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Is there any parameter (in terms of physical quantities such as mass, length, charge...) which can be used to decide when to treat a system quantum mechanically and not classically?

Qmechanic
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Seeker
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    You can treat a system classically if and only if the quantum effects are neglegible for the purpose and precision you are working with. I don't see how this question could ever have another answer in this generality. – ACuriousMind May 01 '16 at 10:39
  • This question is certainly too broad.Some related posts http://physics.stackexchange.com/q/183649/, http://physics.stackexchange.com/q/32112/, http://physics.stackexchange.com/q/57390/, http://physics.stackexchange.com/q/225797/, http://physics.stackexchange.com/q/218523/ – AccidentalFourierTransform May 01 '16 at 11:20
  • and http://physics.stackexchange.com/q/37601/, http://physics.stackexchange.com/q/38221/, http://physics.stackexchange.com/q/247314/, http://physics.stackexchange.com/q/72230/, etc. – AccidentalFourierTransform May 01 '16 at 11:24
  • The system is classical when the action is much greater than $\hbar$. – Robin Ekman May 01 '16 at 15:41

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This is not a simple problem with a single answer, but I'll try to give a few practical examples to demonstrate the thought process involved.

The most basic way to determine "quantumness" is by comparing the length scales in the problem to the de Broglie wavelength: $$ \lambda = \frac{\hbar}{p} $$ where $p$ is the momentum of a particle. As an example, a gas of atoms in an enclosed container, where the average distance between atoms is $d$. If $d$ is much larger than $\lambda$, then the wavefunction overlap between two particles will be small, and the gas can be treated classically. If you either compress the gas to decrease $d$ or cool it down (lower the average momentum) to increase $\lambda$, quantum effects become more important because the wavefunctions of neighboring atoms start to overlap appreciably. Eventually this system will turn either into a Bose-Einstein condensate or a degenerate Fermi gas, depending on the nature of your atoms.

On the other hand, consider an airplane in flight. It has a massive momentum ($\approx 85\times10^6$ kg m for a 747) and so the de Broglie wavelength of this "particle" is going to be $\approx 10^{-42}$ meters, 44 orders of magnitude smaller than the plane itself. If you didn't already know, this makes it painfully obvious that airplanes are not very quantum.

Another way to test for quantumness is to compare energy scales. For a harmonic oscillator, the quantized energy levels differ by $\hbar\omega$, where $\omega$ is the characteristic frequency of the oscillator. Usually the relevant energy of comparison is the thermal energy, $k_BT$ ($k_B$ is Boltzmann's constant). People can actually get macroscopic objects (usually thin circular membranes) to act like quantum harmonic oscillators with discrete energy level spacing, but they have to get it really cold (less than a Kelvin) to get there. At any higher temperature, there is so much thermal energy that an extra $\hbar\omega$ here and there just doesn't matter.

There are a bunch of other possible cases and subtleties, but I'll leave you with one more: individual photons. Those little massless buggers are about as quantum as you can get, so classical mechanics will almost always fail when dealing with just a few of them.

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How about $\Delta p\Delta x$?

From The Physics of Stargates: Parallel Universes, Time Travel and the Enigma of Wormhole Physics, by Enrico Rodrigo:

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Řídící
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I think that the best answer is probably just "no". It's easy to give specific cases where you need/don't need quantum mechanics, and plenty of people attempt to give such a rule of thumb (e.g. relating to de Broglie wavelengths or Plank's constant). However, all of the affirmative answers to this question I've ever heard are either (a) too specific to be useful or (b) have way too many exceptions and caveats to be useful. You'll have to analyze each systems individually.

If you narrowed down your question, e.g. asking about interference effects specifically, precision measurements specifically, particle scattering specifically, etc. you could probably get a better answer.

aquirdturtle
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  • Yes. I agree. The next noble prize is waiting for someone who cooks up such a parameter, I reckon. – Seeker May 06 '16 at 15:48
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One way to see it is to consider isolation as the key factor: when a system is isolated from its environment, it exhibits quantum behavior.

Seen from that angle, one could say that individual particles or atoms have to be described by QM because they are intrinsically isolated due to their small number of degrees of freedom.

See what happens when they interact:

  • When interacting with a complex environment, decoherence happens and they behave classicaly (I'm simplifying a lot here: more exactly, the observables involved in the interaction yield classical values). Such an interaction is considered a measurement.
  • When interacting with another isolated system (for example another particle or atom), both become entangled forming a larger, but still isolated, quantum system.

But much larger and complex systems may need QM to be described, if they are isolated enough.

One way to isolate a mesoscopic system is to cool it down to very low temperatures. This way systems of thousands of atoms can be set in superposition. For pop accounts of such states, see for example Quantum Record! 3,000 Atoms Entangled in Bizarre State or The Same Atoms Exist in Two Places Nearly 2 Feet Apart Simultaneously. Even more impressive: Scientists supersize quantum mechanics where a full mechanical system with excitation modes is put in superposition, again by isolation via low temperature.

For a theoretical perspective laying stress on the isolation factor, see Quantum theory without measurement or state reduction problems (Macdonald).