Schrödinger's Equation and Tunnelling

Beginner

Step 1: energy sets a limit. A ball rolling up a hill trades speed for height. With too little energy to reach the top, it stops partway and rolls back, every time: energy cannot appear from nowhere.

Step 2: tiny things behave like waves. Light acts sometimes as a wave, sometimes as particles (photons), and so do electrons. An electron is described by a wave, written $\psi$ (the Greek letter psi). Where the wave is big, the electron is likely to be found; where it is flat, almost never. In the animation the blue line is the electron's wave and the shaded hump shows where it probably is.

Step 3: a wave can leak through a thin wall. The gold line is the electron's energy and the grey wall is higher, so a ball would always come back. Watch: most of the wave bounces back, but a smaller wave appears beyond the wall and moves on. Each electron is still found whole on one side or the other; here about one in five gets through. This is quantum tunnelling. Light does it too: light trapped in glass by total internal reflection can cross a very thin air gap into a second block.

Check yourself: the wall is made twice as thick. Do more or fewer electrons get through? Answer: far fewer, because the wave dies away quickly inside the wall. A lower wall, though, lets more through.

Step 4: trapped waves have set energies. In the box on the right the wave must fit between the walls like a guitar string, so only whole half-waves are allowed, each with its own energy. Electrons in atoms are trapped too, which is why atoms have energy levels.

Tunnelling lets nuclei fuse inside the Sun and stores photos in flash memory.

Intermediate

The wavefunction and its probabilities

A detector behind a wall higher than the electrons' energy still finds some of them. In quantum mechanics an electron has no definite path. Its state is a wavefunction $\psi (x)$, a wave that is never seen directly: a detector finds the whole electron at one place, and the wavefunction only says how likely each place is.

By the Born rule, $|\psi {|}^2$ is a probability per metre, and the probability of finding the particle between two points is the area under it:
$P=\int |\psi {|}^{2}dx$

In the animation $|\psi {|}^2$ is the purple shading and the blue line is the real part of the wave, $\mathrm{Re}\psi$. The readouts $P(\text{left})$, $P(\text{barrier})$ and $P(\text{right})$ are the areas of the shading in three regions, and over the whole line they add up to 1.

The Schrödinger equation on an energy diagram

For a particle of definite energy $E$ the wavefunction obeys the time-independent Schrödinger equation:
$-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+V(x)\psi =E\psi$
Here $m$ is the mass, $V(x)$ the potential energy and $\hslash =h/2\pi$ the reduced Planck constant. Read it as an energy balance: kinetic plus potential energy equals $E$, and the kinetic part is set by how sharply $\psi$ curves.

The animation is an energy diagram: position runs across, energy runs up. The grey barrier is $V(x)$, zero on both sides and $V_0$ inside, and the gold line is $E$, with the wave drawn about it. Where $E>V$ the wave oscillates with the de Broglie wavelength
$\lambda =\frac{h}{p},p=\sqrt{2m(E-V)}$
as the blue wave does on both sides of the barrier.

Where $E<V$ a classical particle could never be, since its kinetic energy would be negative. The equation still has solutions there, but they are exponentials, not waves, and the decaying one dies away gradually inside the wall.

Inside the barrier: decay and transmission

Inside a thick barrier of height $V_0$ the wave decays as $e^{-\kappa x}$, with the decay constant
$\kappa =\frac{\sqrt{2m(V_0-E)}}{\hslash }$
Over the decay length $1/\kappa$ the amplitude falls by a factor $e\approx 2.72$. A larger deficit $V_0-E$ or a heavier particle makes $\kappa$ larger and the decay faster.

A thin barrier leaves some wave on the far side, with the same wavelength but a smaller amplitude. The fraction of electrons that get through is the transmission probability $T$, the fraction that come back is $R$, and $R+T=1$. For a barrier of width $a$ much larger than $1/\kappa$,
$T\approx 16\left(\frac{E}{V_0}\right)\left(1-\frac{E}{V_0}\right)e^{-2\kappa a}$
The amplitude is multiplied by $e^{-\kappa a}$ and the probability by its square; the prefactor, never above 4, depends only on $E/V_0$.

The width sits in an exponent, so tunnelling is extremely sensitive to it: every extra $1/(2\kappa )$ of thickness divides $T$ by $e$. For barriers a few electronvolts high, only layers at most a few nanometres thick let many electrons through.

