2.11 Barrier penetration and quantum tunnelling

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

A particle of energy $E<V_0$ comes in from the left. Classical mechanics says: it hits the wall, bounces back, and the transmission probability is exactly zero.

The quantum answer is not zero.

Solving region by region

Derivation: the transmission coefficient of a rectangular barrieradvanced~12 min

General solutions in the three regions (for $E<V_0$ ):

\psi_\text{I}=A\ee^{\ii kx}+B\ee^{-\ii kx}\quad(x<0),\qquad k=\frac{\sqrt{2mE}}{\hbar}\tag{2.11.1}

\psi_\text{II}=C\ee^{\kappa x}+D\ee^{-\kappa x}\quad(0<x<a),\qquad \kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}\tag{2.11.2}

\psi_\text{III}=F\ee^{\ii kx}\quad(x>a)\tag{2.11.3}

$A$ is the incident wave, $B$ the reflected wave, $F$ the transmitted wave. Region III has no $\ee^{-\ii kx}$ term because there is no source on the right — that is the only place where physical input enters; everything else is mathematics.

Matching conditions. $V$ is finite, so $\psi$ and $\psi'$ are continuous at both $x=0$ and $x=a$ , giving four equations:

A+B=C+D\tag{2.11.4}

\ii k(A-B)=\kappa(C-D)\tag{2.11.5}

C\ee^{\kappa a}+D\ee^{-\kappa a}=F\ee^{\ii ka}\tag{2.11.6}

\kappa\left(C\ee^{\kappa a}-D\ee^{-\kappa a}\right)=\ii kF\ee^{\ii ka}\tag{2.11.7}

Four equations for four ratios ( $B/A$ , $C/A$ , $D/A$ , $F/A$ ) — solvable.

Elimination. The last two give $C$ and $D$ :

C\ee^{\kappa a}=\frac{F\ee^{\ii ka}}{2}\left(1+\frac{\ii k}{\kappa}\right),\qquad D\ee^{-\kappa a}=\frac{F\ee^{\ii ka}}{2}\left(1-\frac{\ii k}{\kappa}\right)\tag{2.11.8}

Substitute into the first two and eliminate $B$ (multiply the first by $\ii k$ and add the second):

2\ii kA=(\ii k+\kappa)C+(\ii k-\kappa)D\tag{2.11.9}

Insert $C$ and $D$ and tidy up:

\frac{A}{F}=\ee^{\ii ka}\left[\cosh\kappa a+\frac{\ii}{2}\left(\frac{\kappa}{k}-\frac{k}{\kappa}\right)\sinh\kappa a\right]\tag{2.11.10}

Transmission coefficient $T=|F/A|^2$ :

T^{-1}=\cosh^2\kappa a+\frac14\left(\frac{\kappa}{k}-\frac{k}{\kappa}\right)^2\sinh^2\kappa a\tag{2.11.11}

Using $\cosh^2=1+\sinh^2$ and converting $k,\kappa$ back to $E,V_0$ (after some algebra):

\boxed{\ T=\left[1+\frac{V_0^2\sinh^2(\kappa a)}{4E(V_0-E)}\right]^{-1}\ }\tag{2.11.12}

The thick-barrier approximation

For $\kappa a\gg1$ , $\sinh\kappa a\approx\frac12\ee^{\kappa a}$ , so

T\approx\frac{16E(V_0-E)}{V_0^2}\,\ee^{-2\kappa a}\tag{2.11.13}

The exponential factor decides everything

The prefactor lies between 0 and 4 — a number of order one. All the physics is in $\ee^{-2\kappa a}$ .

That makes $T$ exponentially sensitive to the barrier’s width and height:

\frac{\dd\ln T}{\dd a}=-2\kappa\tag{2.11.14}

For an electron with $V_0-E=4$ eV, $\kappa=\sqrt{2m(V_0-E)}/\hbar\approx1.02\ \text{Å}^{-1}$ . Every extra ångström of barrier thickness costs a factor of $\ee^{-2}\approx1/7.4$ in $T$ .

