For everyday objects the action of any reasonable path is enormous compared with $\hbar$, so $e^{iS/\hbar}$ turns through a great many revolutions from one path to its neighbour. Neighbouring paths then cancel each other — except near a path where the action does not change to first order when the path is varied. There the neighbours arrive in step and add up. The condition $\delta S=0$ is Hamilton’s principle of classical mechanics, and it gives Newton’s law, $m\ddot{x}=-V'(x)$. So the classical path is not chosen by nature; it is what survives the interference of all the others as $\hbar\to0$. The same stationary-phase argument, done for a single integral, gives the leading correction: the propagator becomes $e^{iS_{\text{cl}}/\hbar}$ times a smooth factor, and the WKB wavefunction of the approximation chapter is its fixed-energy form.

A crowd clapping out of time makes a noise; the same crowd clapping in unison makes a beat. Paths far from the classical one clap out of time with their neighbours, while those near it clap together, and only the beat is heard.

The real part of $e^{ix^{2}/h}$ with $h=0.25$: near the stationary point $x=0$ the phase hardly changes and the integrand keeps one sign, while farther out it oscillates ever faster, and neighbouring stretches cancel. As shrinks, only a neighbourhood of width about is left.

For an integral of a rapidly turning phase with a single stationary point $x_{0}$, $f'(x_{0})=0$, , as :

• Away from stationary points the phase turns so fast that the integral cancels to higher orders in $\hbar$ (stated).
• For the path integral the stationary paths satisfy $\delta S=0$ with fixed ends: Hamilton’s principle, whose equation is Newton’s $m\ddot{x}=-V'(x)$.

Change the path by a small $\delta x$ that vanishes at the ends. The action changes by the integral of $m\dot{x}\,\delta\dot{x}-V'\,\delta x$. An integration by parts moves the derivative off , the boundary terms vanishing, and the change is the integral of . It vanishes for every only if the bracket does.

The first-order change of $\int(\tfrac12m\dot{x}^{2}-V)\,dt$.

Integrate by parts; the bracket vanishes because $\delta x$ does at the ends.

Collect the terms.

A bracket whose integral against every $\delta x$ vanishes must itself vanish.

Only paths near the stationary one add up; the others cancel.

When the phase turns fast, neighbouring contributions cancel, except near a point where the phase is stationary: there they add in step.

The integral of $e^{ix^{2}}$ over the whole line converges to $\sqrt{\pi}$ times a phase : its size is .

The paths whose action does not change to first order, with the ends held fixed, are exactly the solutions of Newton’s law.

Varying the path, the term $m\dot{x}\delta\dot{x}$ is integrated by parts; the boundary term vanishes because the ends are held fixed.

Great work! You now know how to:

• explain why the classical path dominates as $\hbar\to0$
• derive Newton’s law from the stationary action
• apply the stationary-phase approximation to an integral