Method of Stationary Phase

Derivation

Method of Stationary Phase

D-194 Home PU-205 Threads waves · light Depends on Method of Steepest Descent, Real Integrals by Residues & Jordan's Lemma

Statement

For a real phase \(\phi(t)\in C^\infty\) with a single non-degenerate stationary point \(t_0\in(a,b)\) (\(\phi'(t_0)=0\), \(\phi''(t_0)\neq 0\)) and a smooth amplitude \(g(t)\) of compact support, the oscillatory integral \(\displaystyle I(\lambda)=\int_a^b g(t)\,e^{i\lambda\phi(t)}\,dt\) has the large-parameter asymptotics \(\displaystyle I(\lambda)=g(t_0)\sqrt{\frac{2\pi}{\lambda\,|\phi''(t_0)|}}\;\exp\!\left[i\lambda\phi(t_0)+i\,\frac{\pi}{4}\,\operatorname{sgn}\phi''(t_0)\right]+O(\lambda^{-3/2})\) as \(\lambda\to+\infty\). The factor \(e^{\pm i\pi/4}\) is the Fresnel phase; away from stationary points the integrand self-cancels and contributes only \(O(\lambda^{-\infty})\).

Why it matters

Almost every wave phenomenon in the short-wavelength limit is an oscillatory integral of exactly this form: a Fourier or Fresnel–Kirchhoff propagator carries a rapidly varying phase \(\lambda\phi(t)\) with \(\lambda\) a large frequency, wavenumber, or action-scale \(1/\hbar\). The stationary-phase principle tells you that only the neighbourhoods where the phase is momentarily flat survive the interference; everything else washes out. This is the mathematical engine behind Fermat's principle, the eikonal/geometrical-optics limit, group velocity, and the WKB/semiclassical connection between wave and ray pictures.

The characteristic \(e^{i\pi/4}\) phase slip and the \(\lambda^{-1/2}\) amplitude decay are not decoration — they are the observable signature of a caustic-free ray, the origin of the Gouy phase near a focus, and the Maslov index bookkeeping that keeps semiclassical quantisation self-consistent.

Assumptions

The large parameter \(\lambda\) is real and \(\lambda\to+\infty\).If \(\lambda\) is complex with a growing imaginary part the problem is Laplace-type (steepest descent along a real exponential decay); the oscillatory cancellation argument is replaced by exponential dominance and the \(i\pi/4\) rotation generally does not appear.

The stationary point is non-degenerate: \(\phi''(t_0)\neq 0\).If \(\phi''(t_0)=0\) the local phase is cubic or higher, the Gaussian collapses, and the leading order becomes \(\lambda^{-1/3}\) (Airy) or \(\lambda^{-1/4}\) — a caustic. The formula below diverges there.

The amplitude \(g\) is smooth and \(g(t_0)\neq 0\) is finite.A singular or vanishing amplitude at \(t_0\) changes the power of \(\lambda\); a jump in \(g\) or its derivatives feeds endpoint/edge contributions of their own order.

There is exactly one stationary point in \((a,b)\) and it is interior, with \(g\) of compact support (or decaying) so endpoints do not contribute.Multiple stationary points each add their own term and interfere; a stationary point coinciding with an endpoint halves the contribution; a hard endpoint with \(\phi'\neq0\) there gives a separate \(O(\lambda^{-1})\) boundary term that this leading formula omits.

The phase and amplitude are analytic enough that the local Taylor expansion controls the tail (Morse-lemma change of variables is valid).Without sufficient smoothness the error is no longer \(O(\lambda^{-3/2})\) and the asymptotic series in powers of \(\lambda^{-1}\) may not exist.

