Stationary Phase

Stationary phase is the basic asymptotic mechanism behind many classical-limit statements in quantum dynamics. It estimates oscillatory integrals of the form

$$I(\lambda) = \int dx\, a(x)e^{i\lambda S(x)}$$

when $\lambda$ is large. In quantum mechanics the large parameter is usually an action divided by $\hbar$, so the same expression is written

$$I(\hbar) = \int dx\, a(x) \exp\left[ \frac{i}{\hbar}S(x) \right], \qquad \frac{S_0}{\hbar}\gg1.$$

The leading contributions come from points where the phase is stationary:

$$S'(x_\star)=0.$$

This is why classical stationary-action paths appear in semiclassical propagators and path integrals. It is destructive interference away from stationary phase, not a statement that nonclassical alternatives were never present.

One-Dimensional Formula

Let

$$I(\lambda) = \int_{-\infty}^{\infty} dx\,a(x)e^{i\lambda S(x)}$$

with $\lambda\gg1$. Suppose $x_\star$ is an isolated nondegenerate stationary point:

$$S'(x_\star)=0, \qquad S''(x_\star)\ne0,$$

with real smooth $S$, smooth $a$, and $\lambda\to+\infty$. For example, take $a$ compactly supported, with only this one stationary point on its support. Suppose also $a(x_\star)\neq0$ and no endpoint contribution. Then

$$I(\lambda) \sim a(x_\star) e^{i\lambda S(x_\star)} e^{i\frac{\pi}{4}\operatorname{sgn}S''(x_\star)} \sqrt{ \frac{2\pi} {\lambda\lvert S''(x_\star)\rvert} }.$$

With the quantum normalization $\lambda=1/\hbar$, the same formula is

$$I(\hbar) \sim a(x_\star) \exp\left[ \frac{i}{\hbar}S(x_\star) \right] e^{i\frac{\pi}{4}\operatorname{sgn}S''(x_\star)} \sqrt{ \frac{2\pi\hbar} {\lvert S''(x_\star)\rvert} }.$$

If there are several isolated stationary points, add their contributions:

$$I(\hbar) \sim \sum_{x_\star} a(x_\star) e^{iS(x_\star)/\hbar} e^{i\frac{\pi}{4}\operatorname{sgn}S''(x_\star)} \sqrt{ \frac{2\pi\hbar} {\lvert S''(x_\star)\rvert} }.$$

The sum can show interference between classical branches. Under these smooth compact-support assumptions, the error after the leading term is $O(\lambda^{-3/2})$ in one dimension. If $a(x_\star)=0$ or leading saddle terms cancel, compute the next nonzero term rather than interpreting a zero leading coefficient as an asymptotic equivalence.

Derivation Sketch

Near a stationary point,

$$S(x) = S(x_\star) + \frac{1}{2}S''(x_\star)(x-x_\star)^2 + \cdots,$$

and

$$a(x)=a(x_\star)+\cdots.$$

Keeping the leading terms gives a Gaussian oscillatory integral:

$$I(\lambda) \approx a(x_\star)e^{i\lambda S(x_\star)} \int dy\, \exp\left[ \frac{i\lambda}{2} S''(x_\star)y^2 \right].$$

The regulated Gaussian integral contributes the magnitude

$$\sqrt{ \frac{2\pi} {\lambda\lvert S''(x_\star)\rvert} }$$

and the phase

$$e^{i\frac{\pi}{4}\operatorname{sgn}S''(x_\star)}.$$

The sign of the second derivative matters because real-time quantum amplitudes are oscillatory. The stationary point is not a probability maximum.

Nonstationary Regions

If $S'(x)$ never vanishes on a region and no endpoint contribution is present, repeated integration by parts suppresses that region. A useful identity is

$$e^{i\lambda S(x)} = \frac{1}{i\lambda S'(x)} \frac{d}{dx} e^{i\lambda S(x)}.$$

After integrating by parts, each step introduces a factor of $1/\lambda$ provided $S'(x)$ stays away from zero and the amplitude is smooth. This is the cancellation mechanism behind stationary phase: rapidly varying phases cancel except where the phase becomes insensitive to first-order changes.

