figure the resonances sit at 11.5, 16.0 and 23.5 eV. The deep dive into these spikes — including the Ramsauer–Townsend effect, where noble-gas atoms turn nearly invisible to ~1 eV electrons — belongs to Resonant Transmission and Tunnelling Resonances; the delta-limit cousin lives in Delta-Function Barrier and Scattering.

Limiting cases

case	result	meaning
opaque barrier, $\kappa a\gg1$	$T\approx\dfrac{16E(V_0-E)}{V_0^2}\,e^{-2\kappa a}$	one exponential rules; STM/alpha-decay regime
thin/weak barrier, $\kappa a\ll1$	$T\approx\left[1+\dfrac{mV_0^2a^2}{2\hbar^2E}\right]^{-1}$	nearly transparent; with $V_0\to\infty$ , $V_0a=\alpha$ fixed, recovers the delta-barrier $T=[1+m\alpha^2/2\hbar^2E]^{-1}$
barrier top, $E=V_0$	$T=\left[1+\dfrac{mV_0a^2}{2\hbar^2}\right]^{-1}$	finite, smooth crossover; neither $\sinh$ nor $\sin$ regime
above barrier, $E>V_0$	$T=\left[1+\dfrac{V_0^2\sin^2(k_2a)}{4E(E-V_0)}\right]^{-1}$	partial reflection persists; $T=1$ only at $k_2a=n\pi$
high energy, $E\gg V_0$	$T\to1$	the bump becomes invisible; classical limit recovered
heavy mass or wide barrier	$T\to e^{-2\kappa a}\to 0$	$\kappa\propto\sqrt{m}$ : macroscopic tunnelling is annihilated by the exponent

Common trapTunnelling means the particle briefly borrows energy to hop the wall

No energy is borrowed, and nothing violates conservation. The particle is in an energy eigenstate with sharp, conserved $E<V_0$ throughout; the transmitted wave emerges with exactly the incident energy (same $k$ , same wavelength). What fails inside the barrier is not the energy books but the classical decomposition of energy into kinetic-plus-potential at a point: position and momentum are not simultaneously sharp, so "kinetic energy at $x$ " is not a measurable statement. And do not conclude the particle "was inside the wall with negative KE" — actually localising it within the barrier injects momentum uncertainty $\Delta p\gtrsim\hbar\kappa$ , i.e. $\Delta p^2/2m\gtrsim V_0-E$ , which lifts the paradox.

Worked examples

Example 1 — an electron meets an oxide-scale barrier (confidence builder). An electron of energy $E=5\,\mathrm{eV}$ hits a barrier of height $V_0=10\,\mathrm{eV}$ and width $a=0.5\,\mathrm{nm}$ — numbers typical of a thin insulating film. Find $T$ , then find it again for a film half as thick.

Step 1. Decay rate: $\kappa=\sqrt{2m_e(V_0-E)}/\hbar=\sqrt{2(9.109\times10^{-31})(5)(1.602\times10^{-19})}/(1.055\times10^{-34})=1.15\times10^{10}\,\mathrm{m}^{-1}$ , i.e. $\kappa^{-1}=0.87\,\text{Å}$ : the wave loses amplitude on the scale of a single atom.

Step 2. Opacity: $\kappa a=(1.146\times10^{10})(0.5\times10^{-9})=5.73$ . Comfortably opaque — expect the exponential law to be excellent.

Step 3. Prefactor: with $E=V_0/2$ , $\;16E(V_0-E)/V_0^2=16\cdot\tfrac12\cdot\tfrac12=4$ exactly.

Step 4. Transmission: $T\approx 4\,e^{-2(5.73)}=4\,e^{-11.46}$ . The exact formula gives $\boxed{T=4.2\times10^{-5}}$ and the opaque approximation the same to three digits — one electron in 24 000 crosses.

Step 5. Halve the width, $a=0.25$ nm: $\kappa a=2.86$ and the exact $T=1.3\times10^{-2}$ — a factor of 305 more current from removing a quarter-nanometre of oxide. That is the exponential law in action: $e^{+2\kappa\Delta a}=e^{5.73}\approx 308$ predicts the jump almost perfectly.