Derivation

\[ I(\lambda)=\int_a^b g(t)\,e^{i\lambda\phi(t)}\,dt \]

Starting oscillatory integral; \(\lambda\) large and real, \(\phi,g\) real-analytic. A

\[ \frac{d}{dt}\,e^{i\lambda\phi(t)}=i\lambda\,\phi'(t)\,e^{i\lambda\phi(t)}\quad\Rightarrow\quad e^{i\lambda\phi}=\frac{1}{i\lambda\,\phi'}\frac{d}{dt}e^{i\lambda\phi}\ \ (\phi'\neq0) \]

On any interval free of stationary points \(\phi'\neq0\), so the fast exponential is an exact derivative up to the slowly varying factor \(1/(i\lambda\phi')\). This is the key that makes non-stationary regions negligible. B

\[ \int g\,e^{i\lambda\phi}\,dt=\frac{1}{i\lambda}\int \frac{g}{\phi'}\,\frac{d}{dt}e^{i\lambda\phi}\,dt=\frac{1}{i\lambda}\left[\frac{g}{\phi'}e^{i\lambda\phi}\right]-\frac{1}{i\lambda}\int \frac{d}{dt}\!\left(\frac{g}{\phi'}\right)e^{i\lambda\phi}\,dt \]

Integration by parts. For compactly supported \(g\) the boundary term vanishes and the remaining integral has the same form with an extra \(1/\lambda\); iterating shows any region with \(\phi'\neq0\) contributes \(O(\lambda^{-N})\) for all \(N\). Hence only \(t_0\) matters. B

\[ I(\lambda)\sim \int_{t_0-\delta}^{t_0+\delta} g(t)\,e^{i\lambda\phi(t)}\,dt \qquad(\text{localise to a neighbourhood of }t_0) \]

Multiply by a smooth cutoff equal to \(1\) near \(t_0\); the discarded part is \(O(\lambda^{-\infty})\) by Step 3. Legal because the cutoff difference is supported where \(\phi'\neq0\). A

\[ \phi(t)=\phi(t_0)+\tfrac12\phi''(t_0)(t-t_0)^2+O\!\left((t-t_0)^3\right),\qquad g(t)=g(t_0)+O(t-t_0) \]

Taylor-expand phase and amplitude about the stationary point; the linear phase term is absent because \(\phi'(t_0)=0\). This is where non-degeneracy \(\phi''\neq0\) enters. A

\[ u=\operatorname{sgn}(\phi'')\,\sqrt{\tfrac12|\phi''(t_0)|}\;(t-t_0),\qquad \phi(t)-\phi(t_0)=\operatorname{sgn}(\phi'')\,u^2 \]

Morse lemma: near a non-degenerate critical point a smooth change of variable makes the phase exactly quadratic. To leading order \(dt=\sqrt{2/|\phi''|}\,du\) and higher Taylor terms feed only the \(O(\lambda^{-3/2})\) correction. C

\[ I(\lambda)\sim g(t_0)\,e^{i\lambda\phi(t_0)}\sqrt{\frac{2}{|\phi''(t_0)|}}\int_{-\infty}^{\infty} e^{\,i\lambda\,\sigma\,u^2}\,du,\qquad \sigma\equiv\operatorname{sgn}\phi''(t_0) \]

Insert the leading amplitude \(g(t_0)\), pull out the constant phase \(e^{i\lambda\phi(t_0)}\), and extend the limits to \(\pm\infty\): the Gaussian-oscillatory tail beyond \(\delta\) is exponentially/oscillatorily negligible for large \(\lambda\). B

\[ \int_{-\infty}^{\infty} e^{\,i a u^2}\,du=\sqrt{\frac{\pi}{|a|}}\;e^{\,i\frac{\pi}{4}\operatorname{sgn}a}\qquad(a\ \text{real},\ a\neq0) \]

The Fresnel integral, obtained by rotating the contour \(u\to u\,e^{i\pi/4}\) (for \(a>0\)) onto the steepest-descent line where \(iau^2\to -|a|s^2\), reducing it to \(\int e^{-|a|s^2}\,ds=\sqrt{\pi/|a|}\); the rotation supplies the \(e^{i\pi/4}\). Convergence is in the Abel/regularised sense. C

\[ \int_{-\infty}^{\infty} e^{\,i\lambda\sigma u^2}\,du=\sqrt{\frac{\pi}{\lambda}}\;e^{\,i\frac{\pi}{4}\sigma}\qquad(a=\lambda\sigma,\ |a|=\lambda,\ \operatorname{sgn}a=\sigma) \]

