The WKB approximation

The last three sections all computed energy levels — the system sits quietly in a bound state and we ask what its energy is. This section asks a completely different question, and it starts with a cold case from nuclear physics.

Uranium-238 emits a 4.3 MeV alpha particle with a half-life of 4.5 billion years; polonium-212 also emits alpha particles, at 8.8 MeV, with a half-life of 0.3 microseconds. A factor of 2 in energy, a factor of $10^{23}$ in lifetime. In 1911 Geiger and Nuttall plotted a batch of alpha emitters and found a tidy law linking the logarithm of the lifetime to the energy — but nobody could explain the law, let alone that absurd span.

Worse still, classical mechanics does not even permit the alpha particle to get out. Inside the nucleus it feels the attractive nuclear force; once outside, it faces Coulomb repulsion from the residual nucleus (charge about $90e$), a barrier standing roughly 28 MeV tall at the nuclear surface. A 4.3 MeV particle charging a 28 MeV wall — the classical verdict is unambiguous: it never gets out. And yet the uranium ore sits there, decaying.

A particle showing up under the barrier — we solved tunnelling exactly for the square barrier back in section 2.11. But the Coulomb barrier is not square; no exact solution can be written down, and we need a tool that works for a barrier of any shape. Perturbation theory is no help: the tunnelling probability is an exponentially small quantity of the form $\ee^{-\text{large number}}$, whose dependence on the Hamiltonian is non-analytic — any finite-order power series gives exactly zero. We need an approximation purpose-built for exponentials.

The local plane wave

The stationary equation

$$\frac{\dd^2\psi}{\dd x^2}=-\frac{p^2(x)}{\hbar^2}\psi, \qquad p(x)\equiv\sqrt{2m\left[E-V(x)\right]}\tag{7.4.1}$$

If $V$ is constant, the solutions are plane waves $\ee^{\pm\ii px/\hbar}$ with wavelength $\lambda=h/p$. The core idea of WKB (Wentzel–Kramers–Brillouin) fits in one sentence: if $V(x)$ barely changes over one wavelength, the wavefunction is still approximately a plane wave at every point — only the wavelength adjusts point by point to the local $p(x)$. Like a train of water waves crossing a gentle slope: the slope is gentle, the wave stays a wave, only its spacing gradually changes.

The WKB wavefunction: phase integral and 1/√p amplitudebasic~10 minexpand

Step 1: write the wavefunction as amplitude times phase.

Any complex function can be written

$$\psi(x)=A(x)\,\ee^{\ii\phi(x)}\tag{7.4.2}$$

(with $A,\phi$ real). Substitute into the stationary equation and set real and imaginary parts to zero separately:

$$A''-A\,(\phi')^2=-\frac{p^2}{\hbar^2}A \qquad\text{(real part)}\tag{7.4.3}$$$$2A'\phi'+A\phi''=0 \qquad\text{(imaginary part)}\tag{7.4.4}$$

No approximation yet.

Step 2: the imaginary-part equation solves exactly.

Multiply the imaginary-part equation by $A$ and it is precisely a total derivative, $\left(A^2\phi'\right)'=0$, so

$$A=\frac{C}{\sqrt{\phi'}}\tag{7.4.5}$$

Step 3: make the one and only approximation.

The $A''$ term in the real-part equation is the contribution of the amplitude’s curvature. The “slowly varying potential” assumption translates as: the amplitude changes far more slowly than the phase oscillates. Drop $A''$ (we check this in a moment), and what is left is

$$(\phi')^2=\frac{p^2}{\hbar^2} \quad\Longrightarrow\quad \phi(x)=\pm\frac{1}{\hbar}\int^x p(x')\,\dd x'\tag{7.4.6}$$

Assemble:

$$\psi(x)\approx\frac{C}{\sqrt{p(x)}}\, \exp\!\left[\pm\frac{\ii}{\hbar}\int^x p(x')\,\dd x'\right]\tag{7.4.7}$$

The phase is the integral of the momentum — the plane wave’s $px/\hbar$ generalised into the point-by-point accumulation $\int p\,\dd x/\hbar$.

The $1/\sqrt p$ amplitude has a clean physical reading: the probability density is $|\psi|^2\propto1/p\propto1/v$. A classical particle lingers where it moves slowly — this is the correspondence principle from the end of section 2.7 all over again: WKB is an approximation with the classical limit built in.

Step 4: the classically forbidden region.

Where $E$ falls below $V$, $p$ turns pure imaginary. Writing $\kappa(x)=\sqrt{2m[V(x)-E]}/\hbar$, the oscillating solution becomes an exponential one:

$$\psi(x)\approx\frac{C}{\sqrt{\kappa(x)}}\, \exp\!\left[\pm\int^x\kappa(x')\,\dd x'\right]\tag{7.4.8}$$

— the wavefunction seeping into the forbidden region decays exponentially with the accumulated $\int\kappa\,\dd x$. The seed of tunnelling is planted right here.