Example 2 — why the STM sees single atoms (exam level). In a scanning tunnelling microscope the tip–sample vacuum gap is a barrier whose height is set by the work function, $\phi\approx4.5\,\mathrm{eV}$ for a typical metal. The tunnelling current is proportional to $T\approx T_0\,e^{-2\kappa d}$ at gap $d$ . How sharply does the current respond to the gap?

Step 1. Decay rate at the Fermi level: $\kappa=\sqrt{2m_e\phi}/\hbar=\sqrt{2(9.109\times10^{-31})(4.5)(1.602\times10^{-19})}/(1.055\times10^{-34})=1.09\times10^{10}\,\mathrm{m}^{-1}=1.09\,\text{Å}^{-1}$ .

Step 2. Sensitivity per ångström: the current ratio for $\Delta d=1\,\text{Å}=10^{-10}\,\mathrm{m}$ is $e^{2\kappa\Delta d}=e^{2(1.087)}=\boxed{8.8}$ — nearly an order of magnitude of current per atomic radius. An atom-high surface step announces itself as a ninefold current jump.

Step 3. Sensitivity per ten picometres: $e^{2\kappa(10^{-11})}=1.24$ , a 24% current change for 10 pm — a hundredth of an atomic diameter. Servo the tip to hold the current constant and the feedback voltage traces the surface at picometre vertical resolution — not engineering heroics, just the number $2\kappa\approx2.2\,\text{Å}^{-1}$ doing the work.

The tunnelling century

Origins · 1 minLeo Esaki and how this was discovered

Leo Esaki

1925– · Sony laboratories, Tokyo

Esaki turned tunnelling from an explanation into a device. In 1957, studying heavily doped germanium p–n junctions at what would become Sony, he found a current that decreased as the voltage increased — negative differential resistance, impossible in any classical conduction picture, and exactly the fingerprint of electrons tunnelling through the junction's atomically thin depletion barrier. The Esaki diode was the first device whose operating principle was quantum tunnelling, later joined by the Josephson junction and resonant-tunnelling heterostructures. Esaki shared the 1973 Nobel Prize with Giaever and Josephson — a prize given, in effect, to the formula on this page.

History

1927

— Friedrich Hund finds wavefunctions leaking between the two wells of a molecular double-minimum potential — tunnelling's first appearance

1928

— George Gamow, and independently Gurney and Condon, explain alpha decay as tunnelling through the nuclear Coulomb barrier, cracking the Geiger–Nuttall law

1928

— Fowler and Nordheim describe field emission of electrons from cold metals as tunnelling through a field-tilted surface barrier

1957

— Leo Esaki demonstrates the tunnel diode: negative differential resistance from electrons tunnelling in a heavily doped junction

1962

— Brian Josephson predicts that Cooper pairs tunnel between superconductors, creating a supercurrent with zero applied voltage

1973

— Esaki, Giaever and Josephson share the Nobel Prize for tunnelling phenomena in solids

1981

— Binnig and Rohrer's scanning tunnelling microscope images surfaces atom by atom using the exponential current–gap law

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.6 (the delta and step/barrier potentials) and Problem 2.33 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.5 / scattering in 1D · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 (1D scattering, step and barrier) · Cohen-Tannoudji, Diu, Laloë, Quantum Mechanics, Vol. I, Complement H_I (square barrier) · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 4 · Merzbacher, Quantum Mechanics, 3rd ed., Ch. 6

Provenance. Equations cross-checked against Griffiths 3rd ed. §2.6, Cohen-Tannoudji Vol. I Complement H_I, Shankar Ch. 5, and Merzbacher Ch. 6. A verifier should: (1) re-solve the 4 continuity equations and confirm $T=[1+V_0^2\sinh^2(\kappa a)/(4E(V_0-E))]^{-1}$ ; (2) check the identity $(\kappa/k+k/\kappa)^2=V_0^2/[E(V_0-E)]$ using $k^2=2mE/\hbar^2$ , $\kappa^2=2m(V_0-E)/\hbar^2$ ; (3) confirm the opaque limit prefactor $16E(V_0-E)/V_0^2$ from $\sinh^2\to\tfrac14 e^{2\kappa a}$ ; (4) verify $T+R=1$ via the probability current; (5) confirm resonance energies $E_n=V_0+n^2\pi^2\hbar^2/(2ma^2)$ . All five checks, every worked number, and both limits on this page were re-verified symbolically (sympy) and numerically (independent 4×4 linear solve) for this edition.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Free Particle and Plane Waves · Bound States versus Scattering States · Finite Square Well · Potential Step and Reflection

Leads to: Transmission and Reflection Coefficients · Resonant Transmission and Tunnelling Resonances · Bohr–Sommerfeld and Semiclassical Quantization in 1D · The WKB Approximation · Wave Packets and Their Construction

Does the particle actually pass through the barrier, or does it somehow go around?

