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

Make the Wick rotation $t^{\prime }=-i\tau$, and write $T_E=\beta \hslash$ for the final imaginary time, where $\beta$ is inverse temperature and $\hslash$ is the reduced Planck constant. Then $d{t}^{\prime }=-id\tau$ and $dx/d{t}^{\prime }=idx/d\tau$. Thus the continued action isBoth the kinetic energy and the potential energy enter this Euclidean action with positive signs relative to their real-time definitions. Also $e^{-iH(-i\beta \hslash )/\hslash }=e^{-\beta H}$, so the Euclidean path integral isIts boundary conditions fix the two endpoint positions. They need not coincide: identifying and integrating over them would instead give the thermal trace.

The normalization of $\mathcal{D}x$ can be fixed without continuing a formal measure. Apply the Trotter product formula to $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide $T_E$ into $M$ intervals of size $ϵ=T_E/M$, and integrate the intermediate positions. The resulting time-sliced configuration-space path integral isThis also explains the upper limit $\beta \hslash$ rather than $\beta$ when the reduced Planck constant has not been set to one.

For a free particle, vary the Euclidean action with a variation $h$ obeying $h(0)=h(T_E)=0$. Integration by parts givesThe Euler-Lagrange equation is therefore ${\ddot{x}}_0=0$. The fixed boundary conditions determine its unique solution:This classical path is also the unique minimum among paths with finite Euclidean action: the second variation is $m{\int }_0^{T_E}{\dot{h}}^{2}d\tau >0$ for every nonzero admissible variation. Its Euclidean action is

Write $x=x_0+y$, so the fluctuation has homogeneous Dirichlet boundary conditions $y(0)=y(T_E)=0$. Because ${\dot{x}}_0$ is constant, the cross term in the Euclidean action vanishes exactly:ConsequentlyTranslation of each intermediate integration variable has Jacobian determinant one in the time-sliced configuration-space path integral. This is an exact classical-path factorization of a quadratic path integral, givingThe fluctuation Euclidean action, the homogeneous boundary conditions, and the measure normalization contain neither $x_i$ nor $x_f$. Hence $\mathcal{N}$ is independent of both endpoint positions. It depends on $\beta ,m,\hslash$.

With the standard position-state normalization, the Euclidean free-particle bridge amplitude isFor example, a momentum resolution gives a one-dimensional Gaussian integral,whose evaluation gives the same exponent and prefactor. In particular, the result integrates to one over $x_f$ at fixed $x_i$, as required for its heat kernel interpretation.

The integrated stochastic differential equation readsThe Gaussian white noise has zero expected value, soIts Dirac delta covariance givesThus $\boxed{⟨[y(t)-y(0){]}^2⟩=\Gamma t}$, which is also the variance because the displacement has zero expected value. More generally its covariance is $⟨Y(s)Y(t)⟩=\Gamma \min (s,t)$. Equivalently $Y(t)=\sqrt{\Gamma }{W}_t$ for standard Brownian motion, and the diffusion coefficient is $D=\Gamma /2$.

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.