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If I consider for instance N non interacting particles in a box, I can compute the energy spectrum quantum mechanically, and thus the number of (quantum) microstates corresponding to a total energy between $E_0$ and $E_0 + \delta E$. In the limit of large quantum numbers, the result is well known to coincide with the available volume of the phase space of the corresponding classical system of N newtonian free particles in a box, namely $$ \Omega(E_0,V,N; \delta E)_{\textbf{quantum}} \to \frac{1}{h^N} \int_{E_0<E<E_0 +\delta E} d^{3N}x d^{3N}p $$ in the limit of large quantum numbers.

My question is the following. Is there any proof, besides this specific example of the quantum gas in a box, that the quantum expression is always going to approach the classical one in phase space, for any given physical system (and thus for some generalized coordinates), provided some classical limit is used?

This does not seem a trivial statement to me, and I can't find the proof in textbooks.

Many thanks.

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Well there is a reason in this case of non-interacting particles- it is the so called "Thermodynamic limit". But I can answer this question without invoking the thermodynamic limit.

One very simple way to see this is using $h$. We know $h \ll 1$ so $P = h^{-N}$ for $N \gg 1$ will give you $P \gg 1$. And in some large limit of the number of particles, you can effectively set $ P \rightarrow \infty$ i.e. $h \rightarrow 0$, which happens to give you the classical limit (This is because setting $h = 0$ gives you a classical theory).

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  • $\begingroup$ I quite disagree, even for one particle in a box, the two expression coincide, provided you can count the quantum degeneracy (n^2+p^2+m^2) proportional to E, by a calculation in the continuum, ie if the cell in the wavevector space, $\pi/L$, is very small w.r.t to the radius of the sphere $\propto \sqrt{m E}$. Thus to me it has no link at all with the thermodynamic limit (this is rather about the equivalence of the different statistical ensembles, not about the claissical limit). $\endgroup$ – Jip Apr 1 '16 at 14:49
  • $\begingroup$ @user106422: If you treat $h$ as a dimensionless number in your statement $h\ll 1$, what is the scale you are comparing it with? $\endgroup$ – flaudemus Feb 23 at 11:51

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