University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.3

Stationary phase and the classical limit

This is where the path integral earns its keep. One Fresnel integral, done carefully, explains why a thrown ball follows Newton, where the oscillator propagator's phase and prefactor come from, and why the argument collapses at half a period, the first instant at which every classical path meets again. Each claim is checked against an exact answer.

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The path integral gives every path from x′ to x the same weight, a phasor e^(iS/ħ) of modulus one, and asks for their sum. When the actions involved are many times ħ, a small bend in a path swings its phase through many radians and paths cancel in pairs, except near a path whose action does not change to first order under any bend. That condition, δS = 0, is Hamilton's principle, so the classical trajectory emerges as the one place where neighbouring paths agree, for any smooth potential and not only the quadratic ones for which Ehrenfest's theorem returns Newton's law exactly.

Keeping the second-order change of S turns the neighbours into Gaussian integrals, one for each way of bending the path, and gives K ≈ F exp(iS_cl/ħ): leading order in ħ in general, exact when V is quadratic. The harmonic oscillator tests every part of this. Its exact propagator, quoted, has the classical action as its phase, and a prefactor equal to the free one times √(ωt/sin ωt), the product of how far the potential widens each bending mode's in-phase zone.

The model stops at focal points. At ωt = π every classical path from x′ arrives at −x′, the stationary point is no longer isolated, the prefactor diverges, and the exact propagator is a mirror image; past it the Gaussian form returns carrying an extra phase −π/2.

Simple definition: Stationary phase approximates an integral of e^(iφ/ħ) with φ ≫ ħ by keeping only the neighbourhood of the points where φ is stationary; applied to the path integral, it keeps the classical path and the Gaussian fluctuations around it.
Example: In ∫ exp[iλ(u − 1)²] du with λ = 400 the phase is stationary at u = 1, and only |u − 1| ≲ √(π/λ) = 0.089 adds in phase; the integral is √(π/λ) e^(iπ/4) = 0.0886 e^(iπ/4), exactly, because the phase is quadratic.

Stationary phase in one variable∫ g(u) e^(iφ/ħ) du ≈ g(u₀) √(2πiħ/φ″(u₀)) e^(iφ₀/ħ), where φ′(u₀) = 0 and φ₀ = φ(u₀)

Derived: only a window about √(2πħ/|φ″|) wide around u₀ adds in phase, and the rest cancels. Exact for quadratic φ and constant g; the sign of φ″ sets the phase e^(±iπ/4).

φ an action in J s, ħ in J s, φ″ in J s per unit of u²; the root is a width in units of u. Leading order in ħ.

In-phase width of fluctuation mode nℓₙ² = 4πħt / (m|n²π² − ω²t²|), for a bend aₙ sin(nπs/t)

Derived: bending by aₙ raises the action by m aₙ²(n²π² − ω²t²)/4t, so aₙ = ℓₙ costs phase π. For n = 1 the width diverges at ωt = π.

ħ in J s, t in s, m in kg, ω in rad s⁻¹, ℓₙ in m; s is the time along the path. Free particle: set ω = 0.

Classical action of the oscillator pathS_cl = mω[(x² + x′²) cos ωt − 2xx′] / (2 sin ωt)

Derived by integrating the Lagrangian along the classical path. As ωt → 0 it tends to the free m(x − x′)²/2t; at ωt = π/2 it is −mωxx′.

m in kg, ω in rad s⁻¹, x′ at time 0 and x at time t in m; S_cl in J s.

Oscillator propagator (quoted)K(x, tx′, 0) = √(mω/(2πiħ sin ωt)) exp(iS_cl/ħ), for ωt between 0 and π

Its phase is the classical action, and as t → 0 it becomes the free √(m/2πiħt) exp(im(x − x′)²/2ħt).

K in m⁻¹, with √i = e^(iπ/4). Quoted as Mehler's formula; exact for every x, x′.