Multidimensional Formula

For

$$I(\lambda) = \int_{\mathbb R^n} d^n x\, a(x)e^{i\lambda S(x)},$$

the stationary points satisfy

$$\nabla S(x_\star)=0.$$

Let $H_\star$ be the Hessian matrix at the stationary point:

$$(H_\star)_{ij} = \left. \frac{\partial^2S} {\partial x_i\partial x_j} \right\rvert_{x_\star}.$$

If the stationary point is nondegenerate,

$$\det H_\star\ne0,$$

then the leading contribution is

$$I(\lambda) \sim a(x_\star) e^{i\lambda S(x_\star)} \left( \frac{2\pi}{\lambda} \right)^{n/2} \frac{ e^{i\pi\sigma_\star/4} } {\sqrt{\lvert\det H_\star\rvert}}.$$

Here

$$\sigma_\star = n_+-n_-,$$

where $n_+$ and $n_-$ are the numbers of positive and negative eigenvalues of $H_\star$. This signature phase is the finite-dimensional ancestor of Maslov-type phases in semiclassical mechanics.

With $\lambda=1/\hbar$,

$$\left( \frac{2\pi}{\lambda} \right)^{n/2} = (2\pi\hbar)^{n/2}.$$

Phase and Hessian

The Hessian controls both the size and phase of the leading contribution:

• $\lvert\det H_\star\rvert^{-1/2}$ measures the local spread of nearby phase contours;
• $\sigma_\star$ gives the Gaussian phase;
• zero Hessian eigenvalues signal that ordinary nondegenerate stationary phase is not enough.

Degenerate directions can arise from symmetries, gauge redundancies, continuous families of saddles, or coalescing classical paths. They must be treated separately. Depending on the problem, the repair may involve collective coordinates, gauge fixing, exact integration over a zero mode, Airy functions, or another uniform approximation.

In semiclassical propagators, Hessian determinants become stability determinants such as the Van Vleck determinant. The dedicated method page is Van Vleck Determinant.

Application to Path Integrals

The real-time path integral for a transition amplitude has the schematic form

$$K(q_f,t_f;q_i,t_i) = \int_{q(t_i)=q_i}^{q(t_f)=q_f} \mathcal Dq\, \exp\left[ \frac{i}{\hbar}S[q] \right].$$

Apply stationary phase first to regulated finite time-sliced integrals, and justify the continuum limit separately. The resulting formal expansion is around paths $q_\gamma(t)$ satisfying

$$\delta S[q_\gamma]=0$$

with the endpoint conditions fixed. For an ordinary Lagrangian system, this gives the Euler–Lagrange equations:

$$\frac{d}{dt} \frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0.$$

Write

$$q(t)=q_\gamma(t)+\eta(t), \qquad \eta(t_i)=\eta(t_f)=0.$$

Then

$$S[q] = S[q_\gamma] + \frac{1}{2}\delta^2S_\gamma[\eta,\eta] + \cdots.$$

The linear term vanishes because $q_\gamma$ is stationary. The quadratic fluctuation operator supplies a determinant and a phase; higher terms give semiclassical corrections. In this way, classical paths organize the amplitude without replacing the full quantum theory.

The full propagator formula is developed in Semiclassical Propagator.

Endpoint and Boundary Contributions

Stationary points are not the only possible leading contributions. Endpoints, boundaries, discontinuities, singularities, and poles can dominate an asymptotic integral. For example, an integral over a finite interval may receive leading contributions from an endpoint even when there is no stationary point in the interior.

