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WKB Approximation

The WKB (Wentzel—Kramers—Brillouin) method provides approximate solutions to the one-dimensional Schrodinger equation when the potential varies slowly compared to the de Broglie wavelength.

The ansatz ψ(x)=A(x)eiS(x)/\psi(x) = A(x)e^{iS(x)/\hbar} substituted into 22mψ"+Vψ=Eψ-\frac{\hbar^2}{2m}\psi"' + V\psi = E\psi gives, to leading order in \hbar:

S(x)=±xp(x)dx,p(x)=2m[EV(x)]S(x) = \pm\int^x p(x')\,dx', \quad p(x) = \sqrt{2m[E - V(x)]}

The WKB wavefunctions:

ψ(x)Cp(x)exp ⁣(±ixp(x)dx)(classically allowed, E > V)\psi(x) \approx \frac{C}{\sqrt{p(x)}}\exp\!\left(\pm\frac{i}{\hbar}\int^x p(x')\,dx'\right) \quad \text{(classically allowed, E > V\text{)}}

ψ(x)Cp(x)exp ⁣(±1xp(x)dx)(classically forbidden, E < V)\psi(x) \approx \frac{C}{\sqrt{|p(x)|}}\exp\!\left(\pm\frac{1}{\hbar}\int^x |p(x')|\,dx'\right) \quad \text{(classically forbidden, E < V\text{)}}

At a classical turning point (E=V(x0)E = V(x_0)), the WKB approximation breaks down. The Airy function connects the oscillating and decaying solutions:

2p(x)cos ⁣(1xx0p(x)dxπ4)1p(x)exp ⁣(1x0xp(x)dx)\frac{2}{\sqrt{p(x)}}\cos\!\left(\frac{1}{\hbar}\int_x^{x_0} p(x')\,dx' - \frac{\pi}{4}\right) \longleftrightarrow \frac{1}{\sqrt{|p(x)|}}\exp\!\left(-\frac{1}{\hbar}\int_{x_0}^x |p(x')|\,dx'\right)

The WKB quantisation condition for a bound state in a potential well with turning points aa and bb:

abp(x)dx=(n+12)π,n=0,1,2,\int_a^b p(x)\,dx = \left(n + \frac{1}{2}\right)\pi\hbar, \quad n = 0, 1, 2, \ldots

The factor of 1/21/2 (Maslov index) accounts for the phase loss at each turning point.

Application: Harmonic oscillator. V(x)=12mω2x2V(x) = \frac{1}{2}m\omega^2 x^2. Turning points at x=±2E/(mω2)x = \pm\sqrt{2E/(m\omega^2)}.

AA2mEm2ω2x2dx=πEω=(n+12)π\int_{-A}^{A}\sqrt{2mE - m^2\omega^2 x^2}\,dx = \frac{\pi E}{\omega} = \left(n + \frac{1}{2}\right)\pi\hbar

En=(n+12)ωE_n = \left(n + \frac{1}{2}\right)\hbar\omega

The WKB gives the exact result for the harmonic oscillator --- a fortunate coincidence due to the quadratic potential.

Application: Power-law potential. For V(x)=V0x/aαV(x) = V_0|x/a|^\alpha:

En(n+12)2α/(α+2)E_n \propto \left(n + \frac{1}{2}\right)^{2\alpha/(\alpha+2)}

MethodValidityAccuracyUse case
WKBSlowly varying V(x)V(x)Leading order in \hbarSemiclassical tunnelling
Exact solutionAll potentialsExactSolvable potentials (HO, Coulomb)
Perturbation theorySmall VV' relative to VVϵ2\sim \epsilon^2 errorWeak anharmonic corrections
Variational methodAny (with trial function)Upper bound, depends on trialGround state energies
  • Applying WKB at a turning point. The approximation diverges at x0x_0 where p(x0)=0p(x_0) = 0. Connection formulas using Airy functions are necessary to match solutions across turning points.
  • Forgetting the Maslov index. For bound states, each smooth turning point contributes a phase of π/4\pi/4, giving π/2\pi/2 total (the 1/21/2 in n+1/2n+1/2). Hard walls give different phases.
  • Using WKB for rapidly varying potentials. The condition dλ/dx1|d\lambda/dx| \ll 1 (where λ=h/p\lambda = h/p) must hold; WKB fails at sharp potential steps or barriers.
  • Assuming WKB works for all nn in bound states. The approximation improves for large nn (highly excited states), but is poor for n=0n=0 ground states in shallow wells.
Potential typeTurning pointsQuantisation conditionWKB exact?
Harmonic oscillator2 (smooth)n+1/2n + 1/2Yes
Infinite square well2 (hard wall)nn (no 1/21/2)Yes
Linear potential1 (smooth)Airy zerosNo
Coulomb2 (smooth)n1/2n - 1/2 for l=0l=0Approx.
Quartic double well4 (smooth)Splitting via instantonsNo
  • Nuclear physics: Alpha decay is the canonical example of WKB tunnelling. The Geiger-Nuttall law T1/2exp(1/E)T_{1/2} \propto \exp(1/\sqrt{E}) follows directly from the WKB transmission probability.
  • Quantum chemistry: Tunnelling corrections in reaction rate theory use WKB transmission coefficients for barrier crossing, important for proton transfer and enzyme kinetics.
  • Semiconductor physics: Field emission (Fowler-Nordheim tunnelling) from metal surfaces and tunnelling in MOSFETs are described by WKB transmission through triangular barriers.
  • Cosmology: The WKB method is used in inflationary cosmology to compute the spectrum of primordial density perturbations generated by quantum fluctuations during inflation.
Worked Example 13.1: WKB Tunnelling Through a Barrier

