Path Integral and the Schrödinger Equation

By 1948 quantum mechanics already had two complete languages — Heisenberg's matrices and Schrödinger's wave equation — and they were known to be equivalent. Feynman then wrote down a third, seemingly unrelated, starting point: assign every conceivable trajectory from $x_i$ to $x_f$ the phase $e^{iS[x]/\hbar}$ , where $S=\int L\,\mathrm{d}t$ is the classical action, and add them all up. The amplitude is a sum over histories. Nothing about that prescription looks like a differential equation for a wavefunction. So the honest question is: does "add phases over paths" actually reproduce Schrödinger's $i\hbar\,\partial_t\psi=\hat H\psi$ , term for term — the kinetic operator, the potential, the factor of $i$ , everything? It does, and the proof is one of the most instructive short calculations in the subject. The whole argument lives in a single infinitesimal time step. Because the evolution operator $e^{-i\hat H T/\hbar}$ factorizes into a product of tiny slices, all the physics is hidden in the short-time kernel $K_\varepsilon$ . Over an interval $\varepsilon\to0$ only trajectories near the straight line contribute, the action collapses to a simple function of the displacement, and $K_\varepsilon$ becomes a sharp Gaussian in that displacement times a potential phase. You are ready for this because you already command the Schrödinger equation, Gaussian integrals, and the propagator as a matrix element of the evolution operator. Here we run the machine in reverse: propagate $\psi$ by one short-time kernel, Taylor-expand the convolution to first order in $\varepsilon$ , and watch the Gaussian moments turn into spatial derivatives. The second moment builds the kinetic term; the linear-in- $\varepsilon$ phase builds the potential; the $\varepsilon^0$ term fixes the normalization. Out drops the Schrödinger equation, exactly — the closing of the loop that makes the path integral a genuine formulation of quantum mechanics rather than a mnemonic.

Key takeaways

The path integral $K=\int\mathcal D x\,e^{iS/\hbar}$ and the Schrödinger equation must agree; this article proves the agreement from the short-time kernel.
Slicing $e^{-i\hat H T/\hbar}$ into $N$ factors reduces everything to $K_\varepsilon(x,y)=A\,e^{i m(x-y)^2/2\hbar\varepsilon}\,e^{-i\varepsilon V(x)/\hbar}$ with $A=\sqrt{m/2\pi i\hbar\varepsilon}$ .
Propagating $\psi$ by one slice and expanding to $O(\varepsilon)$ uses just three Gaussian moments: $\langle1\rangle=1$ , $\langle\eta\rangle=0$ , $\langle\eta^2\rangle=i\hbar\varepsilon/m$ .
The $\langle\eta^2\rangle$ moment builds the kinetic operator $-(\hbar^2/2m)\partial_x^2$ ; the phase $e^{-i\varepsilon V/\hbar}$ builds $V$ ; the $\varepsilon^0$ term fixes $A$ .
The result is $i\hbar\,\partial_t\psi=-(\hbar^2/2m)\partial_x^2\psi+V\psi$ — the Schrödinger equation, recovered term by term.
The dominant paths have $(\Delta x)^2\sim\varepsilon$ , so $\Delta x/\Delta t\to\infty$ : quantum paths are continuous but nowhere differentiable, the Wiener signature.
Key idea: the entire Schrödinger equation is encoded in one infinitesimal path-integral step — the second Gaussian moment $\langle\eta^2\rangle=i\hbar\varepsilon/m$ becomes the kinetic operator, and the potential phase becomes $V$ .

The core insight

Evolving the wavefunction by a single short-time path-integral kernel and expanding the convolution to first order in $\varepsilon$ converts the Gaussian moment $\langle\eta^2\rangle=i\hbar\varepsilon/m$ into the kinetic operator and the potential phase into $V$ , reproducing the Schrödinger equation exactly.

Where this lives today

The short-time-kernel argument is not just a consistency check; it is the template for how every path integral is defined and simulated. Lattice quantum field theory discretizes spacetime into slices exactly as here, replaces $e^{iS/\hbar}$ by its Euclidean weight, and integrates the resulting high-dimensional Gaussian-plus-interaction numerically by Monte Carlo — the backbone of modern lattice QCD calculations of hadron masses. In quantum computing, the same Trotter slicing that underlies equation (3) is the standard way to compile a Hamiltonian $\hat H$ into a sequence of short-time gates $e^{-i\hat H\varepsilon}$ ; the error you make truncating at $O(\varepsilon^2)$ is precisely the term we dropped in Step 2. Whenever someone "Trotterizes" a simulation, they are running Feynman's 1948 calculation forward instead of backward.

