TITLE: Intuition behind Harmonic Analysis in Analytic Number Theory
QUESTION [37 upvotes]: As far as I know, in analytic number theory, harmonic analysis appears often. The thing is that I would see the proof of some results where they use harmonic analysis,  and I can follow the argument of the proof and it makes sense, but I have no intuition behind why one would consider using harmonic analysis there (other than that using it works...). 
For example, maybe in a proof one has to estimate a sum of the form $\sum f(n)$ and so they would take the Fourier transform and use Poisson summation formula or something and it works. I would understand the proof, but I just have no idea why it was the "right" thing to do  or why it was a good thing to try (other than of course that it worked out).
I know my question is rather vague, but I would appreciate some explanations if possible! Also I would try to modify the question in a better way if anyone has any suggestion. Thank you very much!

REPLY [5 votes]: Consider the Real numbers modulo the integers. You then get a circle which is of course the group $\mathbb{R}/\mathbb{Z}$, this group is compact and we know that that the dual space is a discrete infinite cyclic  group and thus isomorphic to $\mathbb{Z}$. Consider a continuous function $h:\mathbb{R}/\mathbb{Z} \longrightarrow \mathbb{C}$. We have  by the Plancherel theorem  $$h\left(\bar{0}\right) = \sum_{n\in \mathbb{Z}} \hat{h}(n)$$ Now, there is a canonical map from $C_c(\mathbb{R}) \longrightarrow C_c(\mathbb{R}/\mathbb{Z})$, where $C_C(X)$ is the space of continuous functions of compact support on $X$ with values in the complex numbers. Indeed the map is given by $f\mapsto \bar{f}$, where $$\bar{f}(\bar{i})= \sum_{n\in\mathbb{Z}} f(n+i)$$ in particular $$\sum_{n\in\mathbb{Z}} f(n)=\bar{f}\left(\bar{0}\right) = \sum_{n\in \mathbb{Z}} \hat{\bar{f}}(n)$$ it is not very difficult to show that $$\hat{\bar{f}}(n)= \hat{f}(n)$$ where the $\hat{}$ means their respective fourier transfoms i.e. one for the group $\mathbb{R}/\mathbb{Z}$ and one for $\mathbb{R}$. So this is essentially a short proof of poisson summation formula. 
The question is how can one have an intuition of why the Fourier transfom will be useful in this kind of things. The reason in this case is because the sum $\sum_{n\in\mathbb{Z}} f(n)$ turn out to be a trace and thus can be decomposed into a spectral sum. In more general terms, when can you expect this kind of relationship. I would say every time you work with automorphoic forms. An automorphic form is in general a function that lives in $L^2$ of some topological group say $G$ modulo some discrete subgroup $Q$. In number theory we are interested in the case where $$G(\mathbb{Q})\backslash G(\mathbb{A})$$, where $\mathbb{A}$ are the adeles and $\mathbb{Q}$ are the rationals. In this case we can decompose the space $$L^2(G(\mathbb{Q})\backslash G(\mathbb{A}))$$ into factors related to irreducible automorphic representations. This decomposition relies on a measure and as one would expect a "Fourier transform" with respect to this measure. So in short it is natural to think on Fourier transforms every time you see something like functions on $Q\backslash G$ and in the particular case of number theory this is the study of automorphic forms.