Apply Step 8 with \(a=\lambda\sigma\); since \(\lambda>0\), \(\operatorname{sgn}a=\sigma=\operatorname{sgn}\phi''(t_0)\). Symbols only — no numbers yet. A

\[ I(\lambda)\sim g(t_0)\,e^{i\lambda\phi(t_0)}\sqrt{\frac{2}{|\phi''(t_0)|}}\cdot\sqrt{\frac{\pi}{\lambda}}\;e^{\,i\frac{\pi}{4}\sigma}=g(t_0)\sqrt{\frac{2\pi}{\lambda|\phi''(t_0)|}}\;e^{\,i\lambda\phi(t_0)+i\frac{\pi}{4}\sigma} \]

Combine the two square roots and the two phase factors. This is the leading-order stationary-phase formula. A

Result

\[ \int_a^b g(t)\,e^{i\lambda\phi(t)}\,dt=g(t_0)\sqrt{\frac{2\pi}{\lambda\,|\phi''(t_0)|}}\;\exp\!\left[i\lambda\phi(t_0)+i\,\frac{\pi}{4}\operatorname{sgn}\phi''(t_0)\right]+O\!\left(\lambda^{-3/2}\right) \]

Reading. The whole integral is governed by the single point where the phase stops changing. The amplitude there, \(g(t_0)\), is weighted by a "resonance width" \(\sqrt{2\pi/(\lambda|\phi''|)}\): the flatter the phase (small \(\phi''\)) or the slower the oscillation (small \(\lambda\)), the wider the constructively-adding window and the larger the result. The output oscillates at the stationary phase value \(\lambda\phi(t_0)\), and — crucially — carries an extra fixed \(\pm45^\circ\) phase kick set only by whether \(t_0\) is a phase minimum (\(+\)) or maximum (\(-\)).

Units check. \([g\,dt]\) has units of \(I\). On the right, \(g(t_0)\) carries \([g]\); the phase \(\lambda\phi\) is dimensionless (as it must be in an exponent); and \(\sqrt{2\pi/(\lambda|\phi''|)}\) has units \([\,1/\sqrt{[\lambda][\phi'']}\,]=[\,1/\sqrt{[\lambda\phi]\cdot[t]^{-2}}\,]=[t]\) since \([\lambda\phi]=1\) and \([\phi'']=[\phi][t]^{-2}\Rightarrow[\lambda\phi'']=[t]^{-2}\). Thus the prefactor has units of \(t\), and \([g]\cdot[t]=[g\,dt]=[I]\). Consistent.

Limiting cases

\(\lambda\to\infty\): amplitude \(\propto\lambda^{-1/2}\to0\) — faster oscillation, more cancellation, smaller integral. The leading term becomes exact relative to corrections.
\(\phi''\to0\) (approach to a caustic): the prefactor diverges as \(|\phi''|^{-1/2}\); the formula signals its own breakdown and must be replaced by an Airy (\(\lambda^{-1/3}\)) uniform approximation.
\(\phi''>0\) vs \(\phi''<0\): only the sign of the Fresnel phase flips (\(+\pi/4\) for a minimum, \(-\pi/4\) for a maximum); the magnitude is identical.
Several isolated stationary points: sum one such term per point; their relative phases \(e^{i\lambda\phi(t_k)}\) produce the interference fringes (e.g. two-slit / two-ray beating).
Higher orders: retaining the next Taylor terms of \(\phi,g\) gives an asymptotic series \(I\sim\lambda^{-1/2}\sum_{n\ge0}c_n\lambda^{-n}\) — divergent but with exponentially small optimal truncation error.

