Principles of Quantum Mechanics · Problem 21.17

Path Integrals—II — R. Shankar

Physics — Quantum Mechanics · path-integrals-phase-space

Problem

Derive the phase-space path integral ⟨x,t∣x′,t′⟩=∫exp⁡(iℏ∫t′t[px˙−H(p,x)]dt′′)D[x(t′′)]D[p(t′′)]\langle x, t | x', t'\rangle = \int \exp\left(\frac{i}{\hbar}\int_{t'}^{t} \left[p\dot{x} - H(p,x)\right] dt''\right) \mathcal{D}[x(t'')] \mathcal{D}[p(t'')] starting from the position-space path integral, and show how integrating out the momentum gives back the Lagrangian path integral for a particle with kinetic energy p2/2mp^2/2m.
Intuitive explanation

The phase-space path integral is the most general form—it treats position and momentum as independent variables evolving through time. Starting from the position-space path integral, you'll insert complete sets of momentum eigenstates at each time slice to "promote" momentum to a dynamical degree of freedom; this transforms the Hamiltonian constraint into an explicit integral over all momentum paths. The key insight is recognizing that the exponent px˙−H(p,x)p\dot{x} - H(p,x) is the Legendre transform connecting phase space back to configuration space—when you Gaussian-integrate out pp, its equation of motion x˙=∂H/∂p\dot{x} = \partial H/\partial p automatically emerges, and you recover the Lagrangian L=px˙−HL = p\dot{x} - H.

Intuitive explanation

The phase-space path integral is the most general form—it treats position and momentum as independent variables evolving through time. Starting from the position-space path integral, you'll insert complete sets of momentum eigenstates at each time slice to "promote" momentum to a dynamical degree of freedom; this transforms the Hamiltonian constraint into an explicit integral over all momentum paths. The key insight is recognizing that the exponent px˙−H(p,x)p\dot{x} - H(p,x) is the Legendre transform connecting phase space back to configuration space—when you Gaussian-integrate out pp, its equation of motion x˙=∂H/∂p\dot{x} = \partial H/\partial p automatically emerges, and you recover the Lagrangian L=px˙−HL = p\dot{x} - H.

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