This question is about the Uncertainty Principle

$$\sigma_x \sigma_p ~\ge ~\frac{\hbar}{2}.$$

Looking at the maths, I understant why the uncertainty in the poistion increases as the uncertainty in the momentum decreases and vice versa, but I don't understand it in terms of an experiment.

For starter, how do you measure a particle's position or momentum? What is stopping me from making two simultaneous measurements, one for position and one for momentum?

(I understand that the Uncertainty Principle is a natural phenomenon, not a limitation of our technology.)

  • 1
    $\begingroup$ The crux of the problem is that in order to measure the position accurately, you need to bounce low wavelength light of the particle. But to measure momentum accurately, you need low frequency light instead. $\endgroup$
    – David H
    Mar 24, 2014 at 9:35
  • $\begingroup$ You could also have a look at how predicting a particle's position works. I think I read a great article about it on medium: dreti.medium.com/…. $\endgroup$ Jan 5, 2021 at 13:58

1 Answer 1


There is nothing like bubble chamber photos to clarify such issues.

Here is an anti Lamda:

antilamdaenter image description here

The part relevant to your question is that we can measure with an accuracy of microns the vertex of the decaying antilamda, and with an accuracy of an MeV/c the momenta of the particles it decays into, an antiproton and a pion. The antiproton identifiable by ionisation and the interaction into five pions . The Heisenberg Uncertainty Principle ,HUP, is fulfilled because the experimental accuracies are much larger than the constraints of the uncertainty principle. This holds true for all measurements in particle physics detectors.

So nothing is keeping you except the crudeness of real measurements with respect to the HUP.

For testing the HUP special experimental setups have to be devised .

This link shows how using the HUP one can estimate energy levels in nuclei and atoms.

  • $\begingroup$ Fantastic! I have actually never seen the HUP expressed so directly in terms of experiment before. $\endgroup$ Mar 25, 2014 at 0:14

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.