Prefactor from the Gaussian fluctuations|K_osc| / |K_free| = ∏ₙ 1/√(1 − (ωt/nπ)²) = √(ωt / sin ωt)

The potential softens every bending mode and widens its in-phase zone. The ratio is 1.2533 at ωt = π/2 and tends to 1 as t → 0.

Dimensionless; same m, t and endpoints; for ωt between 0 and π, and past π with |1 − (ωt/π)²| and |sin ωt|. The product is derived mode by mode; Euler's product for sin, which closes it, is quoted.

Half a period: the focal pointK(x, π/ωx′, 0) = e^(−iπ/2) δ(x + x′)

Every classical path from x′ reaches −x′, so no stationary point is isolated and the Gaussian formula diverges. Past ωt = π the semiclassical K gains a factor e^(−iπ/2).

Exact; derived from K = Σ e^{(−iEₙt/ħ)} ψₙ(x)ψₙ(x′) with parity ψₙ(−x) = (−1)ⁿψₙ(x). δ in m⁻¹.

Neighbouring paths cancel unless the action is stationary

In the time-sliced path integral every path carries a unit phasor e^(iS/ħ). None is weighted more than another; what differs is how fast the phase turns as a path is deformed. Take one variable first. In ∫ e^(iφ/ħ) du, with φ a function of u, the phase advances by 2π each time u moves 2πħ/|φ′|, and consecutive half-cycles cancel, so only the neighbourhood of a point with φ′(u₀) = 0 survives. Expanding φ to second order there leaves a Fresnel integral of width √(2πħ/|φ″|), the first card, correct to leading order in ħ. For paths, bend x(s) into x(s) + η(s), with s the time along the path and η(0) = η(t) = 0. Integrating by parts, the first-order change is δS = −∫ [mẍ + V′(x)] η ds, which vanishes for every η only if mẍ = −V′(x). The path whose neighbours all add in phase is the Newtonian one: Hamilton's principle, derived here as a statement about interference. How far the in-phase zone reaches is set by ħ. A free path bent by a₁ sin(πs/t) stays within phase π of the straight one out to a₁ = ℓ₁ = √(4ħt/πm).

A classical limit that needs no quadratic potential

Unit 3 reached classical motion through Ehrenfest's theorem, d⟨p⟩/dt = −⟨V′(x)⟩, which follows Newton's law only when ⟨V′(x)⟩ = V′(⟨x⟩), exactly so for V at most quadratic. Stationary phase makes no such demand. Expand the action about the classical path: S[x_cl + η] = S_cl + ∫ (½mη̇² − ½V″(x_cl)η²) ds + O(η³). There is no linear term, because the path is classical, so any smooth V gives K ≈ F exp(iS_cl/ħ), with F the Gaussian integral over η. The cubic and higher terms matter only as far as η reaches, about √(ħt/m), so they correct F at relative order ħ: the result is leading order in ħ for an anharmonic V and exact for a quadratic one, where those terms vanish. Two features have no Ehrenfest analogue. If several classical paths join x′ to x in time t, each is a separate stationary point and their contributions interfere with relative phase (S₁ − S₂)/ħ, shifted by −π/2 for each focal point one path has passed and the other has not (the Maslov phase, met below). And the argument needs S to change by many ħ across the region where V″ varies; it says nothing where that fails, or where two stationary points merge.

The oscillator's phase is its classical action

For V = ½mω²x² the classical path from x′ at s = 0 to x at s = t is x_cl(s) = [x′ sin ω(t − s) + x sin ωs]/sin ωt. On it the Lagrangian collapses: ½mẋ² − ½mω²x² = ½m d(xẋ)/ds − ½mx(ẍ + ω²x), and the bracket vanishes by the equation of motion, so S_cl = ½m[x ẋ(t) − x′ ẋ(0)]. With ẋ(t) = ω(x cos ωt − x′)/sin ωt and ẋ(0) = ω(x − x′ cos ωt)/sin ωt this is the third card, S_cl = mω[(x² + x′²) cos ωt − 2xx′]/(2 sin ωt), derived. The oscillator propagator, quoted in closed form as Mehler's formula, has exactly this as its phase. Checks: mωx² is kg s⁻¹ m² = J s, an action. As ωt → 0, expanding cos and sin gives S_cl ≈ m(x − x′)²/2t − (mω²t/6)(x² + xx′ + x′²): the free action of the propagator topic, less t times the potential energy averaged along the straight path. At ωt = π/2 it is simply −mωxx′, positive when the path crosses the trap centre.

