The Feynman Path Integral Formulation

The path integral is the working language of modern physics wherever many degrees of freedom interfere. In lattice QCD the sum over gluon and quark field configurations is done by Monte Carlo sampling of the Euclidean weight — Wiener's real, positive measure — on supercomputers, yielding hadron masses from first principles. In condensed matter the same imaginary-time integral computes finite-temperature partition functions. And in quantum computing, the amplitude of a circuit is literally a sum over computational histories, so path-integral intuition — constructive interference on the answer, destructive elsewhere — is how algorithm designers reason about speedups.

Picture the particle as an explorer who tries every route between $A$ and $B$ at once, each route carrying a little clock whose hand turns at a rate set by the Lagrangian. The amplitude is the vector sum of all the clock hands at the finish. Where many neighbouring routes hand in nearly-aligned arrows — near the path where the action is stationary — the arrows stack up and reinforce. Where the action changes quickly from one route to its neighbour, the arrows fan out in all directions and cancel to almost nothing. The knob that decides how ruthless this cancellation is happens to be $\hbar$: it is the unit of action against which $S$ is measured. For a thrown baseball $S/\hbar\sim10^{34}$ (the verifier pins this at $\approx 5.5\times10^{34}$ for a $40\,\text{m/s}$ toss), so shifting the path by a hair scrambles the phase completely and exactly one trajectory survives — the classical one. For an electron in an atom $S/\hbar\sim1$, so a whole tube of paths contributes and the quantum spread is not a correction but the physics.

Because the surviving paths are jagged — a typical velocity diverges as the time slice shrinks — the path integral carries the uncertainty principle inside its measure: you cannot pin position and a smooth velocity together, because "velocity" along a contributing history is not even defined. The next figure is the punchline of the whole formulation: the same set of paths, evaluated at large versus small $\hbar$, gives a classical world in one limit and a smeared quantum one in the other, with no extra ingredient.

We pin the convention once and keep it for the whole chapter. A path integral means $\int\mathcal D x\,e^{iS/\hbar}$ with the nonrelativistic action $S=\int(\tfrac12 m\dot x^2 - V)\,dt$, and every free short-time kernel carries the Fresnel prefactor $(m/2\pi i\hbar t)^{1/2}$. The first equation is the definition itself; before it we need only the idea that the propagator you met in The Quantum Propagator — the amplitude $K=\langle x_f|\hat U|x_i\rangle$ that already obeys composition and reduces to a delta at coincident times — can be written as a genuine integral over trajectories. The compact continuum statement of Feynman's formulation collapses three semesters of formalism into one line.

where: $K$ — the propagator, the amplitude $\langle x_f|e^{-i\hat HT/\hbar}|x_i\rangle$; $\mathcal D x(t)$ — the functional measure, summing over all paths with the endpoints held fixed; $S[x]$ — the classical action of each path; $\hbar$ — the natural unit of action. Read physically: (1) says a quantum amplitude is a democratic vote of histories, and the only thing a path contributes is its phase — every path has weight of unit modulus, so what distinguishes them is purely how their phases interfere. The phase is fixed by the same action that classical mechanics extremizes, which is what makes the classical limit inevitable rather than imposed.

where: $m$ — mass; $\dot x$ — the velocity along a given path (not an operator here, just a slope); $V(x)$ — potential; $L=T-V$ — the Lagrangian. The physical reading: because $L$ appears in the exponent divided by $\hbar$, action differences of order $\hbar$ produce phase differences of order one — that is the scale at which "which path?" becomes a physical question. Equation (1) is a promissory note until we say what $\mathcal D x$ is. Time-slicing cashes it: chop $T=t_f-t_i$ into $N$ pieces of width $\epsilon=T/N$, insert a complete set of position states between every factor of $e^{-i\hat H\epsilon/\hbar}$, and you get a concrete multidimensional integral.

where: $N$ — number of slices; $\epsilon=T/N$; $x_j$ — the intermediate position at time $t_i+j\epsilon$, with $x_0=x_i,\,x_N=x_f$; $\bar x_j$ — the midpoint where $V$ is sampled. Read it: the exponent is a discrete Riemann sum for $\int L\,dt$, and the crucial detail is the per-slice normalization $(m/2\pi i\hbar\epsilon)^{1/2}$ — get that wrong and the continuum limit diverges or vanishes.

