Back to the on-screen lesson ·
Slowly varying potentials: local wavelengths, tunneling through barriers of any shape with $e^{-2\int\kappa\,dx}$, and quantization by $\int p\,dx = (n + \tfrac{1}{2})\pi\hbar$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to estimate tunneling probabilities through barriers of any shape, find approximate energy levels from the WKB quantization rule, and judge where the approximation can be trusted.
You know that in a region of constant potential the wave function is $e^{\pm ikx}$ where $E > V$ and $e^{\pm\kappa x}$ where $E < V$, and that a rectangular barrier transmits about $e^{-2\kappa L}$. You also know that the Bohr model quantized angular momentum by fiat. This lesson generalizes both to potentials of any shape, provided they change slowly.
| Term | What it means |
|---|---|
| WKB approximation | Named for Wentzel, Kramers and Brillouin: treat the potential as locally constant. |
| Local momentum | $p(x) = \sqrt{2m(E - V(x))}$, the classical momentum at each point. |
| Classical turning point | Where $E = V(x)$, so $p = 0$: the edge of the allowed region. |
| Connection formulas | Rules that join WKB solutions across a turning point, where WKB itself fails. |
| Bohr–Sommerfeld quantization | $\int p\,dx = (n + \tfrac{1}{2})\pi\hbar$ between two smooth turning points. |
| Gamow factor | $\gamma = \tfrac{1}{\hbar}\int\sqrt{2m(V - E)}\,dx$ through a barrier; tunneling probability $e^{-2\gamma}$. |
| Field emission | Electrons tunneling out of a metal through a barrier tilted by a strong electric field. |
If $V$ were constant, the wave function would be $e^{\pm ipx/\hbar}$. If $V$ changes slowly, it is natural to let the momentum depend on position, $p(x) = \sqrt{2m(E - V(x))}$, and accumulate the phase:
$$\psi(x) \approx \frac{C}{\sqrt{p(x)}}\exp\left(\pm\frac{i}{\hbar}\int^x p(x')\,dx'\right).$$
The amplitude factor $1/\sqrt{p}$ has a classical meaning: the particle spends more time where it moves slowly, so it is more likely to be found there. In a forbidden region, $p$ becomes imaginary and the solution becomes growing and decaying exponentials with $\kappa(x) = \sqrt{2m(V - E)}/\hbar$.
Two powerful results follow. For tunneling through a barrier of any shape,
$$T \approx \exp\left(-2\int_{x_1}^{x_2}\kappa(x)\,dx\right),$$
the rectangular result with $\kappa L$ replaced by an integral. For bound states in a well with two smooth turning points, the wave must fit with the right phase:
$$\int_{x_1}^{x_2}p(x)\,dx = \left(n + \tfrac{1}{2}\right)\pi\hbar.$$
The approximation holds when the potential changes little over a wavelength, $|d\lambda/dx| \ll 1$. It fails at the turning points, where $p \to 0$, and connection formulas derived from the exact solution of a linear potential patch it there; they produce the $\tfrac{1}{2}$ in the quantization rule.
Another way: picture
Picture a wave traveling over a gently sloping seabed. Where the water is deep the wave moves fast and its crests are far apart; where it is shallow they bunch up and the wave grows taller. The wave never reflects because nothing changes abruptly. WKB describes quantum waves in the same way: the local wavelength follows the local kinetic energy, and the amplitude adjusts to keep the flow of probability constant.
Another way: steps
Checks. For a box with two hard walls, WKB gives $E_n = n^2\pi^2\hbar^2/2ma^2$, exact. For the oscillator it gives $(n + \tfrac{1}{2})\hbar\omega$, also exact. For a rectangular barrier it reproduces $e^{-2\kappa L}$. A tunneling exponent must increase with the barrier's width, its height above $E$ and the particle's mass.
Divide a barrier of any shape into thin slices, each nearly rectangular with its own $\kappa(x)$. Through each slice of width $dx$ the amplitude falls by $e^{-\kappa(x)\,dx}$, so through the whole barrier by $\exp(-\int\kappa\,dx)$, and the probability by the square of that. The approximation ignores reflections at each slice, which is justified when $\kappa$ changes slowly.
The integral makes the barrier's shape matter. A triangular barrier, like the one a strong field creates at a metal surface, has an exponent two-thirds of what a rectangle of the same height and width would give, because it narrows toward the top. A Coulomb barrier, tall near the nucleus but with a long tail, gives an exponent dominated by its thickness at the particle's energy. In each case the exponential makes the result extraordinarily sensitive to the parameters, which is what explains the enormous range of alpha-decay half-lives.
