Lecture Notes#

The Stationary-Phase Formula#

Many quantum-mechanical calculations reduce to an oscillatory 1D integral of the form

(93)#\[ I = \int A(x)\,\mathrm{e}^{\mathrm{i}S(x)/\hbar}\,\mathrm{d}x, \]

with \(S(x)\) a real “action-like” phase function and \(A(x)\) a slowly varying amplitude. (Calling the phase function \(S\) — not a separate symbol like \(\Phi\) — is intentional: the path integral has the same form, only with \(x\) replaced by an entire path.) When \(\hbar\) is small, the phase \(S(x)/\hbar\) oscillates rapidly and contributions cancel everywhere except near points where the phase is stationary, \(S'(x_{0}) = 0\).

Stationary-phase approximation

For an isolated stationary point \(x_{0}\) where \(S'(x_{0}) = 0\), the leading contribution as \(\hbar\to 0\) is

(94)#\[ I \;\approx\; A(x_{0})\,\sqrt{\frac{2\pi\mathrm{i}\hbar}{S''(x_{0})}}\,\mathrm{e}^{\mathrm{i}S(x_{0})/\hbar}, \]

with the principal-branch square root: \(\sqrt{2\pi\mathrm{i}\hbar/S''} = \sqrt{2\pi\hbar/\vert S''\vert}\,\mathrm{e}^{\mathrm{i}\pi/4\,\mathrm{sgn}(S'')}\). For multiple isolated saddles, sum their contributions.

Validity: each saddle is isolated, \(A\) varies slowly compared with \(\mathrm{e}^{\mathrm{i}S/\hbar}\), and \(\vert S''(x_{0})\vert\ne 0\).

Derivation: SPA from a Taylor expansion

Near \(x_{0}\), expand \(S(x) \approx S(x_{0}) + \tfrac{1}{2}S''(x_{0})(x - x_{0})^{2}\) — the linear term vanishes by stationarity. Replace the slowly varying \(A(x)\) by \(A(x_{0})\). The integral becomes Gaussian:

\[ I \;\approx\; A(x_{0})\,\mathrm{e}^{\mathrm{i}S(x_{0})/\hbar}\int\mathrm{e}^{\mathrm{i}\,S''(x_{0})\,(x - x_{0})^{2}/(2\hbar)}\,\mathrm{d}x. \]

Use the Fresnel integral \(\int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\alpha y^{2}}\,\mathrm{d}y = \sqrt{\mathrm{i}\pi/\alpha}\) (for real \(\alpha\neq 0\), with \(\sqrt{\mathrm{i}} = \mathrm{e}^{\mathrm{i}\pi/4}\)) at \(\alpha = S''(x_{0})/(2\hbar)\) to get \(\sqrt{2\pi\mathrm{i}\hbar/S''(x_{0})}\), completing (94).

The width of the contributing region around \(x_{0}\) is \(\Delta x \sim \sqrt{\hbar/\vert S''(x_{0})\vert}\) — it shrinks to zero as \(\hbar\to 0\), leaving only the saddle.

Worked Example: A Mexican-Hat Action#

To see SPA in action — and to see when it breaks down — consider the Mexican-hat phase

(95)#\[ S(x) = (x^{2} - 1)^{2},\qquad A(x) = \mathrm{e}^{\mathrm{i}kx}, \]

so the SPA integral is the Fourier transform of \(\mathrm{e}^{\mathrm{i}S(x)/\hbar}\) at wavevector \(k\):

(96)#\[ I(k) = \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}S(x)/\hbar}\,\mathrm{e}^{\mathrm{i}kx}\,\mathrm{d}x. \]

The saddles are \(S'(x_{0}) = 4x_{0}(x_{0}^{2} - 1) = 0\), giving three points: \(x_{0} = -1, 0, +1\), with \(S(\pm 1) = 0\), \(S(0) = 1\), \(S''(\pm 1) = 8\), \(S''(0) = -4\) (Fig. 14). Applying (94) to each and summing,

(97)#\[ I_{\mathrm{SPA}}(k) = \sum_{x_{0}\in\{-1, 0, +1\}} \mathrm{e}^{\mathrm{i}kx_{0}}\,\sqrt{\frac{2\pi\mathrm{i}\hbar}{S''(x_{0})}}\,\mathrm{e}^{\mathrm{i}S(x_{0})/\hbar}. \]

The SPA result is a sum of three plane waves \(\mathrm{e}^{\mathrm{i}kx_{0}}\), one per saddle. Inverse-Fourier-transforming back to position space yields three delta peaks located precisely at \(x_{0} = -1, 0, +1\) — the SPA literally reads off the saddle positions from the \(k\)-space oscillation pattern.