There is no "around" in one dimension, and no hidden path. The question "where was the particle while crossing?" has no measurement-independent answer in quantum mechanics. What the theory predicts — and experiment confirms — is arrival statistics: a fraction $T$ of identically prepared particles is detected beyond the barrier, with unchanged energy. If you do measure position inside the barrier you will sometimes find the particle there, but localising it that tightly transfers enough momentum to make its subsequent motion classically allowed. The paradox lives only in unperformed measurements.

Is tunnelling instantaneous? Does the particle cross faster than light?

No superluminal signalling occurs, but "how long does tunnelling take" is genuinely subtle and still researched. Stationary states have no clock in them — they describe steady fluxes, not journeys. A transmitted wave-packet peak can appear to cross anomalously fast (the Hartman effect), but the peak of a reshaped, exponentially filtered packet is not a signal, and causality survives. Attoclock experiments bound the tunnelling delay in strong-field ionisation at tens of attoseconds or below. The clean treatment needs Wave Packets and Their Construction.

If the wavefunction decays as e^(−κx) inside, why is the transmission e^(−2κa)?

Amplitude versus probability. The wave's amplitude is attenuated by roughly $e^{-\kappa a}$ across the width. Detection probabilities are amplitudes squared, so the transmitted probability carries $(e^{-\kappa a})^2=e^{-2\kappa a}$ . The same factor-of-two bookkeeping appears everywhere in quantum mechanics — it is the reason STM sensitivity quotes $2\kappa\approx2.2\,\text{Å}^{-1}$ rather than $\kappa$ .

Could a tennis ball tunnel through a brick wall if I threw it forever?

Formally yes; physically never. A 58 g ball with a 1 J energy deficit against a 10 cm wall has $\kappa\approx3.2\times10^{33}\,\mathrm{m}^{-1}$ , so $2\kappa a\approx6.5\times10^{32}$ and $T\sim10^{-2.8\times10^{32}}$ — a probability whose decimal form needs more zeros than there are atoms in the observable universe. Throwing once per Planck time for the age of the universe changes nothing. Tunnelling is a nanoscale privilege.

Alpha decay half-lives range from microseconds to billions of years. How can one mechanism span 20-plus orders of magnitude?

Because the lifetime is an exponential of an integral. The Gamow exponent $2\int\kappa(x)\,\mathrm{d}x$ across the Coulomb barrier varies only modestly between nuclei — but it sits in an exponent, so a factor-two spread in alpha energies (4–9 MeV) becomes twenty-plus orders of magnitude in $T$ , and hence in the decay rate $\lambda\approx f\,T$ ( $f\sim10^{21}\,\mathrm{s}^{-1}$ is the knock frequency). The straight-line Geiger–Nuttall plot of $\log\lambda$ against $1/\sqrt{E}$ is the exponential law wearing experimental clothes.

Does the rectangular-barrier formula apply to real, smooth barriers?

The exact $\sinh^2$ formula is special to the rectangle, but its opaque-limit structure — order-one prefactor times $e^{-2\kappa a}$ — generalises: for a smooth barrier replace $\kappa a$ by $\int\kappa(x)\,\mathrm{d}x$ over the forbidden stretch (the WKB/Gamow prescription, made honest in The WKB Approximation). The rectangle is the calibration case where the semiclassical answer is checked against an exact one.

What happens exactly at E = V₀? Both formulas seem to break.

Neither breaks — they meet. From below, $\kappa\to0$ and $\sinh(\kappa a)\approx\kappa a$ ; from above, $k_2\to0$ and $\sin(k_2a)\approx k_2a$ . Both give the same finite value $T=[1+mV_0a^2/2\hbar^2]^{-1}$ — about 0.058 for the worked 10 eV, 0.5 nm barrier. The crossover is smooth because the exact solution is analytic in energy; the apparent singularity is an artefact of the $\kappa$ / $k_2$ notation.