In 1928 George Gamow, and independently Ronald Gurney and Edward Condon, explained alpha decay as tunneling. Inside the nucleus an alpha particle is held by the strong force; outside, it is repelled by the daughter's charge. The Coulomb barrier at the nuclear surface is about $30$ to $40$ MeV high, while the alpha's energy is only $4$ to $9$ MeV, so classically it could never escape.
WKB gives the tunneling probability $e^{-2\gamma}$ with $\gamma \approx 1.980Z/\sqrt{E} - 1.485\sqrt{Zr_1}$, with $E$ in MeV and $r_1$ in fm. The alpha rattles inside the nucleus about $10^{21}$ times per second, and the half-life is roughly its inverse times $e^{2\gamma}$. Because $\gamma$ depends on $1/\sqrt{E}$ inside an exponential, doubling the energy from $4.3$ MeV for uranium-238 to $9.0$ MeV for polonium-212 changes $2\gamma$ from $100$ to $41$, and the half-life from $4.5$ billion years to $0.3$ microseconds — the Geiger–Nuttall law, explained.
A metal's electrons are held in by a barrier one work function high, about $4$ to $5$ eV. A strong electric field outside tilts the potential, turning the step into a triangle of width $L = \phi/eE$. At fields of a few volts per nanometer — billions of volts per meter — $L$ is about a nanometer and electrons tunnel out, even at room temperature. The WKB exponent is $6.83\,\phi^{3/2}/E$, with $\phi$ in eV and $E$ in V/nm; Ralph Fowler and Lothar Nordheim derived the current in 1928.
Such fields are reached at the tips of very sharp needles, where field lines concentrate. Cold field-emission guns in electron microscopes use a tungsten tip a few tens of nanometers across and produce the brightest, most coherent electron beams available. Field emission is also a hazard: it limits how high a voltage vacuum equipment, particle-accelerator cavities and high-voltage switches can hold before sharp spots on their surfaces begin to leak current.
The old quantum theory of Bohr and Sommerfeld demanded $\oint p\,dx = nh$ around a classical orbit. WKB derives the correct version, $\oint p\,dx = (n + \tfrac{1}{2})h$ for two soft turning points, and explains the extra half as the phase the wave loses when it reflects from each turning point. The old rule missed the half, which is why Bohr's oscillator lacked zero-point energy.
The quantization rule has a vivid meaning: each quantum state occupies an area $h$ in phase space, the plane of $x$ and $p$. The $n$-th state's classical orbit encloses $(n + \tfrac{1}{2})h$. Counting states by phase-space area is how statistical mechanics counts quantum states in a box, how the density of states of a gas is found, and how semiclassical methods estimate energy levels of complicated systems such as highly excited molecules.
WKB is a short-wavelength approximation, formally an expansion in powers of $\hbar$. It improves as the quantum number increases, because the wavelength becomes short compared with the distance over which the potential changes. For the lowest states it can be off by several percent, except in lucky cases like the oscillator where the errors cancel exactly.
For the bouncing neutron, WKB puts the ground state at $13.62$ μm; the exact solution, an Airy function, gives $13.72$ μm, less than one percent higher, and the agreement improves for higher levels. For tunneling, WKB gets the exponent right but not the prefactor, which is fine for estimates since the exponent dominates. Near the top of a barrier, where the turning points merge, WKB fails and a parabolic-barrier formula takes over.
Geiger and Nuttall noticed in 1911 that alpha emitters with higher energies have enormously shorter half-lives, but no one could explain why until tunneling. With WKB, the explanation is the Gamow factor: the barrier's thickness at the alpha's energy shrinks as the energy rises, and the tunneling probability, exponential in that thickness, rises steeply. Alpha energies from $4$ to $9$ MeV span half-lives from $10^{10}$ years to $10^{-7}$ seconds.
This matters in practice. Americium-241, with a $5.49$ MeV alpha and a $432$-year half-life, powers household smoke detectors: long-lived enough to last, active enough to ionize the air in the detector's chamber. Plutonium-238, with a $5.59$ MeV alpha and an $88$-year half-life, heats the radioisotope generators that power NASA's Curiosity and Perseverance rovers and the Voyager probes, still running after nearly fifty years. Its half-life, set by tunneling, is what makes it suitable.
