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This question already has an answer here:

I quote from the Wikipedia article on Planck length:

According to the generalized uncertainty principle, the Planck length is in principle, within a factor of order unity, the shortest measurable length – and no improvements in measurement instruments could change that.

I also link to an answer provided to this Phys.SE question:

The reason there is only a finite number of ways is that we assume separations smaller than the Planck length can't be distinguished.

I have hear this claimed every once in a while, but I never learned this formally at university. Can it be proven using quantum mechanics/uncertainty principle that the Planck length is the smallest measurable quantity, or is this still an area of debate and opinion in modern physics?

If you have a simple proof, could you please provide.

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marked as duplicate by Ben Crowell, Qmechanic Apr 23 '13 at 20:29

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.

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    $\begingroup$ Related: physics.stackexchange.com/q/28720/17609 $\endgroup$ – Keep these mind Apr 23 '13 at 11:30
  • $\begingroup$ I added "measurable" to your title because, while the Planck length or something of that order may be the shortest measurable length, it is not the shortest length. $\endgroup$ – David Z Apr 23 '13 at 18:37
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    $\begingroup$ What you want to look for are references on the Bekenstein bound or the covariant entropy bound; these say that in quantum gravity theories, there are finitely many degrees of freedom in any bounded region of spacetime, and hence physics cannot be precisely local. Operationally, what happens is that if you try to measure smaller distances, you make a black hole. $\endgroup$ – Matt Reece Apr 23 '13 at 19:05
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    $\begingroup$ Voted to close as a duplicate of the question identified by Gugg. $\endgroup$ – Ben Crowell Apr 23 '13 at 19:39