Breaks when

Degenerate / coalescing stationary points (\(\phi''(t_0)=0\)): two stationary points merge at a caustic; the Gaussian model fails, \(|\phi''|^{-1/2}\) blows up, and the true scaling is \(\lambda^{-1/3}\) with Airy-function structure. Uniform (Chester–Friedman–Ursell) methods are required.
Stationary point at or near an endpoint: if \(t_0\) sits at \(a\) or \(b\), only half the Gaussian is integrated and the coefficient halves; if \(t_0\) is within \(O(\lambda^{-1/2})\) of the endpoint the leading and boundary terms overlap and neither isolated formula is valid.
Small or non-real \(\lambda\): for moderate \(\lambda\) the neglected \(O(\lambda^{-3/2})\) terms are not small and the single-term formula is quantitatively wrong; for \(\lambda\) with a real part (Laplace-type) the whole oscillatory-cancellation logic is replaced by steepest descent.
Non-smooth amplitude or phase: a kink, pole, or endpoint discontinuity in \(g\) or \(\phi\) injects contributions the local Taylor expansion cannot see, corrupting both the power of \(\lambda\) and the phase.

Failure modes

Dropping the \(e^{i\pi/4}\): keeping only \(|I|\) and forgetting the Fresnel phase — fatal when several stationary points interfere, because the relative \(\pi/4\)'s (and Maslov jumps) set the fringe positions.
Sign confusion in \(\operatorname{sgn}\phi''\): writing \(+\pi/4\) universally. A phase maximum gives \(-\pi/4\); the sign is the Morse index and is physical (Gouy phase, Maslov index).
Using \(\phi''(t_0)\) with the wrong argument: evaluating \(\phi''\) at a generic \(t\) or at \(t=0\) instead of at the stationary point \(t_0\).
Forgetting to check \(\phi'(t_0)=0\) has a real solution in \((a,b)\): if there is none, the integral is \(O(\lambda^{-\infty})\) — not the formula's value; students often plug in a boundary or a complex root and misuse the real formula.
Factor-of-2 / \(\sqrt2\) slips: mishandling \(\tfrac12\phi''\) in the Taylor step, giving \(\sqrt{\pi/(\lambda|\phi''|)}\) instead of \(\sqrt{2\pi/(\lambda|\phi''|)}\).
Applying it at a caustic: using the formula when \(\phi''\approx0\) and reporting the divergent prefactor as a physical intensity instead of switching to the Airy result.

Discussion

Stationary phase is the wave-optics statement of Fermat's principle. Write a wave amplitude as a sum over paths, each contributing \(e^{ik\,L[\text{path}]}\) with \(L\) the optical path length and \(k\) large. Only paths for which \(L\) is stationary — the classical rays — add coherently; all others interfere destructively. The method thus derives geometrical optics from wave optics, and the same argument with \(\lambda=1/\hbar\) turns Feynman's path integral into classical mechanics, the stationary point being the trajectory that extremises the action. The condition \(\phi'(t_0)=0\) is the Euler–Lagrange / ray equation in disguise.

The \(\lambda^{-1/2}\) falloff and the \(e^{i\pi/4}\) are not cosmetic. In a focusing wave the number of stationary points changes as you cross a caustic, and each time a ray touches a caustic its stationary point passes through degeneracy; the Morse index \(\operatorname{sgn}\phi''\) jumps and the field picks up an extra \(-\pi/2\). Accumulated, these are the Maslov indices, and they are exactly what makes the Gouy phase shift of a focused Gaussian beam \(\pi\) (in 2D) as it passes through a waist. The bookkeeping that looks like a mathematical technicality is a measurable phase.

Group velocity and wave-packet propagation are the same computation. A packet \(\int A(k)e^{i(kx-\omega(k)t)}dk\) has phase \(\phi(k)=kx-\omega(k)t\); the stationary point \(\phi'(k)=0\) gives \(x=\omega'(k)t\), i.e. the packet centre travels at the group velocity \(d\omega/dk\), and the \(\lambda^{-1/2}\) (here \(t^{-1/2}\)) prefactor is the spreading and \(1/\sqrt t\) amplitude decay of a dispersing packet. Dispersion, group velocity, and packet spreading all fall out of Steps 5–10.

