From the path integral to the Schrödinger equation

Last section we computed the free-particle propagator and checked that it satisfies the Schrödinger equation — but that was only one example. Now we settle a much larger debt: in chapter 2 we accepted the Schrödinger equation as a postulate; this section proves that once you accept the single rule “sum over all paths, each contributing $\ee^{\ii S/\hbar}$”, the Schrödinger equation is no longer a postulate but a consequence. This is the central calculation of Feynman’s 1948 paper, and every step uses nothing beyond first-year calculus.

Strategy: look at just one short interval

Tackling the finite-time path integral head-on is hard (for a general potential it cannot be done in closed form). But a differential equation only ever talks about “what happens in the next instant” — so all we need is the short-time behaviour.

Take a short interval $\varepsilon$. In the slicing definition, this interval contains just a single step: the particle goes in a straight line from $x'$ to $x$. By the zigzag rule the action of this step is

$$S_\varepsilon\approx\frac{m(x-x')^2}{2\varepsilon}-\varepsilon V\!\Bigl(\frac{x+x'}{2}\Bigr)\tag{11.3.1}$$

The kinetic term uses the average velocity $(x-x')/\varepsilon$; the potential term uses the midpoint value times the duration (to first order in $\varepsilon$, midpoint, endpoint or average all give the same result — see the warning box at the end). The short-time propagator is therefore

$$K_\varepsilon(x,x')=\sqrt{\frac{m}{2\pi\ii\hbar\varepsilon}}\, \exp\!\left[\frac{\ii m(x-x')^2}{2\hbar\varepsilon}-\frac{\ii\varepsilon}{\hbar}V\!\Bigl(\frac{x+x'}{2}\Bigr)\right]\tag{11.3.2}$$

The evolution formula for this one small step becomes an integral equation:

$$\Psi(x,t+\varepsilon)=\int_{-\infty}^{\infty}K_\varepsilon(x,x')\,\Psi(x',t)\,\dd x'\tag{11.3.3}$$

From here the whole game is: expand both sides to first order in $\varepsilon$ and see what equation gets forced out.

Which x′ actually contribute?

The derivation

From the short-time propagator to the Schrödinger equationadvanced~12 minexpand

Step 1: change variables. Let $\eta=x'-x$ (the displacement of this step); the integral equation becomes

$$\Psi(x,t+\varepsilon)=\sqrt{\frac{m}{2\pi\ii\hbar\varepsilon}} \int\dd\eta\;\ee^{\ii m\eta^2/2\hbar\varepsilon}\; \ee^{-\ii\varepsilon V(x+\eta/2)/\hbar}\;\Psi(x+\eta,t)\tag{11.3.7}$$

Step 2: count the orders of each factor. The effective $\eta\sim\sqrt{\hbar\varepsilon/m}$, so $\eta^2\sim\varepsilon$: to keep first order in $\varepsilon$, $\Psi$ must be expanded to second order in $\eta$ — the single most important judgement in the whole derivation (and the origin of the second spatial derivative):

$$\Psi(x+\eta,t)=\Psi+\eta\,\partial_x\Psi+\frac{\eta^2}2\,\partial_x^2\Psi+O(\eta^3)\tag{11.3.8}$$

In the potential factor, $\varepsilon V$ is already first-order small, so $V(x+\eta/2)\to V(x)$ suffices:

$$\ee^{-\ii\varepsilon V/\hbar}=1-\frac{\ii\varepsilon}{\hbar}V(x)+O(\varepsilon^2)\tag{11.3.9}$$

Step 3: do the Gaussian integrals term by term. Three moments are needed (all standard Gaussian integrals, with $\alpha\equiv m/2\hbar\varepsilon$):

$$\int\ee^{\ii\alpha\eta^2}\dd\eta=\sqrt{\frac{\pi\ii}{\alpha}},\qquad \int\eta\,\ee^{\ii\alpha\eta^2}\dd\eta=0,\qquad \int\eta^2\,\ee^{\ii\alpha\eta^2}\dd\eta=\frac{\ii}{2\alpha}\sqrt{\frac{\pi\ii}{\alpha}}\tag{11.3.10}$$

(The second vanishes because the integrand is odd; the third follows from the first by differentiating with respect to $\alpha$.) Insert $\alpha=m/2\hbar\varepsilon$: the zeroth moment exactly cancels the normalisation prefactor $\sqrt{m/2\pi\ii\hbar\varepsilon}$ — total coefficient 1 — and the second moment brings an extra factor $\ii\hbar\varepsilon/m$.

