Endpoint phases quantize a closed orbit
Derive the half-integer shift for two soft turning points and separate it from hard-wall quantization.
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The question for today
What you’ll understand
- Relate a closed-orbit action to semiclassical eigenenergies.
- Count the phase contribution from two simple turning points.
- Explain why hard-wall endpoints change the quantization offset.
A bound wave must reproduce itself after propagation across a well and reflection from its endpoints. WKB supplies the accumulated action, while local matching supplies the reflection phases. Keeping only the action misses the ground-state offset in a smooth well. We will derive the two-turning-point condition, express it with a Maslov index, and use a harmonic orbit as a check before examining why an infinite square well uses a different endpoint rule.
A global standing wave needs consistent local matches
Consider a smooth one-dimensional confining potential with exactly two simple turning points x₁ and x₂ at the energy of interest. Between them the momentum is real; outside them a normalizable bound state must decay. Starting from each decaying tail gives two allowed-region sine forms with quarter-cycle shifts. They describe the same state only when the phase accumulated between the turning points satisfies a discrete condition. The resulting half-integer offset is not an arbitrary correction fitted to a known spectrum.
Integrating momentum from the left turning point to the right gives half the classical closed-orbit action. The two Airy matches require that integral to be πħ times n plus one half, where n is a nonnegative integer counting the bound states in this simple well. The formula is a leading semiclassical approximation in general. It is most naturally controlled when the wavelength is short on the potential scale; its exact success for every harmonic-oscillator level is a special property of that potential.
The Maslov index records the missing phase
Write the full classical action as an integral around one closed orbit in phase space. For the standard one-dimensional soft well, each turning-point caustic contributes one to the Maslov index, so the total index is two. This notation packages the matching phase into the closed-orbit quantization rule. It does not remove the need to inspect the geometry: multi-dimensional caustics, singular coordinate reductions, and imposed boundary conditions can change the phase bookkeeping.
A useful independent check comes from differentiating action with respect to energy. For a one-dimensional bound orbit this derivative equals its classical period. Neighboring quantized actions differ by 2πħ, so the leading level spacing is approximately 2πħ divided by that period. Long-period motion therefore produces more densely spaced semiclassical levels. Near a separatrix the period can become large and the simple approximation can require uniform treatment; one should not extend a regular isolated-orbit estimate through every critical energy.
A hard wall supplies its own boundary condition
An infinite hard wall imposes a wavefunction zero directly. There is no narrow smooth region where a linear potential and Airy matching describe its reflection. Two Dirichlet walls give pL=Nπħ for a constant-potential interior, with positive integer N. The N=0 sine is the identically zero function and is not a state. Reusing the soft-well half-integer rule would therefore predict the wrong square-well ground energy. A mixed hard-wall and soft-turning-point problem has yet another offset determined by their combined phases.
The oscillator lab checks the classical period that enters the action derivative. Reset and set both controls to damping ratio zero and natural frequency one; predict a period 2π in its time units. Reset again and set damping zero and frequency two; predict period π. The model has unit mass and starts at displacement one with zero velocity. It computes a classical oscillator trajectory, not a quantum spectrum, a wavefunction, or a Maslov index. Its role here is to test the classical period relation used in the semiclassical reasoning.
Work it through
Quantize the harmonic phase-space ellipse
For a one-dimensional harmonic oscillator, use the classical orbit semiaxes x_max=√(2E/(mω²)) and p_max=√(2mE). Find its closed-orbit action and apply the two-soft-turning-point rule. Then give energies for n=0 and n=2 when ħω=2 energy units.
The enclosed phase-space ellipse has area πx_max p_max. Multiplying the two semiaxes gives 2E/ω, so J(E)=2πE/ω. This is also the closed-orbit integral of p dx.
Set this action equal to 2πħ(n+1/2). Cancel 2π and solve for E to obtain E_n=ħω(n+1/2). The offset arises from the two soft endpoints.
For ħω=2, the n=0 energy is one and the n=2 energy is five. Consecutive levels differ by two, consistent with dJ/dE=2π/ω and the classical period.
The action is 2πE/ω, giving E_n=ħω(n+1/2); the requested energies are 1 and 5 energy units.
The living notebook
Explore the model
Interactive lab · oscillator
Test the period behind the action spacing
Reset and set BOTH controls to ζ=0 and ω₀=1; predict period 2π. Reset and set ζ=0 and ω₀=2; predict period π. This unit-mass model starts at x=1,v=0 and contains no quantum levels.
- 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 forThe classical period controls the derivative of action; endpoint phases must still be supplied separately.
Pause & reason
Check your understanding
Your turn
Try it yourself
An infinite square well has width L=π, scaled ħ=1, and mass m=1/2. Use the exact Dirichlet condition pL=Nπħ to find its lowest nonzero momentum and ground energy. What energy would incorrectly reusing n+1/2 at n=0 give?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The interior energy is p²/(2m)=p² in these units; N must be positive.
Compare with the solution
The condition gives p=N, so the smallest nonzero momentum is one and the ground energy is one. Replacing the positive integer N with n+1/2 would give p=1/2 at n=0 and energy 1/4. That incorrect value comes from importing the soft-turning-point phase into a problem with hard Dirichlet walls.
Keep the idea
Take it with you
- Action accumulation and endpoint phases jointly determine semiclassical quantization.
- Two simple soft turning points give μ=2 and the half-integer shift.
- Hard walls require their boundary phases, and the classical period checks level spacing.
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