[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.

Best Answer

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.