Two lessons from this example:

1. SPA picks out saddles. The complex-plane traces in Fig. 15 (left column) make the geometric meaning of “rapid-oscillation cancellation” concrete: the integral spirals only where the phase is stationary; everywhere else it stalls. The SPA sum (97) is exactly the analytic prediction for these jumps.

2. SPA breaks down when saddles overlap. The contributing width around each saddle is \(\sqrt{\hbar/\vert S''\vert}\). When \(\hbar\) becomes large enough that adjacent saddles’ contributing regions overlap (compare the saddle separation \(\sim 1\) here with \(\sqrt{\hbar/\vert S''\vert}\sim\sqrt{\hbar/4}\)), neighbouring saddles cease to be isolated and SPA fails (Fig. 15, lower-right).

Homework#

1. Gaussian check. Apply the stationary-phase approximation to the pure Gaussian integral \(I = \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\alpha x^{2}/\hbar}\,\mathrm{d}x\) with \(\alpha > 0\) real.

(a) Find the stationary point \(x_{0}\) where \(S'(x_{0}) = 0\) for \(S(x) = \alpha x^{2}\).

(b) Evaluate the SPA formula (94) and compare with the exact Fresnel result \(\sqrt{\pi\hbar/(-\mathrm{i}\alpha)} = \sqrt{\pi\hbar/\alpha}\,\mathrm{e}^{\mathrm{i}\pi/4}\). Show that SPA is exact here. Why?

2. Cubic phase. Consider \(I = \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}(x^{2} + \epsilon x^{3})/\hbar}\,\mathrm{d}x\) with \(\epsilon\) small and positive.

(a) Find all stationary points of \(S(x) = x^{2} + \epsilon x^{3}\).

(b) For the stationary point near \(x = 0\), apply the SPA formula and write the leading approximation to \(I\).

(c) Explain why the other stationary point (at \(x = -2/(3\epsilon)\)) gives a subdominant contribution when \(\epsilon\) is small.

3. Stationary-region width. The SPA integral receives its dominant contribution from a region of width \(\Delta x\) around each stationary point.

(a) Using \(\vert S''(x_{0})\vert(x - x_{0})^{2}/(2\hbar)\sim 1\) as the criterion for significant phase variation, show \(\Delta x \sim \sqrt{\hbar/\vert S''(x_{0})\vert}\).

(b) For the free-particle classical action \(S_\text{cl} = m(\Delta x)^{2}/(2t)\), identify \(S'' = m/t\) and estimate \(\Delta x\) in position space. How does it scale with \(m\) and \(t\)?

(c) Evaluate this width for an electron (\(m \approx 10^{-30}\,\text{kg}\), \(t = 10^{-15}\,\text{s}\)) and a baseball (\(m = 0.15\,\text{kg}\), \(t = 1\,\text{s}\)). Why is the baseball’s quantum uncertainty negligible?

4. Second variation. For a particle in a potential \(V(x)\), the second variation of the action around the classical path is \(\delta^{2}S = \int_{0}^{T}\delta x(t)\,\hat{D}\,\delta x(t)\,\mathrm{d}t\) with \(\hat{D} = -m\,\mathrm{d}^{2}/\mathrm{d}t^{2} - V''(x_\mathrm{cl})\).

(a) For the harmonic oscillator \(V = \tfrac{1}{2}m\omega^{2}x^{2}\), show that \(\hat{D} = -m(\mathrm{d}^{2}/\mathrm{d}t^{2} + \omega^{2})\).

