5
$\begingroup$

Suppose the Fourier transform $\hat{f}(k)$ (with $k \in \mathbb{R}^d$) is given, and one intends to get some information about its position-space counterpart $f(x)$. When the analytical calculation of the inverse Fourier transform of $\hat{f}(k)$ is not possible, one may still be able to extract useful information by specializing to specific regions of $k$ space; for instance, in statistical physics, it is often customary to study the "macroscopic" properties of, e.g., correlation functions, by examining the $k\to 0$ limit of their Fourier transforms. It appears to me that such a process is somewhat analogous to looking at the Taylor series of a Fourier transform, i.e., \begin{equation} \hat{f}(k) = \hat{f}\big\rvert_{k=0} + k \partial_k\hat{f}\big\rvert_{k=0} + \ldots \end{equation} If one truncates this series and then tries to perform on it the inverse Fourier transformation, $$ \int \frac{dk}{2\pi} e^{ikx} \hat{f}_{\rm trunc}(k), $$ in some cases one might find that the result diverges as $k\to\infty$. However, in many theories, and especially in field theories, there is an upper cutoff for $k$ which determines the range of validity of that theory; such a cutoff often resolves the possible divergence of the inverse Fourier transform.

Question Does the position-space function that is obtained from the inverse transformation of the truncated Taylor series $\hat{f}_{\rm trunc}$, with some cutoff $\Lambda$, approximate the original function $f(x)$ in any sense? otherwise, is there a systematic way of obtaining such an approximate form from its Fourier transform $\hat{f}(k)$?

$\endgroup$
0
$\begingroup$

When you truncate the Taylor expansion around $0$, you are saying that you are interested in modes with long wavelength. These are often the modes that are long-lived, so that for long times they will approximately describe your system. In spirit, it is like doing a coarse graining: you forget about the fast microscopic dynamics and retain only macroscopoic information. In a more rigorous sense, one has $|| \mathcal{F}^{-1} [\hat f_{trunc}](x) - f(x) ||_2 = || \hat f_{trunc}(k) - \hat f (k) ||_2$, so if the approximation of your fourier transform is good in the $L^2$ sense so it will be the approximation of the position space $f(x)$.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. I understand your point about the long-lived modes (having a background in stat phys). In fact, I think my question can be cast as when does the approximation in the Fourier space gets better by including more terms in the Taylor expansion. I also don't have an understanding of 'a better approximation in $L^2$ sense'. $\endgroup$ – SaMaSo Jan 15 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.