# [Physics] Why are position and momentum space examples of Pontryagin duality

dualityfourier transformheisenberg-uncertainty-principlequantum mechanics

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.

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.