The Propagator from Path Integrals

You now have two descriptions of the same object. The Quantum Propagator introduced $K(x_f,t_f;x_i,t_i)$ as the kernel of the evolution operator and handed you its properties — composition, the spectral sum over energy eigenstates, the delta function at coincident times — as axioms, verified numerically but taken as given. The Feynman Path Integral Formulation built $K$ from the ground up as $\int\mathcal D x\,e^{iS/\hbar}$ . This article is where the two meet: we take the sum-over-paths definition and show that the propagator's headline property, the composition (Chapman–Kolmogorov) law, is not an axiom at all but a one-line consequence of how paths concatenate. The idea is almost embarrassingly simple once you see it. A path from $i$ to $f$ passing through some intermediate time $t_m$ must occupy some position $x_m$ at that instant. Grouping the sum over all paths by their value at $t_m$ , then summing over $x_m$ , splits the integral into "paths $i\to m$ " times "paths $m\to f$ " — which is exactly $K(f,i)=\int dx_m\,K(f,m)K(m,i)$ . No operator algebra, no completeness relation, just the geometry of trajectories. The delta-function limit is equally transparent: as the elapsed time shrinks, the free kernel becomes a spike of width $\sqrt{\hbar t/m}$ . What is this good for? It is the practical bridge that lets a wavefunction be propagated: $\psi(x_f,t_f)=\int dx_i\,K(x_f,t_f;x_i,t_i)\,\psi(x_i,t_i)$ . Feed in a state, integrate against the kernel, and out comes the evolved state — no differential equation solved. You are ready because you know the free kernel, Gaussian integrals, and the position/momentum bases; here we assemble them into the machine that turns "sum over histories" into a working propagator.

Key takeaways

The path integral gives the propagator that evolves any state: $\psi(x_f,t_f)=\displaystyle\int dx_i\,K(x_f,t_f;x_i,t_i)\,\psi(x_i,t_i)$ .
The composition law $K(f,i)=\int dx_m\,K(f,m)K(m,i)$ is derived from grouping paths by their position at an intermediate time — the Chapman–Kolmogorov identity, which The Quantum Propagator states as an axiom.
Composition is exactly Huygens' principle for amplitudes: every intermediate point is a source of secondary "wavelets."
At coincident times the propagator collapses to a delta: $K(x_f,t;x_i,t)=\delta(x_f-x_i)$ , the identity operator's kernel.
The free kernel is a nascent delta of width $\sqrt{\hbar t/m}$ — it acts as $\delta$ as $t\to0$ and reproduces free Gaussian spreading when you propagate a real state.
KEY IDEA: the propagator's composition law is not postulated in the path-integral picture — it is what "a path is a path-to-a-midpoint followed by a path-from-the-midpoint" says, summed over the midpoint.

The core insight

Grouping the sum over paths by their value at an intermediate time turns the functional integral into a product of propagators integrated over the midpoint — the composition law, derived rather than assumed.

Where this lives today

Propagators are the atoms of every scattering calculation. In quantum field theory each internal line of a Feynman diagram is a propagator, and the composition law derived here is the reason diagrams can be built by gluing sub-diagrams at internal vertices. Numerically, split-operator and Chebyshev methods for the time-dependent Schrödinger equation exploit exactly (2): they compose many short-time propagators (each easy to apply) to evolve a wavepacket over a long time — the workhorse of attosecond and strong-field physics simulations, where an electron's wavefunction is marched through an intense laser pulse one short kernel at a time.

The physical picture

Think of the amplitude as light — not the metaphor, the real thing. Huygens taught that every point a wavefront reaches becomes a source of secondary spherical wavelets, and the next wavefront is their envelope. The path integral says the same thing about quantum amplitude: the amplitude to reach $x_f$ is built by treating every intermediate position $x_m$ as a fresh source, weighting it by the amplitude $K(m,i)$ to have arrived there, and adding up the onward amplitudes $K(f,m)$ . Summing over $x_m$ is Huygens' envelope. This is why The Quantum Propagator calls composition "Chapman–Kolmogorov for waves" and ties it to Huygens' three-century-old picture.

Figure

Composition as Huygens' principle: every midpoint re-emits

The delta limit is the other end of the same intuition. As the elapsed time $t\to0$ the free kernel is a Gaussian whose width $\sqrt{\hbar t/m}$ collapses to nothing: the particle has had no time to move, so the amplitude to be anywhere but its starting point vanishes. That collapsing spike is the identity operator written in the position basis, $K(x_f,t;x_i,t)=\delta(x_f-x_i)$ . The punchline figure shows the kernel narrowing into a delta as $t$ shrinks — the visual signature that the propagator starts life as the identity.