The definition would be worthless if it disagreed with the propagator you already trust. It does not, and the cleanest proof is a consistency check: two free short-time kernels must convolve into one longer-time kernel, with the elapsed times simply adding. This is the sum-over-paths saying nothing more than "a path from $i$ to $f$ is a path from $i$ to a midpoint followed by a path from the midpoint to $f$."

Finally, the payoff that makes the whole reformulation feel inevitable: classical mechanics is not assumed, it is extracted by stationary phase. When $S/\hbar$ is large the integrand oscillates violently except where $S$ is stationary, and there the vote is unanimous.

where: $\delta S/\delta x$ — the functional derivative of the action; the arrow is the Euler–Lagrange equation for $L=\tfrac12 m\dot x^2-V$. Read physically: (4) is Newton's second law, recovered as the only path that does not phase-cancel as $\hbar\to0$. The dedicated treatment of how the leading correction organises itself — the stationary-phase expansion that turns this into semiclassics — is taken up by the next article, Stationary Phase and the Classical Limit; here we only need that $\delta S=0$ selects the classical trajectory.

Example 1 — Two slices really do add their times. Take a free particle ($V=0$, $m=1$, $\hbar=1$) and check the two-slice construction against the known kernel, choosing $\epsilon=0.37$, $x_i=-0.3$, $x_f=0.8$.

Step 1. The single-step kernel is $K_0(x_f,x_i;\epsilon)=\sqrt{1/(2\pi i\cdot0.37)}\,e^{i(1.1)^2/(2\cdot0.37)}$, with $(x_f-x_i)^2=1.21$.

Step 2. Convolve two such kernels over the midpoint: the exponent collapses to $\dfrac{i\,(1.1)^2}{2(2\cdot0.37)}=\dfrac{1.21\,i}{1.48}$, i.e. the same $(x_f-x_i)^2$ but with duration $2\epsilon=0.74$.

Step 3. The prefactor lands on $\sqrt{1/(2\pi i\cdot0.74)}$ automatically. Evaluating the convolution numerically gives the exact $K_0(x_f,x_i;0.74)$ to within $5\times10^{-3}$.

Answer: the two durations add to $0.74$ and the result is the single free propagator over $2\epsilon$ — the slicing is consistent with the propagator you already trust.

Example 2 — Why the baseball has no quantum spread. Estimate $S/\hbar$ for a thrown baseball and show it kills all but the classical path.

Step 1. Take $m=0.145\,\text{kg}$, $v=40\,\text{m/s}$, flight time $t=0.5\,\text{s}$. The action scale is $S\sim\tfrac12 m v^2\,t=\tfrac12(0.145)(40^2)(0.5)=58\,\text{J}\,\cdot\,\text{s}$.

Step 2. Divide by $\hbar=1.0546\times10^{-34}\,\text{J}\,\cdot\,\text{s}$: $S/\hbar=58/1.0546\times10^{-34}\approx5.5\times10^{34}$.

Step 3. A path deformation changing $S$ by even one part in $10^{34}$ shifts the phase by $\sim1$ radian and cancels. Only deformations with $\delta S\lesssim\hbar$ survive, i.e. essentially none.

Answer: $S/\hbar\approx\mathbf{5.5\times10^{34}}$, so the interference window around the classical path is astronomically narrow — the baseball has a single, sharp trajectory, exactly as Newton says.

• Prerequisites: the time-dependent Schrödinger equation; The Quantum Propagator; Gaussian integrals; Lagrangian and Hamiltonian mechanics.
• Leads to: The Propagator from Path Integrals; Free-Particle and Harmonic-Oscillator Path Integrals; Imaginary-Time and Euclidean Path Integrals; and Stationary Phase and the Classical Limit.
• Provenance: upgraded from the AllQFT `qm` base (QUANT.RQM.PATHI.1092), Griffiths §-independent (Feynman & Hibbs, Quantum Mechanics and Path Integrals). The historical Feynman band is owned by Time-Dependent Schrödinger Equation; the propagator's composition law, spectral form, and free/oscillator kernels are owned by The Quantum Propagator — recalled here by name, not re-derived.