In 2002, a team at the Institut Laue–Langevin in Grenoble observed that neutrons slow enough to be held by gravity above a horizontal mirror cannot bounce at any height: they occupy discrete quantum states. A neutron's potential is $mgz$ above the mirror and a hard wall at it, and WKB gives the turning heights $z_n = 5.868\,[\tfrac{3\pi}{2}(n - \tfrac{1}{4})]^{2/3}$ μm: $13.6$, $24.0$, $32.4$ μm, and so on. The experiment detected neutrons passing through a slit above the mirror and saw the transmission rise in steps as the slit opened past each height.
The energies are tiny, about $1.4$ peV for the ground state, so transitions between levels can be driven by gently vibrating the mirror at a few hundred hertz. The qBounce collaboration uses this gravity resonance spectroscopy to test Newton's law of gravity at micrometer distances and to search for hypothetical new forces and dark-energy fields, placing limits that no other experiment reaches at those scales.
Because WKB reproduces exact results for the box and the oscillator, it is tempting to treat it as exact. It is an approximation that assumes the potential changes slowly on the scale of the wavelength. Near a turning point the wavelength becomes infinite and the WKB wave function blows up; the connection formulas bridge the gap but are themselves approximations. For the lowest states of a general potential, WKB energies can be off by several percent.
A second misconception is that tunneling probability depends only on the barrier's peak height. The exponent is an integral over the whole barrier, so a barrier that is tall but narrow at the particle's energy can be crossed far more easily than a lower one that is thick — exactly why a strong field, which thins the barrier, releases electrons from a metal without lowering its peak much.
Write the local momentum inside the well.
$p(x) = \sqrt{2mE}$
Constant, since $V = 0$ inside.
Identify the boundary conditions.
$\psi(0) = \psi(a) = 0$
Two hard walls: no phase loss correction beyond $\pi$ per half-wave.
Write the quantization condition.
$\displaystyle\int_0^a p\,dx = n\pi\hbar$
For two hard walls.
Evaluate the integral.
$\sqrt{2mE}\,a = n\pi\hbar$
The momentum is constant.
Solve for the energies.
$E_n = \dfrac{n^2\pi^2\hbar^2}{2ma^2}$
Exact: WKB is exact when the potential is flat between hard walls.
Tungsten has $\phi = 4.5$ eV in a field of $5$ V/nm. Find the barrier width.
$L = \dfrac{\phi}{eE} = \dfrac{4.5\ \text{eV}}{5\ \text{eV/nm}} = 0.90\ \text{nm}$
Where the tilted potential drops back to the Fermi level.
Find the maximum decay constant.
$\kappa_{\max} = 5.123\sqrt{4.5} = 10.87\ \text{nm}^{-1}$
At the metal surface, where the barrier is highest.
Integrate across the triangle.
$\displaystyle\int_0^L\kappa\,dx = \tfrac{2}{3}\kappa_{\max}L = \tfrac{2}{3} \times 10.87 \times 0.90 = 6.52$
The average of $\sqrt{\text{linear}}$ over the triangle is two-thirds of its maximum.
Find the tunneling probability.
$T \approx e^{-2 \times 6.52} = e^{-13.0} = 2.2 \times 10^{-6}$
Per attempt at the barrier.
Compare with a rectangular barrier.
$e^{-2 \times 10.87 \times 0.90} = e^{-19.6} = 3.2 \times 10^{-9}$
The triangle is far easier to cross.
Find the effect of doubling the field.
$L = 0.45\ \text{nm}, \quad T \approx e^{-6.5} = 1.5 \times 10^{-3}$
A thousandfold increase: field emission switches on sharply.
Uranium-238 emits a $4.27$ MeV alpha, leaving thorium-234 ($Z = 90$). Find the outer turning point.
$r_2 = \dfrac{2Z \times 1.44\ \text{MeV fm}}{E} = \dfrac{2 \times 90 \times 1.44}{4.27} = 60.7\ \text{fm}$
Using $e^2/4\pi\varepsilon_0 = 1.44$ MeV fm.
Find the nuclear radius.
$r_1 = 1.07 \times 234^{1/3} = 6.59\ \text{fm}$
The alpha starts at the daughter's surface.
Find the barrier height at the surface.
$V(r_1) = \dfrac{2 \times 90 \times 1.44}{6.59} = 39.3\ \text{MeV}$
Nine times the alpha's energy.
Evaluate the first term of $\gamma$.
$1.980 \times \dfrac{90}{\sqrt{4.27}} = 86.2$
The thick-barrier limit.
