As I understand it, if $S \gg h$ then we are in the classical realm, whereas if $S \leq h$ we are in the quantum realm. My question is what happens somewhere in between those 2 limits? Are we quantum and classical at the same time?
3 Answers
The heuristic that compares the action $S$ to Planck's constant is vaguely useful as an initial criterion, but the limit from quantum to classical mechanics is rather more subtle, in ways that make the simplistic comparison close to useless in practice.
As a pair of counter-examples:
If you prepare a harmonic oscillator in a coherent state, then in practice it will be indistinguishable from something you could model as a classical harmonic oscillator with some added shot noise, and this happens regardless of the mean number of excitation states or of the ratio $S/h$.
On the other hand, it is technologically challenging but in-principle possible to prepare an $n$-photon Fock state with an arbitrarily high but well-defined photon number $n$, and this will exhibit clearly quantum behaviour even for arbitrarily large $S/h$.
Thus the limit from quantum to classical mechanics needs to be done more carefully, and a simple heuristic will never suffice beyond serving as a fuzzy qualifier.

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I really love it how many people come here with dichotomic questions, just to be shown there is a fuzzy continuum! Keep up the good work… – Jul 07 '17 at 13:11
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Can you elaborate on what you mean by the second example clearly exhibiting quantum behavior? In particular what quantum behavior do you have in mind? – Borun Chowdhury Jul 08 '17 at 07:31
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In your second example, the system would indeed behave classically. Nonetheless, the initial configuration would be a classic probability distribution that is not concentrated in one point of the phase-space. It is possible to interpret this as a "quantum residue", however let me stress again that the description is perfectly classical (even if statistical in some sense). – yuggib Nov 23 '17 at 07:47
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By the way, I could write explicitly what the classical probability distribution corresponding to that state is ;-) – yuggib Nov 23 '17 at 07:48
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@yuggib That's completely incorrect. Fock states are nonclassical, and the physics of quantum optics goes far beyond the quadrature probability distribution. – Emilio Pisanty Nov 23 '17 at 08:45
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I can prove that fock states with a very large number of photons behave like probability distributions, and that the expectation of quantum observables converges to the average w.r.t. the probability distribution of the classical observables. I could give you the references if you want. – yuggib Nov 23 '17 at 08:51
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Then your proof is either wrong or is explicitly restricting itself to a strict subset of the possible experiments you can do. This is well-trod textbook material and I'm not interested in clarifying the point; if you're still confused, ask separately. – Emilio Pisanty Nov 23 '17 at 09:26
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@EmilioPisanty Your arrogance is completely uncalled for, and you should be careful in making bold statements. I am saying that if you have a quantum Fock state with many many photons, then it behaves like a classical probability distribution. With that I mean that the error that is made in approximating quantum averages with classical averages is small (but of course not zero). It is actually possible to give bounds on how small it is, bounds that have been confirmed by numerical simulations, and I am pretty sure would be (or have been, I am not an expert) confirmed by experiments as well. – yuggib Nov 23 '17 at 09:39
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And what I'm saying is that that argument is meaningless unless you specify the averages of which class of quantum measurement you're considering. A simple example is an interference experiment between your state and a photon-substracted version of the state; this gives nonzero contrast for the classical approximation to the Fock state and absolutely no interference for a true Fock state, regardless of $N$. But again, unless you say what class of experiments you're considering, your argument isn't even an argument. – Emilio Pisanty Nov 23 '17 at 09:44
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I am quite sure you do not even know what the "classical approximation" of a Fock state is, so I strongly doubt that what you are saying is meaningful. Anyways, I don't know which experimental apparatus is feasible or not, but if you are able e.g. to have cavity radiation in a Fock state interacting with particles (also in a prescribed initial state, if they are large and heavy it is easier to predict the outcome, but this is not necessary), – yuggib Nov 23 '17 at 09:56
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and you could measure some dynamical property of such particles, you would indeed see that their motion is driven by the interaction with a classical radiation field, in a precise initial statistical configuration. Anyways this is just an example, and it is possible to make many others. – yuggib Nov 23 '17 at 09:57
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1@yuggib You seem to have missed the point. The existence of a class of experiments where Fock states can be seen as classical is entirely irrelevant; I'm not sure why you keep bringing it up. The issue at stake is the existence of any experiment on high-$N$ Fock states that cannot be explained by the classical limit; your only nontrivial relevant claim is that that class is empty, despite being given an explicit example in that class. Anything you say that is not directed at refuting that counterexample is irrelevant. – Emilio Pisanty Nov 23 '17 at 10:13
