3.2.1 Path Integral Formulation#
Prompts
Define the propagator \(K(x, t; x', t')\). Why is it the central object in quantum mechanics? How does it determine the wavefunction at any later time?
Feynman’s path integral says \(K = \int \mathcal{D}[x] \, \mathrm{e}^{\mathrm{i}S/\hbar}\). Why does each path contribute with the same amplitude but different phase? What happens in the \(\hbar \to 0\) limit?
What does “time-slicing” mean? How does inserting complete position bases at each time step turn the functional integral into ordinary integrals?
Connect the path integral to Huygens’ principle (§3.1.2). How are they saying the same thing about waves?
Lecture Notes#
Overview#
Section 3.1.3 ended with the core idea: quantization means action becomes the phase exponent, and quantum mechanics is a sum over all paths. This section makes that precise. We introduce the propagator — the fundamental transition amplitude — express it as Feynman’s path integral (sum over paths weighted by \(\mathrm{e}^{\mathrm{i}S/\hbar}\)), and decode the functional integral via time-slicing into concrete integrals over single time steps. The result: a complete, equivalent reformulation of quantum mechanics that makes the classical limit transparent.
The Propagator: Transition Amplitude#
Propagator Definition
The propagator \(K(x, t; x', t')\) is the quantum amplitude for a particle at position \(x'\) at time \(t'\) to be found at position \(x\) at time \(t > t'\):
where \(\hat{U}(t, t') = \mathrm{e}^{-\mathrm{i}\hat{H}(t-t')/\hbar}\) is the time-evolution operator.
It determines everything: Given any initial state \(\psi(x', t')\), the state at any later time is
\[ \psi(x, t) = \int K(x, t; x', t') \, \psi(x', t') \, \mathrm{d}x' \]The propagator also satisfies a composition property: you can split a path at any intermediate time \(t_m\):
\[ K(x, t; x', t') = \int K(x, t; x_m, t_m) \, K(x_m, t_m; x', t') \, \mathrm{d}x_m \]This follows from inserting the identity \(\int \vert x_m \rangle \langle x_m \vert \, \mathrm{d}x_m\) into the product \(\hat{U}(t, t') = \hat{U}(t, t_m) \hat{U}(t_m, t')\).
Feynman’s Path Integral#
The Path Integral
The propagator equals a functional integral over all paths from \((x', t')\) to \((x, t)\):
where \(S[x] = \int_{t'}^{t} L(x, \dot{x}, \tau) \, \mathrm{d}\tau\) is the action along path \(x(\tau)\).
Each path contributes with unit probability weight \(|\mathrm{e}^{\mathrm{i}S/\hbar}|=1\) but different phase \(S/\hbar\). Paths with similar action interfere constructively; paths with wildly different action oscillate and cancel.
This is the stationary-phase principle from §3.1.3 realized exactly: the classical path (where \(\delta S = 0\)) is surrounded by neighboring paths with nearly equal phase, so they all add coherently. Off the classical path, the phase varies rapidly and neighboring paths cancel. Thus the classical path dominates, and quantum effects emerge from the sum of near-classical paths.
Discussion: Is the Functional Integral Real?
Summing over uncountably many paths seems ill-defined. Yet the path integral must give the same results as the Schrödinger equation—both describe the same physics. The resolution is time-slicing: discretize time, perform finitely many ordinary integrals, then take the continuum limit carefully. The mathematical details are subtle, but the result is rigorous and equivalent to Schrödinger.
The Infinitesimal Propagator: One Time Slice#
For a single small time step \(\delta t\), we focus on the short-time propagator \(K_{\delta t}(x, x')\) from \(x'\) to \(x\) in time \(\delta t\).