When does the approximation hold? Go back and check the discarded $A''$ term; the condition tidies up to

$$\left|\frac{\dd\lambda}{\dd x}\right|\ll1, \qquad \lambda(x)=\frac{2\pi\hbar}{p(x)}\tag{7.4.9}$$

The wavelength’s change over one wavelength must be far smaller than the wavelength itself — the precise version of “the potential varies slowly”. It holds well in two situations: a gentle potential, or high energy (large $p$, short $\lambda$ — relative to a short enough wavelength, even an ordinary potential counts as “gentle”). Hence WKB’s other name, the semiclassical approximation: $\lambda\to0$ is exactly the limit where quantum degenerates into classical.

First deliverable: the quantisation condition

For a bound state oscillating back and forth between two turning points $x_1,x_2$, the connection formulas require the phase integral to satisfy

$$\int_{x_1}^{x_2}p(x)\,\dd x=\left(n+\frac12\right)\pi\hbar, \qquad n=0,1,2,\dots\tag{7.4.10}$$

This is the Bohr–Sommerfeld quantisation condition — the rule Bohr and Sommerfeld once wrote down by guesswork in the old quantum theory (section 1.4), here derived properly from the Schrödinger equation, plus a $\tfrac12$ they could never have guessed (it comes from the wavefunction’s exponential tails beyond the turning points — each “soft bounce” steals $\pi/2$ of phase).

Test: the harmonic oscillator. With $V=\tfrac12m\omega^2x^2$, the phase integral is the area of an ellipse and comes out to $\pi E/\omega$. Insert into the condition:

$$\frac{\pi E}{\omega}=\left(n+\frac12\right)\pi\hbar \quad\Longrightarrow\quad E_n=\left(n+\frac12\right)\hbar\omega\tag{7.4.11}$$

Identical to the exact result, word for word — even the zero-point energy is right (a fluke of the harmonic oscillator; no other potential gets that treatment, though at large $n$ the WKB levels always approach the exact ones). An approximation that hands you an entire spectrum for free — astonishing value for money.

Second deliverable: the tunnelling factor

For a wide barrier (with entry and exit turning points $x_1,x_2$), the connection formulas give the penetration probability

$$T\approx\ee^{-2\gamma},\qquad \gamma=\frac{1}{\hbar}\int_{x_1}^{x_2}\kappa(x)\,\dd x =\frac{1}{\hbar}\int_{x_1}^{x_2}\sqrt{2m\left[V(x)-E\right]}\,\dd x\tag{7.4.12}$$

How to read it: the wavefunction decays by $\ee^{-\gamma}$ crossing the forbidden region, and the probability is the amplitude squared. The exponent is an area-like integral over the whole barrier — the taller and wider the barrier, the heavier the particle, the more $\gamma$ grows linearly and $T$ collapses exponentially.

Plug in a threshold-level number first: an electron hitting a square barrier standing $1\ \text{eV}$ above its energy and $0.5\ \text{nm}$ wide.

$$\kappa=\frac{\sqrt{2mc^2\cdot(1\,\text{eV})}}{\hbar c} =\frac{\sqrt{2\times511000}}{197.3}\approx5.1\ \text{nm}^{-1}, \qquad T\approx\ee^{-2\times5.1\times0.5}\approx6\times10^{-3}\tag{7.4.13}$$

One success in about one hundred and sixty attempts — and the scanning tunnelling microscope earns its living on exactly this: $\gamma$‘s exponential sensitivity to width makes the tunnelling current change by nearly an order of magnitude for every 0.1 nm of tip height, which is why an STM can “feel out” the bumps of individual atoms.

Alpha decay: Gamow unlocks thirty orders of magnitudeadvanced~10 minexpand

In 1928 Gamow (and, independently, Gurney and Condon) aimed the tunnelling factor at uranium. The model: the alpha particle rattles inside the nucleus (radius $r_0\approx9\ \text{fm}$), and at each wall collision escapes through the Coulomb barrier with probability $T$

$$V(r)=\frac{2(Z-2)e^2}{4\pi\varepsilon_0 r}\qquad(r>r_0)\tag{7.4.14}$$

(The $2$ is the alpha particle’s charge number, $Z-2$ the residual nucleus. Uranium decays to thorium, $Z-2=90$, $V(r_0)\approx\dfrac{2\times90\times1.44\ \text{MeV}\cdot\text{fm}}{9\ \text{fm}}\approx29\ \text{MeV}$, far above $E_\alpha\approx4.3\ \text{MeV}$ — a classically forbidden region worthy of the name.)