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Yeah, I am really saying that the behavior of high-$N$ Fock states is always well-approximated by classical objects. To be precise, a state $\lvert \phi^{\otimes N}\rangle\langle \phi^{\otimes N}\rvert$, where $\phi\in \mathfrak{h}$ is a one-particle (i.e. classical) configuration of radiation, converges (in a suitable topology) to the probability distribution $\int_{0}^{2\pi}\frac{d\theta}{2\pi} \delta(,\cdot,-e^{i\theta}\phi)$, as a probability measure on $\mathfrak{h}$. – yuggib Nov 23 '17 at 10:23
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Take this probability measure, and compute the predicted classical interference pattern of your example, mediating w.r.t. this probability. If $N$ is large enough, the prediction will be very close to the true interference pattern given by a fixed $N$. Was this the classical approximation of a Fock state you had in mind, or was it different? Because the one above is the correct one (of course you should interpret the quantum state in the Fock space, with only non-zero component in the $N$-particle sector given by $\phi^{\otimes N}$). – yuggib Nov 23 '17 at 10:24
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@yuggib Yes, that is the classical limit I had in mind, and no, photon subtraction doesn't particularly make sense in that regime. (In case you missed it - that's a class of experiments the limit doesn't reproduce well.) The approximation you give has nonzero contrast (perfect contrast in the $N\to\infty$ limit) in the fringes; the quantum-mechanical result has no interference whatsoever. – Emilio Pisanty Nov 23 '17 at 10:40
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Are large $n$ Fock states not more succeptible to decoherence, i.e. more likely to become more classical? It's surely true that large GHZ states decohere easily, but I guess that's not the same thing... – DanielSank Dec 16 '17 at 20:55
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1@DanielSank Indeed they are, but that just mean that the quantumness is more fragile not that it's not there. – Emilio Pisanty Dec 16 '17 at 21:01
In Feynman path integral formalism, which is the context of your question as I understand it, the classical domain is recovered for $S \gg \hbar$. If $S$ is on the contrary of the order of $\hbar$, then the system will exhibit quantum behaviours of one sort or another, regardless of whether $S \ge \hbar$ or $S \le \hbar$. Now as @Countto10 and @Emilio Pisanty correctly stated, the devil is in the details and this statement of mine is full of caveats. But I guess you just wanted the gist of it.
There are two different ways in which this question can be asked:
1) ISOLATED SYSTEM: If you are asking about an isolated system then recall that the classical equations of motion are derived by minimizing the action. If $S>> \hbar$ then the saddle point approximation to the path integral
$$ \int [dx] e^{i S(x)/\hbar} \approx e^{i S(x_{cl}) /\hbar} $$
is a good approximation as fluctuations around the classical path cancel out. This implies that all the correlators will be peaked around their classical values.
Let me give a trivial example. Suppose you are interested in the transition amplitude from $x_0$ at time $t_0$ to $x_1$ at $t_1$. This is give by
$$ \mathcal A = \langle x_1| e^{-i H (t_1-t_0)}|x_0 \rangle $$
Using the Hamiltonian for a free particle we get
$$ \mathcal A = \int_{-\infty}^\infty dp e^{- i (\frac{p^2}{2m} \delta t - p \delta x)} $$
Being a Gaussian integral its trivial to solve but solving it will defeat the purpose. What we want to observe is that if $\delta x,\delta t \gg 1$ ($\hbar$ in natural units) we can approximate the integral by the value at the saddle point $p= m \frac{\delta x}{\delta t}$. You'll recognize this as the classical definition of momentum.
This method is often called 'sum over all paths' but the different paths are coming only because the initial and final states are not eigenstates of the Hamiltonian. For instance if one were to take a Harmonic oscillator in the $n^{th}$ state we would get
$$ \langle m,t| n,0 \rangle = e^{-i n t} \delta_{mn} $$
that is unless the final state is exactly the same as the initial the amplitude is 0. One can also start off the harmonic oscillator in a position eigenstate or coherent state and see how it leads to sum over all paths and can also play with why the evolution of a coherent state seems to track a classical path even when $S\sim \hbar$ as mentioned in another answer above.
2) OPEN SYSTEM: If you are interested in an open system then decoherence from interaction with the environment puts the system in a impure state or density matrix where the density matrix is diagonal in the so called pointer basis (environmental superselection). For large enough ''classical objects'' this basis is usually position and for small enough ''quantum objects'' this basis is usually energy. However, in a lab setting this decoherence can be controlled by tuning the interaction with the environment to be something else. For instance a very trivial example is making the pointer basis up-down or left-right trajectory of a beam of electrons by rotating the Stern-Gerlach apparatus. [ref: http://www.springer.com/gp/book/9783540357735]

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At energies of LHC the momentum of the beams is extremely large. Does this make S, the action, very large compared with h? The difference is 10 orders of magnitude. With that value of S are we back in the classical realm ? – Majid Malik Jan 18 '18 at 09:49
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@MajidMalik look at Emilio Pisanty's second 'counter-example' above to understand why at LHC scales you still have 'quantum' behavior. This is also covered at the end of my point about isolated systems. – Borun Chowdhury Jan 18 '18 at 11:51
Very confusing!!!!!
– Majid Malik Dec 30 '17 at 11:07