Worked example: the barrier in the animation

The electron has $E=3\mathrm{eV}$ and meets a barrier $V_0=4\mathrm{eV}$ high and $a=0.25\mathrm{nm}$ wide, the values written in the scene. The top of the barrier is 1 eV, $1.60\times 10^{-19}\mathrm{J}$, above the energy, so
$\kappa =\frac{\sqrt{2\cdot 9.11\times 10^{-31}\cdot 1.60\times 10^{-19}}}{1.055\times 10^{-34}}\approx 5.12\times 10^9{\mathrm{m}}^{-1}=5.12\mathrm{n}{\mathrm{m}}^{-1}$
and the decay length $1/\kappa =0.195\mathrm{nm}$ is a little shorter than the barrier.

Then $2\kappa a=2\cdot 5.12\cdot 0.25=2.56$ and $e^{-2.56}\approx 0.077$. The prefactor is $16\cdot 0.75\cdot 0.25=3$, so
$T\approx 3\cdot 0.077\approx 0.23$
The exact formula of the advanced level gives 0.214, and the moving packet, which holds a small spread of energies, 0.218. That is where the readouts settle: $P(\text{right})=0.22$ and $P(\text{left})=0.78$.

Outside the barrier $\lambda =h/\sqrt{2mE}=0.708\mathrm{nm}$, and the transmitted wave has the same wavelength but only about $\sqrt{0.22}\approx 0.47$ of the incoming amplitude. In front of the barrier the incoming and reflected waves overlap in tall fringes $\lambda /2=0.354\mathrm{nm}$ apart. About 4 s in, the readouts show 0.80 on the left, 0.10 in the barrier and 0.10 on the right: for a moment a tenth of the probability sits inside the wall.

Worked example: a scanning tunnelling microscope

A sharp metal tip is held a fraction of a nanometre above a conducting surface. For electrons near the top of the filled levels the gap is a barrier whose top lies about one work function above them, here $V_0-E=4.5\mathrm{eV}$, and the current is proportional to the transmission, $I\propto e^{-2\kappa d}$, with $d$ the width of the gap. By what factor does it change when the tip moves 0.1 nm closer?

Since $\kappa$ grows as the square root of the deficit, $\kappa =5.12\cdot \sqrt{4.5}\approx 10.9\mathrm{n}{\mathrm{m}}^{-1}$, and the prefactor cancels in a ratio: $I_2/I_1=e^{2\kappa \Delta d}=e^{2\cdot 10.9\cdot 0.1}=e^{2.17}\approx 8.8$. With the rounder $\kappa \approx 10\mathrm{n}{\mathrm{m}}^{-1}$ it is $e^2\approx 7$: a shift of less than half an atom's diameter multiplies the current several times.

A bump only 0.01 nm high changes the current by $e^{0.217}\approx 1.24$, a 24 % rise, while an atomic step 0.2 nm high changes it about 77 times. A feedback loop moves the tip to keep the current constant as it scans, and the tip's height then traces the surface atom by atom.

Trapped waves: energy levels in a box

Now trap the electron in a box of width $L$ with $V=0$ inside and walls it cannot pass. The wavefunction must vanish at both walls, like a guitar string fixed at both ends, so only whole numbers of half-wavelengths fit, $L=n\lambda /2$. Then $p=h/\lambda =nh/(2L)$, and the kinetic energy $p^2/2m$ gives the allowed energies
$E_n=\frac{n^2{h}^2}{8m{L}^2}$
with $n=1,2,3,\dots$ counting the half-waves; $n=0$ would mean no wave at all, so the lowest energy is never zero.

The levels grow as $n^2$, in the ratio 1 : 4 : 9, as in the box of the animation, labelled $E_1$, $4E_1$ and $9E_1$. Each standing wave keeps a fixed shading $|{\psi }_n{|}^2$: a state of definite energy is stationary. For $L=1\mathrm{nm}$, $E_1=(6.626\times 10^{-34}{)}^2/(8\cdot 9.11\times 10^{-31}\cdot 10^{-18})\approx 6.0\times 10^{-20}\mathrm{J}=0.38\mathrm{eV}$, so $E_2=1.50\mathrm{eV}$ and $E_3=3.38\mathrm{eV}$. A jump from $n=3$ to $n=2$ releases $5E_1=1.88\mathrm{eV}$, a red photon of about 660 nm. A smaller box spaces the levels further apart, as $1/L^2$.

