Computing friction tensor for a colloidal particle in (moving) harmonic potential

Question

I am having difficulties reproducing the result for the friction tensor presented in equation (19) of this article.

Here is my understanding: Consider the over-damped Langevin dynamics of a particle diffusing in a one-dimensional harmonic potential, with a Cartesian friction coefficient $\zeta^{\text{c}}$ and control parameters $x_0$ and $\omega$. The over-damped Hamiltonian is given by $V(x|\lambda) = \omega (x-x_0)^2 / 2$, where $\lambda$ represents the control parameters, with $\lambda = (\lambda^i)_{i=1,2}$, $\lambda^1(t) = x_0(t)$, and $\lambda^2(t)=\omega(t)$. The equation of motion is described by the Langevin equation: \begin{equation} \dot{x} = -\frac{1}{\zeta^{\text{c}}}\frac{\partial}{\partial x} V(x|\lambda) + \xi = -\frac{1}{\zeta^{\text{c}}}\omega (x - x_0) + \xi, \end{equation} where $\xi$ is a Gaussian white noise with $\langle \xi(t) \rangle = 0$ and $\langle \xi(t)\xi(t')\rangle \propto\delta(t-t')$.

To compute the mean excess work over a finite-duration protocol, the article employs linear-response theory and introduces the friction tensor $\zeta_{ij}(\lambda(t))$. This tensor is defined as: \begin{equation} \zeta_{ij}(\lambda(t)) = \beta \int_0^{\infty} dt' \langle \delta X_j(0) \delta X_i(t')\rangle_{\lambda(t)}, \end{equation} where $\delta X_i(t') = X_i(t')-\langle X_i\rangle_{\lambda(t')}$ represents the difference between the instantaneous conjugate force and its equilibrium mean. The correlation function $\langle \delta X_j(0) \delta X_i(t')\rangle_{\lambda(t)}$ describes the temporal correlation of forces $X_i$ and $X_j$, which are conjugate to control parameters $\lambda^i$ and $\lambda^j$, respectively. These averages are computed with respect to the canonical distribution of microstates $x$ at $\lambda(t)$, as detailed in equation (1) of the same article.

In the article, the friction tensor is a $2\times2$ matrix given by: $$ \zeta(\lambda) = \begin{pmatrix} \zeta^{\text{c}} & 0 \\ 0 & \frac{\zeta^{\text{c}}}{4\beta\omega^3} \end{pmatrix} $$

which I have been unable to reproduce. I tried computing the conjugate forces $X_1 = \omega (x - x_0)$ and $X_2=-\frac{1}{2}(x - x_0)^2$, and substituting them into the equation for the friction tensor. However, I am not able to solve the resulting time integral.

Could someone guide me through the correct computation of the friction tensor? Any help would be greatly appreciated.

Answer 1

First, for ease of notation let's note that the underlying Langevin equation can be written in terms of $u=x-x_0$. Then, we have

\begin{equation} \dot{u} = -\frac{\omega}{\zeta^c}u + \xi(t), \end{equation} where $\langle \xi(t) \rangle = 0$ and $\langle \xi(t)\xi(t^\prime) \rangle=2 \frac{\zeta^c}{\beta}\delta(t-t^\prime)$.

Then, we can formally write the solution for $u(t)$: \begin{equation} u(t) = \frac{1}{\zeta^c} \int\limits_{-\infty}^t dt^\prime \exp(-(t-t^\prime)/\tau) \xi(t^\prime), \end{equation} where $\tau=\frac{\zeta^c}{\omega}$.

Here, we can immediately note that $\langle x(t) \rangle = 0$ since $\langle \xi(t) \rangle = 0$. Therefore, $\langle X_1 \rangle = 0$ and so $\delta X_1(t) = \omega u(t)$.

