Why are position and momentum space examples of Pontryagin duality? https://en.wikipedia.org/wiki/Position_and_momentum_space
https://en.wikipedia.org/wiki/Pontryagin_duality
I am trying to understand logic behind the uncertainity principle. And as far as I understand, it follows mathematically if we assume that wave function in momentum space is Fourier transform of the wave function in position space. I tried to dig in and find out why they should be related so, and the only explanation I could find out was Pontryagin duality.
 A: Practically speaking, the full machinery of Pontryagin duality is way more advanced than physicists need to understand the uncertainty principle. There are several ways to "derive" that the momentum-space wavefunction is the Fourier transform of the position-space wavefunction, which depend somewhat on your choice of starting postulates. Here's one common path:
One common starting fundamental postulate is the commutation relation $[\hat{x}, \hat{p}] = i \hbar.$ The most common position-space representation of this commutation relation is $\hat{x} \to x,\ \hat{p} \to -i \hbar \frac{\partial}{\partial x}$. In this representation, taking the inner product of $\langle x |$ and the eigenvalue equation $\hat{p} |p\rangle = p | p \rangle$ gives the differential equation
$$-i \hbar \frac{d\, \psi_p(x)}{dx} = p\, \psi_p(x),$$
which has solution $\psi_p(x) = \langle x | p \rangle \propto e^{(i p x)/\hbar}$. Then to express an arbitrary state $| \psi \rangle$ in the momentum basis, we can use the resolution of the identity
$$ \psi(p) = \langle p | \psi \rangle = \int dx\ \langle p | x \rangle \langle x | \psi \rangle \propto \int dx\ e^{-ipx/\hbar} \psi(x),$$
which is just the Fourier transform. This generalizes straightforwardly into higher dimensions.
BTW, the fact that position-space and momentum-space wavefunctions are Fourier transforms of each other (or more precisely, can be chosen to be Fourier transforms of each other) gives some nice intuition for the uncertainty relation but isn't actually necessary to derive it. All you need is the commutation relation, as I explain here.
A: Seems that one may have a bit of a chicken-and-egg situation here. What comes first, quantum mechanics or the Fourier transform? According to tparker's answer it seem that one should take quantum mechanics as more fundamental and the Fourier transform then follows from it. However, I suspect it is the other way around.
The Fourier properties were more or less imposed at the point where Planck discovered the relationship between energy and frequency $E=\hbar\omega$, which was later extended to a relationship between momentum and the propagation vector ${\bf p}=\hbar{\bf k}$. Due to these relationships, the fathers of quantum mechanics expanded everything in terms of plane waves. Well, plane waves form an orthogonal basis. Hence, such an expansion comes down to a Fourier analysis. As a consequence, one then obtains the Heisenberg uncertainty relationship.
However, one also finds in quantum mechanics some Heisenberg-type uncertainty relations that does not seem to follow directly from a Fourier relationship. For example, consider the uncertainty relationship associated with spin. This begs the question, what is the underlying principle that leads to an uncertainty relationship, which is shared by Fourier analysis?
This underlying principle, in my view, is the notion of mutually unbiased bases. Any inner product between elements from the respective mutually unbiased bases $\langle x|k\rangle$ gives a constant magnitude, independent of the choice of elements (the phase could be different). Any state with a particular representation in one basis will have a representation in a mutually unbiased basis that obeys a Heisenberg-type uncertainty relationship; the width in terms of one representation would be inverse proportional to the width in the other representation.
What does this have to do with Fourier analysis? Well, a Fourier transform is a link between representations in two mutually unbiased bases. This follows from the fact that for these bases $\langle x|k\rangle=\exp(-ixk)$, which means that $|\langle x|k\rangle|=constant$. This property, ultimately leads to the uncertainty relationship as we know it. 
A: I will try to adress this question from a point of view that is useful when building intuition. Thus, see other peoples answers, and references in Wikipedia, to fill in technical details. 