The physical picture

Picture evolving $\psi$ one tiny step $\varepsilon$ at a time. The short-time kernel is a bell-shaped Gaussian centred on $x=y$ with width $\sqrt{\hbar\varepsilon/m}$ — but a Gaussian with an imaginary variance. Convolving $\psi$ with it is like applying a diffusion operator with an imaginary diffusion constant, and that single $i$ is what turns ordinary spreading into the wavelike $\partial_t\psi\sim i\,\partial_x^2\psi$ of free quantum motion. The kinetic energy is not written into the kernel as a term; it is hidden in the shape of the Gaussian, and it emerges only when you ask how the kernel reshapes $\psi$ over its width.

Figure

One short-time step: a narrow Gaussian smears ψ, and the potential adds a phase

The potential does its job in a completely different way. It enters as a pure phase $e^{-iV(x)\varepsilon/\hbar}\approx 1-iV\varepsilon/\hbar$ — first order in $\varepsilon$ , zeroth order in derivatives. So kinetic energy and potential energy, which sit on the same footing in $\hat H=\hat p^2/2m+V$ , arise from utterly different parts of the kernel: one from the second moment of the Gaussian, the other from a scalar phase. Seeing that asymmetry is the whole content of the calculation, and it is why the derivation is worth doing by hand once.

Figure

The three moments that do all the work

Because the relevant displacements scale as $\Delta x\sim\sqrt{\varepsilon}$ , the trajectories that dominate have velocity $\Delta x/\Delta t\sim 1/\sqrt{\varepsilon}\to\infty$ : continuous but nowhere differentiable, exactly the analytic character of Brownian (Wiener) motion rotated to real time. The Schrödinger equation is, in this light, the imaginary-diffusion-constant cousin of the heat equation — an observation we make precise in Imaginary-Time and Euclidean Path Integrals.

The core equations

Everything rests on two statements that must be shown to agree. The first is the path-integral definition of the propagator, the amplitude to travel from one spacetime point to another; the second is its operator definition. We pin the conventions here and use them throughout: the action is $S=\int L\,\mathrm{d}t$ with $L=\tfrac12 m\dot x^2-V(x)$ , and the free propagator carries the prefactor $\sqrt{m/2\pi i\hbar t}$ .

K(x_f,t_f;x_i,t_i)=\int\mathcal D[x(t)]\;e^{\,iS[x(t)]/\hbar},\qquad S=\int_{t_i}^{t_f}\!\Big(\tfrac12 m\dot x^2-V\Big)\mathrm{d}t \qquad (1)

where: $K$ — the propagator, i.e. the transition amplitude from $(x_i,t_i)$ to $(x_f,t_f)$ ; $S$ — classical action along a path; $L$ — Lagrangian; $\mathcal D[x]$ — the functional measure summing over all paths with fixed endpoints. Physically, the construction of this object from time-slicing is owned by The Feynman Path Integral Formulation; here we take it as given and ask what differential law it obeys. The operator definition is the anchor that ties $K$ to ordinary quantum mechanics. It is the position matrix element of the evolution operator.

K(x_f,t_f;x_i,t_i)=\langle x_f|\,e^{-i\hat H (t_f-t_i)/\hbar}\,|x_i\rangle,\qquad \hat H=\frac{\hat p^2}{2m}+V \qquad (2)

where: $\hat H$ — Hamiltonian; $\hbar$ — reduced Planck constant. Reading it physically: $K$ is the kernel that propagates any state, $\psi(x_f,t_f)=\int K\,\psi(x_i,t_i)\,\mathrm dx_i$ — the property established in The Propagator from Path Integrals — so if we can find the equation $K$ satisfies over an infinitesimal step, we have found the equation $\psi$ satisfies. The short-time kernel is where the two definitions meet. For a single slice $\varepsilon$ , equation (1) has only the straight-line path to leading order, and equation (2) is expanded to first order in $\varepsilon$ ; both give the same Gaussian.