The outer turning point sits where $V(r_2)=E$. Integrate $\kappa$ along the Coulomb tail (an elementary substitution does it); in the limit $r_2\gg r_0$ the leading term is

$$2\gamma\approx\frac{2\pi\cdot2(Z-2)e^2}{4\pi\varepsilon_0\hbar}\sqrt{\frac{m_\alpha}{2E}} \ \propto\ \frac{Z-2}{\sqrt{E}}\tag{7.4.15}$$

The logarithm of the decay rate is proportional to $-1/\sqrt E$ — exactly the Geiger–Nuttall empirical line, derived for the first time seventeen years after its discovery.

The order-of-magnitude ledger: plugging in numbers for uranium-238 gives $2\gamma\approx90$, i.e. $T\sim\ee^{-90}\sim10^{-39}$. The alpha particle hits the wall at a rate of about $v/2r_0\sim10^{21}\ \text{s}^{-1}$, so the lifetime is

$$\tau\sim\frac{1}{10^{21}\times10^{-39}}\ \text{s}=10^{18}\ \text{s}\approx3\times10^{10}\ \text{yr}\tag{7.4.16}$$

The same order of magnitude as the measured 4.5 billion years — for an estimate carrying a 90 in its exponent, that is a resounding success.

The answer to the thirty-order riddle: as the energy rises from 4.3 to 8.8 MeV, $1/\sqrt E$ shrinks by only about a third — but it rides an exponent of order 90: $2\gamma$ drops from $\sim90$ to $\sim50$, and with the barrier details stacked on, $T$ swings by tens of orders of magnitude. The absurd span of lifetimes is just the exponential function going about its daily business. The other end of the same exponential lever is the stars: protons in the solar core ignite fusion by tunnelling, and the tunnelling probability’s extreme sensitivity to temperature sets the burn rate of every star.

∎Key formulas

• WKB wavefunction

  $$\psi\approx\frac{C}{\sqrt{p(x)}}\exp\left[\pm\frac{\ii}{\hbar}\int^x p\,\dd x'\right]$$

  In the forbidden region p→iħκ, oscillation becomes exponential decay; |ψ|²∝1/v is the classical dwell time

• Validity condition

  $$\left|\frac{\dd\lambda}{\dd x}\right|\ll1$$

  The wavelength barely changes over one wavelength; maximally violated at turning points

• Quantisation condition

  $$\int_{x_1}^{x_2}p\,\dd x=\left(n+\tfrac12\right)\pi\hbar$$

  Bohr-Sommerfeld + half-integer correction; exact for the harmonic oscillator

• Tunnelling factor

  $$T\approx\ee^{-2\gamma},\quad\gamma=\frac{1}{\hbar}\int_{x_1}^{x_2}\sqrt{2m(V-E)}\,\dd x$$

  Alpha decay: 2γ∝(Z−2)/√E, i.e. the Geiger-Nuttall law

?Self-check4 questions

1. 1.

   The validity condition of the WKB approximation is:

   • The de Broglie wavelength barely changes over a distance of one wavelength
   • The potential energy is far smaller than the total energy
   • The perturbation matrix elements are far smaller than the level spacing
   • The particle moves at nearly the speed of light

   Submit
2. 2.

   The physical interpretation of the WKB amplitude factor 1/√p(x) is:

   • The probability density goes as 1/v — the classical particle’s dwell time at each point
   • A mathematical requirement of normalisation, with no physical meaning
   • The wavefunction must diverge at the turning points
   • Larger momentum means a larger wavefunction

   Submit
3. 3.

   Why can a factor of 2 in alpha-decay energy produce a factor of 10²³ in lifetime?

   • The decay rate goes as e^(−2γ) with 2γ ∝ 1/√E a large number of order tens; a moderate change in the exponent gets amplified into tens of orders of magnitude
   • The higher-energy alpha particle hits the nuclear wall 10²³ times more often
   • The Coulomb barrier of a heavy nucleus grows exponentially with energy
   • It is experimental error; the real lifetimes differ little

   Submit
4. 4.

   An electron tunnels through a square barrier 1.0 eV above its energy and 0.50 nm wide. Estimate the penetration probability with T ≈ e^(−2κa). (Hint: κ = √(2mc²·ΔE)/ħc, mc² = 511000 eV, ħc = 197.3 eV·nm)

   0% relative tolerance

   Submit

What comes next

So far, every Hamiltonian in this chapter has been time-independent: the levels are eternal, and a stationary state, once moved in, never moves out. Yet in the real world atoms absorb and emit light, and excited states rarely live beyond a few nanoseconds. Where is the contradiction? In the assumption “the Hamiltonian does not depend on time” — when a light wave hits an atom, what the atom feels is an electric field oscillating in time. The next section bolts a time axis onto perturbation theory, to compute one of the most useful classes of numbers in quantum mechanics: transition probabilities.