Figure

The free kernel narrows into a delta as t → 0

The core equations

We fix the convention (unchanged from the last article): $K=\int\mathcal D x\,e^{iS/\hbar}$ , free kernel $K_0=\sqrt{m/2\pi i\hbar t}\,e^{im(\Delta x)^2/2\hbar t}$ . The first equation is what makes the propagator useful: it evolves states. If you know the amplitude everywhere at $t_i$ , the amplitude at $t_f$ is the propagator-weighted sum over starting points, because each starting point launches its own bundle of paths.

\psi(x_f,t_f)=\int_{-\infty}^{\infty}\!dx_i\;K(x_f,t_f;x_i,t_i)\,\psi(x_i,t_i) \qquad (1)

where: $K(x_f,t_f;x_i,t_i)=\int\mathcal D x\,e^{iS/\hbar}$ — the sum over paths landing at $x_f$ ; $\psi(x_i,t_i)$ — the initial amplitude, the weight each launch point carries. Read physically: (1) is the Green's-function statement — the propagator is the response to a point source, and a general state is a superposition of point sources, so you convolve. The structural heart of the article is the composition law. In the operator picture it follows from $\hat U(t_f,t_i)=\hat U(t_f,t_m)\hat U(t_m,t_i)$ and inserting completeness; The Quantum Propagator takes that route and calls the result an axiom of the kernel. The path integral derives it geometrically, with no operators at all.

K(x_f,t_f;x_i,t_i)=\int_{-\infty}^{\infty}\!dx_m\;K(x_f,t_f;x_m,t_m)\,K(x_m,t_m;x_i,t_i) \qquad (2)

where: $t_m$ — any intermediate time, $t_i<t_m<t_f$ ; $x_m$ — the (integrated) position at that time. Read it: (2) says amplitudes for consecutive stages multiply and intermediate positions are summed — the quantum analogue of the classical Chapman–Kolmogorov equation for transition probabilities, but with amplitudes (which can interfere) instead of probabilities (which cannot).

Derivation AThe composition law from path concatenation

The goal is to prove (2) directly from the sum-over-paths definition, exhibiting the composition law as a theorem rather than a postulate. The strategy is to slice the path sum at a single intermediate time $t_m$ : every path from $i$ to $f$ passes through exactly one position $x_m$ at $t_m$ , so we can organise the sum by first fixing $x_m$ , summing all paths on each side, and then integrating over $x_m$ . The one thing to watch is that the action is additive over the split, which is what lets the two path sums factor.

Step 1. Split the action along any path at $t_m$ : $S[x]=\int_{t_i}^{t_f}L\,dt=\int_{t_i}^{t_m}L\,dt+\int_{t_m}^{t_f}L\,dt=S_{i\to m}[x]+S_{m\to f}[x]$ . Why: the Lagrangian is local in time, so the action of the whole path is the sum of the actions of its two halves — this additivity is the entire engine of the proof.

Step 2. A path from $i$ to $f$ is uniquely a first leg $i\to(x_m,t_m)$ glued to a second leg $(x_m,t_m)\to f$ . So $\int_{i\to f}\mathcal D x=\int dx_m\int_{i\to(x_m,t_m)}\!\!\mathcal D x_1\int_{(x_m,t_m)\to f}\!\!\mathcal D x_2$ . Why: the measure over all paths factorises into "paths on the left" $\times$ "paths on the right," with the shared endpoint $x_m$ integrated over — geometry, not algebra.

Step 3. Because $e^{iS/\hbar}=e^{iS_{i\to m}/\hbar}e^{iS_{m\to f}/\hbar}$ , the integrand factors too, giving $K(f,i)=\int dx_m\big[\int\mathcal D x_2\,e^{iS_{m\to f}/\hbar}\big]\big[\int\mathcal D x_1\,e^{iS_{i\to m}/\hbar}\big]$ . Why: the exponential of a sum is a product, so the additive action of Step 1 turns into a product of two independent path integrals — one for each leg.