Common mistakes

• The electron does not lose energy in the barrier. Beyond it the wavelength is still 0.708 nm, so a 3 eV electron comes out with 3 eV; only the amplitude is smaller.
• The electron does not split. $P(\text{right})=0.22$ means that about 22 of every 100 electrons are found beyond the barrier, each one whole.
• Transmission is not inversely proportional to width: from 0.25 nm to 0.5 nm, $T$ falls from 0.214 to 0.018, about twelve times, because doubling the width squares the exponential factor.
• Mass matters as much as energy. A proton with the same 3 eV at the same barrier has a $\kappa$ larger by $\sqrt{1836}\approx 43$, so $2\kappa a\approx 110$ and $T\approx 6\times 10^{-48}$.

Where you meet it

Flash memory stores each bit as charge on an isolated gate, written and erased by driving electrons through an insulating layer a few nanometres thick. Tunnel diodes, scanning tunnelling microscopes and the escape of alpha particles from nuclei follow the rule of this lesson, and in the Sun's core protons fuse only thanks to tunnelling.

The same wave rules govern light in Reflection and Refraction, where a totally reflected beam can cross a thin gap. Nuclei tunnel in Radioactivity and Half-Life and in Fission, Fusion and E = mc²; Semiconductors, Diodes and LEDs builds energy bands from electron waves; and Atomic Models: the Bohr Atom has relatives of the box levels.

Advanced

One frequency, many wavelengths

Tunnelling is a wave effect, so start with the wave of a particle that has a definite energy and ask how it changes as the potential energy changes along its path. Two rules settle almost everything that follows: the frequency is the same everywhere, and the wavelength adjusts to the local kinetic energy. Where the kinetic energy would be negative, the wave does not stop dead; it fades.

Legend

$E$	Energy of the state: one value for the whole wave.	$\mathrm{J}$
$V(x)$	Potential energy at position $x$, drawn in grey: $0$, $\frac{3}{4}E$ and $\frac{5}{4}E$.	$\mathrm{J}$
$\psi$	Wave function: its real part at one instant.	${\mathrm{m}}^{-1/2}$
$x$	Position along the line of motion.	$\mathrm{m}$
${\lambda }_1$	Wavelength on the lowest level, where $E-V$ is largest.	$\mathrm{m}$
${\lambda }_2$	Wavelength on the middle level, where $E-V$ is a quarter as large: ${\lambda }_2=2{\lambda }_1$.	$\mathrm{m}$
${\kappa }^{-1}$	Decay length where $V>E$: over it $\psi$ falls to $1/e$ of its value at the edge.	$\mathrm{m}$

Figure 1 shows one state of energy $E$ (the purple line) crossing three flat levels of the potential energy $V(x)$ (grey). Underneath, the blue curve is the wave function $\psi$ at one instant. On the lowest level $V=0$, and the wave oscillates with wavelength ${\lambda }_1$. On the middle level $V=\frac{3}{4}E$, so the kinetic energy $E-V$ is only a quarter of what it was: the wave stretches to ${\lambda }_2=2{\lambda }_1$, and its crests grow taller. On the highest level $V=\frac{5}{4}E$ lies above $E$. There the wave stops oscillating and dies away, falling to $1/e$ of its value at the edge over the short length ${\kappa }^{-1}$ marked at the right. Look also at the two steps: the curve is continuous there and has no kink.

Why the frequency is shared. When $V$ does not depend on time, a state of definite energy has the form $\psi (x)e^{-i\omega t}$: a fixed shape in space times a single oscillating factor, with the same angular frequency $\omega =E/\hslash$ at every $x$. Nothing in the potential changes in time, so nothing can drive one part of the wave faster than another. Light entering glass behaves in the same way: it keeps its frequency and changes only its wavelength. One frequency for the whole wave means one energy for the whole motion. This is energy conservation, in the language of waves.

Why the wavelength changes. The fixed energy is shared differently between kinetic and potential energy from place to place. Writing $\hslash \omega =p^2/2m+V(x)$ and solving for the momentum gives the local momentum $p(x)=\sqrt{2m(E-V(x))}$, and with it the local wavelength $h/p(x)$. Where $V$ dips, $p$ is larger and the waves crowd together; where $V$ rises, they stretch. On the middle level of Figure 1 the kinetic energy is a quarter as large, so $p$ is halved and the wavelength doubles from ${\lambda }_1$ to ${\lambda }_2$. The crests are taller there because a slower particle spends more time on each stretch of its path, so it is more likely to be found there.

What happens where $V>E$. A classical particle can never be there, since its kinetic energy $E-V$ would be negative: this is the classically forbidden region. In the formula, the square root of a negative number is imaginary, $p=i\hslash \kappa$, with the real decay constant
$\kappa =\frac{\sqrt{2m(V-E)}}{\hslash }$
Put into the plane wave $e^{ipx/\hslash }$, this turns it into the real exponential $e^{-\kappa x}$. The equation also allows the growing solution $e^{+\kappa x}$, but in a forbidden region that goes on without end it would make $\psi$ grow without limit, so it is ruled out. The amplitude therefore does enter the forbidden region, but it decays there, by a factor $e$ over every length $1/\kappa$.