Rigorously, the leading term is the \(n=0\) coefficient of a genuine asymptotic (Poincaré) expansion generated by the Morse lemma and Watson's lemma applied to the transformed integral; the full series \(I(\lambda)\sim e^{i\lambda\phi(t_0)}\sum_n a_n\lambda^{-(n+1/2)}\) is generically divergent but Borel-summable, with the least term of order \(e^{-c\lambda}\) — the same exponentially small scale that hides the Stokes phenomenon connecting stationary-phase to steepest-descent contributions. Hörmander's theorem generalises all of this to \(N\)-dimensional oscillatory integrals, where \(\sqrt{2\pi/(\lambda|\phi''|)}\) becomes \((2\pi/\lambda)^{N/2}|\det\operatorname{Hess}\phi|^{-1/2}\) and the single \(\pi/4\) becomes \((\pi/4)\times(\text{signature of the Hessian})\).

Common misconceptions. The method does not claim the integrand is negligible except at \(t_0\) pointwise — the integrand has the same magnitude everywhere; it is the integral of the fast-oscillating part that cancels. Nor is stationary phase the same as steepest descent: they are the real-oscillatory and real-exponential faces of the same saddle-point idea, and the \(e^{i\pi/4}\) is precisely the rotation from one contour to the other.

Worked examples

Bessel function at large argument: \(\displaystyle J_0(\lambda)=\frac{1}{\pi}\int_0^\pi \cos(\lambda\sin t)\,dt=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{\,i\lambda\sin t}\,dt.\)

Identify \(\phi(t)=\sin t\), \(g(t)=1/(2\pi)\), large parameter \(\lambda\). A

\[ \phi'(t)=\cos t=0\ \Rightarrow\ t_0=\pm\tfrac{\pi}{2};\qquad \phi''(t)=-\sin t\ \Rightarrow\ \phi''(\pm\tfrac{\pi}{2})=\mp1. \]

Two non-degenerate stationary points in \((-\pi,\pi)\); \(\phi(\pm\pi/2)=\pm1\), \(|\phi''|=1\), \(\operatorname{sgn}\phi''=\mp1\). A

\[ J_0(\lambda)\sim \frac{1}{2\pi}\sqrt{\frac{2\pi}{\lambda\cdot1}}\Big[e^{\,i\lambda-i\pi/4}+e^{-i\lambda+i\pi/4}\Big]=\frac{1}{\sqrt{2\pi\lambda}}\cdot2\cos\!\left(\lambda-\frac{\pi}{4}\right). \]

Sum the two stationary-point terms (Fresnel signs \(\mp\pi/4\)); they combine into a cosine. Numbers now: with \(\lambda=20\), \(\sqrt{2\pi\lambda}=\sqrt{125.66}=11.21\), and \(\lambda-\pi/4=20-0.7854=19.215\ \text{rad}\); reducing modulo \(2\pi\), \(19.215-6\pi=0.365\ \text{rad}\), so \(\cos(19.215)=\cos(0.365)=0.934\). B

\[ J_0(\lambda)\sim\sqrt{\frac{2}{\pi\lambda}}\cos\!\left(\lambda-\frac{\pi}{4}\right);\qquad J_0(20)\approx\frac{2(0.934)}{11.21}=0.1667. \]

Reading. The true value is \(J_0(20)=0.16702\). The leading asymptotic gives \(\sqrt{2/(\pi\lambda)}=0.1784\) times \(\cos(19.215)=0.934\), i.e. \(0.1667\) — already correct to about \(0.2\%\). The single stationary-phase formula reproduces both the amplitude envelope \(\sqrt{2/(\pi\lambda)}\) and the \(-\pi/4\) phase for \(\lambda\gtrsim10\). Units: \(J_0\) and all factors are dimensionless.

Fresnel diffraction at a straight edge / near-field intensity: field \(\displaystyle U(\lambda)=\int_{-\infty}^{0} e^{\,i\lambda\,t^2}\,dt\) with \(\lambda=\dfrac{\pi}{2}\cdot\dfrac{1}{\text{(Fresnel scale)}}\), here take \(\phi(t)=t^2\), \(g=1\), and a stationary point at the interior edge of the aperture.

