Initial equilibrium at $a_I$ gives
$$
\langle p_{\mathbf q,i}(t_1)p_{-\mathbf q,j}(t_1)\rangle
=\delta_{ij}\frac{k_BT}{a_I+\kappa q^2}.
$$
The initial field and later noise are independent, so their cross terms vanish. Define
$$
\Delta t=t_2-t_1.
$$
With the supplied <Gaussian white noise> covariance, the noise contribution is
$$
2k_BT\Gamma\delta_{ij}
\int_{t_1}^{t_2}ds\,e^{-2r(q)(t_2-s)}
=\delta_{ij}\frac{k_BT}{a_F+\kappa q^2}
\left(1-e^{-2r(q)\Delta t}\right),
$$
Using the decay rate found in part (c), it follows that
$$
\langle p_{\mathbf q,i}(t_2)p_{-\mathbf q,j}(t_2)\rangle
=\delta_{ij}\left[
\frac{k_BT}{a_I+\kappa q^2}e^{-2r(q)\Delta t}
+\frac{k_BT}{a_F+\kappa q^2}
\left(1-e^{-2r(q)\Delta t}\right)
\right].
$$
This is the covariance interpolation in a <Gaussian Model A quench>: each mode forgets its initial equilibrium with relaxation time $1/r(q)$ and approaches the final equilibrium variance. Larger-$q$ modes relax faster.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-344/3/d/solution