For a nonrelativistic particle with Lagrangian \(L = \tfrac{1}{2}m\dot{x}^2 - V(x)\), the action over one slice (midpoint rule for the potential) is
One-step propagator (1D)
The Gaussian integration over momentum yields the short-time propagator for one step \(\delta t\):
For a free particle (\(V=0\)), this reduces to
Derivation: Short-Time Propagator
Over an infinitesimal time step, \(\hat{U}(\delta t) = \mathrm{e}^{-\mathrm{i}\hat{H}\delta t/\hbar}\) with \(\hat{H} = \tfrac{\hat{p}^2}{2m} + V(\hat{x})\). Using the Trotter-Suzuki formula:
Insert momentum and position resolutions of identity: \(\int \frac{\mathrm{d}p}{2\pi\hbar}\) and \(\int \vert x_n \rangle \langle x_n \vert \mathrm{d}x_n\). The momentum integral is Gaussian:
Completing the Gaussian integral over \(p\) gives the one-step propagator (64). The prefactor \((m/2\pi\mathrm{i}\hbar\,\delta t)^{1/2}\) per slice ensures proper normalization across the full time evolution.
Time-Slicing: Turning Paths into Integrals#
Divide the total time interval from \(t'\) to \(t\) into \(N\) small steps of duration \(\delta t = (t-t')/N\). Insert complete sets of position eigenstates at each intermediate time \(t_0 < t_1 < \cdots < t_N\):
The full propagator factorizes:
Each factor \(K_{\delta t}\) is the one-step kernel (64). Multiplying the exponentials builds up the total phase along a discretized path; in the limit \(N \to \infty\) (\(\delta t \to 0\)), this becomes the continuum path integral.
Connection to Huygens’ Principle
Compare this to Huygens’ principle (§3.1.2): each point \(x_n\) at time \(t_n\) acts as a source of matter wavelets that propagate to \(x_{n+1}\), with phase proportional to the action \(S_{\text{slice}}\). The time-sliced path integral is exactly this: a chain of wavelets, each with phase \(\mathrm{e}^{\mathrm{i}S_{\text{slice}}/\hbar}\), summed over all paths.
Summary#
The propagator \(K(x, t; x', t') = \langle x \vert \hat{U} \vert x' \rangle\) is the central transition amplitude; it encodes the full time evolution.
Feynman’s path integral expresses \(K\) as a sum over all histories, each weighted by \(\mathrm{e}^{\mathrm{i}S/\hbar}\). The classical path dominates via stationary phase; quantum paths contribute through interference.
The one-step propagator \(K_{\delta t}\) is given explicitly in (64) (free-particle limit and \(d\)-dimensional prefactor noted there).
Time-slicing converts the functional integral into a sequence of ordinary integrals over intermediate positions, which converges to the continuum path integral as \(N \to \infty\).
Homework#
1. From the definition \(K(x, t; x', t') = \langle x \vert \hat{U}(t, t') \vert x' \rangle\), derive the wavefunction evolution formula:
2. Prove the composition property by inserting a complete set of position states. If \(t' < t_m < t\), show that
3. For a free particle (\(V=0\)) in one dimension, the classical path from \((x', t')\) to \((x, t)\) is a straight line: \(x_{\text{cl}}(\tau) = x' + v(\tau - t')\) where \(v = (x - x')/(t - t')\). Compute the action \(S_{\text{cl}}\) along this path and express it in terms of \(m\), \(\Delta x = x - x'\), and \(\Delta t = t - t'\).
4. Two nearby paths have actions \(S_1\) and \(S_2\) with \(S_1 - S_2 = \delta S\). (a) For what condition on \(\delta S / \hbar\) do these paths interfere constructively? (b) Explain why the classical path \(\delta S = 0\) collects the most constructive interference. (c) Why does the limit \(\hbar \to 0\) select only the classical path?
5. In the time-sliced path integral, why does the normalization factor \((m / 2\pi\mathrm{i}\hbar\,\delta t)^{1/2}\) appear per time slice? What would go wrong if this factor were missing?
6. Explain why the path integral is a third formulation of quantum mechanics, distinct from the Schrödinger and Heisenberg pictures. What new physical insight does it provide? Give one concrete example.