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I'm given a system represented by a Hilbert space $L^2([0, 1])$. In this space, the momentum basis is $\frac{1}{\sqrt{L}} \exp{\frac{2 \pi i nx}{L}}$. It seems like Heisenberg uncertainty would be violated for any particle in a momentum eigenstate. So I reasoned that this is somehow equivalent to the particle in an infinite box. But the problem seems to still exist there:

If we have a particle in an infinite box of width 1. We measure its momentum. Now it's in an eigenstate of momentum, such that $\Delta p$ is 0. And $\Delta x \leq 1$ (since the particle must be inside the box). Obviously something is wrong with this reasoning. What's wrong?


marked as duplicate by Dvij Mankad, Jon Custer, Kyle Kanos, John Rennie quantum-mechanics May 23 at 6:13

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