Find correlation function $\langle R^2(t_1) R^2(t_2)\rangle$ for 2d stochastic dynamics of polymer

Question

The problem:

Consider 2d stochastic dynamics \begin{equation} \partial_t R_\alpha = \sigma_{\alpha \beta} R_\beta \end{equation} \begin{equation} \langle \sigma_{\alpha \beta}(t_1) \sigma_{\mu \nu}(t_2) \rangle = A (3 \delta_{\alpha \mu} \delta_{\beta \nu} - \delta_{\alpha \beta}\delta_{\mu \nu} - \delta_{\alpha \nu}\delta_{\beta \mu})\delta(t_1-t_2) \end{equation} Find correlation function $\langle R^2(t_1) R^2(t_2)\rangle$ , $t_2 > t_1 $, $R_{\alpha}(0) = R_0 n_\alpha$, where $n_\alpha$ is unit vector

A solution attempt:

I was able only calculate same time correlation functions by solving equation exactly \begin{equation} R_\alpha = U_{\alpha \beta} R_0 n_\beta \qquad U = \hat{T} exp[\int_0^t \hat{\sigma}(t') dt'], \end{equation} where $\hat{T} exp$ means ordered exponentional. I then decompose $\langle (R^2(t+ \Delta t))^2 \rangle$ , leaving only first order of $\Delta t$.

I've obtained differential equation for same time correlation function. Solution is $R_0^4 \exp[2 4 A t]$. However, i am confused what to do, when $t_1 \not = t_2$.

I also managed to find distribution for $\rho = \ln[ R(t) /R_0]$. Here $R(t)$ means modulus of vector $\overline{R}$. \begin{equation} \mathcal{P}(\rho) = \frac{1}{\sqrt{4 \pi A t}} \exp[- \frac{(\rho - 2 A t)^2}{4 A t}] \end{equation}

Answer 1

I think that you probably already know everything necessary to obtain a solution. The form of the equations for $R_\alpha$ and the fact that $\sigma$ is $\delta$-correlated imply that the random process is Markov and homogeneous in the following sense: for $t_2>t_1$ the ratio $$ \frac{R(t_2)}{R(t_1)} \equiv e^\rho $$ does not depend on $R(t_1)$ and on the evolution for $t<t_1$. Then the following formula is valid for the required average $$ \langle R^2(t_1)R^2(t_2)\rangle = \langle R^4(t_1)\left(\frac{R(t_2)}{R(t_1)}\right)^2\rangle = \langle R^4(t_1)\rangle\langle e^{2\rho}\rangle $$ Now, if you have the correct formulas for $\langle R^4(t_1)\rangle$ and ${\cal P}(\rho)$, you only need to calculate the following integral $$ \langle e^{2\rho}\rangle = \int\limits_{-\infty}^\infty\frac1{\sqrt{4\pi At}}\exp\left(-\frac{(\rho-2At)^2}{4At}+2\rho \right)\ d\rho $$ for $t = t_2-t_1$.

Update. Xian-Zu, you did not show how you arrived at the formulas for $\langle R^4(t)\rangle$ and ${\cal P}(\rho)$. So unless there is a guarantee that this result is absolutely correct, I would assume that there is an error in your calculations. My analysis showed that there are the following time dependencies $$ \langle R^2(t)\rangle = R_0^2e^{4At},\quad \langle R^4(t)\rangle = R_0^4 e^{12At},\quad \langle \ln(R(t))\rangle = \ln(R_0) + At, $$ which are consistent with $$ {\cal P}(\rho) = \frac1{\sqrt{2\pi At}}\exp\left(-\frac{(\rho-At)^2}{2At}\right). $$ I also derived the following equation $$ \frac{\partial}{\partial t_2}\langle R^2(t_2) R^2(t_1)\rangle = 4A \langle R^2(t_2) R^2(t_1)\rangle $$ for $t_2>t_1$, from which we get $$ \langle R^2(t_2) R^2(t_1)\rangle = e^{4A(t_2-t_1)}\langle R^2(t_1) R^2(t_1)\rangle = e^{4A(t_2-t_1)}e^{12At_1}R_0^4 $$ The last formula is consistent with what I wrote earlier, it can be written as $$ \langle R^2(t_2) R^2(t_1)\rangle = \langle R^4(t_1)\rangle \langle e^{2\rho(t)}\rangle_{t = t_2-t_1}. $$

Addition 1. Now I see that my and your formulas differ by a factor of 2. Perhaps you took the following equality $$ \int_{t_0}^{t_1} \delta (t_1 - t)dt = 1, $$ and I think that the correct one is: $$ \int_{t_0}^{t_1} \delta (t_1 - t)dt = \frac12. $$

Addition 2. According to Xian-Tzu's commentary, when calculating $\partial\langle R^2(t)\rangle/\partial t$, it is necessary to use the correlation $\langle \sigma_{\alpha\rho}(t_1)\sigma_{\rho\beta}(t_2)$ . This expression, however, implies the sum $$ \sum_{\rho = 1}^2\langle \sigma_{\alpha\rho}(t_1)\sigma_{\rho\beta}(t_2)\rangle $$ which is equal to zero by virtue of the definition of $\sigma$ correlations in the problem statement. We have $$ \sum_{\rho = 1}^2\langle \sigma_{\alpha\rho}(t_1)\sigma_{\rho\beta}(t_2)\rangle = A\delta(t_2-t_1)\sum_{\rho=1}^2\left(3\delta_{\alpha\rho}\delta_{\rho\beta}-\delta_{\alpha\rho}\delta_{\rho\beta}-\delta_{\alpha\beta}\delta_{\rho\rho}\right) = $$ $$ =A\delta(t_2-t_1)\left(3\delta_{\alpha\beta}-\delta_{\alpha\beta}-\delta_{\alpha\beta}\sum_{\rho=1}^21\right) = 0. $$ Therefore, the equation for $\langle R^2(t)\rangle$ takes the form $$ \frac{\partial\langle R^2(t)\rangle}{\partial t} = 4A\langle R^2(t)\rangle. $$ I think that when solving this problem it is better to use the following forms of equalities $$ \dot{R}_\alpha(t) = \sum_\beta\sigma_{\alpha\beta}(t)R_\beta(t), \quad R^2 = \sum_\alpha R_\alpha R_\alpha $$