This is exactly how the scanning tunnelling microscope works: pull the tip 1 Å further from the surface and the tunnelling current drops by almost an order of magnitude. That extreme distance sensitivity is what lets an STM resolve height differences of single atoms.

Orders of magnitude: three real examplesbasic~6 min

1. Scanning tunnelling microscope. A metal work function $\phi\approx4$ eV and a tip–sample gap $a\approx5$ Å:

\kappa=\frac{\sqrt{2m\phi}}{\hbar}=\frac{\sqrt{2\times511000\times4}}{197.3}\ \text{nm}^{-1}=10.2\ \text{nm}^{-1}\tag{2.11.15}

2\kappa a=2\times10.2\times0.5=10.2\quad\Longrightarrow\quad \ee^{-10.2}\approx3.7\times10^{-5}\tag{2.11.16}

A transmission probability around $10^{-4}$ – $10^{-5}$ . That sounds tiny, but around $10^{15}$ electrons try every second, so the tunnelling current is in the nanoamp range — comfortably measurable.

2. Alpha decay. An alpha particle is bound by the nuclear force inside a well of radius about 7 fm and has to get through the Coulomb barrier (peaking near 25 MeV, while the alpha’s energy is only 4–9 MeV). Approximating the Coulomb barrier as a stack of rectangular barriers and summing (this is the WKB approximation) gives

\ln T\propto-\frac{Z}{\sqrt{E}}\tag{2.11.17}

This is the Gamow factor, and it explains the Geiger–Nuttall law: a small change in alpha energy sweeps the half-life across 24 orders of magnitude (14 billion years for $^{232}$ Th versus 0.3 microseconds for $^{212}$ Po). Exponential sensitivity taken to its extreme.

Gamow’s 1928 calculation of alpha decay was one of the earliest major successes of quantum mechanics.

3. Flash memory. Writing to NAND flash pushes electrons by tunnelling through an oxide layer about 8 nm thick into the floating gate. On read, the same oxide is thick enough to keep the electrons in place (ten-year data retention). Writing raises $T$ by applying a high voltage that “tilts” the barrier (Fowler–Nordheim tunnelling).

Every single bit in your phone’s storage chip is one controlled act of tunnelling.

E > V₀: there is still reflection

The formula above stays valid for $E>V_0$ ; simply set $\kappa\to\ii k_2$ (with $k_2=\sqrt{2m(E-V_0)}/\hbar$ ), which turns $\sinh$ into $\ii\sin$ :

T=\left[1+\frac{V_0^2\sin^2(k_2a)}{4E(E-V_0)}\right]^{-1}\tag{2.11.18}

Classically the particle slows down, crosses, and transmission is 100%. Quantum mechanically $T<1$ : there is reflection.

The picture

This should not really be a surprise. Waves partially reflect at any abrupt change of medium: light going from air into glass reflects about 4%, and sound going from air into water reflects almost completely. An abrupt change in potential does exactly the same thing to a matter wave.

Reflection comes from impedance mismatch, and has nothing to do with whether the particle has enough energy to get over the top.

The mathematics

Resonant transmission. When

k_2a=n\pi\quad(n=1,2,\dots)\tag{2.11.19}

we get $\sin(k_2a)=0$ and $T=1$ — the barrier becomes perfectly transparent.

The condition says the barrier width holds an exact whole number of half-wavelengths, so the reflections from the two interfaces cancel — exactly like an anti-reflection coating in optics.

Wave-packet evolution and scattering

A Gaussian packet hits a rectangular barrier. The time-dependent Schrödinger equation is solved live by the split-operator method, with ħ = m = 1.

E / V₀: 0.80
Reflection R: 0.000
Transmission T: 0.000
Packet width Δx: 3.00

Press play to send the packet into the barrier. R and T lock automatically once the packet has completely left the barrier region. The barrier width on the grid is a = 1.500.