Now, we just need to find correlation functions of $u$ and we have the matrix. For $\zeta_{11}$, we need $\langle u(0) u(t) \rangle$: \begin{equation} \begin{split} \langle u(0) u(t) \rangle & = \frac{1}{\left(\zeta^c\right)^2} \int\limits_{-\infty}^t dt_1 \int\limits_{-\infty}^0 dt_2 \exp(-(t-t_1-t_2)/\tau)\langle \xi(t_1) \xi(t_2) \rangle \\ & = \frac{1}{\left(\zeta^c\right)^2} \int\limits_{-\infty}^t dt_1 \int\limits_{-\infty}^0 dt_2 \exp(-(t-t_1-t_2)/\tau) 2 \frac{\zeta^c}{\beta}\delta(t_1-t_2) & = \frac{2}{\beta \zeta^c} \int\limits_{-\infty}^0 dt_2 \exp(-(t-2t_2)/\tau) \\ & = \frac{1}{\beta \zeta^c}\tau \exp(-t/\tau). \end{split} \end{equation}

Then, we can write down our expression for $\zeta_{11}$: \begin{equation} \zeta_{11} = \beta \int\limits_0^\infty dt^\prime \omega^2 \frac{\tau}{\beta \zeta^c} \exp(-t^\prime/\tau) = \frac{\omega^2 \tau^2}{\zeta^c} = \zeta^c. \end{equation} By Isserlis' theorem, the off-diagonal elements (which contain a product of an odd number of $xi$) will be 0. Therefore, it remains to find $\zeta_{22}.$

We can also decompose expectation value of the product of four $\xi$ into the known expectation values of two $\xi$: \begin{equation} \langle \xi(t_1)\xi(t_2)\xi(t_3)\xi(t_4) \rangle = \langle \xi(t_1)\xi(t_2) \rangle \langle \xi(t_3)\xi(t_4) \rangle + \langle \xi(t_1)\xi(t_3) \rangle \langle \xi(t_2)\xi(t_4) \rangle + \langle \xi(t_1)\xi(t_4) \rangle \langle \xi(t_2)\xi(t_3) \rangle. \end{equation}

This will allow us to compute $\langle u(0)^2 u(t)^2 \rangle$ in a similar way to above, but more tedious (you can fill in some of the missing steps, they are straightforward): \begin{equation} \begin{split} \langle u(0)^2 u(t)^2 \rangle & = \frac{4}{\beta^2 \left(\xi^c\right)^2} \int\limits_{-\infty}^t dt_1 \int\limits_{-\infty}^t dt_2 \int\limits_{-\infty}^0 dt_3 \int\limits_{-\infty}^0 dt_4 \exp\Big(-(2t - t_1 -t_2 - t_3 -t_4)\Big) \times \Big(\delta(t_1-t_2)\delta(t_3-t_4)+(\delta(t_1-t_3)\delta(t_2-t_4)+(\delta(t_1-t_4)\delta(t_2-t_3)\Big) \\ & = \frac{4}{\beta^2 \left(\xi^c\right)^2} \left(\frac{\tau^2}{4} + \frac{\tau^2 \exp(-2t/\tau)}{4} + \frac{\tau^2 \exp(-2t/\tau)}{4} \right) \\ & = \frac{2}{\beta^2 \omega^2}\exp(-2t/\tau) + \frac{1}{\beta^2 \omega^2}. \end{split} \end{equation} Note that $\langle u^2(t) \rangle = \frac{1}{\beta \omega}$, which means $\langle X_2 \rangle = -\frac{1}{2 \beta \omega}$, and so one can show \begin{equation} \langle X_2(0) X_2(t) \rangle = \left\langle \frac{1}{4}u(0)^2 u(t)^2 \right\rangle - \frac{1}{4} \frac{1}{\beta^2 \omega^2} = \frac{1}{2 \beta^2 \omega^2}\exp(-2t/\tau). \end{equation} Substituting this expression into the integral for the friction coefficient, we have \begin{equation} \zeta_{22} = \beta \int\limits_0^\infty dt^\prime \frac{1}{2 \beta^2 \omega^2}\exp(-2t^\prime/\tau) = \frac{1}{2 \beta \omega^2} \frac{\tau}{2} = \frac{\xi^c}{4 \beta \omega^3}, \end{equation} which is the desired expression.