Comparing the top and middle levels shows how short this decay length is. In Figure 1, $V-E$ on the top level equals $E-V$ on the middle level, so $\kappa$ equals the middle level's wavenumber $2\pi /{\lambda }_2$ and the decay length is ${\kappa }^{-1}={\lambda }_2/2\pi$: the wave is gone within about a sixth of a wavelength.

Light fades in the same way, and the effect can be seen on a laboratory bench. A ray travels in glass of refractive index $n$, at an angle $\theta$ to the normal of a surface with air beyond it. The wave pattern along the surface must match on both sides, so the component of the wavevector along the surface, $k_y=n\omega \sin \theta /c$, is the same in the air. In air, however, the whole wavevector is only $\omega /c$ long. Beyond the critical angle, where $n\sin \theta >1$, the component along the surface is already longer than that on its own, so the square of the component across the gap is negative:
$k_x^2=(\omega /c{)}^2-k_y^2=(\omega /c{)}^2(1-n^2{\sin }^2\theta )$
Then $k_x=i\kappa$ is imaginary, and the field in the air decays as $e^{-\kappa x}$. For $n=1.5$ at 60°, $\kappa {\lambda }_0=2\pi \sqrt{n^2{\sin }^2\theta -1}\approx 5.2$, where ${\lambda }_0$ is the vacuum wavelength. Green light of 550 nm therefore decays over about 0.1 µm. Bring a second glass block that close, and part of the light crosses the gap into it. This is frustrated total internal reflection: barrier penetration for photons.

Matching at the edges: the exact transmission

Now make the forbidden region finite: a barrier with allowed ground on both sides. The decaying wave reaches the far side with a reduced but non-zero amplitude, and there it starts a travelling wave again. To find how much gets through, solve the wave equation in each region and join the pieces at the edges.

Legend

$V_0$	Height of the barrier.	$\mathrm{J}$
$E$	Energy of the wave, below the top of the barrier.	$\mathrm{J}$
$a$	Width of the barrier.	$\mathrm{m}$
$\psi$	Wave function: its real part at one instant.	${\mathrm{m}}^{-1/2}$
$x$	Position along the line of motion.	$\mathrm{m}$
$r$	Amplitude of the reflected wave, for an incident wave of amplitude 1.
$t$	Amplitude of the transmitted wave behind the barrier, on the same scale.

Figure 2 shows a wave of unit amplitude arriving from the left at a barrier of height $V_0$ and width $a$. It is drawn for $E=\frac{3}{4}{V}_0$ and $\kappa a=1.6$, a barrier just under half a wavelength thick. In front of the barrier the incident wave (arrow 1) and the reflected wave (arrow $r$) overlap and form a nearly standing wave. Inside, the curve bends away from the axis and sinks. Behind the barrier a single wave of amplitude $t$ runs on to the right, and nothing comes back from there. The arrow lengths are the amplitudes: here $|r|=0.94$ and $|t|=0.34$. The two dots mark the edges, where $\psi$ and its slope ${\psi }^{\prime }$ must have the same values on both sides.

Why must they agree? Rearrange the Schrödinger equation into a statement about curvature:
$\frac{d^2\psi }{d{x}^2}=\frac{2m}{{\hslash }^2}(V-E)\psi$
Where $V<E$, the curvature has the opposite sign to $\psi$, so the curve always bends back towards the axis and oscillates. Where $V>E$, it has the same sign, so the curve bends away from the axis, like $e^{\pm \kappa x}$. At an edge $V$ jumps by a finite amount, so the curvature jumps too, but it stays finite. A finite curvature acting over a vanishing distance cannot change the slope, so ${\psi }^{\prime }$ is continuous; and a curve whose slope is finite cannot jump, so $\psi$ is continuous as well. That is why the curve in Figure 2 passes through the dots without a kink.

Now the calculation. Put the barrier on $0<x<a$, with $V=0$ outside it. Outside, the wavenumber is $k=\sqrt{2mE}/\hslash$; inside, the decay constant is $\kappa =\sqrt{2m(V_0-E)}/\hslash$. On the left, $\psi =e^{ikx}+r{e}^{-ikx}$: the incident wave of unit amplitude plus the reflected wave. Inside, $\psi =C{e}^{\kappa x}+D{e}^{-\kappa x}$; the growing term is allowed this time, because the barrier ends before it can grow without limit. On the right, $\psi =t{e}^{ikx}$ alone, because nothing comes in from that side.

