1
$\begingroup$

So we know that for the really small world we have quantum mechanical behavior and for big things we have classical behavior. But what is the boundary that differentiates the two? If we make a thought experiment with a maze getting smaller and smaller,when can we say that it goes from the classical world to the quantum mechanical and does it has to do only with the mass and size of an object?

$\endgroup$
1

1 Answer 1

2
$\begingroup$

Quantum mechanics, excluding gravity, is valid at all scales. In principle, you can do classical mechanics via quantum mechanics, but this is just very hard because of the number of particles involved and decoherence. Therefore, you switch to classical mechanics, which describes an effective theory of large scales (and small energies), because most of the "quantum" effects are so small due to e.g. decoherence, they don't matter at all. The idea is that if the number of particles is infinite (and we take similar limits also for other parameters), we obtain classical mechanics, which means that classical mechanics is never truly valid in real life, but since all the differences are well below experimental precision, we don't care.

There are, however, certain effects that you can never describe with classical mechanics in a satisfactory way, such as the photoelectric effect.

In this sense, there is no boundary. In practice, you'll need to include quantum mechanical effects if the size of the objects approaches something like one nanometre (+- a couple of orders of magnitude) and/or the number of particles is well below the Avogadro constant. But this is just a rough estimate. One thread discussing this experimental boundary to some degree is this one. There is no real boundary, because it might be that certain objects do not show any "quantum" behaviour and some much larger objects do, just because the latter are more coherent in some way.

$\endgroup$
8
  • 1
    $\begingroup$ Just as a remark, there are quantitative theoretical bounds on the error that we do in approximating the true dynamics with the effective (classical or mean field) one, at least in some simple situation. For QM systems with regular enough interactions we can provide semiclassical expansions of the time evolved observables that are mathematically precise (Egorov-type theorems). $\endgroup$
    – yuggib
    Commented Mar 19, 2015 at 13:47
  • $\begingroup$ @yuggib good point. I know that the limit quantum --> classical is not simple in most situations but can often be done, so I suspected this exists, but I don't have a lot of background there. Can you provide a/some references for my own curiosity? $\endgroup$
    – Martin
    Commented Mar 19, 2015 at 13:55
  • $\begingroup$ Yes of course, but I have to say that the references I know are quite mathematical, so the language may not be so familiar. There is a sharp distinction between results for QM classical limit ("finite degrees of freedom"), that are more rich and accurate; and for QFT/mean field ("infinite degrees of freedom") where less can be said. For the QM limit, a result in quite physical terms using coherent states is due to Hepp; the general theory (with more quantitative bounds) ... $\endgroup$
    – yuggib
    Commented Mar 19, 2015 at 15:09
  • $\begingroup$ uses heavily the techniques of the semiclassical calculus à la Weyl-Hörmander and is well outlined in this book by Martinez. For the QFT/mean field limits, there is a result extending the work of Hepp by Ginibre and Velo to systems of non-relativistic bosons (with recent quantitative bounds in the mean field setting by Rodnianski-Schlein and Schlein-Chen-Lee). For systems where particles can be destroyed or created... $\endgroup$
    – yuggib
    Commented Mar 19, 2015 at 15:13
  • $\begingroup$ almost anything is known, except for the Nelson model of non relativistic particles interacting with a scalar field (Falconi and Ammari-Falconi). Anyways, it is a very active research field in mathematical physics...and sorry for the long digression ;-) $\endgroup$
    – yuggib
    Commented Mar 19, 2015 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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