The prefactor counts the paths that add in phase

Bend the classical path by η(s) = Σ aₙ sin(nπs/t), which vanishes at both ends. For a quadratic V the expansion above is exact and the modes decouple: mode n raises the action by m aₙ²(n²π² − ω²t²)/4t, and n²π² − ω²t² is its stiffness. Each aₙ integral is a Fresnel integral whose size is the in-phase width ℓₙ of the second card, so up to a normalisation shared with the free particle the prefactor is the product of the ℓₙ. Dividing by the free case, where each stiffness is n²π², gives |K_osc|/|K_free| = ∏ 1/√(1 − (ωt/nπ)²), and Euler's product formula for sin, quoted from analysis, closes it to √(ωt/sin ωt). The potential softens every mode, so the ratio exceeds 1. At ωt = π/2 the n = 1 factor is 1.155, n = 2 lifts the product to 1.193, n = 5 to 1.225, and the whole product is 1.2533, the ratio the quoted √(mω/2πiħ sin ωt) gives. As t → 0 it tends to 1 and K_osc to the free √(m/2πiħt). The Van Vleck formula, quoted, agrees: |K| = √(|∂²S_cl/∂x∂x′|/2πħ), and here ∂²S_cl/∂x∂x′ = −mω/sin ωt.

Half a period: a focal point

Now let ωt approach π. Every classical path leaving x′ is x′ cos ωs + (v₀/ω) sin ωs, and at ωs = π all of them pass through −x′, whatever v₀: the trap is a perfect lens. The failure shows up two ways. The n = 1 stiffness π² − ω²t² goes to zero, because sin(πs/t) is then exactly the difference between two classical paths, so bending along it costs no action and ℓ₁ grows without bound. And in the Van Vleck form the prefactor is 1/√(2πħ|∂x/∂p₀|), set by how many classical paths, labelled by their initial momentum p₀, land per unit length near x; with ∂x/∂p₀ = sin ωt/mω the whole fan lands on one point. The stationary point is no longer isolated, and the Gaussian formula, which needs one, fails. The exact answer is easy anyway. From the spectral sum with Eₙ = (n + ½)ħω and parity ψₙ(−x) = (−1)ⁿψₙ(x), K(x, π/ω; x′, 0) = e^(−iπ/2) Σ ψₙ(−x)ψₙ(x′) = e^(−iπ/2) δ(x + x′), derived: any initial state arrives at half a period as its own mirror image, multiplied by −i.

Past the focal point: the Maslov phase and the ½

Between ωt = π and 2π the n = 1 stiffness is negative, so the classical path is a saddle of S, not a minimum, yet it still dominates. A Fresnel integral of negative stiffness carries e^(−iπ/4) where a positive one carried e^(+iπ/4), so past the focal point the semiclassical propagator gains a factor e^(−iπ/2), the Maslov phase, and becomes √(mω/2πħ|sin ωt|) exp(iS_cl/ħ) e^(−i3π/4). A full period passes two focal points, and K(x, 2π/ω; x′, 0) = e^(−iπ) δ(x − x′). The spectral sum demands e^{(−iEₙT/ħ)} = −1 for every n, which with spacing ħω forces Eₙ = (n + ½)ħω up to whole quanta: the zero-point ½ counted as two Maslov phases. Both are measured. The paraxial wave equation of optics is the free Schrödinger equation with distance for time, and light through a focus gains Gouy's measured phase of π, π/2 per transverse direction. H₂ and D₂ share one potential curve, and their measured dissociation energies differ by 0.078 eV, against ½ħ[ω(H₂) − ω(D₂)] = 0.080 eV from quoted vibrational frequencies. WKB, next, runs this argument at fixed energy, with turning points as the focal points.

