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.

When calculating in a rest frame, doesn't one assume both, definite velocity (zero) and position (origin)? Why is Heisenberg okay with that?

Edit: E.g. For a decay we can do calculations in which we say that the particle is truly at rest with 0 (three vector) momentum. Wouldn't that automatically mean that the particle could be literally everywhere in space? Isn't it overall a little bit troublesome, since nothing can actually really be truly at rest? So is that center of mass frame merely a good enough tool for calculation with no strict real meaning?

share|improve this question
    
" everywhere in space? " but it is in the microcosm. It just means we are not interested where within the measurement uncertainties, both for the momenta defining the rest frame and the space definition of the point of interaction, the rest frame is. It is within the measurement errors which are huge with respect to any nailing down on the scale where the HUP becomes important. Its meaning is within the measurement errors. –  anna v Jun 30 '13 at 6:46
1  
If atoms had consciousness then they would feel very uncertain about their rest frame :) –  anna v Jun 30 '13 at 6:48
1  
have a look at youtube.com/watch?v=oSCX78-8-q0&feature=player_embedded . You can see the uncertainty in position of the atoms from the fuzziness. –  anna v Jun 30 '13 at 6:53
    
Haha, okay, thank you! –  Jack Jun 30 '13 at 12:08

2 Answers 2

up vote 1 down vote accepted

The HUP holds for elementary particle frameworks. A lamda goes into a pion and a proton, and when we calculate the rest frame we never define an (x,y,z,t), we are interested in the momentum and energy four vectors of the produced particles.

lamda

When we say a Kaon hit a proton in the target, and assume the target at rest, the magnitudes of spatial uncertainty of the target, microns in this picture for the target and velocity the proton has due to the temperature of the target, plus the measurement errors on the momentum of the Kaon, a few MeV/c, are below our experimental discrimination and fulfill inevitably the HUP.

share|improve this answer
    
I am sorry, I totally forgot about this question because I was thinking about the second part of your answer - that one couldn't measure that anyway - and then it kinda stopped bothering me any longer. And yes, I guess you are right, the position isn't actually really defined in the first place in these kind of calculations. Thank you! –  Jack Jun 30 '13 at 5:22
    
Hm, on the other hand... Now when I think about it again, unfortunately I am still not really that much satisfied, so I have now edited my question. It would be awesome if you could also clear these doubts. –  Jack Jun 30 '13 at 5:37

I think the answer lies in wave-particle duality.

So, any particle is made up of waves, lots and lots of tiny waves, so it doesn't make sense to say $x = 0$ for all $t$. The closest you can get to a particle-like solution is a Gaussian wave packet that peaks at $x = 0$.

And for that, when you solve for $p$ (the Fourier Transform of $x$) and then approximate for $\delta x$ and $\delta p$, you get the uncertainty relation.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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