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.


1 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.

  • 1
    $\begingroup$ Thank you! I appreciate your clear explanation, and it solved my problems. Just to clarify, I suppose you are assuming a fixed value of the control parameters at time $\lambda(s)$, since the correlations are all computed at this fixed value along the protocol. Is that correct? Indeed, in my question, I made a typo, and the difference between the instantaneous conjugate force and its equilibrium mean should have been $\delta X_i(t') = X_i(t')-\langle X_i\rangle_{\lambda(s)}$. (Here, I am using $s$ instead of $t$ for better clarity). $\endgroup$
    – cidrolin
    Jun 12, 2023 at 14:45
  • $\begingroup$ @cidrolin Yes, I have assumed a fix value of the control parameters in this calculation. This is based on the equation (12b) from the paper (the average is taken at $\mathbf{\lambda}(t_0)$. The assumption they make is that "control parameter velocities change on timescales slower than the relaxation time of the system’s force fluctuations." Indeed, the resulting friction matrix doesn't explicitly depend on time, but does depend on one of the control parameters $\omega$. $\endgroup$ Jun 12, 2023 at 17:05

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