K_\varepsilon(x,y)=\sqrt{\frac{m}{2\pi i\hbar\varepsilon}}\;\exp\!\Big[\frac{i m (x-y)^2}{2\hbar\varepsilon}\Big]\;\exp\!\Big[-\frac{i\varepsilon V(x)}{\hbar}\Big] \qquad (3)

where: $x-y=\eta$ is the displacement in one slice; the first exponential is the free (kinetic) part, the second the potential phase. The normalization $\sqrt{m/2\pi i\hbar\varepsilon}$ is not an input — it is fixed by demanding $\int K_\varepsilon\,\mathrm dy=1+O(\varepsilon)$ , i.e. that a static wavefunction stays normalized. That self-consistency is the $\varepsilon^0$ moment below.

Derivation AThe three Gaussian moments of the short-time kernel

Our goal is the three integrals that carry the entire calculation: the zeroth moment (which fixes the normalization $A$ ), the first moment (which vanishes by symmetry), and the second moment (which will become the kinetic operator). The subtlety is that the Gaussian is oscillatory — its variance is imaginary — so the integrals are Fresnel integrals, evaluated by analytic continuation of the ordinary Gaussian formula. We keep the endpoint $x$ fixed and integrate over the displacement $\eta=y-x$ , using the shorthand $a=m/2\hbar\varepsilon$ .

Step 1. Write the free-kernel weight as $w(\eta)=A\,e^{i a\eta^2}$ with $A=\sqrt{m/2\pi i\hbar\varepsilon}$ and $a=m/2\hbar\varepsilon$ . Why: stripping the potential phase (which multiplies everything by a constant $e^{-i\varepsilon V/\hbar}$ handled separately) isolates the pure Gaussian in the displacement.

Step 2. Zeroth moment. The Fresnel integral $\int_{-\infty}^{\infty}e^{i a\eta^2}\mathrm d\eta=\sqrt{\pi/a}\,e^{i\pi/4}=\sqrt{i\pi/a}$ , so $A\!\int e^{i a\eta^2}\mathrm d\eta=1$ . Why: this is exactly the condition that fixes $A$ ; the $e^{i\pi/4}$ Fresnel phase is what supplies the $\sqrt i$ in the denominator of $A$ , so the normalization is consistent, not imposed twice.

Step 3. First moment. $\langle\eta\rangle=A\!\int\eta\,e^{ia\eta^2}\mathrm d\eta=0$ because the integrand is odd. Why: no drift — the kernel is symmetric about $x=y$ , so to leading order it neither pushes $\psi$ left nor right; the physics is pure spreading.

Step 4. Second moment. Differentiating the Fresnel integral with respect to the parameter gives $\langle\eta^2\rangle=A\!\int\eta^2 e^{ia\eta^2}\mathrm d\eta=\dfrac{i}{2a}=\dfrac{i\hbar\varepsilon}{m}$ . Why: this single imaginary number is the seed of the entire kinetic operator. Its magnitude $\hbar\varepsilon/m$ is the mean-square displacement per step (the Wiener scaling $\langle\eta^2\rangle\sim\varepsilon$ ), and its phase $i$ is what will make the kinetic term appear with the right factor of $i$ .

Go deeper · Derivation B · 1 minAssembling the Schrödinger equation at order ε

Derivation BAssembling the Schrödinger equation at order ε

Now we propagate the wavefunction by one slice and match powers of $\varepsilon$ . The strategy: substitute the short-time kernel into $\psi(x,t+\varepsilon)=\int K_\varepsilon(x,y)\psi(y,t)\,\mathrm dy$ , Taylor-expand $\psi(y)$ around $x$ and the potential phase in $\varepsilon$ , then use the three moments. Watch that the $\eta^2$ moment is itself $O(\varepsilon)$ , so a second-derivative term survives at first order even though it looks higher order.

Step 1. Substitute and expand: with $\eta=y-x$ , $\psi(y)=\psi+\eta\,\psi_x+\tfrac12\eta^2\psi_{xx}+\cdots$ and $e^{-i\varepsilon V/\hbar}=1-i\varepsilon V/\hbar+\cdots$ , so

\psi(x,t+\varepsilon)=\Big(1-\tfrac{i\varepsilon V}{\hbar}\Big)\Big[\langle1\rangle\psi+\langle\eta\rangle\psi_x+\tfrac12\langle\eta^2\rangle\psi_{xx}\Big]+O(\varepsilon^2).

