5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/114133/2451 , physics.stackexchange.com/q/24068/2451 , physics.stackexchange.com/q/54184/2451 and links therein. $\endgroup$ – Qmechanic Sep 22 '16 at 8:18
  • $\begingroup$ @Qmechanic Thank you for the link. It doesn't answer my question though. $\endgroup$ – Deep Sep 22 '16 at 8:45
  • $\begingroup$ $\textit{Exact}$ duplicate? The link provided at top is related but doesn't answer my question. $\endgroup$ – Deep Sep 23 '16 at 4:54
  • $\begingroup$ @ACuriousMind "...even a perfect experiment free of all real-world imprecisions has these uncertainties...". That is exactly my point when I say you can make either $\Delta x$ or $\Delta p$ arbitrarily small until uncertainty principle becomes manifest in the macroscopic world. $\endgroup$ – Deep Sep 23 '16 at 4:59
  • $\begingroup$ This post is not a duplicate. See discussion in chat (make sure you turn on mathjax). $\endgroup$ – DanielSank Sep 23 '16 at 8:35
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +1 to you. I read both links you provided but much of it was beyond my level. So in summary, what you are saying is, invocation of decoherence puts a stop to an uncritical application of uncertainty principle (just at the point where it would become manifest to the macroscopic world). $\endgroup$ – Deep Sep 24 '16 at 4:48
  • $\begingroup$ I'm not clear on what you mean by uncritical application. The uncertainty principle applies to all physical systems all the time. But if you have a state that is very broad in momentum or position, it can be written as a superposition of narrower states and you'll see one of the narrower states. Both the broad state and the narrower state will respect the uncertainty principle. $\endgroup$ – alanf Sep 24 '16 at 9:52
  • $\begingroup$ I thought when you said "Decoherence would select a set of states that are narrow in position and momentum on a macroscopic scale..." you meant that when decoherence sets in, uncertainty principle becomes inapplicable and quantum objects begin to behave like classical ones. $\endgroup$ – Deep Sep 24 '16 at 10:39
  • $\begingroup$ No. Decoherence is a consequence of quantum mechanics. It doesn't contradict quantum mechanics. $\endgroup$ – alanf Sep 24 '16 at 11:37
  • $\begingroup$ So you are saying uncertainty principle will hold even for macroscopic objects, say a football. $\endgroup$ – Deep Sep 25 '16 at 7:28
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +1 to you. So you are saying measurements more precise than a certain threshold are impossible to obtain in practice. $\endgroup$ – Deep Sep 24 '16 at 4:09
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your answer, although I am afraid I am not competent enough to judge it. $\endgroup$ – Deep Sep 26 '16 at 8:54

protected by Qmechanic Sep 23 '16 at 12:58

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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