Incident momentum k₀2.00

E = k₀²/2 = 2.00

Barrier height V₀2.5

positive = a barrier

Barrier width a1.5

T decays exponentially with a: T ~ e^(−2κa)

Initial packet width σ3.0

larger σ means better-defined momentum, closer to a plane wave

View

Show Re ψ

Speed8×

Try this

At the default parameters E ≈ 2.0 < V₀ = 2.5, a classical particle bounces back 100% of the time. Press play: part of the packet gets through. That is tunnelling — the principle behind the scanning tunnelling microscope and alpha decay.
Take a from 1.5 to 4 and replay. The transmission collapses — T ~ e^(−2κa) is exponential, so widening the barrier a little costs several orders of magnitude.
Set V₀ to 1.0 (now E > V₀, so classically transmission should be 100%) and replay: a clear reflected peak still runs back to the left. A wave partially reflects at any abrupt change of potential, exactly as light does at a glass surface.
Switch to the transmission curve and make V₀ negative (a well): at certain energies the curve returns to T = 1. That is resonant transmission, historically seen as the Ramsauer–Townsend effect, where noble gases are almost transparent to slow electrons.
Set σ to 8 and scatter again: the measured T lands much closer to the open circle on the analytic curve — the nearer the packet is to a plane wave, the better the single-energy approximation.

This is an interactive simulation (wave-packet-scattering) and needs JavaScript to run. The derivations and conclusions in the text do not depend on it, so feel free to read on.

How long does tunnelling take?

The question looks simple and has been argued over for seventy years without a fully settled answer.

The difficulty is that quantum mechanics has no time operator, so “how long the particle spent inside the barrier” cannot be defined directly. Different definitions (Wigner phase time, Büttiker–Landauer time, the Larmor clock, and others) give different answers, some of them apparently “superluminal” and independent of barrier width — though it can be shown that none of them can be used to send information, so causality is safe.

Around 2020 several ultracold-atom and attosecond experiments produced measured values, but the argument over what “tunnelling time” even means is not over. It is one of the few genuinely foundational questions still actively debated; see section 13.7.

Key formulas

Rectangular-barrier transmission (E < V₀)
$T=\left[1+\frac{V_0^2\sinh^2(\kappa a)}{4E(V_0-E)}\right]^{-1},\quad\kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}$
Exact, no approximation
Thick-barrier approximation
$T\approx\frac{16E(V_0-E)}{V_0^2}\ee^{-2\kappa a}$
For κa ≫ 1; the exponential dominates everything
The E > V₀ case
$T=\left[1+\frac{V_0^2\sin^2(k_2a)}{4E(E-V_0)}\right]^{-1},\quad k_2=\frac{\sqrt{2m(E-V_0)}}{\hbar}$
Generally T < 1; resonant transmission T = 1 when k₂a = nπ
Gamow factor
$\ln T\propto-\frac{Z}{\sqrt{E}}$
The origin of the 24 orders of magnitude in alpha-decay half-lives

Self-check4 questions

End of the chapter

Look back at what one complex-valued function and one equation accomplished:

Derived energy quantisation (infinite well, harmonic oscillator) with no extra assumptions
Explained why atoms are stable (stationary states do not radiate)
Predicted a phenomenon classically forbidden outright (tunnelling), which now supports an entire field of technology
Absorbed classical mechanics as a limiting case (Ehrenfest’s theorem, the correspondence principle)

It also left a pile of things unexplained: why do observables correspond to operators? What actually happens in a measurement? What entitles $|c_n|^2$ to be a probability?

Answering those requires a different language. Chapter 3 abstracts the wavefunction into a vector in Hilbert space — after which you will find that most of the long integral derivations of this chapter take two or three lines.

Barrier penetration and quantum tunnelling

Recommended first

After this section you should be able to

Solving region by region

The thick-barrier approximation

E > V₀: there is still reflection

The picture

The mathematics

Wave-packet evolution and scattering

Key formulas

Self-check4 questions

End of the chapter

Solving region by region

The thick-barrier approximation

E > V₀: there is still reflection

◑The picture

∑The mathematics

∎Key formulas

End of the chapter

The picture

The mathematics

Key formulas