Why: the Gaussian kills the odd term and weights the even terms by their moments; the potential factors out because $V(x)$ is evaluated at the fixed endpoint.

Step 2. Insert the moments $\langle1\rangle=1,\ \langle\eta\rangle=0,\ \langle\eta^2\rangle=i\hbar\varepsilon/m$ :

\psi(x,t+\varepsilon)=\Big(1-\tfrac{i\varepsilon V}{\hbar}\Big)\Big[\psi+\tfrac{i\hbar\varepsilon}{2m}\psi_{xx}\Big]+O(\varepsilon^2)=\psi+\varepsilon\Big[\tfrac{i\hbar}{2m}\psi_{xx}-\tfrac{i}{\hbar}V\psi\Big]+O(\varepsilon^2).

Why: the $\varepsilon^0$ term reproduces $\psi$ (consistency), and the cross term $(i\varepsilon V/\hbar)(i\hbar\varepsilon/2m)\psi_{xx}$ is $O(\varepsilon^2)$ and dropped.

Step 3. Match to $\psi(x,t+\varepsilon)=\psi+\varepsilon\,\psi_t+O(\varepsilon^2)$ and read off the $O(\varepsilon^1)$ coefficient:

\psi_t=\frac{i\hbar}{2m}\psi_{xx}-\frac{i}{\hbar}V\psi.

Why: the left side is just the definition of the time derivative; equating first-order terms is the whole matching.

Step 4. Multiply by $i\hbar$ :

i\hbar\,\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V\psi.

Why: the kinetic operator carries the sign and factor set entirely by $\langle\eta^2\rangle$ ; the potential is the scalar phase. The Schrödinger equation has appeared with no extra assumptions.

Limiting cases

case	result	meaning
$V=0$	free short-time kernel, $\langle\eta^2\rangle=i\hbar\varepsilon/m$ only	pure diffusion with imaginary constant → free Schrödinger equation
$\hbar\to0$	Gaussian width $\sqrt{\hbar\varepsilon/m}\to0$	kernel collapses to the classical straight-line path; stationary phase (see Stationary Phase and the Classical Limit)
$t\to-i\tau$	$\langle\eta^2\rangle\to\hbar\varepsilon/m$ (real)	imaginary-time rotation turns the wave kernel into the real diffusion (heat) kernel
finite $T$	composition of $N$ slices	the one-step equation guarantees the full-time evolution by concatenation

Common trapThe kinetic term does not come from a "kinetic phase" in the kernel

A common misreading is to expect the kinetic energy to enter the short-time kernel the same way the potential does — as an explicit $-i\varepsilon (p^2/2m)/\hbar$ phase. It does not. The kinetic operator is never written into $K_\varepsilon$ ; it is generated by the shape of the Gaussian, specifically by the second moment $\langle\eta^2\rangle=i\hbar\varepsilon/m$ . That is why the potential appears at first order in $\varepsilon$ and zeroth order in derivatives, while the kinetic term appears at first order in $\varepsilon$ but second order in derivatives. Treating them symmetrically double-counts the kinetic energy and produces a wrong prefactor.

Worked examples

Example 1 — the second moment, numerically. Compute $\langle\eta^2\rangle$ for the short-time kernel with $m=1$ , $\hbar=1$ , and slice $\varepsilon=0.01$ (natural units), and confirm it is the seed of the kinetic term.

Step 1. The moment formula from the derivation is $\langle\eta^2\rangle=i\hbar\varepsilon/m$ .

Step 2. Substitute: $\langle\eta^2\rangle=i\,(1)(0.01)/(1)=\mathbf{0.01\,i}$ .

Step 3. Interpret. The mean-square displacement per step is $|\langle\eta^2\rangle|=0.01$ , so the typical hop is $\sqrt{0.01}=0.1$ — the Wiener scaling $\sqrt{\hbar\varepsilon/m}$ . The pure imaginary value is what makes the kinetic operator $\tfrac{i\hbar}{2m}\langle\eta^2\rangle/\varepsilon\cdot\partial_x^2=\tfrac{i\hbar}{2m}\cdot\tfrac{i\hbar}{m}\partial_x^2$ … wait, more directly: the coefficient of $\psi_{xx}$ in $\psi_t$ is $\tfrac12\langle\eta^2\rangle/\varepsilon=\tfrac12(i\hbar/m)=i\hbar/2m$ , giving after $\times i\hbar$ the term $-\hbar^2/2m$ . The single number $0.01i$ carries the sign, the factor $i$ , and the magnitude of the kinetic operator.