For example, $\int_0^1 e^{i\lambda x}\,dx=(e^{i\lambda}-1)/(i\lambda)$ is entirely an endpoint contribution despite having no stationary point. In dynamics, endpoint conditions are part of the problem. A propagator path integral fixes $q_i$ and $q_f$, so variations vanish at the endpoints. A trace path integral imposes periodic boundary conditions. A transition amplitude with coherent states uses different boundary data. The stationary-phase equations and fluctuation determinants depend on those choices.

Common Mistakes

• Treating stationary phase as a probability maximum.
• Applying the nondegenerate formula when $S''(x_\star)=0$ or $\det H_\star=0$.
• Ignoring endpoint contributions.
• Dropping all but one saddle without checking whether several saddles have comparable magnitude.
• Forgetting the Hessian signature phase.
• Saying that the classical path is the only path in the exact path integral.
• Using stationary phase without identifying the large dimensionless parameter.

References

• C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers, Springer, 1999.

• N. Bleistein and R. A. Handelsman, Asymptotic Expansions of Integrals, Dover, 1986.

• M. C. Gutzwiller, Chaos in Classical and Quantum Mechanics, Springer, 1990.

• R. G. Littlejohn, “The semiclassical evolution of wave packets,” Physics Reports 138, 193, 1986.

• NIST Digital Library of Mathematical Functions, §2.3(iv), Method of Stationary Phase.

• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.

• R. Wong, Asymptotic Approximations of Integrals, SIAM, 2001.

Exercises

1. Evaluate $I(\lambda)=\int_{\mathbb R}e^{-x^2/2}e^{i\lambda x^2/2}\,dx$ and compare its leading large-positive-$\lambda$ behavior with stationary phase.

Solution

The convergent Gaussian gives $I(\lambda)=\sqrt{2\pi}(1-i\lambda)^{-1/2}$, on the branch continuous from $\lambda=0$. Thus

$$I(\lambda)=e^{i\pi/4}\sqrt{\frac{2\pi}{\lambda}}\,[1+O(\lambda^{-1})].$$

The stationary point is zero, with $a(0)=1$ and $S''(0)=1$. Both the $\lambda^{-1/2}$ magnitude and Fresnel phase agree with the formula.

2. Why are nonstationary regions suppressed?

Solution

If $S'(x)$ is nonzero on a region, then

$$e^{i\lambda S(x)} = \frac{1}{i\lambda S'(x)} \frac{d}{dx} e^{i\lambda S(x)}.$$

Integrating by parts transfers the derivative onto the slowly varying prefactor. Under suitable smoothness and boundary assumptions, each integration by parts produces a factor of $1/\lambda$. Thus the contribution is suppressed for large $\lambda$.

3. For a three-dimensional saddle with Hessian eigenvalue signs $(+,+,-)$, what is the signature phase?

Solution

Here $n_+=2$ and $n_-=1$, so

$$\sigma=n_+-n_-=1.$$

The multidimensional stationary-phase factor contains

$$e^{i\pi\sigma/4} = e^{i\pi/4}.$$

4. What goes wrong when the Hessian determinant vanishes?

Solution

The nondegenerate formula contains

$$\frac{1}{\sqrt{\lvert\det H_\star\rvert}}.$$

If $\det H_\star=0$, this expression is singular and the quadratic approximation does not control all directions. The saddle may be degenerate, part of a continuous family, affected by a symmetry, or near a caustic where saddles coalesce. One must use a problem-specific repair such as a collective coordinate, gauge fixing, or a uniform approximation.

5. Explain why stationary phase gives the Euler–Lagrange equations in a path integral.

Solution

The path-integral phase is

$$\exp\left[ \frac{i}{\hbar}S[q] \right].$$

Stationary phase requires the first variation to vanish:

$$\delta S[q_\gamma]=0$$

for variations satisfying the endpoint conditions. For an action

$$S[q]=\int L(q,\dot q,t)\,dt,$$

fixed-endpoint variation gives

$$\frac{d}{dt} \frac{\partial L}{\partial\dot q} - \frac{\partial L}{\partial q} = 0.$$

Thus the leading saddle paths are classical Euler–Lagrange trajectories.