That makes four unknowns, $r$, $C$, $D$ and $t$, and the edges supply four conditions: $\psi$ and ${\psi }^{\prime }$ continuous at $x=0$ and at $x=a$. These are four linear equations, and eliminating $C$ and $D$ leaves
$t=\frac{e^{-ika}}{\cosh \kappa a+i\frac{{\kappa }^2-k^2}{2k\kappa }\sinh \kappa a}$
The transmission probability is its squared size, one over the squared size of the denominator:
$|t{|}^2={\left[{\cosh }^2\kappa a+{\left(\frac{{\kappa }^2-k^2}{2k\kappa }\right)}^2{\sinh }^2\kappa a\right]}^{-1}$
Replace ${\cosh }^2$ by $1+{\sinh }^2$. The two ${\sinh }^2$ terms then combine, with the coefficient $1+({\kappa }^2-k^2{)}^2/(2k\kappa {)}^2=({\kappa }^2+k^2{)}^2/(2k\kappa {)}^2$. Since ${\kappa }^2+k^2=2m{V}_0/{\hslash }^2$, this coefficient 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 is $|t{|}^2$ the fraction that gets through? The probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ measures how much probability flows past a point per second. In a stationary state probability piles up nowhere, so $j$ is the same at every $x$. On the left it is $(\hslash k/m)(1-|r{|}^2)$, the incident flow minus the reflected flow; on the right it is $(\hslash k/m)|t{|}^2$. Setting them equal gives $|r{|}^2+|t{|}^2=1$: every particle is either reflected or transmitted. The speed $\hslash k/m$ is the same on both sides, so $T=|t{|}^2$ is the transmitted fraction.

The arrows and the formula both give a numerical check of this result. In Figure 2 the arrows give $|r{|}^2=0.88$ and $|t{|}^2=0.12$, which add up to 1, and the 12 % that gets through is $T$. The formula agrees: with $E=\frac{3}{4}{V}_0$ the factor $V_0^2/(4E(V_0-E))$ is $4/3$, $\sinh 1.6=2.38$, and $T=1/(1+\frac{4}{3}\cdot 5.65)=0.12$.

For a thick barrier, $\kappa a\gg 1$, two simplifications apply. First, $\sinh \kappa a=\frac{1}{2}(e^{\kappa a}-e^{-\kappa a})\approx \frac{1}{2}{e}^{\kappa a}$, because the second exponential is tiny. Second, the ${\sinh }^2$ term is then huge, so the 1 in the bracket is negligible and the bracket can be inverted directly:
$T\approx 16\left(\frac{E}{V_0}\right)\left(1-\frac{E}{V_0}\right)e^{-2\kappa a}$
The prefactor is never more than 4, its value at $E=\frac{1}{2}{V}_0$, so the exponential decides: each extra decay length $1/\kappa$ of width divides $T$ by $e^2\approx 7.4$.

How transmission falls with width

The exact formula is hard to read at a glance. A graph of $T$ against the width shows its two regimes: a gentle start for thin barriers and a pure exponential for thick ones.

Legend

$\ln T$	Natural logarithm of the transmission probability; 0 means $T=1$.
$a$	Width of the barrier.	$\mathrm{m}$
$T$	Exact transmission probability for a barrier of width $a$.
$e^{-2\kappa a}$	Thick-barrier law, here $T\approx 3e^{-2\kappa a}$: the dashed straight line.
$\kappa a$	Width in decay lengths $1/\kappa$; the dashed marker at $\kappa a=2$ is where the straight line takes over.
$a_0$	Width of the animation's barrier, 0.25 nm.	$\mathrm{m}$
$\Delta a$	A step in the barrier width.	$\mathrm{m}$
$\Delta (\ln T)$	Change of $\ln T$ over that step, $-2\kappa \Delta a$: $T$ is multiplied by $e^{-2\kappa \Delta a}$.

Figure 3 plots the exact $T(a)$ for $E=3\mathrm{eV}$ and $V_0=4\mathrm{eV}$, up to $a=0.6\mathrm{nm}$; the Advanced layer of the animation shows the same curve. Here it is drawn on a logarithmic scale, as $\ln T$ against $a$, because on that scale an exponential is a straight line. For an electron with $V_0-E=1\mathrm{eV}$, $\kappa =5.12\mathrm{n}{\mathrm{m}}^{-1}$, so one decay length is about 0.2 nm.