For a potential barrier V(x)=V0(1x2/a2)V(x) = V_0(1 - x^2/a^2) for x<a|x| < aWith E<V0E < V_0The WKB transmission probability is:

Texp ⁣(2a0a02m(V0(1x2/a2)E)dx)T \approx \exp\!\left(-\frac{2}{\hbar}\int_{-a_0}^{a_0}\sqrt{2m(V_0(1 - x^2/a^2) - E)}\,dx\right)

Where a0=a1E/V0a_0 = a\sqrt{1 - E/V_0} is the classical turning point.

Texp ⁣(22mV0a0a01E/V0x2/a2dx)T \approx \exp\!\left(-\frac{2}{\hbar}\sqrt{2mV_0}\int_{-a_0}^{a_0}\sqrt{1 - E/V_0 - x^2/a^2}\,dx\right)

=exp ⁣(22mV0πa22a(1E/V0))= \exp\!\left(-\frac{2}{\hbar}\sqrt{2mV_0}\cdot\frac{\pi a^2}{2a}(1 - E/V_0)\right)

=exp ⁣(πa2mV0(1EV0))= \exp\!\left(-\frac{\pi a}{\hbar}\sqrt{2mV_0}\left(1 - \frac{E}{V_0}\right)\right)

For alpha decay (V025V_0 \approx 25 MeV, a30a \approx 30 fm, E=5E = 5 MeV, m=4×931.5m = 4 \times 931.5 MeV/c2c^2):

\frac{\pi a}{\hbar c}\sqrt{2mc^2 V_0}\left(1 - \frac{E}{V_0}\right) = \frac{\pi \times 30\,\text{fm}{197\,\text{MeV}\cdot\text{fm}\sqrt{2 \times 3726 \times 25}\times 0.8}}

=0.479×432.6×0.8=165.7= 0.479 \times 432.6 \times 0.8 = 165.7

Te165.75×1073T \approx e^{-165.7} \approx 5 \times 10^{-73}

This extremely small probability explains the enormously long half-lives of alpha-emitting nuclei (109\sim 10^9 years for 238^{238}U). The Geiger—Nuttall law relates logT1/2\log T_{1/2} to E1/2E^{-1/2}Consistent with the WKB exponential dependence.

13.8 Worked Example: WKB for Quartic Oscillator

Section titled “13.8 Worked Example: WKB for Quartic Oscillator”

Problem. Estimate the ground state energy of the quartic oscillator V(x)=14mω2a2(x/a)4V(x) = \frac{1}{4}m\omega^2 a^2 (x/a)^4 using the WKB quantisation condition.

Solution. For V(x)=αx4V(x) = \alpha x^4 with α=mω2/(4a2)\alpha = m\omega^2/(4a^2), the turning points are at x=±(E/α)1/4x = \pm (E/\alpha)^{1/4}. The WKB integral: AA2m(Eαx4)dx=(n+1/2)π\int_{-A}^{A} \sqrt{2m(E - \alpha x^4)}\,dx = (n + 1/2)\pi\hbar.

For n=0n = 0: AA2mα(A4x4)dx=π/2\int_{-A}^{A} \sqrt{2m\alpha(A^4 - x^4)}\,dx = \pi\hbar/2 where A=(E/α)1/4A = (E/\alpha)^{1/4}.

The integral evaluates to 43A22mαA4Γ(1/4)24π1.748A22mαA2\frac{4}{3}A^2\sqrt{2m\alpha A^4} \cdot \frac{\Gamma(1/4)^2}{4\sqrt{\pi}} \approx 1.748 A^2\sqrt{2m\alpha A^2}.

Solving: E0(3π8Γ(1/4)2πα2m)4/31.022α/ma2E_0 \approx \left(\frac{3\pi\hbar}{8} \frac{\Gamma(1/4)^2}{\sqrt{\pi}} \sqrt{\frac{\alpha}{2m}}\right)^{4/3} \approx 1.022 \hbar \sqrt{\alpha/m} a^2.

The exact result is E0=1.060α/ma2E_0 = 1.060 \hbar \sqrt{\alpha/m} a^2, so WKB is within 4% for the ground state and improves for higher nn. \blacksquare