Change one variable at a time

Make the relationship visible.

Interactive model

End time ωt90 °

Action scale mωL²/ħ10

Drag ωt from 90° towards 174°: the fan closes on the focal point and the shaded band widens past the frame. Step to 186° and the dashed Im curve drops below the axis: the lowest mode's stiffness has turned negative, the source of the Maslov phase. Raise mωL²/ħ and the band shrinks onto the classical path at A = 0.

IN-PHASE WIDTH ℓ₁0.52 L

PREFACTOR |K_osc|/|K_free|1.25

MASLOV PHASE0 °

PHASE δS/ħ AT A = L11.8 rad

Live interpretationIN-PHASE WIDTH ℓ₁: 0.52 L. PREFACTOR |K_osc|/|K_free|: 1.25. MASLOV PHASE: 0 °. PHASE δS/ħ AT A = L: 11.8 rad

Catch the common trap

Explain before calculating.

A particle of mass m in a harmonic well travels for a quarter period, ωt = π/2. Compared with a free particle of the same mass between the same endpoints in the same time, how large is the modulus of its propagator, and why?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA laser-cooled ⁸⁷Rb atom (m = 1.443 × 10⁻²⁵ kg) and a 5.00 g steel ball bearing each move freely for t = 0.250 s. For each, find ℓ₁, the amplitude of the lowest bend sin(πs/t) at which a neighbouring path is π out of phase with the classical one, and compare it with a length that matters for that object.

For a free particle the lowest bend raises the action by mπ²a₁²/4t, with no linear term because the unbent path is classical. Setting this equal to πħ gives ℓ₁ = √(4ħt/πm).
Atom: 4ħt/πm = 4 × 1.0546 × 10⁻³⁴ × 0.250 ÷ (π × 1.443 × 10⁻²⁵) = 2.326 × 10⁻¹⁰ m², so ℓ₁ = 1.53 × 10⁻⁵ m = 15.3 μm, about 20 times the measured 780 nm wavelength of the light that images the atom: paths a camera can tell apart still add in phase. The atom's own size, about 0.5 nm, does not enter; what matters is ℓ₁ against the scale a measurement resolves.
Ball: the same formula with m = 5.00 × 10⁻³ kg gives ℓ₁ = 8.2 × 10⁻¹⁷ m, a tenth of the proton's measured charge radius, 0.84 fm, far below any displacement of a ball that can be measured.
Checks: ħt/m has units J s × s ÷ kg = m², so ℓ₁ is a length; and 15.3 μm ÷ 8.2 × 10⁻¹⁷ m = 1.9 × 10¹¹ = √(5.00 × 10⁻³ ÷ 1.443 × 10⁻²⁵), as ℓ₁ ∝ 1/√m at fixed t requires.

Answerℓ₁ = 15.3 μm for the atom, about 20 imaging wavelengths, so non-classical paths contribute on a resolvable scale and its motion is wave-like; ℓ₁ = 8.2 × 10⁻¹⁷ m for the ball, a tenth of a proton radius, so only its classical path survives.

MediumA ⁴⁰Ca⁺ ion (m = 6.64 × 10⁻²⁶ kg) moves in a trap with ω = 2π × 1.00 MHz. Find the classical action and the modulus of the propagator for travel from x′ = +30 nm to x = −20 nm in a quarter period, and compare both with a free ion joining the same endpoints in the same time.

