Past exam of the field theories and collective phenomena in condensed matter course of the University of Oxford / 2025 / 1 / e

The Brownian motion displacement has a normal distribution with variance $\Gamma t$. Its transition probability density function is thereforeComparison with the free particle heat kernel obtained in part (c) gives the required matching:This matches whole path distributions after a linear reparametrization, not merely endpoint variances. To see this, divide the Brownian duration $t$ into $M$ slices of size $\delta t$. The independent increments have normal distributions with variance $\Gamma \delta t$, so their joint density contributesFor the Euclidean path integral, the corresponding expression isThe matching condition makes $\Gamma \delta t=\hslash \delta \tau /m$, so both the exponents and all normalization factors coincide. In particular, identifying Brownian time with imaginary time, $t=\beta \hslash$, requires $\Gamma =\hslash /m$, or diffusion coefficient $D=\hslash /(2m)$.

With both endpoints fixed, the normalized path law is a Brownian bridge. The Euclidean path integral itself is the endpoint density times that conditional path law. In the normalized time coordinate $s=\tau /(\beta \hslash )$, its expected value and covariance areThe classical path is therefore the bridge's expected value, while individual Brownian bridge paths fluctuate about it. Typical Brownian motion paths are continuous but nowhere differentiable: the kinetic Euclidean action is interpreted through the time-sliced diffusion path integral, rather than by assuming that a typical path has an ordinary velocity.