Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question already has an answer here:

This is probably a stupid and simple question, but does the heisenberg uncertainty principle set this upper bound? That knowledge of the momentum is limited, so it can't reach a very low value and thus have a very large (visible) debroglie wavelength? I am figuring if you wanted to test this on a macroscopic particle you would need it frozen to near-absolute zero, but I assume QM comes in and makes it impossible to see visible wavelengths with the naked eye. Thanks!

share|improve this question

marked as duplicate by John Rennie, user1504, Manishearth Jun 29 '13 at 10:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
    
Interesting, one of the comments leads off to the question I pose: "You are correct, but I don't know if it is due to the present limitations in knowledge or technology or there is a fundamental theoretical limitation of decoherence. In the first case it might be possible in the future to increase the experimental bounds where interference is observable." I suppose my question is better framed in terms of decoherence. –  walczyk Jun 23 '13 at 17:35
add comment

1 Answer 1

No there isn't.

QM imposes limitation on the accuracy of the "measured" momentum, but that doesn't mean at all that in some moment of time it can't be smaller than what Heisenberg principle tells us, so de Broglie wavelength can be any value (maybe up to universe' scale, but that another question), indeed, we can generate photons of any wave length.

BUT don't mix photon's wave length with de Broglie wave length: Photons are actually exception because they are bosons and massless, in other words, deBroglie wave length is not a physical thing that can be measured by it self, it's actually wave function's wave length, i.e it describes probability wave length, and you can't measure it directly, maybe only indirectly by determining how big/small the objects that particular particle will interact with at highest probability (roughly speaking).

share|improve this answer
add comment

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