Follow the solid curve $T$ from the left. It starts at $T=1$, that is $\ln T=0$, since no barrier means no reflection. It leaves that level gently: for small $a$, $\sinh \kappa a\approx \kappa a$, so $T\approx [1+1.33(\kappa a{)}^2{]}^{-1}$, where $1.33=V_0^2/(4E(V_0-E))$. The exact values are still 0.92 at 0.05 nm and 0.72 at 0.1 nm. Beyond the dashed marker at $\kappa a=2$ the curve is practically a straight line. The triangle reads off its slope: a step $\Delta a$ lowers $\ln T$ by $\Delta (\ln T)=-2\kappa \Delta a$, so each further 0.1 nm multiplies $T$ by $e^{-1.02}\approx 0.36$. The ring marks the animation's barrier, $a_0=0.25\mathrm{nm}$, where $T=0.214$.

The dashed line is the thick-barrier law. For these energies $16(E/V_0)(1-E/V_0)=16\cdot \frac{3}{4}\cdot \frac{1}{4}=3$, so $T\approx 3e^{-2\kappa a}$: a straight line that starts at $\ln 3\approx 1.1$, and that the exact curve joins. For thin barriers the law is poor. At $a=0$ it gives 3, more than any probability can be, and at 0.25 nm it gives 0.232, 8 % too high. From 0.4 nm on it agrees within 2 %.

The animation sends a wave packet, not a wave of a single energy: its energies are spread by 0.34 eV around 3 eV. Because $T$ rises steeply with $E$ (a higher energy means a smaller $\kappa$), the faster components get through more easily. The packet's transmission, 0.218, is therefore slightly above the 0.214 of a 3 eV wave, and the transmitted electrons average 3.08 eV against 2.99 eV for the reflected ones.

What if the energy is above the barrier, $E>V_0$? Classically every particle would pass. In the formula $V_0-E$ is now negative, so $\kappa$ becomes imaginary: $\kappa =iq$, where $q=\sqrt{2m(E-V_0)}/\hslash$ is the real wavenumber over the barrier. Then $\sinh \kappa a$ becomes $i\sin qa$, and both ${\sinh }^2\kappa a$ and $V_0-E$ change sign, so
$T={\left[1+\frac{V_0^2{\sin }^2(qa)}{4E(E-V_0)}\right]}^{-1}$
Now $T$ oscillates with the width instead of falling. It reaches 1 only at the resonances $qa=n\pi$, where the barrier holds a whole number of half-wavelengths and the waves reflected at its two edges cancel. At every other width some of the wave is reflected, even though the particle has energy to spare. For a 3 eV electron over a barrier 2 eV high, $q=5.12\mathrm{n}{\mathrm{m}}^{-1}$: as the width grows, $T$ dips to 0.75 and returns to 1 at $a=0.61\mathrm{nm}$, 1.23 nm and so on.

Smooth barriers and alpha decay

Real barriers are rarely rectangular: their height changes gradually along the path. The thick-barrier law extends to them slice by slice, and its best-known use explains why some nuclei live for billions of years and others for less than a microsecond.

Legend

$V(r)$	Potential energy of the alpha particle at distance $r$ from the centre: a well inside the nucleus, electric repulsion outside.	$\mathrm{J}$
$Q$	Energy released in the decay: the alpha particle's energy, the $E$ of the formulas.	$\mathrm{J}$
$\alpha$	The alpha particle, trapped in the well.
$r$	Distance from the centre of the nucleus.	$\mathrm{m}$
$r_1$	Nuclear surface, where the barrier begins.	$\mathrm{m}$
$r_2$	Outer turning point, where $V=Q$ again and the barrier ends.	$\mathrm{m}$
$\kappa (r)$	Decay constant $\sqrt{2m(V-Q)}/\hslash$ at distance $r$; it falls to zero at $r_2$.	${\mathrm{m}}^{-1}$
$dr$	Width of one thin slice of the barrier; the slice multiplies the amplitude by $e^{-\kappa dr}$.	$\mathrm{m}$
$\int \kappa dr$	Area under $\kappa (r)$ from $r_1$ to $r_2$: the exponent in $T\approx e^{-2\int \kappa dr}$.