Evaluate the correction.
$1.485 \times \sqrt{90 \times 6.59} = 36.2$
From starting at $r_1$ instead of $0$.
Find Gamow's factor.
$\gamma = 86.2 - 36.2 = 50.1, \qquad 2\gamma = 100.1$
The tunneling probability is $e^{-100}$.
Estimate the lifetime.
$\tau \approx \dfrac{2r_1}{v}e^{2\gamma} = \dfrac{2 \times 6.59 \times 10^{-15}}{1.4 \times 10^{7}}e^{100.1} \approx 3 \times 10^{22}\ \text{s}$
With the alpha's speed $v = \sqrt{2E/m} = 1.4 \times 10^{7}$ m/s.
Compare with the measured lifetime.
$\tau_{\text{measured}} = \dfrac{4.5 \times 10^{9}\ \text{yr}}{\ln 2} = 2 \times 10^{17}\ \text{s}$
Off by about $10^5$, yet across a range of $10^{24}$ the formula tracks every emitter.
Recall how the exponent depends on the field.
$X \propto \dfrac{1}{E}$
The barrier width is $\phi/eE$.
Divide by the factor of four.
$X = \dfrac{20}{4}$
Four times the field.
Evaluate the exponent.
The WKB approximation treats the wave function as a plane wave whose wavelength changes slowly. Where does it fail?
Complete the worked solution: at a certain field, the field-emission exponent $2\int\kappa\,dx$ is $48$. Find it at twice and three times that field, and the difference between those two.
Halve the exponent for twice the field.
$X_{2E} = \dfrac{48}{2} =$ a
The barrier is half as wide.
Divide the exponent by three for three times the field.
$X_{3E} = \dfrac{48}{3} =$ b
A third as wide.
Subtract the two exponents.
$X_{2E} - X_{3E} =$ c
The current rises by $e$ to this power between the two fields.
Match each WKB result to its expression.
| $\frac{C}{\sqrt{p}}e^{\pm i\int p\,dx/\hbar}$ | $\frac{C}{\sqrt{|p|}}e^{\pm\int\kappa\,dx}$ | $e^{-2\int\kappa\,dx}$ | $\int p\,dx = (n + \tfrac{1}{2})\pi\hbar$ | |
|---|---|---|---|---|
| the allowed-region wave function | ||||
| the forbidden-region wave function | ||||
| the tunneling probability | ||||
| the quantization condition |
Apply WKB quantization to an oscillator with $\hbar\omega = 48$ meV. Fill in the energy, in meV, and the classical turning point, in units of $\sqrt{\hbar/m\omega}$, for $n = 0$, $4$ and $12$.
| energy (meV) | turning point | |
|---|---|---|
| $n = 0$ | ||
| $n = 4$ | ||
| $n = 12$ |
A strong electric field $E = 3$ V/nm at a metal surface with work function $\phi = 4.5$ eV tilts the barrier into a triangle. Find the WKB exponent $2\int\kappa\,dx$ for an electron at the Fermi level to tunnel out.
Answer:
The nucleus uranium-238 decays by emitting an alpha particle of energy $4.27$ MeV, leaving a daughter nucleus of charge $Z = 90$ and radius $r_1 = 6.59$ fm. Using the WKB approximation, find Gamow's factor $\gamma$, where the tunneling probability is $e^{-2\gamma}$.
Answer:
In experiments at the Institut Laue–Langevin, very slow neutrons bounce on a horizontal mirror in Earth's gravity, a potential $V = mgz$ with a hard wall at $z = 0$. WKB gives the heights of the turning points as $z_n = z_0\left[\tfrac{3\pi}{2}\left(n - \tfrac{1}{4}\right)\right]^{2/3}$ with $z_0 = 5.868$ μm. How high does a neutron in level $n = 3$ rise, in μm?
Answer: μm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Apply WKB quantization to an oscillator with $\hbar\omega = 66$ meV. Fill in the energy, in meV, and the classical turning point, in units of $\sqrt{\hbar/m\omega}$, for $n = 0$, $4$ and $12$.
| energy (meV) | turning point | |
|---|---|---|
| $n = 0$ | ||
| $n = 4$ | ||
| $n = 12$ |
You can apply the WKB approximation. Explain to someone why alpha-decay half-lives range from microseconds to billions of years.
17. Your turn: a field-emission exponent is $X = 20$ at a certain field. What is it at four times that field?, step 3
$X = 5$
The tunneling probability rises by $e^{15}$, about three million.