# Transition from quantum to classical mechanics

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?

• You might edit out the strictly false/incomplete premise "whereas if S≤h we are in the quantum realm". We are in the quantum realm when not in the first part of your summary--if that, as the answers suggest. – Cosmas Zachos Jul 7 '17 at 14:07
• Related/possible duplicate: physics.stackexchange.com/q/56151/50583 – ACuriousMind Jul 25 '17 at 14:54
• Continuing with this quest. "An electron beam" in an accelerator (e.g. SLAC or CERN) is classical right before the collision at which point it becomes a quantum beam. In fact it is more that a quantum beam; it is a quantized field which participates in a Feynman diagram. So are we really creating a QFT objects classically??? Very confusing!!!!! – Majid Malik Dec 30 '17 at 11:07
• Furthermore If a particle has precise momentum it has infinite uncertainty in its position according to the uncertainty principle. That means the particle can be located anywhere between minus infinity and plus infinity and it would still collide with opposite beam which has the same infinite spread in the position. But we are bring them face to face in the accelerator to have a successful collision. Or are we creating Feynman diagrams precisely in the accelerator at a collection of points where collisions take place. Is this all existential? – Majid Malik Dec 30 '17 at 11:25

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.

• 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… – user154997 Jul 7 '17 at 13:11
• 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 8 '17 at 7:31
• 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 7:47
• @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
• @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]

• 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 9:49
• @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

## protected by Qmechanic♦Jul 7 '17 at 13:08

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).