Step 4. Each bracket is itself a propagator: $K(f,i)=\int dx_m\,K(x_f,t_f;x_m,t_m)K(x_m,t_m;x_i,t_i)$ , which is (2). Why: this is Chapman–Kolmogorov for amplitudes. Nothing was assumed about $t_m$ except that it lies between $t_i$ and $t_f$ , so the identity holds for every intermediate time — and iterating it $N$ times regenerates the time-sliced path integral of the previous article, closing the loop. *Verified: the composition law is confirmed symbolically for the free kernel — the residual exponent completes to $(x_f-x_i)^2/(t_1+t_2)$ (durations add) and the prefactors combine to $\sqrt{m/2\pi i\hbar(t_1+t_2)}$ — and numerically to $<5\times10^{-3}$ at the asymmetric split $t_1=0.4,\,t_2=0.6$ .*

The remaining structural fact is the initial condition. A propagator must reduce to "nothing has happened yet" when no time has elapsed, and in the position basis "nothing has happened" is the delta function.

Go deeper · Derivation B · 1 minThe delta-function limit and how the kernel evolves a state

Derivation BThe delta-function limit and how the kernel evolves a state

Here we establish that $K$ starts life as the identity and that (1) genuinely evolves a state. The plan: show the free kernel is a nascent delta of shrinking width, so at $t\to0$ integrating it against any smooth $\psi$ returns $\psi$ unchanged; then confirm the full (1) reproduces a known solution — the free spreading of a Gaussian — so the propagator really is the evolution operator's kernel. The subtlety is that $K_0$ is oscillatory, not a decaying bump, so "delta" here means it acts as a delta under integration, not that $|K_0|$ is peaked.

Step 1. Write the free kernel $K_0(x_f,x_i;t)=\sqrt{m/2\pi i\hbar t}\,e^{im(x_f-x_i)^2/2\hbar t}$ ; its phase oscillates on a length scale $\ell(t)=\sqrt{2\pi\hbar t/m}\to0$ as $t\to0$ . Why: as the oscillation wavelength collapses, the integral of $K_0$ against a smooth test function is dominated by the stationary point $x_f=x_i$ — the defining behaviour of a nascent delta.

Step 2. Consequently $\displaystyle\lim_{t\to0}\int dx_i\,K_0(x_f,x_i;t)\,\psi(x_i)=\psi(x_f)$ , i.e. $K(x_f,t;x_i,t)=\delta(x_f-x_i)$ . Why: this is exactly the initial condition (1) needs — at $t_f=t_i$ the evolved state equals the initial state. The verifier checks this by convolving $K_0$ with a Gaussian test function and watching $(K_0*f)(x)\to f(x)$ as $t\to0.01$ .

Step 3. For finite $t$ , feed an initial Gaussian $\psi_0(x)=(\pi s^2)^{-1/4}e^{-x^2/2s^2}$ into (1). The Gaussian $x_i$ -integral gives $\psi(x,t)=(\pi)^{-1/4}\sqrt{s/\alpha}\,e^{-x^2/2\alpha}$ with $\alpha=s^2+i\hbar t/m$ . Why: the width $\alpha$ acquires an imaginary part $\propto t$ — the packet spreads and its phase curves, precisely the free spreading you know. The verifier reproduces $\psi(0.6,0.5)$ from the path-integral convolution to $<3\times10^{-3}$ . *Verified: the delta limit and the free-Gaussian spreading are both confirmed numerically; the composition law is confirmed symbolically and numerically.*

Limiting cases

case	result	meaning
$t_f\to t_i$	$K\to\delta(x_f-x_i)$	propagator is the identity's kernel
split at $t_m$	$K(f,i)=\int dx_m K(f,m)K(m,i)$	composition = Chapman–Kolmogorov
free particle	$K_0=\sqrt{m/2\pi i\hbar t}\,e^{im\Delta x^2/2\hbar t}$	width $\sqrt{\hbar t/m}$
$\psi_0$ a Gaussian	spreads to $\alpha=s^2+i\hbar t/m$	propagation = Gaussian broadening
stationary spectrum known	$K=\sum_n\psi_n(x_f)\psi_n^*(x_i)e^{-iE_n(t)/\hbar}$	spectral form (owned by the propagator article)
many splits	iterate (2) $N$ times	recovers the time-sliced path integral

Worked examples

Example 1 — Composition with unequal legs. Verify $K_0(f,i;t_1+t_2)=\int dx_m K_0(f,m;t_1)K_0(m,i;t_2)$ for $t_1=0.4,\,t_2=0.6$ , $x_i=-0.5$ , $x_f=1.2$ ( $m=\hbar=1$ ).

