Is ${e^{ikx}} \to {x^2}$ when $k \to 0$

Question

I have a question regarding some information from an online video lecture (around 36min, where the exactly statement is at 36min33secs).

Suppose we have a system of $N$ particles, $\left\{ {{{\vec r}_i}(t)} \right\}_{i = 1, \ldots, N}$ are the position vectors of the particles. I was told in the lecture that the so-called self intermediate scattering function is defined as

$${F_s}(k,t) = \frac{1}{N}\left\langle {\sum\limits_{i = 1}^N {{e^{i\vec k \cdot [{{\vec r}_i}(t) - {{\vec r}_i}(0)]}}} } \right\rangle$$

(for homogeneous system, it only depends on the absolute value of $\vec k$). Here, $\left\langle {} \right \rangle$ is the ensemble average.

Furthermore, it is said by the lecturer that when $k \to 0$

$${F_s}(k,t) \to \frac{1}{N}\left\langle {\sum\limits_{i = 1}^N {{{[{{\vec r}_i}(t) - {{\vec r}_i}(0)]}^2}} } \right\rangle$$

but I can't see why. Could anybody give me some help on it?

Answer 1

Your title question's answer is no; the correct asymptotic behaviour is $1$, as you can see by substituting $k=0$. (Oh, by the way, I won't make vectors bold or give them arrows or anything like that.) We can obtain the next-order correction to this approximation, viz.$$F_s(k,\,t)=\frac1N\left\langle\sum_{j=1}^N\left(1+ik\cdot[r_j(t)-r_j(0)]-\frac12\left(k\cdot[r_j(t)-r_j(0)]\right)^2+o(k^2)\right)\right\rangle\\=1+\frac1N ik\cdot\underbrace{\left\langle\sum_j^N[r_j(t)-r_j(0)]\right\rangle}_0-\frac{1}{2N}\left\langle\sum_j^N\left(k\cdot[r_j(t)-r_j(0)]\right)^2\right\rangle+o(k^2)\\=1-\frac{1}{2N}\left\langle\sum_j^N\left(k\cdot[r_j(t)-r_j(0)]\right)^2\right\rangle+o(k^2).$$You can hear Berthier rattle off, without making the coefficients explicit, the existence of the $1,\,k,\,k^2$ terms from 36:36-36:45, although he doesn't mention the $k$ term's coefficient is a vanishing ensemble average. Then, by Cauchy-Schwarz,$$\frac{2|F_s-1|}{k^2}\le\color{blue}{\frac{1}{N}\left\langle\sum_j\left([r_j(t)-r_j(0)]^2\right)\right\rangle}+o(k^2).$$Berthier wrote the blue term on the board, denoting it $\Delta^2$, and noted it's the mean squared displacement. However, he never wrote or otherwise implied this blue term was a $k\to0$ approximation of $F_s$ itself, which is what you misunderstood him as saying. His aim was to discuss how quickly $F_s$ varies around its value of $1$ at $k=0$, by quantifying an upper bound on the $k^2$ coefficient (up to an unimportant multiplicative constant) in the variation.