# particle accelerators and Heisenberg uncertainty principle

In accelerators we shoot very high momentum particles at each other to probe their structure at very small length scales. Has that anything to do with the HUP that addresses the spread of momentum and space?

Related, when we accelerate a proton at exactly, say, 1 GeV, then we know exactly its momentum. But for high momentum particles the Broglie wave length also shrinks, the particles position becomes more precise. But that would violate HUP.

What goes on with high momentum particles and their momentum spread?

thanks

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Good question. My curiosity with this is the role that interaction (via EM) with the accelerator machine itself plays in the spread of the wavefunction. It seems that within a certain limit, you couldn't reliably "shoot" particles from the accelerator at a target due to quantum limits. That would require releasing it into an empty field (I think) in the first place so one role that the babysitting of the particle's trajectory by EM fields plays would be to keep the particle's wave not much larger than the quntum limit. – Alan Rominger Jul 22 '12 at 18:30

The reason accelerators increase in energy is in order to be able to probe smaller distances, the smaller the wavelength the more details as with optics.

The de Broglie wavelength does not describe the location of the particle in space time, that is the function of the wave packet, as elucidated in this note of Hans de Vries: the de Broglie wavelength is a consequence of the HUP.

Those protons in the accelerator are wave packets and do not have a unique frequency and because of HUP, it is not possible to have exactly 1GeV energy.

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Thanks for all the answers! After studying them and some more thinking on my own part, here is my answer to my questions. (Again, any responses highly welcome.)

1. Particles with higher momentum have a shorter de Broglie wave length. The de Broglie wavelength has nothing to do with HUP. The spacing between the wavecrests are shorter for particles with higher momentum. You get a higher resolution, so to speak. But if you have a precise momentum, no matter how high, the probabilties to find the particle in space are represented by sine waves. They are only shorter/denser for higher momentum. (And lower, so that total probability is one.)

2. To get loacalized wavepackets you need a superpostion of many sine waves, i.e. a great uncertainty in your momentum. But how great? According to HUP deviations at the order of h are enough. For particles with large momentum as in particle accelerators, for momenta that are many orders higher than h, already small deviations will localize the particle in space. So the reasoning in point 1. does not matter much. The high resolution waves only work when the momentum is very precise, precise at the order of h.

3. Due to relativity and the speed limit c, and the connection of space and momentum measurement through HUP, there must be also a limit in the localization and narrowing of particles. If momentum reaches relativistic limits, particle creation takes place.

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The thing is,

$\Delta p \Delta x = \frac{\hbar}{2}$ (Check out the size of $h$ versus $1 \text{ GeV}$)

Quick answer: it is a order of magnitude problem

When you have a particle accelerator you don't accelerate your proton to exactly one GeV. Because at one GeV, $\hbar$ becomes insignificantly small and the change to the velocity is even more insignificant due to relativistic effects (at velocities near $c$, particles don't get only a little bit faster, for a lot of energy). Besides that Proton (assuming a ring accelerator) is constantly loosing energy due to it being a charged particle in an Magnetic field. And a Proton consists out of a quark-gluon mixture so the definition to where it is would be very interesting. Electrons have currently no lower bound in their size are a better subject for this kind of question.
Notice that the Heisenberg uncertainty relationship involves only the precision with which the two quantities are known, not their magnitude. In $$\Delta p \Delta x \ge \frac{\hbar}{2}$$ we see $\Delta p$ but not $p$.