Step 1. The exponent of the product is $\dfrac{i}{2}\big[\dfrac{(x_f-x_m)^2}{t_1}+\dfrac{(x_m-x_i)^2}{t_2}\big]$ ; the stationary $x_m$ is $x_m^\star=\dfrac{x_f/t_1+x_i/t_2}{1/t_1+1/t_2}$ .

Step 2. Completing the square, the residual exponent is $\dfrac{i(x_f-x_i)^2}{2(t_1+t_2)}=\dfrac{i(1.7)^2}{2(1.0)}=\dfrac{2.89\,i}{2}$ — the durations have added to $t_1+t_2=1.0$ .

Step 3. The Gaussian $x_m$ -integral converts the two prefactors $\sqrt{1/2\pi i\,0.4}\,\sqrt{1/2\pi i\,0.6}$ into $\sqrt{1/2\pi i\,1.0}$ . Numerically the whole convolution matches $K_0(x_f,x_i;1.0)$ to $<5\times10^{-3}$ .

Answer: composition holds exactly — durations add, prefactor tracks $t_1+t_2$ — with no operator manipulation, straight from path concatenation.

Example 2 — Propagate a Gaussian and watch it spread. Evolve $\psi_0(x)=(\pi)^{-1/4}e^{-x^2/2}$ ( $s=1$ ) for $t=0.5$ using (1), and check against the analytic free-spreading law.

Step 1. Insert the free kernel: $\psi(x_f,t)=\int dx_i\,\sqrt{1/2\pi i\,t}\,e^{i(x_f-x_i)^2/2t}\,(\pi)^{-1/4}e^{-x_i^2/2}$ .

Step 2. The $x_i$ -integral is Gaussian; the result is $\psi(x_f,t)=(\pi)^{-1/4}\sqrt{1/\alpha}\,e^{-x_f^2/2\alpha}$ with $\alpha=1+i\hbar t/m=1+0.5i$ .

Step 3. At $x_f=0.6$ : $\alpha=1+0.5i$ , so $\psi=(\pi)^{-1/4}(1+0.5i)^{-1/2}e^{-0.36/(2(1+0.5i))}$ . The path-integral convolution reproduces this to $<3\times10^{-3}$ .

Answer: the packet's width becomes complex, $\alpha=\mathbf{1+0.5\,i}$ — it spreads and its phase curves, exactly the free-particle spreading, confirming (1) is genuine time evolution.

Origins · 1 minpropagator and how this was discovered

propagator

— · the kernel of quantum evolution

The propagator is the amplitude to go from one spacetime point to another — the Green's function of the Schrödinger equation and the object the entire path-integral chapter is built to compute. Its name captures its job: it propagates a wavefunction forward in time by convolution. Two features make it the quantum workhorse. It composes — the Chapman–Kolmogorov / Huygens law derived above — so long evolutions factor into short ones. And it diagonalizes over energy, $K=\sum_n\psi_n(x_f)\psi_n^*(x_i)e^{-iE_n(t_f-t_i)/\hbar}$ , turning any solved spectrum into a dynamics machine. In relativistic theory the same object, made causal, becomes the Feynman propagator of quantum field theory.

History

1690

— Huygens' principle: each wavefront point re-emits secondary wavelets

1933

— Dirac links the short-time amplitude to

e^{iS/\hbar}

1948

— Feynman writes the propagator as a sum over paths and derives its composition

1949

— Feynman introduces the causal propagator for relativistic perturbation theory

Going further

Prerequisites: The Feynman Path Integral Formulation; The Quantum Propagator; the time-dependent Schrödinger equation; Gaussian integrals; completeness and the position/momentum bases.
Leads to: Free-Particle and Harmonic-Oscillator Path Integrals; Path Integral and the Schrödinger Equation; Imaginary-Time and Euclidean Path Integrals.
Provenance: upgraded from the AllQFT `qm` base (QUANT.RQM.PIPRP.1093), after Feynman & Hibbs. The kernel's composition law, spectral form, δ-limit, and the free/oscillator kernels are owned by The Quantum Propagator (which states composition as an axiom and machine-verifies it to $10^{-14}$ ); this article derives the composition law from path concatenation. The historical Feynman band belongs to Time-Dependent Schrödinger Equation.

Isn't the composition law the same thing the operator formalism proves with completeness?

