# Fourier Transform and Lattices

I've been always confused Fourier transforms on lattices because some times a continuous version is used and others a discrete version is used. I don't understand well when should use one or the other. What is the reason for using one or the other in each situation, e.g. when there are periodic boundary conditions or when the system is infinite, or if the system is finite?

As an example in the book "Quantum Field Theory approach to condensed matter physics" in the first chapter the Fourier transform of the density distribution is shown as

$$f (X) = \sum_q f(q) \exp(i q X) \, ,$$

yet, when they show the inverse Fourier transform a few equations later they show

$$f(Q) = \int_V f(X) \exp(-iQX) d^3X \, .$$

Why is it a continuous integral for the inverse but not for the first one? Any help with this question would be greatly appreciated or any source explaining this for several distinct cases would be helpful.

$f(X)$ is a periodic function that has a Fourier series. In other words, $f(X)$ is a periodic function and so its Fourier transform has a spectrum at only discrete values of q. Still the transformed output $\tilde f(q)$ is potentially infinite and non-periodic in q-space. Electrons are not discrete in space but form a spread out cloud. It is assumed the cloud exists on an infinite periodic lattice and hence $f(x)$ can be represented by Fourier series. Here is a table of Fourier transforms. Periodic can also be interpreted as "finite" in some contexts. \begin{align*} f(x)& & \tilde f(q) & & \text{example}\\ \hline\text{continuous},\,\text{infinite} & & \text{continuous},\,\text{infinite} & &\text{FT}\\ \text{continuous},\,\text{periodic} & & \text{discrete},\,\text{infinite} & & \text{F.Series}\\ \text{discrete},\,\text{infinite} & & \text{continuous},\,\text{periodic } & &\text{DTFT}\\ \text{discrete},\,\text{periodic } & & \text{discrete},\,\text{periodic } & &\text{DFT}\\ \end{align*}
We are dealing with the 2nd line on this table. If you transform discrete $f(q)$ you will get a periodic $f(x)$. Only in the case were $\tilde f(q)$ is also periodic will one get $f(x)$ being discrete though. If $\tilde f(q)$ is discrete and doesn't vary from point to point, that is called a Dirac comb. A Dirac comb transforms into another Dirac comb. However if you give the individual points on the comb a different scaling factor then it just transforms into a general periodic function.
On a lattice, the momentum $q$ can only take discrete values (multiples of some minimal momentum), this is why you can sum over it but not integrate. However, $x$ is is a continuous variable, so you integrate over it rather than summing.
• Thank you for your answer, as I understand $q$ takes discrete values because There are only discrete positions for the atoms in the system, why not make the inverse FT discrete also?. But anyway my doubt is more general than that particular example, if the atoms where in a ring I think both the FT and the inverse are discrete (I'm not sure though) why is it discrete in that case and not in the particular example? What if the number of atoms where finite but not in a ring (only in a line for example) how could I know if the FT is discrete or continuous? – Pam Feb 5 '18 at 18:49