First a comment on theory-building. Much of theoretical physics begins as someones hunch, or educated guess. You then see if you can formalise what you feel should be the case, and use the new framework to make predictions that are tested with experiment (ideally). Iterations will then strengthen your framework, or prove it to be unfeasible. Inevitably, theories will contain some axioms (statements that you assume are true without proving them). Note that these axioms are not necessarily unique, and what you choose to be an axiom, and what is a theorem (consequence of the axioms) is somewhat ambiguous; the distinction can dependent on your personal tastes (though there is commonly some consensus in a field).
One possible axiom for Quantum Mechanics is that: a state of a system can be described as a "wave-function" over the spacial coordinates, like x y z. (This axiom is usually generalised and expressed in terms of Hilbert spaces instead.) 
Then, if a state of a physical system (like a particle) is modelled with a function you might wonder what you can do with this function, and what that can tell you. One obvious thing you can do with functions is decomposing it as sums. Like consider for instance the functions $f$, $g$ and $h$.
$$
f(x) = x^2 + 5x
\hspace{10pt};\hspace{10pt}
g(x) = x
\hspace{10pt};\hspace{10pt}
h(x) = x^2
$$
Clearly, $f$ can be expressed was a sum of $g$ and $h$; if you take 5 of $g$ and 1 of $h$. But there are much more sophisticated ways to do the exact same thing. Like on a interval $x \in[a,b]$, any function can be expressed as an infinite sum of sines and cosines. This is the idea behind Fourier series. Then you can ask yourself if the same thing is possible on an infinite interval, $x\in[- \infty, \infty]$, and it turns out that it is. This is the idea of Fourier transforms. Clearly when you do this you will pick up at lot of technicalities, which are important when actually dealing with Fourier transforms. But in spirit you're doing the same thing as in the trivial example above.
When you do a Fourier transform, how to mix the functions (like 1 of $g$ and 5 of $h$ in the example) is summarised in a second function, called the "transform". This function cannot be a function of $x$ (clearly), so it is a function of something else, let's call the variable $k$ and the function $\tilde{f}(k)$. You can then go back to your physical theory and ask if $\tilde{f}(k)$ has any kind of physical significance. I mean, if it tells you anything useful abut your system.
It turns out that it does; $k$ is actually associated with the momentum $p$ of your system. Where $p = \hbar k$ gives you the correct predictions when compared to experiment. On top of that it ties nicely into other existing theories, and other ideas you hold for true.
Your friends doing the experiments then come back and tell you that they cannot seem to simultaneously nail down the momentum and position of a particle. And you respond that you know why. That it is a consequence of the fact that particles are best modelled by functions, and a well localised function in $x$ becomes a non localised function in $k$ (which is basically $p$), and vise versa. Then doing some further investigations into the theory of Fourier transforms you find that the function that is simultaneously "the most localised" in both $x$ and $k$ is the gaussian function, and from that function you conclude that $\Delta x \Delta p \geq \frac{\hbar}{2}$, where $\Delta x$ is the standard deviation in $x$, and $\Delta p$ is the standard deviation in $p$.
You then make a cup of coffee, and wait for a phone call from Stockholm.
[Edit below:]
The idea that $p = \hbar k$ is called the de Broglie hypothesis, and an intuition about why anyone (like de Broglie) would suggest this hypothesis can be gained by considering special relativity. Basically: If you have a stationary wave alternating in sync, and you travel by it at some (high) speed, the loss of simultaneity will mean that the wave gains a spacial frequency, as well as momentum (relative to you).
See this resource for some pretty animations and further explanations: 3blue1brown about the uncertainty principle.
A: Each uncertainty is a reflection of an existing fundamental symmetry of nature. For example, per the Noether theorem, the time symmetry leads to energy conservation while the space translation symmetry leads to conservation of momentum. The quantum reflection of these symmetries is the time/energy and position/momentum uncertainty. While the Noether theorem is limited to only a few symmetries for technical reasons, it reveals the connection between symmetries and conservation laws and gives an insight on the nature of uncertainty.