Figure 4 shows the barrier that holds an alpha particle inside a uranium-238 nucleus. The coordinate along the motion is now the distance $r$ from the centre, in the role of $x$. The upper panel shows the potential energy $V(r)$ and the energy $Q$ released in the decay, which plays the part of $E$. Inside the nuclear surface $r_1$ the strong force makes a deep well (only sketched), and the alpha particle $\alpha$ is trapped in it. Just outside $r_1$ the electric repulsion puts $V$ far above $Q$. It falls off as $1/r$ and meets $Q$ again at the outer turning point $r_2$, where the barrier ends. The lower panel plots the decay constant $\kappa (r)=\sqrt{2m(V(r)-Q)}/\hslash$ over the same range: it is largest at $r_1$ and falls to zero at $r_2$.

To handle a barrier whose height varies, cut it into thin slices of width $dr$, like the gold strip. If $\kappa$ changes little over one decay length, that is $|d\kappa /dr|\ll {\kappa }^2$, each slice behaves like a thin piece of uniform barrier and multiplies the amplitude by $e^{-\kappa dr}$. Multiplying the factors of all the slices adds their exponents, so the amplitude falls by $e^{-\int \kappa dr}$, where $\int \kappa dr$ is the shaded area under the curve. The probability is the square of the amplitude, which doubles the exponent:
$T\approx \exp \left(-2{\int }_{r_1}^{r_2}\kappa (r)dr\right)$
This is the WKB estimate, named after Wentzel, Kramers and Brillouin. It drops a prefactor of order 1, like the 3 of the rectangular barrier, because the exponent decides. For a rectangular barrier the area is simply $\kappa a$, and the estimate gives back $e^{-2\kappa a}$.

George Gamow used this in 1928 to explain alpha decay. Beyond the reach of the strong force, the daughter nucleus, of charge $Ze$, repels the alpha particle, of charge $2e$, with
$V=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}r}$
the curve of Figure 4. For uranium-238 the daughter is thorium, $Z=90$. The barrier peaks near 28 MeV at the nuclear surface, about 9 fm from the centre, while the energy released is only $Q=4.27\mathrm{MeV}$. The alpha particle must therefore tunnel out to where $V$ has fallen to $Q$, about 61 fm from the centre. The integral gives $2\int \kappa dr\approx 88$, so $T\approx e^{-88}\approx 6\times 10^{-39}$.

Turning this into a lifetime uses the same idea as the worked problem below. Inside the nucleus the alpha particle strikes the wall roughly $10^{21}$ times per second, with the chance $T$ of escaping at each strike. The escape rate is about $10^{21}\cdot 6\times 10^{-39}=6\times 10^{-18}$ per second, so it escapes after some $10^{17}$ s: a few billion years. That is the right order for the measured half-life of 4.5 billion years, a real success for an estimate whose exponent is 88.

The exponent depends steeply on energy. Polonium-212 releases 8.95 MeV. Its outer turning point moves in to 26 fm, the barrier becomes thinner and lower, the exponent falls to about 33, and the half-life is only 0.3 µs. A drop of 55 in the exponent is a factor $e^{55}\approx 10^{24}$: roughly doubling the energy, from 4.3 to 9 MeV, shortens the half-life about $10^{24}$ times. This steep trend had already been measured, as the empirical Geiger–Nuttall rule; tunnelling explains it.

Bound states and zero-point energy

The same matching conditions do more than count transmitted particles. Applied to a particle trapped between walls, they produce its energy levels.

Start with a box whose walls are infinitely high. Then $\kappa$ is infinite in the walls, the wave cannot enter them at all, and $\psi$ must vanish at $x=0$ and $x=L$. Inside, where $V=0$, the solutions are $\sin kx$ and $\cos kx$. The condition at $x=0$ keeps only $\sin kx$, and the condition at $x=L$ requires $kL=n\pi$, with $n$ a positive whole number: a whole number of half-wavelengths must fit in the box. This leaves
${\psi }_n=\sqrt{2/L}\sin (n\pi x/L)$
where the factor $\sqrt{2/L}$ makes the total probability 1. The energy is the kinetic energy ${\hslash }^2{k}_n^2/2m$ with $k_n=n\pi /L$:
$E_n=\frac{n^2{h}^2}{8m{L}^2}$

In a well of finite depth the walls are not perfectly hard. Inside, the solution oscillates; outside, it must decay on both sides; and at each wall the two must join with the same value and slope. For most energies the inside wave reaches a wall with the wrong slope, and the outside piece would have to grow without limit. Only special energies let both joins work at once: quantisation is a boundary condition, not an extra postulate. The tails also let the wave spread a little beyond the walls, as if the box were slightly wider, so the wavelength is longer and the energy lower. For a well 1 nm wide and 4 eV deep the lowest level is 0.26 eV, not the 0.38 eV of the infinite box.

