WKB states and Bohr–Sommerfeld quantization
Separate eikonal and transport equations, match through simple turning points, and retain the Maslov correction and regular-energy scope.
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The question for today
What you’ll understand
- Derive the one-dimensional eikonal and leading amplitude equations.
- State a two-simple-turning-point quantization rule with its Maslov correction.
- Check the harmonic oscillator and distinguish regular-energy asymptotics from a well-bottom limit.
A WKB state separates a rapidly changing phase from a slowly varying amplitude. Substitution into the Schrödinger equation first determines a classical momentum and then an amplitude transported along that momentum. This representation breaks down at a turning point where the momentum vanishes. Matching with an Airy model repairs the representation and contributes the phase shift responsible for the half-integer quantization rule in a smooth one-dimensional well.
The phase and amplitude solve different equations
Use P_h=−h²∂ₓ²/2+V and an ansatz u=a exp(iS/h), with real phase in the allowed region. Expanding the derivatives separates the order-one eikonal equation from the order-h transport equation. Writing p(x)=S′(x), the positive branch is sqrt(2(E−V(x))) where E>V. The leading amplitude is a constant times p^(−1/2) on a branch without turning points. It is not constant when the classical particle speed changes.
The approximation is local and requires bounds away from p=0 plus smooth coefficients and an asymptotic remainder construction. Merely solving the first two formal equations does not establish a globally small residual to every order. At a simple turning point V(x)=E with V′ nonzero, the p^(−1/2) amplitude diverges and the oscillatory coordinate chart fails. The exact eigenfunction need not diverge there. An Airy approximation supplies a different local representation that connects oscillatory and decaying regions.
Two soft turning points give a half-integer shift
For a precise regular-energy setting, take a smooth real potential equal to a positive quadratic outside a compact set, and a fixed compact energy interval on which exactly two simple turning points bound one allowed well. There are no competing wells at those energies and the turning points stay separated and nondegenerate. The self-adjoint operator has discrete eigenvalues there. Airy matching at both ends gives the closed action rule below, with total Maslov index two and no order-h subprincipal term for this Schrödinger operator.
The closed action is twice the integral of positive momentum between the endpoints. Each simple soft turning point contributes to the phase matching, producing the shift n+1/2 rather than n. Hard boundaries, degenerate turning points and multiple-well matching require different arguments; the same formula cannot simply be copied into them. The O(h²) action remainder is uniform in the specified regular energy window. It is not a uniform assertion at a coalescing pair of turning points as the energy approaches the bottom with h.
Classical action calibrates the quantum label
For a harmonic well V=ω²x²/2, an energy curve is an ellipse with position radius sqrt(2E)/ω and momentum radius sqrt(2E). Its area is J(E)=2πE/ω, and its period is the action derivative 2π/ω. Inserting this action into the rule gives hω(n+1/2). The exact oscillator spectrum has precisely those values, but its exact solvability is stronger than the general WKB construction and should not be inferred for an anharmonic well.
At a fixed positive energy, the state index grows like one over h. Holding n fixed instead drives oscillator energy toward the minimum and lies in a different asymptotic regime. The harmonic formula remains exact there by direct solution, while a general potential requires a well-bottom harmonic approximation and its corrections. Keeping these regimes distinct prevents an error estimate proved near a noncritical energy from being advertised as a uniform description of every low-lying eigenvalue.
Work it through
The harmonic action and the zero-point shift
For V=x²/2, compute the closed action at energy E and recover the harmonic eigenvalues from the Maslov-corrected rule. Identify what is special about the result.
The energy curve is x²+ξ²=2E, a circle of radius √(2E). Its enclosed phase-space area, equal to the closed action, is π·2E=2πE.
The quantization condition gives 2πE=2πh(n+1/2), hence E=h(n+1/2). Omitting the Maslov term would incorrectly give zero energy for n=0.
For this quadratic operator the formula is exact, as the Gaussian ground state and creation-operator spectrum confirm. Its exact ground-state value h/2 does not extend the general regular-energy O(h²) assertion to arbitrary wells whose turning points coalesce at their minimum.
The harmonic oscillator has exact levels h(n+1/2); the half-integer shift records two soft turning points and their matching phases.
The living notebook
Explore the model
Interactive lab · oscillator
Period and action of a quadratic well
Pause and Reset, then set ζ = 0 and ω₀ = 1. The orbit from (x,ξ)=(1,0) has energy 1/2, period 2π, and action π. Reset and set both ζ = 0 and ω₀ = 2: the same initial point has energy two, period π, and action 2π. Use J=2πE/ω₀ to check both predictions. The instrument shows classical position and velocity, not quantized energy levels or Airy matching, and it has no h control.
- Displacement x(t)
- 1
- Velocity v(t)
- 0
- Regime
- underdamped
Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forClassical action and period can be checked independently of the additional phase-matching condition that selects quantum energies.
Pause & reason
Check your understanding
Your turn
Try it yourself
For P_h=−h²∂ₓ²/2+2x², use h=0.1 and n=3 to find the harmonic energy, both turning points, and the closed action. Verify the quantization equation.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Here ω=2. Solve 2x²=E for the turning points and use J(E)=2πE/ω.
Compare with the solution
The exact harmonic energy is hω(n+1/2)=0.1·2·3.5=0.7. The turning points are , and J(E)=πE=0.7π. The right side of the action rule is 2π·0.1·3.5=0.7π, in exact agreement. Both turning derivatives V′=4x are nonzero at these points. The calculation uses a single quadratic well with two separated soft turning points; it does not establish the same spectrum for a potential with extra wells or a nonsmooth boundary.
Keep the idea
Take it with you
- WKB phase and amplitude solve separate eikonal and transport equations away from turning points.
- Two uniformly simple soft turning points yield a Maslov shift of one half in a regular-energy window.
- An exact harmonic spectrum does not remove regime restrictions from general quantization theorems.
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