Step 4: collect to first order in ε. Multiply the pieces together and keep only $\varepsilon^1$:

$$\Psi(x,t+\varepsilon)=\Psi(x,t) +\frac{\ii\hbar\varepsilon}{2m}\,\partial_x^2\Psi -\frac{\ii\varepsilon}{\hbar}V(x)\,\Psi +O(\varepsilon^2)\tag{11.3.11}$$

(Cross terms such as $\varepsilon V\times\eta^2\partial_x^2\Psi$ are of order $\varepsilon^2$ and get discarded.)

Step 5: take the limit. Write the left side as $\Psi(x,t)+\varepsilon\,\partial_t\Psi+O(\varepsilon^2)$, cancel $\Psi(x,t)$ from both sides, divide by $\varepsilon$, let $\varepsilon\to0$, and multiply through by $\ii\hbar$:

$$\boxed{\;\ii\hbar\,\frac{\partial\Psi}{\partial t} =-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V(x)\,\Psi\;}\tag{11.3.12}$$

The Schrödinger equation. Chapter 2’s postulate has just become a theorem.

At this point the three formulations formally join forces: Heisenberg’s matrices (1925), Schrödinger’s wave equation (1926), Feynman’s sum over paths (1948) — three formal languages, one quantum mechanics. Chapter 3 proved the first two equivalent; this section supplies the third piece of the puzzle.

∎Key formulas

• Short-time propagator

  $$K_\varepsilon=\sqrt{\frac{m}{2\pi\ii\hbar\varepsilon}}\,\exp\Bigl[\frac{\ii m(x-x')^2}{2\hbar\varepsilon}-\frac{\ii\varepsilon}{\hbar}V\Bigr]$$

  Exact to first order in ε; everything expands from here

• Effective displacement range

  $$|x-x'|\sim\sqrt{\hbar\varepsilon/m}$$

  Gaussian oscillation suppresses distant contributions; displacement ∝ √ε is diffusive scaling

• Key Gaussian moment

  $$\int\eta^2\ee^{\ii\alpha\eta^2}\dd\eta=\frac{\ii}{2\alpha}\sqrt{\frac{\pi\ii}{\alpha}}$$

  The η² moment → the kinetic term −(ħ²/2m)∂ₓ²

• Conclusion of the derivation

  $$\ii\hbar\,\partial_t\Psi=-\frac{\hbar^2}{2m}\partial_x^2\Psi+V\Psi$$

  The Schrödinger equation: demoted from postulate to consequence of the path integral

?Self-check4 questions

1. 1.

   Why must Ψ(x+η) be expanded to second order in η in the derivation, rather than first order?

   • Because the effective η satisfies η²∼ε, so keeping first order in ε requires second order in η
   • Because the first-order term is non-zero and needs a second-order correction
   • Because Gaussian integrals can only handle even powers
   • Because second order is always more accurate than first — the more the better

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2. 2.

   Which are properties of typical paths in the path integral? (Select all that apply.)

   Select all that apply

   • The typical displacement per step ε scales as √ε — the same scaling as Brownian motion
   • They are everywhere continuous but nowhere differentiable; instantaneous velocity is undefined
   • They tend to smooth classical trajectories as ε→0
   • This jaggedness is consistent with the uncertainty principle: finer time slicing means larger momentum fluctuations

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3. 3.

   In the path-integral derivation, the kinetic term −(ħ²/2m)∂ₓ²Ψ in the Schrödinger equation originates from:

   • The second Gaussian moment of the kinetic phase factor e^{imη²/2ħε}
   • The first-order expansion of the potential factor e^{−iεV/ħ}
   • The choice of normalisation constant A
   • The Taylor expansion in the time direction

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4. 4.

   Regarding "evaluate the potential at the midpoint or the endpoint", the correct statement is:

   • For an ordinary potential V(x) the two differ only at higher order and give the same result; for velocity-dependent potentials (such as magnetic coupling) the midpoint rule is mandatory
   • The midpoint must be used in all cases, or the derivation fails
   • Any choice works in all cases — it is purely a matter of notation
   • The endpoint rule corresponds to the correct operator ordering; the midpoint rule is an approximation

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What comes next

With the equivalence proven, the path integral might seem to have merely “re-derived what we already knew”. The next section lets it show off a trick the wavefunction language finds very hard: rotate time onto the imaginary axis, and tunnelling becomes a classical mechanics problem — hiding beneath the barrier is a classical path called the “instanton”.