Example 2 — assembling the operator by hand. Show that the three parts of the short-time kernel map onto the three pieces of $\hat H\psi$ and confirm the term-by-term correspondence for a definite potential, say $V(x)=\tfrac12 m\omega^2 x^2$ .

Step 1. The $\varepsilon^0$ piece: $\langle1\rangle\psi=\psi$ . This is the identity — it says $\psi$ does not change at zeroth order, and it fixed $A=\sqrt{m/2\pi i\hbar\varepsilon}$ .

Step 2. The $\langle\eta^2\rangle$ piece: $\tfrac12\langle\eta^2\rangle\psi_{xx}=\tfrac12(i\hbar\varepsilon/m)\psi_{xx}$ . Dividing by $\varepsilon$ and multiplying by $i\hbar$ gives $-\tfrac{\hbar^2}{2m}\psi_{xx}$ — the kinetic operator.

Step 3. The potential-phase piece: $-\tfrac{i\varepsilon V}{\hbar}\psi=-\tfrac{i\varepsilon}{\hbar}\big(\tfrac12 m\omega^2 x^2\big)\psi$ . Dividing by $\varepsilon$ and multiplying by $i\hbar$ gives $+\tfrac12 m\omega^2 x^2\,\psi=V\psi$ — the potential operator, here the oscillator potential.

Step 4. Sum. $i\hbar\psi_t=-\tfrac{\hbar^2}{2m}\psi_{xx}+\tfrac12 m\omega^2 x^2\psi$ — the oscillator Schrödinger equation, reconstructed entirely from the short-time path-integral kernel. Every piece of $\hat H$ has a distinct kernel origin: identity → $\varepsilon^0$ , kinetic → second moment, potential → linear phase.

Origins · 1 minclassical action and how this was discovered

classical action

central object S = ∫L dt · The action functional whose stationary value gives the classical trajectory (Hamilton's principle) and whose exponential e^{iS/ℏ} weights each path in Feynman's sum over histories — the bridge from the Lagrangian world to quantum amplitudes

The classical action is the object this whole formulation is built on: Hamilton's principle $\delta S=0$ picks the classical path, and Feynman's insight was that quantum mechanics keeps all paths but weights them by $e^{iS/\hbar}$ . In the derivation above, $S$ appears in the short-time kernel as $m\eta^2/2\varepsilon$ (the free part) minus $\varepsilon V$ (the potential part) — the discretized action of a single slice. That the Schrödinger equation drops out of summing these phases is the precise statement that "extremize $S$ classically" and "sum $e^{iS/\hbar}$ quantumly" are two faces of one dynamics.

History

1923

— Wiener constructs a rigorous measure on continuous, nowhere-differentiable paths — the imaginary-time ancestor of the path integral

1933

— Dirac notes that e^{iS/ℏ} "corresponds to" the quantum transformation function over short times

1942

— Feynman's Princeton thesis recasts quantum mechanics as a sum over spacetime histories

1948

— Feynman publishes the path integral and proves its equivalence to the Schrödinger equation by exactly this short-time argument

1949

— Kac turns the imaginary-time version into the rigorous Feynman-Kac formula

Going further

This article closes the loop opened by The Feynman Path Integral Formulation: that article builds the propagator $K=\int\mathcal D x\,e^{iS/\hbar}$ from time-slicing; here we show the same $K$ , applied over one slice, satisfies the Schrödinger equation. The kernel's role as the object that evolves $\psi$ is owned by The Propagator from Path Integrals, and the historical Feynman-vs-Schrödinger story is told in the Time-Dependent Schrödinger Equation. The imaginary-time rotation flagged in the limiting-cases table is developed in Imaginary-Time and Euclidean Path Integrals, and the $\hbar\to0$ collapse to the classical path is taken up by Stationary Phase and the Classical Limit.

Prerequisites. Lagrangian and Hamiltonian mechanics · the Time-Dependent Schrödinger Equation · Gaussian integrals and moments · The Quantum Propagator. Leads to. Imaginary-Time and Euclidean Path Integrals · Stationary Phase and the Classical Limit · lattice field theory.

Why is the second moment imaginary while an ordinary diffusion variance is real?