Model a quadratic phase (paraxial propagator). \(\phi'(t)=2t=0\Rightarrow t_0=0\); \(\phi''=2>0\), \(\operatorname{sgn}\phi''=+1\), \(\phi(0)=0\). Here \(t_0\) is at the endpoint, so use half the Gaussian. A

\[ \int_{-\infty}^{\infty} e^{\,i\lambda t^2}\,dt=\sqrt{\frac{\pi}{\lambda}}\,e^{\,i\pi/4};\qquad \int_{-\infty}^{0}=\tfrac12\sqrt{\frac{\pi}{\lambda}}\,e^{\,i\pi/4}\ \ (\text{stationary point at the endpoint}). \]

The full Fresnel integral is the stationary-phase result with \(g=1\), \(|\phi''|=2\): \(\sqrt{2\pi/(2\lambda)}=\sqrt{\pi/\lambda}\). An endpoint-located \(t_0\) contributes half. B

\[ |U|^2=\tfrac14\,\frac{\pi}{\lambda}. \]

Take modulus squared for intensity. Numbers: for \(\lambda=\pi\) (i.e. one Fresnel unit), \(|U|^2=\tfrac14\cdot\frac{\pi}{\pi}=0.25\); the field at the geometric shadow edge is one-half the amplitude and one-quarter the intensity of the fully illuminated region \(|U_\infty|^2=\pi/\lambda=1\). B

\[ \frac{I_{\text{edge}}}{I_{\text{open}}}=\frac{|U|^2}{|U_\infty|^2}=\frac{\tfrac14\pi/\lambda}{\pi/\lambda}=\frac14. \]

Reading. A stationary point sitting exactly on the boundary (the shadow edge of a straight diffracting edge) delivers half the amplitude of an interior one, hence the textbook result that the intensity at the geometrical shadow boundary is one-quarter of the unobstructed intensity. Units: taking \(g,\phi\) dimensionless, \(|U|^2\) is a pure number here; in a physical problem \(\lambda\) carries \([t]^{-2}\) and \(|U|^2\) inherits \([t]\).

