# Why are short distances synonymous with high energies?

Context: I'm thinking about QFT here and why introducing a momentum cutoff means ignoring short distances.

This has been asked before (Why are high energies equivalent to short distances?) but I didn't find any of the answers satisfying. I understand that taking units of $\hbar = c = 1$ gives energy units equal to inverse space units, but that doesn't give me any $physical$ intuition for why short distances is the same as high energy. Another answer given on that thread (the accepted answer, in fact) was about how high energy corresponds to short $wavelengths$, which is true for the massive and massless particles, but I still feel like that is missing a real link to the physical intuition:'stuff' still happens at 'long' wavelengths, and I can't see why distance and energy should be considered the same in this context.

Can somebody help me find the physical intuition here?

## 2 Answers

Think of a one dimensional continuous medium whose behavior we would like to approximate by a collection of linear coupled oscillators. Clearly, the more oscillators we have the better our approximation and the more minute details we can capture. In particular the range of frequencies we can talk about will be very large. But having more oscillators means the distances between each oscillator is really small. So if I decide to reduce the number of oscillators I am reducing the range of frequencies I can consider.

Picking a momentum cut off therefore is another way of picking how many oscillators will be in your approximation and therefore the distance between each oscillator i.e the lattice spacing.

the simplest explanation I have come across is as follows: to probe what's happening on extremely short distance scales requires the use of a wavelength of light that is smaller than that distance scale- otherwise, the scattering of that light will not give you any detailed information about what's going on down there. so for example, the accelerator tube for a benchtop electron microscope might be 3 feet long and the resulting beam energy (~ 1/wavelength) is sufficient to resolve ~10 nanometer-sized features- but to resolve the interior of a proton requires an accelerator tube ~10,000 feet long to get the energy high enough so as to obtain a wavelength short enough.

• Thanks for your response. This idea is fairly good, but bothers me a bit because I used to do scanning probe experiments, and we could see on the Angstrom scale with very low energies. Also, this kind of explanation bothers me because we aren't necessarily trying to "see" anything in the sense of having photons (or electrons) bounce off things into some detector. Like in some Feynman diagram calculation, if I'm trying to imagine what the Feynman diagram represents, there's nothing in there to say that short distances mean high energy, and I'm not trying to image the process. Commented Dec 11, 2017 at 2:47
• an electron can be thought of as having a de broglie wavelength inversely proportional to its energy. increase its energy and you shorten its wavelength. if you ramp up the incident energy of an electron aimed at a proton, at low energies it scatters off the proton as if it were a little sphere but as you increase the beam energy it begins to resolve the interior structure of the proton i.e., you start scattering off the quarks inside... see the book "hunting the quark" by riordan for the full explanation. Commented Dec 11, 2017 at 5:13