Yes — it is the same identity, reached two ways. The operator route writes $\hat U(t_f,t_i)=\hat U(t_f,t_m)\hat U(t_m,t_i)$ , sandwiches it between position states, and inserts $\int|x_m\rangle\langle x_m|dx_m=\mathbb 1$ . The path-integral route splits each path at $t_m$ and uses the additivity of the action. They agree because the completeness relation and the path-concatenation geometry are the same statement in different clothes. The value of the path-integral derivation is that it needs no Hilbert-space machinery — it is pure trajectory bookkeeping, which is why it generalizes so cleanly to fields.

|K_0|

is the same everywhere, how can it be a delta function?

Because "delta" is about how $K_0$ acts under integration, not about its pointwise size. The free kernel has constant magnitude $\sqrt{m/2\pi\hbar t}$ but a phase that oscillates with wavelength $\sqrt{2\pi\hbar t/m}$ . As $t\to0$ that wavelength collapses, so integrating $K_0$ against any smooth function leaves only the stationary-phase point $x_f=x_i$ surviving — everything else cancels. That is exactly the defining property of $\delta(x_f-x_i)$ . A decaying-bump nascent delta and an oscillatory one both do the job; the free kernel is the oscillatory kind.

Does the intermediate time

t_m

have to be between

t_i

and

t_f

For the composition law as written, yes — $t_i<t_m<t_f$ — because we split a forward evolution into two forward legs. You can extend it to $t_m$ outside the interval using the propagator's group property $\hat U(t_a,t_b)\hat U(t_b,t_c)=\hat U(t_a,t_c)$ with reversed-time kernels, and $K(x_f,t_f;x_i,t_i)=K^*(x_i,t_i;x_f,t_f)$ handles running backwards. But the clean "sum over paths through $x_m$ " picture is the in-between case; the reversal identity is owned by The Quantum Propagator.

Can I use this to actually compute an evolution, or is it just formal?

It is genuinely computational, and it is how many time-dependent codes work. Split-operator methods write the short-time propagator as $e^{-iV\epsilon/2\hbar}e^{-iT\epsilon/\hbar}e^{-iV\epsilon/2\hbar}$ (the potential part is diagonal in $x$ , the kinetic part in $k$ ), apply it via FFTs, and compose many such short kernels to march $\psi$ forward — exactly iterating (2). The composition law is what guarantees that chaining short, easy steps reproduces the exact long-time evolution up to the controlled $O(\epsilon^2)$ Trotter error.

Why does propagating a Gaussian give a complex width

\alpha=s^2+i\hbar t/m

The imaginary part is the whole point: it encodes both spreading and phase curvature. The real part of $1/\alpha$ controls the (broadening) probability width, and the imaginary part puts a position-dependent phase on the packet — the "chirp" of a spreading wave. You can read it off directly: $|\psi|^2\propto\exp(-x^2\,\mathrm{Re}(1/\alpha))$ widens as $t$ grows, while the phase $\propto x^2\,\mathrm{Im}(1/\alpha)$ is what makes a free packet's different momentum components separate. It is the same $\alpha$ that appears in free wave-packet spreading.

How does the spectral form

K=\sum_n\psi_n\psi_n^* e^{-iE_n t/\hbar}

relate to the path integral?

They are the same $K$ expanded in two different bases: the spectral form uses energy eigenstates, the path integral uses position paths. For an exactly solvable spectrum (oscillator, box) the spectral sum is the fastest route; for a general or field-theoretic problem the path integral is. They must agree, and checking that they do — e.g. resumming the oscillator's path-integral kernel into $\sum_n e^{-i(n+\frac12)\omega t}\psi_n\psi_n^*$ — is a standard consistency test, carried out in the next article. The spectral form itself is owned by The Quantum Propagator.

Does the derivation assume the paths on the two legs are independent?

It assumes they are independently summed, which is correct, not that they are dynamically independent. When you fix the junction point $x_m$ at $t_m$ , a left leg $i\to(x_m,t_m)$ and a right leg $(x_m,t_m)\to f$ can be chosen freely and separately — any left path pairs with any right path, as long as they meet at $x_m$ . That is what makes the measure factor: $\int_{i\to f}\mathcal D x=\int dx_m\big(\int_{i\to m}\mathcal D x_1\big)\big(\int_{m\to f}\mathcal D x_2\big)$ . The coupling between the legs is restored precisely by the integral over the shared $x_m$ , which is where interference between "how you got to $x_m$ " and "where you go next" lives. Drop that integral and you would be fixing the intermediate position — a measurement — which is a different physical situation.