The lowest level is never zero. Taking $n=0$ would give $\psi =0$ everywhere, which is no particle at all. The uncertainty principle says the same. In the ground state the position spread is $\Delta x=0.18L$, and the momentum, of size $\pi \hslash /L$ in either direction, has the spread $\Delta p=\pi \hslash /L$. Their product is $\Delta x\Delta p\approx 0.57\hslash$, just above the minimum $\hslash /2$ that the principle allows. Confining a particle to a region of size $L$ therefore costs kinetic energy of order ${\hslash }^2/(m{L}^2)$, the zero-point energy. It keeps atoms from collapsing: pulling an electron closer to the nucleus lowers its potential energy but raises this kinetic cost.

Worked problem: a leaky trap

Place two of the animation's barriers, each 4 eV high and 0.25 nm thick, a distance $L=2\mathrm{nm}$ apart, with an electron of about 3 eV between them. How long does the electron stay trapped, and how does that depend on the thickness of the walls?

Treat the electron as a particle bouncing to and fro at the speed $v=\sqrt{2E/m}=1.03\times 10^{6}m/s$, which is 1.03 nm per femtosecond, with a chance $T$ of escaping each time it hits a wall. Between hits it crosses the gap once, so it hits a wall every $L/v=1.95\mathrm{fs}$. The chance of escape per unit time, the escape rate, is therefore $\Gamma =Tv/L$. A constant chance per unit time means that the number still inside decays as $e^{-\Gamma t}$, with mean lifetime $1/\Gamma$:
$\tau =\frac{L}{vT}=\frac{1.95\mathrm{fs}}{T}$
With the animation's 0.25 nm walls, $T=0.214$ and $\tau \approx 9\mathrm{fs}$. With walls 1 nm thick, $T=1.06\times 10^{-4}$ and $\tau \approx 18\mathrm{ps}$. With 2 nm walls, $T=3.8\times 10^{-9}$ and $\tau \approx 0.5\mu \mathrm{s}$.

Notice the pattern. By the thick-barrier law, each extra nanometre of wall divides $T$, and so multiplies the lifetime, by $e^{2\kappa \cdot 1\mathrm{nm}}=e^{10.25}$, about 28,000: exactly the step from 18 ps to 0.5 µs.

A proton in the same trap would stay trapped far longer. Its mass is 1836 times larger, so at the same energy it moves 43 times more slowly and faces a $\kappa$ 43 times larger, since both scale with $\sqrt{m}$. Even with 0.25 nm walls, $T$ falls to about $6\times 10^{-48}$ and $\tau$ grows to some $10^{34}$ s, far longer than the age of the Universe. The bouncing picture is crude for $T=0.214$, where the electron is gone within a few bounces, and accurate when $T$ is small, as for alpha decay.

Limits of validity

Everything here is non-relativistic. A 3 eV electron moves at 0.34 % of the speed of light, and its energy is $6\times 10^{-6}$ of its rest energy, 511 keV, so the Schrödinger equation describes it excellently. The rectangular barrier and the infinite box are idealisations. In a tunnel junction, for instance, the electron induces charge in the nearby metal and is attracted to it, and this attraction rounds and lowers the barrier. That raises $T$ but keeps the exponential dependence on the width.

The model is also one-dimensional. In three dimensions only the motion across the barrier counts, so an electron arriving at an angle, with less of its energy in that direction, tunnels less. Inside a solid the electron responds to forces as if it had a different mass, its effective mass, which replaces $m$. In gallium arsenide it is 0.067 of the free mass, which cuts $\kappa$ about fourfold and lets electrons tunnel through layers a few nanometres thick in resonant-tunnelling diodes. Finally, the potential was taken as fixed and the electron as alone. In very small junctions one electron's charge changes the potential enough to block the next one.

Where this leads

Wave–Particle Duality introduced the amplitude and its probabilities; this lesson gave it an equation. Radioactivity and Half-Life shows the exponential decay law that a leaky trap produces, and Fission, Fusion and E = mc² the tunnelling that lets nuclei fuse in stars. Semiconductors, Diodes and LEDs applies the same matching conditions to a periodic potential, where they give energy bands.

Reflection and Refraction treats total internal reflection, whose frustrated form is the optical twin of tunnelling. The orbits of Atomic Models: the Bohr Atom give way to the three-dimensional Schrödinger equation, whose solution for hydrogen yields the same $-13.6\mathrm{eV}/n^2$ levels and the shapes of the orbitals, and Particles and Forces: the Standard Model makes the same wave mechanics relativistic.