Does limit $\hbar \rightarrow 0$ in Quantum Mechanics mean anything? [duplicate]

Assuming that I learn Quantum Mechanics first, and then I approach Classical Mechanics as a special case of Quantum Mechanics, I will definitely find the relationship between Quantum Mechanics and Classical Mechanics very confusing. I don't know how to make sense of what happens when $\hbar \rightarrow 0$.

So, does limit $\hbar \rightarrow 0$ in Quantum Mechanics mean anything? How should we interpret it? Or does the above contradiction reveal yet another flaw in the fundamentals of Quantum Mechanics?

• Non-relativistic limits are limits with $c \to \infty$, not $\hbar \to 0$. Mar 5 '14 at 10:53
• What you you mean by "yet another flaw in the fundamentals of Quantum Mechanics"? Mar 5 '14 at 11:22
• Classical mechanics is not a special case of Qunatum Mechanics. It emerges from the underlying framework of QM and is valid in situations where h_bar can be taken to zero. ON how classical fields emerge from quantum fields have a look motls.blogspot.com/2011/11/… . It aint simple. Mar 5 '14 at 11:29
• possible duplicate of $\hbar \rightarrow 0$ in quantum mechanics Sep 28 '15 at 20:18

The question duplicates $\hbar \rightarrow 0$ in QM where it has been soundly answered. The most direct bridge is through deformation quantization, the phase-space formulation of QM, where operator observables are mapped injectively onto their Wigner transforms, c-number phase-space functions, just like their classical counterparts. It is then evident that QM laws are deformations of the classical laws, and include those in their $O(\hbar^0)$ pieces. So, for example, the Wigner transforms of the density matrix, the Wigner functions, go to Dirac deltas in phase space in that $\hbar \rightarrow 0$ limit. These δs, in turn, transcribe to the Liouville theorem for single particles, and hence Hamilton's dynamics, out of Heisenberg's equations of motion. The Ehrenfest theorem structures of course replicate and parameterize this injection. Note this is a mere shift in point of view: $\hbar \rightarrow 0$ means considering (macroscopic) phenomena at much larger scales, which thus dwarf the scale of $\hbar$ and make it insignificant. It is mere sloppy shorthand of the emergence of our classical world out of QM at large scales (the correspondence principle).
For the limit to make sense, however, you must always consider angular momenta and actions S much larger than $\hbar$: it makes no sense to consider $\hbar \rightarrow 0$ limits for a single particle of small actions and spins---$\hbar$ is a dimensionful quantity, and its scale matters. I gather there is no interest in microscopic physics without $\hbar$.
The least tortured limit then, is, as above, consideration of the $O(\hbar^0)$ part of macroscopic quantities. For details, see, e.g., Ref. 1. For instance, for a Freshman lab oscillator with maximum oscillation amp 10cm, m=10g, and ω=2Hz, the characteristic action is S=E/ω= $10^{-4}$ Js and so $\hbar / S= 10^{-30}$, suppressing any and all quantum terms and validating the limit. (One might be interested in thinking about the Wigner function for the nth excited state for n ~ $10^{30}$, a very spikey cookie-cutter function, indeed!)