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Do large objects behave quantum mechanically?

As part of my series on the Higgs boson, I mentioned that it used to be that we thought of objects in the macroscopic (‘classical’) world as consisting of either particles or waves while in the microscopic (‘quantum’) world, we had wave-particle duality, where entities had both properties. Both worlds were governed by different laws.

The more modern idea that everything at every size consists of quantum fields eliminates that problem. There is no ‘wave-particle duality’ (as formerly understood) and this implies that there is no quantum to classical transition. But it also implies that even macroscopic objects must obey the laws governing quantum fields, such as having a non-zero ground state energy because of having to obey the Heisenberg uncertainty principle. So the search has been on to test this idea for larger and larger objects.

Here is an ingenious experiment that shows that a small square mirror of size about 10 micrometers (which is very large when compared to atoms or elementary particles and would normally be considered to be of a size where classical laws applied) behaves according to the laws of quantum mechanics.

(If the embedded video doesn’t work for some reason, you can see it here. Via Sean Carroll.)

Comments

  1. says

    I have trouble watching videos at work, so I haven’t seen it yet. That said….

    I have long heard that the effects of quantum uncertainty “average out”. A single neutron can tunnel, as it requires only that three very tightly bound and effectively interchangeable quarks be in the same quantum state at the same time. A hydrogen atom, on the other hand, requires that the three quarks in the proton and an unrelated electron be in the same quantum state, which is more difficult. A helium atom requires that four sets of quarks, plus both electrons, be in the same state at the same time, and so on. As you add more and more particles, the quantum state of the whole coalesces into the definitive boundaries we detect day-to-day. Sure, there are still individual fluctuations, but they are as measurable as half a fart on the main floor of a national political convention during the opening speeches.

  2. MarkF says

    Isn’t each photon reflected by the electrons surrounding just a few atoms? If I recall my physics correctly, the reflection exhibited my mirrors comes from the interaction of electrons with the d-electrons of the silver atoms.

    How do the authors make the claim that their results reflect the behaviour of the entire mirror rather than just small clusters of atoms? Or is that even a meaningful distinction? How do you distinguish between a macroscopic object behaving non-classically and its atoms behaving non-classically?

  3. slc1 says

    Although I haven’t done the exercise, one can set up Schrodinger’s Equation for, say, the earth/moon system, with the “charges” being the masses of the earth and the moon, M and m, and the universal gravitational constant, G,replacing episilon(0), the permittivity of the vacuum. One doesn’t even need to solve the 2nd order differential equation as the solution is the same as for the hydrogen atom, with the above replacements. One will then get quantized energy and angular momentum levels. The problem is that the energy levels and the angular momentum levels are so close together that they are indistinguishable from a continuum. Thus, no useful information arises of such an exercise which is why it is not done in the books on quantum mechanics.

    In other words, the quantum mechanical effects of the earth/moon system are completely unobservable.

  4. MNb says

    “there is no quantum to classical transition”
    What my teachers taught me (and me being stubborn back then as I am now it took one teacher quite a long time) is that every single classical event accurately can be described by QM. Thus classical theory is a simplification of QM. I clearly remember I had to do the calculation connected to this with the aid of the correspondence principle.
    So the transition to me always has meant the grey area where classical theory gradually begins to become inaccurate. Therefor this

    “even macroscopic objects must obey the laws governing quantum fields” doesn’t surprise me at all. We only continue to use classical physics because its calculations are so much easier.

  5. Jenora Feuer says

    Of course, super-cooling Helium can lead to all sorts of fun.

    Via de Broglie, the wavelength is inversely proportional to the momentum, so if you cool helium enough, the wavelengths of the entire atom can get large enough to overlap other atoms.

    He4 has a total integral spin, so it acts as a boson, allowing the overlap. This results in the odd ‘superfluid’ state of He4 where it can start siphoning itself out of any non-sealed container, reached at about 2 Kelvin.

    He3 has a total non-integral spin, so it acts as a fermion, and He3 will still act as a normal liquid until the temperature gets so low (about 2 milliKelvin) that we get bound Cooper pairs forming which collectively act as bosons. Only then does it shift into superfluid mode.

  6. Rob Grigjanis says

    Gregory, a free neutron decays because it has two d quarks and one u quark, and a d quark can decay to a u via the weak interaction. A proton (with or without the electron which makes it a hydrogen atom) is stable because there is no lower energy state that it can decay to (not in the Standard Model, anyway).

    A neutron in a nucleus (and therefore the nucleus itself) is more stable because decaying to a proton so close to other protons is not energetically favourable.

  7. MarkF says

    The problem is that the energy levels and the angular momentum levels are so close together that they are indistinguishable from a continuum. Thus, no useful information arises of such an exercise which is why it is not done in the books on quantum mechanics.

    Really? That sounds like a great calculation to do in a quantum mechanics textbook. No practical use, but very informative about why we see continuums in nature. But then I don’t do calculations of that kind, so my opinion is of limited value. :)

  8. slc1 says

    Of course, the other problem is that, as we sit here today, we can’t do relativistic quantum gravity calculations because there is no theory of quantum gravity. In addition to which any quantum gravity theory that involved the exchange of massless spin 2 gravitons is unrenormalizable so it won’t fit in with the current standard model (all field theories involving particles with spin higher then 1 are unrenormalizable).

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