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The uncertainty principle says that $\Delta x\Delta p\geq \frac{\hbar}{2}$. The uncertainty principle is to be viewed as a fundamental fact of nature herself, and the principle has nothing to do with measurement limitations. If such an uncertainty does not show itself in the macroscopic world, it is explained, it is because of the smallness of $\hbar$.

But if one accepts this, then it means that in principle one could make as precise measurements as one wishes by making either $\Delta x$ or $\Delta p$ smaller without bounds. The other will rise without bounds, and at some point must be capable of observation in the macroscopic world, no matter how small $\hbar$ is.

If such a large uncertainty has not been observed in the macroscopic world, how can this be reconciled with the idea described above?

ACuriousMind
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Deep
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3 Answers3

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In order to test whether a system is in a state with some wide uncertainty in position or momentum, you would have to do an interference experiment. The smallness of $\hbar$ does not explain why you can't do such an experiment with a macroscopic system.

Rather, what happens is that the system interacts with its environment and undergoes decoherence. Decoherence would select a set of states that are narrow in position and momentum on a macroscopic scale. The system would then exist in each of those states, but the states would be unable to undergo interference as a result of the decoherence. As a result, each version of you would see the system in one of the allowed states, none of which is wide in position and momentum. For more explanation of decoherence see

https://arxiv.org/abs/quant-ph/0306072

https://arxiv.org/abs/1212.3245

and references therein.

alanf
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for theoretical purposes it usually suffices that $\hbar >0$ but let's look at the numbers for once.

using SI units (I'm lazy) $\hbar \approx 10^{-34}$, lets say a nice macroscopic effect would be roughly $1$ in SI units. This means that we would need a measurement of the complementary quantity (position or momentum) thats precise to $34$ digits. The concept of an macroscopic "object" (say a chair, a human, a bowl of lemon ice, ..) becomes nonsensical way before that. (typical atoms are $\approx 10^{-10}$ so you still need $24$ orders). The only feasible way to (directly) see effects of the uncertainty relation is to make both errors small and for that you have to leave the macroscopic realm.

side note: there are of course effects that you can easily observe that are at least in part related to uncertainty. For example the finite thickness (in space as in frequency) of spectral lines.

Bort
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I'm not sure that I've correctly understood your point. In Nature, usual quantum systems are such that the uncertainity is shared in a rather "equal" way between one variable and its conjugate. ($x$ and $p_x$, or $L_z$ and $\varphi$, etc).

Indeed there are cases where the uncertainity of one variable goes to $+\infty$ while the uncertainity of the conjugate variable goes to $0$: these are the "squeezed" states. A typical example is the order parameter (a sort of macroscopic wavefunction) of Bose-Einstein condensates. Generally speaking, a wavefunction can always be written in the form: \begin{equation*} \psi=\sqrt{\rho} e^{i\theta} \end{equation*} where $\rho$ corresponds to the density (number of bosons per unit volume) and $\theta$ is the phase of the wavefunction describing the condensate. If you fragment the condensate in many sites (thanks to an optical lattice), the system is described in terms of the Bose-Hubbard Model. To keep things simple and clear, there is a quantum phase transition, according to the parameters you set: In the Mott insulator phase the uncertainity about $\rho$ goes to zero (i.e. the populations of each site is exactly defined) but the uncertainity on the phases goes to $+\infty$. The opposite happens in the Superfluid phase.
The two different phases exhibit a macroscopic different behaviour.

AndreaPaco
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