Problems

Compute the leading stationary-phase asymptotics of \(\displaystyle I(\lambda)=\int_{-\infty}^{\infty} e^{-t^2}\,e^{\,i\lambda t^2}\,dt\) as \(\lambda\to\infty\), and comment on why this one is actually exact.
Solution
This is a Gaussian and can be done exactly: \(\int e^{-(1-i\lambda)t^2}dt=\sqrt{\pi/(1-i\lambda)}\). For large \(\lambda\), \(1-i\lambda\approx-i\lambda=\lambda e^{-i\pi/2}\), so \(\sqrt{\pi/(1-i\lambda)}\approx\sqrt{\pi/\lambda}\,e^{\,i\pi/4}\). This matches stationary phase with \(g(0)=1\), \(\phi(t)=t^2\), \(\phi''=2\), \(t_0=0\): \(g(t_0)\sqrt{2\pi/(\lambda\cdot2)}\,e^{i\pi/4}=\sqrt{\pi/\lambda}\,e^{i\pi/4}\). The amplitude \(e^{-t^2}\) plays the role of \(g\); because it is smooth and the phase purely quadratic, the leading term already captures the exact large-\(\lambda\) behaviour, with corrections \(O(\lambda^{-3/2})\) from the \(1\) in \(1-i\lambda\).
For \(\displaystyle I(\lambda)=\int_0^{2\pi} e^{\,i\lambda\cos t}\,dt\), find all stationary points, their \(\phi''\), and the leading asymptotic form. Relate to \(J_0\).
Solution
\(\phi=\cos t\), \(\phi'=-\sin t=0\Rightarrow t_0=0,\pi\). \(\phi''=-\cos t\): at \(t=0\), \(\phi''=-1\) (\(\operatorname{sgn}=-1\), \(\phi=+1\)); at \(t=\pi\), \(\phi''=+1\) (\(\operatorname{sgn}=+1\), \(\phi=-1\)). Sum: \(\sqrt{2\pi/\lambda}\,[e^{i\lambda-i\pi/4}+e^{-i\lambda+i\pi/4}]=\sqrt{2\pi/\lambda}\cdot2\cos(\lambda-\pi/4)\). Since \(J_0(\lambda)=\frac{1}{2\pi}\int_0^{2\pi}e^{i\lambda\cos t}dt\), this gives \(J_0(\lambda)\sim\sqrt{2/(\pi\lambda)}\cos(\lambda-\pi/4)\), the standard large-argument Bessel asymptotic.
A wave packet \(\psi(x,t)=\int_{-\infty}^{\infty}A(k)\,e^{\,i(kx-\omega(k)t)}dk\) with \(\omega(k)=\hbar k^2/2m\). Use stationary phase to find where the packet is centred at time \(t\) and the amplitude's time-decay.
Solution
Phase \(\phi(k)=kx-\omega(k)t\) with large parameter effectively \(t\). Stationary: \(\phi'(k)=x-\omega'(k)t=x-(\hbar k_0/m)t=0\Rightarrow k_0=mx/(\hbar t)\), i.e. the packet centre moves at group velocity \(v_g=\omega'(k_0)=\hbar k_0/m=x/t\). \(\phi''(k)=-\omega''(k)t=-(\hbar/m)t\), \(|\phi''|=\hbar t/m\). Amplitude \(\propto\sqrt{2\pi/|\phi''|}=\sqrt{2\pi m/(\hbar t)}\propto t^{-1/2}\): the free-particle packet spreads and its peak amplitude decays as \(t^{-1/2}\), with Fresnel phase \(e^{-i\pi/4}\) (since \(\phi''<0\)). This is the standard \(1/\sqrt t\) spreading of a dispersing Schrödinger packet.
Evaluate the endpoint-plus-stationary contributions of \(\displaystyle I(\lambda)=\int_0^{\infty} e^{\,i\lambda(t^3/3-t)}\,dt\) at leading order, and identify what happens when the two stationary points coalesce (they don't here, but discuss the nearby case \(\int e^{i\lambda(t^3/3+xt)}dt\)).
Solution
\(\phi=t^3/3-t\), \(\phi'=t^2-1=0\Rightarrow t=\pm1\); only \(t_0=1\) lies in \((0,\infty)\). \(\phi''=2t\Rightarrow\phi''(1)=2>0\), \(\phi(1)=1/3-1=-2/3\). Contribution: \(\sqrt{2\pi/(2\lambda)}\,e^{i\lambda(-2/3)+i\pi/4}=\sqrt{\pi/\lambda}\,e^{-2i\lambda/3+i\pi/4}\), plus an \(O(\lambda^{-1})\) endpoint term from \(t=0\) where \(\phi'(0)=-1\neq0\): \(\frac{-1}{i\lambda\phi'(0)}e^{i\lambda\phi(0)}=\frac{-1}{i\lambda(-1)}=\frac{1}{i\lambda}\), i.e. \(-i/\lambda\). For the Airy case \(\int e^{i\lambda(t^3/3+xt)}dt\), the two stationary points \(t=\pm\sqrt{-x}\) merge at \(x=0\); there \(\phi''=0\), stationary phase fails, and the integral is \(2\pi\,\mathrm{Ai}\) with scaling \(\lambda^{-1/3}\) — the caustic/Airy transition.
Two well-separated stationary points give \(I\sim c_1 e^{i\lambda\phi_1}+c_2 e^{i\lambda\phi_2}\). Show \(|I|^2\) exhibits fringes and find their spacing in \(\lambda\). Take \(c_1=c_2=c\) real, \(\phi_1-\phi_2=\Delta\).
Solution
\(|I|^2=|c|^2|e^{i\lambda\phi_1}+e^{i\lambda\phi_2}|^2=2|c|^2[1+\cos(\lambda(\phi_1-\phi_2))]=2|c|^2[1+\cos(\lambda\Delta)]=4|c|^2\cos^2(\lambda\Delta/2)\). This oscillates between \(0\) and \(4|c|^2\) — full-contrast interference fringes. Adjacent maxima occur when \(\lambda\Delta\) changes by \(2\pi\), so the fringe spacing in \(\lambda\) is \(\delta\lambda=2\pi/\Delta=2\pi/|\phi_1-\phi_2|\). Physically this is two-ray interference (e.g. two-slit or a Newton's-rings-type beating): the fringe period is inversely proportional to the optical-path difference \(\Delta\) between the two stationary (ray) paths, and the \(\pm\pi/4\) Fresnel phases shift the pattern but do not change the spacing.