Because the exponent of the free kernel is $im\eta^2/2\hbar\varepsilon$ , not $-\eta^2/2\sigma^2$ . The variance is $\sigma^2\to i\hbar\varepsilon/m$ , an imaginary "diffusion constant." That single factor of $i$ is the difference between a heat equation (real diffusion, spreading probability) and the Schrödinger equation (imaginary diffusion, spreading amplitude with interference). Rotate time to $t=-i\tau$ and the $i$ disappears — you get genuine diffusion, which is exactly the Euclidean path integral.

Why can I Taylor-expand ψ(y) but not the potential's spatial dependence?

You expand $\psi(y)$ around $x$ because the Gaussian confines $y$ to within $\sqrt{\hbar\varepsilon/m}$ of $x$ , so only a few derivatives of $\psi$ survive at each order in $\varepsilon$ . The potential is evaluated at the fixed endpoint $x$ (a standard mid-point/endpoint prescription for a smooth $V$ ); the difference $V(y)-V(x)$ is $O(\eta)$ , and multiplied by the already- $O(\varepsilon)$ potential phase it is $O(\varepsilon^{3/2})$ — beyond the order we keep. For singular potentials the ordering prescription matters and must be handled with care.

Doesn't a second-derivative term look like it should be higher order than first order in ε?

It would be, if $\langle\eta^2\rangle$ were $O(1)$ . But it is $O(\varepsilon)$ : $\langle\eta^2\rangle=i\hbar\varepsilon/m$ . So $\tfrac12\langle\eta^2\rangle\psi_{xx}$ is genuinely first order in $\varepsilon$ even though it carries two spatial derivatives. This is the crux of the calculation and the reason a naive power counting fails: in the path integral, one factor of $\varepsilon$ "costs" two factors of $\Delta x$ , the Wiener scaling $(\Delta x)^2\sim\varepsilon$ .

Where does the factor of i in front of ∂ₜψ actually come from?

From the second moment. The coefficient of $\psi_{xx}$ in $\psi_t$ is $\tfrac12\langle\eta^2\rangle/\varepsilon=i\hbar/2m$ , which is imaginary. Multiplying through by $i\hbar$ to clear it produces $i\hbar\,\psi_t$ on the left and the real operator $-\hbar^2/2m\,\psi_{xx}$ on the right. So the $i$ multiplying the time derivative is inherited directly from the imaginary variance of the free kernel — no separate assumption of complex time evolution is needed.

Is this derivation exact, or only leading order?

The equation it derives is exact — the Schrödinger equation is recovered with no approximation. What is "leading order" is the single-slice kernel: it agrees with the true short-time propagator only to $O(\varepsilon)$ , with $O(\varepsilon^2)$ corrections (operator-ordering and $V^2$ terms). But the differential equation lives at $O(\varepsilon^1)$ , and the neglected $O(\varepsilon^2)$ terms vanish in the continuum limit $\varepsilon\to0$ . This is exactly the Trotter error in Hamiltonian simulation, controlled by taking more, thinner slices.

How does this relate to the operator-ordering ambiguity in quantum mechanics?

The endpoint at which you evaluate $V$ (initial point, final point, or midpoint) corresponds to different operator orderings of $\hat p$ and $\hat x$ in $\hat H$ . For a potential $V(x)$ with no momentum dependence, all choices agree at $O(\varepsilon)$ and give the same Schrödinger equation. For Hamiltonians with $x$ -dependent kinetic terms (a curved configuration space, or $p\,f(x)\,p$ ) the choice does not drop out, and the path integral must be defined with a specific ordering — the midpoint (Weyl) prescription is the standard one. This subtlety is invisible for the ordinary $\hat p^2/2m+V$ treated here.

Why call the paths "Brownian" if there is no randomness in quantum mechanics?

The analogy is analytic, not probabilistic. Quantum paths and Brownian paths share the same short-distance scaling $\langle(\Delta x)^2\rangle\propto\Delta t$ , so both are continuous and nowhere differentiable, and both are described by a Gaussian short-time kernel. The difference is the factor of $i$ : Brownian motion has a real, positive transition density (a genuine probability), while the quantum kernel is complex and interferes. Wick-rotating time turns the quantum path integral into a true Wiener (Brownian) integral, which is why Wiener's 1923 measure is the rigorous foundation of the Euclidean theory.