(b) Fluctuations \(\delta x(t)\) vanish at the endpoints (paths have fixed boundary conditions), so \(\hat{D}\,\delta x = \lambda\,\delta x\) has Dirichlet boundary conditions \(\delta x(0) = \delta x(T) = 0\), with eigenfunctions \(\sin(n\pi t/T)\) for \(n = 1, 2, \ldots\). Find the eigenvalues, identify the values of \(T\) at which a zero eigenvalue appears, and explain what goes wrong with the semiclassical propagator at those special times.

5. Validity breakdown. Apply the SPA naively to \(I = \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}x^{4}/\hbar}\,\mathrm{d}x\).

(a) Find the stationary points of \(S(x) = x^{4}\) and compute \(S''\) at each.

(b) Why does the SPA formula (94) fail here? Which condition of the standard SPA is violated?

(c) The correct leading behaviour is \(I \sim \hbar^{1/4}\), not \(\hbar^{1/2}\). Argue on dimensional grounds why a degenerate stationary point (\(S'' = 0\) but \(S^{(4)}\neq 0\)) changes the \(\hbar\)-scaling.

6. Mexican-hat saddles. For the example (95)–(96) with \(S(x) = (x^{2}-1)^{2}\) and \(A(x) = \mathrm{e}^{\mathrm{i}kx}\):

(a) Verify that \(S'(x_{0}) = 0\) has the three solutions \(x_{0} = -1, 0, +1\) and compute \(S(x_{0})\) and \(S''(x_{0})\) at each.

(b) Apply (94) to write down \(I_{\mathrm{SPA}}(k)\) as a sum of three plane waves in \(k\).

(c) Compute the inverse Fourier transform \(\tilde{I}(x) = (1/2\pi)\int I_{\mathrm{SPA}}(k)\,\mathrm{e}^{-\mathrm{i}kx}\,\mathrm{d}k\). Show that it is a sum of three delta peaks at the saddle positions, with relative weights set by \(1/\sqrt{\vert S''(x_{0})\vert}\).

(d) From the criterion \(\sqrt{\hbar/\vert S''\vert}\ll\) saddle separation, estimate the value of \(\hbar\) at which adjacent saddles begin to overlap and SPA breaks down. Compare with Fig. 15.

7. Free particle from SPA. The free-particle action \(S[x] = \int_{0}^{t}\tfrac{1}{2}m\dot{x}^{2}\,\mathrm{d}\tau\) is purely quadratic in the path, so SPA applied to the path integral is exact.

(a) Compute the classical action \(S_\text{cl}(x,t;x_{0},0) = m(x - x_{0})^{2}/(2t)\) along the straight-line trajectory.

(b) Compute the mixed derivative \(-\partial^{2}S_\text{cl}/(\partial x\,\partial x_{0})\) and show it equals \(m/t\).

(c) The semiclassical propagator from a quadratic action is \(K \approx \sqrt{-\partial^{2}S_\text{cl}/(\partial x\,\partial x_{0})/(2\pi\mathrm{i}\hbar)}\,\exp(\mathrm{i}S_\text{cl}/\hbar)\) (the Van Vleck formula — the SPA result for a path integral with quadratic action). Verify that this reproduces the exact free propagator from §3.2.3.

8. Correspondence principle. The ratio \(S_\text{cl}/\hbar\) controls whether a system is quantum or classical.

(a) For a free particle of mass \(m\) traversing distance \(d\) in time \(t\), \(S_\text{cl} = md^{2}/(2t)\). Compute \(S_\text{cl}/\hbar\) for an electron (\(m\approx 10^{-30}\,\text{kg}\)) moving \(1\,\text{nm}\) in \(1\,\text{ns}\), and for a ball (\(m = 0.1\,\text{kg}\)) moving \(1\,\text{m}\) in \(1\,\text{s}\).

(b) For the ball, the SPA selects one dominant path. Estimate the number of \(2\pi\) phase oscillations as the endpoint \(x\) varies over \(1\,\text{mm}\). Why does this make the propagator effectively a delta function on macroscopic scales?

(c) Explain how this quantitative argument embodies the correspondence principle: classical mechanics emerges from quantum mechanics when \(S_\text{cl}\gg\hbar\).