A quarter period is t = π/2ω = 2.50 × 10⁻⁷ s, so cos ωt = 0 and sin ωt = 1. The classical path x′ cos ωs + (v₀/ω) sin ωs reaches v₀/ω at t, so v₀ = ωx = −0.126 m s⁻¹.
S_cl = mω[(x² + x′²) cos ωt − 2xx′]/(2 sin ωt) reduces to −mωxx′ = −(4.172 × 10⁻¹⁹ kg s⁻¹)(−20 × 10⁻⁹ m)(30 × 10⁻⁹ m) = 2.50 × 10⁻³⁴ J s, a phase S_cl/ħ = 2.37 rad. Integrating ½mẋ² − ½mω²x² along the path numerically returns the same value.
Free ion: S = m(x − x′)²/2t = 6.64 × 10⁻²⁶ × (5.0 × 10⁻⁸)² ÷ (5.0 × 10⁻⁷) = 3.32 × 10⁻³⁴ J s, or 3.15 rad. The difference, −0.77 rad, is within 7% of the short-time estimate −(mω²t/6)(x² + xx′ + x′²)/ħ = −0.73 rad, even though ωt = π/2 is not small.
Prefactors: |K_free| = √(m/2πħt) = 2.00 × 10⁷ m⁻¹ and |K_osc| = √(mω/2πħ sin ωt) = 2.51 × 10⁷ m⁻¹. The ratio, 1.253, is √(ωt/sin ωt) = √(π/2): the potential has widened every fluctuation mode's in-phase zone.
Units: mωx² is kg s⁻¹ m² = J s, an action, and √(mω/ħ) is m⁻¹, as a one-dimensional propagator must be.

AnswerS_cl = 2.50 × 10⁻³⁴ J s (2.37 rad) against 3.32 × 10⁻³⁴ J s (3.15 rad) free; |K_osc| = 2.51 × 10⁷ m⁻¹, 1.25 times the free 2.00 × 10⁷ m⁻¹.

HardAn ⁸⁷Rb atom (m = 1.443 × 10⁻²⁵ kg) moves in a magnetic trap with ω = 2π × 50.0 Hz that is harmonic only for |x| up to about 30 μm, so the Gaussian treatment of the real trap is trusted only while ℓ₁ ≤ 30 μm. Find ℓ₁ at ωt = 0.9π, the time window around T/2 in which the treatment fails, the prefactor ratio at the window's edge, and what K does at and just after T/2.

Length scale: x₀² = ħ/mω = 1.0546 × 10⁻³⁴ ÷ (1.443 × 10⁻²⁵ × 314.2) = 2.326 × 10⁻¹² m², so x₀ = 1.53 μm. Since ħt/m = x₀²ωt, the second card reads ℓ₁² = 4πθx₀²/|π² − θ²| with θ = ωt.
At θ = 0.9π: ℓ₁² = 4 × 0.9 x₀²/(1 − 0.81) = 18.9x₀², so ℓ₁ = 4.35x₀ = 6.64 μm, well inside 30 μm: the Gaussian treatment is sound.
Near θ = π write θ = π ∓ δ. Then |π² − θ²| ≈ 2πδ and ℓ₁² ≈ 2πx₀²/δ, so ℓ₁ reaches 30 μm at δ = 2πx₀²/(30 μm)² = 2π × 2.326 × 10⁻¹² ÷ 9.00 × 10⁻¹⁰ = 1.62 × 10⁻² rad, a time δ/ω = 51.7 μs from T/2 on either side; solved without the approximation, 51.6 μs before and 51.8 μs after.
The treatment is therefore unsafe for about 103 μs around T/2 = 10.0 ms, 1.0% of the half period. At the window's edge sin θ ≈ δ, so |K_osc|/|K_free| = √(θ/sin θ) = √(π/0.0162) = 13.9, on its way to diverging.
At T/2 itself the exact harmonic result is K = e^(−iπ/2) δ(x + x′): a packet centred at +20 μm arrives centred at −20 μm, reflected and multiplied by −i. Just after, the n = 1 stiffness is negative, its Fresnel integral carries e^(−iπ/4) in place of e^(+iπ/4), and K has gained the Maslov factor e^(−iπ/2).

Answerℓ₁ = 6.6 μm at ωt = 0.9π; the treatment fails for about ±52 μs around T/2 = 10.0 ms, where the prefactor ratio has reached 14; K is a mirror image at T/2 and carries an extra phase −π/2 after it.