Is $ \hbar \rightarrow 0 $ in path integral merely technical thing? Is there any justification? (especially in open quantum theory)

Question

I wonder whether the limit $ \hbar \rightarrow 0 $ in path integral is merely technical stuff when we explain the classical limit of quantum mechanics, or there may be any physical meaning or sense behind it.

When we deal with path integral, it's usually said that we obtain the classical Lagrange equation if we take a limit $ \hbar \rightarrow 0 $. But what on earth is the condition that we can take such a limit? I have found some discussion in here, but I couldn't find any satisfying one. I found this one gives some clue, but I think it's a bit mathematical content and does not help answer my question directly. Rather, I want to find a more physics-oriented answer, especially in the context of open quantum systems.

The reason why I have the open system in my mind is the following: when we do experiments, we always think of accuracy problems. For example, when the experiment considers small scale or high energy, we may think of quantum correction to the classical result. But this consideration is pertaining not only to the observed phenomena but also to the external environment such as experiment apparatus. Moreover, I prefer to solve this problem in path integral formalism because I think this approach may have more benefit when discussing open QFT.

In other areas of physics, one assumes that one quantity is extremely small relative to the other quantity when we say such a limit. For example, in relativity, we take a limit $ v/c \rightarrow 0 $ if the system's speed is very low relative to the speed of light. So, I wonder if we may compare $ \hbar $ with something in quantum mechanics too.

Especially, I wonder if this idea may be related to some concepts like decoherence, which frequently appears in open quantum systems. I mean, if we consider an open system in path integral formalism, can we extract some physical quantity $ X $ (maybe related to decoherence rate) and compare it with $ \hbar $ so that we justify when to take a limit $ \hbar \rightarrow 0 $, that is, $ \hbar / X \rightarrow 0 $. Or perhaps there would be an 'effective $\hbar$' depending on the total system? (so that it provides a reason for the argument 'when the scale of the action is much larger than $\hbar$ we can take a limit $\hbar \rightarrow 0 $.')

I know that open quantum system and open QFT is a relatively new subject in physics, but I wonder if there are some results on my question.

Answer 1

Basic Meaning

• As mentioned in the comments, the meaning of $\hbar\to 0$ is really $\hbar/S\to 0$. The limit $\hbar\to 0$ doesn't make much sense because it is a dimensionful quantity. If you make it dimensionless by setting $\hbar=1$ then the limit is really $S>>1$ and not $\hbar\to0$ because $\hbar$ has been set to $1$. Anyway, one way or the other, the meaning of the limit is that the action is large compared to Planck's constant.

Propagator and the large $S$ limit

• When you take this kind of a large action limit, in non-weird systems (see the post linked by OP for a more nuanced take), what happens is that the propagator becomes classical as can be seen using the saddle point approximation.
• The basic idea is that the contribution of each path in the Feynman integral for the propagator has the weight (which is a pure phase) of $\exp[iS/\hbar]$, thus, if $S/\hbar$ is very large then the fluctuations in the phase of different paths is so wild that they all cancel each other out. The only path that contributes is the one that corresponds to a stationary point of the action functional because the variations in the functional around a stational point will not be wild and thus, its contribution would remain. You do the calculation to find out as to which path will be the stationary point of the action functional and voila, you get the Euler-Lagrange equation.

Decoherence

• However, crucially, this does not mean that you have obtained the classical limit. In particular, the propagator only describes the evolution of the quantum system. If the initial condition of a quantum state is characteristically quantum then even with a large action, the system would behave quantum mechanically. As far as my understanding goes (kindly correct me if you know better), this is where decoherence comes in. Decoherence explains that when a quantum system is open, it entangles with the environment and loses the information to the environment, and becomes un-coherent or decohered. And thus, it is unlikely that we would just find an object that is in a characteristically quantum state, to begin with.
• The main reason as to why this is relevant is because it is quite hard to perfectly isolate big objects -- in other words, to make them non-open. This, combined with the large-action behavior of the propagator, makes big objects behave classically. However, with the advancements in technology, the objects that we can isolate are getting bigger and bigger, and thus, consequently, we are able to see characteristically quantum behavior in bigger and bigger objects.