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The uncertainty principle states that $\Delta x\Delta p \geq \frac{\hbar}{2}$. So if you know momentum to high precision, you can't know position to high precision.

In the context of accelerator physics though, it's stated as meaning that a beam with impulse $p\geq \frac{\hbar}{\Delta x}$ is required to probe particles with spatial dimensions of around $\Delta x$, or interactions taking place over that kind of length scale. I don't really see why the latter statement follows from the uncertainty principle. As in, if you have a high momentum beam, does that mean the uncertainty in momentum is large and because of that the uncertainty in its position is small? But that's still the beam position that has small uncertainty so why does that help you probe small particles? I have a feeling the answer should be clear to me because it's not really explained anywhere, but I really can't see it.

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    $\begingroup$ Possible duplicate of particle accelerators and Heisenberg uncertainty principle $\endgroup$ – Hal Hollis Oct 12 '18 at 14:40
  • $\begingroup$ The way that it's been explained to me in the past doesn't actually involve the uncertainty principle at all. Instead, the length scale probed by a given beam is on the order of the de Broglie wavelength of a particle with momentum $p$. $\endgroup$ – probably_someone Oct 12 '18 at 15:00

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