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 Incoming links: Time-sliced diffusion path integral

Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 2 / b / Solution 2026-10-20

Take

τ = t_{f} - t_{i} > 0

and split it into

M

intervals of width

Δ t = τ / M

. The markov property and Gaussian convolution give

K = \int_{R^{M - 1}} \prod_{j = 1}^{M - 1} d x_{j} \prod_{j = 0}^{M - 1} \frac{e x p [ - ( x _{j + 1} - x _{j} ) ^{2} / ( 4 D Δ t )]}{4 π D Δ t}, x_{0} = x_{i}, x_{M} = x_{f} .

This finite-dimensional identity is exact for every positive

M

. Define the path-integral measure by these normalized time slices:

\int D x (\dots) = lim_{M \to \infty} (4 π D Δ t)^{- M /2} \int \prod_{j = 1}^{M - 1} d x_{j} (\dots) .

In the time-sliced diffusion path integral, the discrete exponent is

\sum_{j} [(x_{j + 1} - x_{j}) /Δ t]^{2} Δ t / (4 D)

. Its conventional continuum notation is therefore

K (x_{f}, t_{f} ∣ x_{i}, t_{i}) = \int_{x_{i}}^{x_{f}} D x exp [- \int_{t_{i}}^{t_{f}} \frac{x ˙ ^{2}}{4 D} d t] .

The time-slice normalization is essential, not an arbitrary multiplicative constant. Brownian paths are almost surely nondifferentiable, so the displayed action is shorthand for the Gaussian limiting measure rather than an ordinary integral of the derivative of each typical path.

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