Consider an ensemble of identically prepared pencils balanced on its tip. If a pencil is precisely vertical, ##x=0##, and precisely at rest, ##p=0##, then it will never fall. But some pencils in the ensemble would fall, because according to the uncertainty principle, the standard deviation ##\sigma_x## of position of the ensemble and the standard deviation ##\sigma_p## of momentum of the ensemble is related by ##\sigma_x\sigma_p\geq\hbar/2##. From this inequality, it can be shown that the mean time ##\tau## of fall for an ensemble is at most 4s. Also, the smaller the product ##\sigma_x\sigma_p##, the longer the mean time ##\tau##.(adsbygoogle = window.adsbygoogle || []).push({});

If we set up a large number of pencils and wait sufficiently long, the set of pencils still standing would have a big value of ##\tau##. This would imply that for this set of pencils, ##\sigma_x\sigma_p## must be smaller than ##\hbar/2##, violating the uncertainty principle.

Next consider an ensemble of identically prepared radioactive uranium. Suppose its half life is 4s. If we started off with 100 uranium atoms, then after 8s, we would have roughly 25 atoms left. For this set of 25 atoms, the mean time ##\tau## of decay is greater than 8s. This set of undecayed uranium atoms must somehow be inherently more stable than those 75 that have already decayed. Then the half life of these 25 remaining uranium atoms must be more than 4s. But this contradicts experimental observations. Why is this so?

It seems that even though the mean time ##\tau## of decay is greater than 8s for the set of 25 remaining uranium atoms, they are not more stable than those 75 that have already decayed. Similarly, even though the mean time ##\tau## of fall is greater for the set of remaining pencils, their ##\sigma_x\sigma_p## is still greater than ##\hbar/2##. But why?

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# I Radioactive decay, falling pencils and the uncertainty principle

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