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.


2 Answers 2


$f(X)$ 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.

*It's common to model the electrons as Bloch waves. They consist of a periodic part times a plane wave.

  • $\begingroup$ Thank you for your answer! this is exactly what I was looking for. Just a small question. Could you tell me why in the third case DTFT is used? Is there a simple physical example where this is used? $\endgroup$
    – Apo
    Feb 6, 2018 at 5:52
  • $\begingroup$ It is dictated by the conditions. You can also think of row 3 as being like Fourier series only with the x and q roles swapped. Re-interpret a discrete signal as a continuous signal that is an infinite weighted pulse train and apply the FT you will get a similar situation. DTFT is is related to the Z-transform and the Ordinary Generating Function. Both of these are also on row 3 just like DTFT. Sometimes when we are on row 3 we use Z-transform instead of DTFT. Some examples would be a recurrence relation or a birth death process. It is important as a discrete signal processing tool. $\endgroup$
    – WhatIAm
    Feb 9, 2018 at 4:50

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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Apo
    Feb 5, 2018 at 18:49
  • $\begingroup$ Sorry for the late reply, I assume that the question is already answered. :) $\endgroup$
    – Photon
    Feb 6, 2018 at 17:48
  • $\begingroup$ I am also confused at the fourier transform in finite size case. In textbook(solid state physics or others), it seems that they always assumed period (i.e. translation symmetry), then momentum (or wave number ) q takes 2Pi*n/L. How to understand q in finite system, for example, look at the numerical research (QMC, DMRG,...), they produce some spectra (such as dynamic structure factor S(q,omega)) of finite system. Another example that confuses me is q_y component along rung in two-leg (spin) ladder system, there are only two atoms along the rung. $\endgroup$
    – ZJX
    Jan 9, 2019 at 2:25

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