3
$\begingroup$

It's easy to think of ways to reduce the state of a particle to a position eigenfunction (or at least a narrow spread in position space), whether by trapping the particle in a potential well or by striking it with a probe photon, and we can calculate the precision $\Delta x$ necessary to produce an experimentally significant $\Delta p$ through the [x,p] uncertainty relation. However, despite how much emphasis there is in QM texts on position and momentum eigenfunctions and the [x,p] uncertainty relation, I have not come across an example of an actual experiment in which a momentum measurement could reduce a state to a momentum eigenfunction, with corresponding spread in the position state in accordance with the uncertainty relation. For example in HEP momentum is measured through sequential position measurements in order to establish radius of curvature, but clearly you cannot produce momentum eigenstates by measuring position! Similarly for time-of-flight measurements, or anything else I can think of. Diffraction could be used to establish the expectation value of wavelength (and thus momentum) with an ensemble of identically prepared states, but you can't measure a diffraction pattern with a single particle!

Can someone point out any experiment that measures particle momentum with the result of leaving the state in a momentum eigenfunction?

$\endgroup$
2
$\begingroup$

Use a collimator to filter out particles with wrong direction, and then use a narrow band-pass filter. Such a filter can be made using interference to reflect particles with wavevector outside of the passband. For photons this is typically implemented in photonic crystals, for electrons a superlattice can be made.

$\endgroup$
  • $\begingroup$ Doesn't collimation produce diffraction by constraining position in exactly the kind of way that would make it impossible to end up with a momentum eigenstate? $\endgroup$ – user1247 Feb 21 at 14:46
  • $\begingroup$ @user1247 not necessarily so. If the collimator's aperture is much wider than wavelength of the particle, then diffraction on the output can be negligible. You just have to make sure that the length of the collimator is still sufficient to let the collimator absorb the off-axis particles. $\endgroup$ – Ruslan Feb 21 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.