In Dirac's book Principles of quantum mechanics ([4th ed., p. 87-88]), he seems to give a very elementary argument as to how the commutator $[X,P]$ reduces to the Poisson brackets ${x,p}$ in the limit $\hbar\to 0$. However, I don't understand the argument that he is making. Could someone please explain this?
[Physics] Classical Limit of Commutator
classical-mechanicscommutatorpoisson-bracketsquantum mechanics
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The subtlety is that an arbitrary wavefunction doesn't reduce to a point of the classical phase space in the limit $\hbar \to 0$ (thinking about phase space makes more sense since in the classical limit one should have definite coordinates and momenta).
So one could ask, which wavefunctions do. And the answer is that the classical limit is built on the so-called coherent states -- the states that minimize the uncertainty relation (though I don't know any mathematical theorem proving that it's always true in the general case, but in all known examples it is indeed so). States close to the coherent ones can be thought of as some "quantum fuzz", corresponding to the quasiclassical corrections of higher orders in $\hbar$.
Example of this for the harmonic oscillator can be found in Landau Lifshits.
Regarding the fluid argument. About your remark (1): the $|\psi|^2$ for the stationary state is indeed stationary, but it still satisfies the continuity equation since the current is zero for such states. Your remarks (2) and (3) are quite right because, as I already said, the classical limit can't be sensibly taken for arbitrary states, it is built from coherent states.
And also I must admit that the given fluid argument indeed doesn't provide any classical-limit manifestation. It's just an illustration that "everything behaves reasonably well" to convince readers that everything is OK and to presumably drive their attention away from the hard and subtle point -- it often happens in $\it{physics}$ books, probably unintentionally :). The problem of a nice classical limit description is actually an open one (though often underestimated), leading to rather deep questions, like the systematic way to obtain the symplectic geometry from the classical limit. In my opinion it is also connected to the problem of quantum reduction (known also as the "wave function collapse").
I do not know about deep questions. And people seem to give pretty deep answers here. My contribution is to show
$$ \lim_{\hbar \to \infty} \frac{1}{i\hbar}[ F(p,x) , G(p,x)] = \{F(p,x), G(p,x)\}_{P.B.} $$
where $ [ F, G] = F G - G F $ and
$$ \{ F(p,x), G(p,x) \} = \frac{\partial F}{\partial x} \frac{\partial G}{\partial p} - \frac{\partial G}{\partial x} \frac{\partial F}{\partial p}. $$
Preliminars.
With $[x, p] = i \hbar$, you can show the following two equalities:
$$ [x, f(p) ] = i \hbar \frac{\partial f}{\partial p} $$
and
$$ [p , g(x) ] = - i \hbar \frac{\partial g}{\partial x}. $$
I think this is almost mandatory for every QM course, so I will skip this derivation. In any case, the standard route is to consider the commutator of x with increasing powers of p; then use induction when developing $f(p)$ as a Taylor series.
A more illustrative example is the following:
$$ [x^{2} , f(p) ] = [x ,f(p) ] x + x [x, f(p)] \\ = i \hbar f'(p)\, x + i \hbar x \, f'(p) = 2 i\hbar x f'(p) - i\hbar [x , f'(p)]\\ = 2 i \hbar x f'(p) - (i \hbar)^{2} f''(p) $$
where I have introduced the pretty useful notation $ f'(p) = d f /dp $.
By now you can see the fun is in arbitrary powers of $x$. You should more or less be able to guess the result and prove it by induction.
Lemma.
$$ [x^{n} , f(p) ] = \sum_{j=1}^{n} (-)^{j+1} \binom{n}{k} \, (i \hbar)^{j} x^{n-j} \, f^{(j)}(p) $$
Proof: You do it. Use induction. It should be more or less straightforward. By the way, $\binom{n}{k}$ denotes the binomial coefficient.
Moment of truth.
The previous argument can be used to include an analytic function of $x$. Consider
$$ [ g(x) , f(p)] = \Biggl[ \, \sum_{k=0}^{\infty} \frac{1}{k!} g^{(k)} (0) x^{k}, \, f(p) \Biggr] = \sum_{k=0}^{\infty} \frac{1}{k!} g^{(k)} (0) \Biggl[ x^{k}, f(p) \Biggr] \\ = \sum_{k=0}^{\infty} \frac{1}{k!} g^{(k)} (0) \sum_{j=1}^{k} (-)^{j+1} C^{k}_{j} \, (i \hbar)^{j} x^{k-j} \, f^{(j)}(p) \\ = \sum_{j=1}^{\infty} (-)^{j+1} \, (i \hbar)^{j} g^{(j)}(x) \, f^{(j)}(p). $$
The trick in the fourth equality is to switch the sums (and then expand $C^{k}_{j}$... everything fits nicely).
It is interesting to notice that the double summations collapsed into one. This somehow makes sense by dimensional analysis, powers of x and p decrease together so that $\hbar$ appears.
The final part is the most subtle point. A general $f(x,p)$ is tricky because $x$ and $p$ do not commute. So you would have problems with "hermiticity" and ordering. I will choose every $p$ to be the left of every $x$. Once this is agreed, a general $F(p,x)$ can be written as
$$ F(p, x) = \sum_{n=0}^{\infty} \alpha_{n} (p) \,\, f_{n} (x). $$
Now, we can compute
$$ \Biggl[ F(p,x) , G(p,x) \Biggr] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \Biggl[ \alpha_{n} (p) \,\,f_{n} (x), \, \beta_{m} (p) \, \, g_{m} (x) \Biggr] \\ = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \alpha_{n} (p) \biggl[ \, f_{n} (x) , \beta_{m} ( p) \biggr] g_{m} (x) + \beta_{m} (p) \, \biggl[ \, \alpha_{n} ( p) , g_{m} (x) \biggr] \, f_{n} (x) \\ = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \, \alpha_{n} (p) \, \biggl( \sum_{j=1}^{\infty} (-)^{j+1} \, (i \hbar)^{j} f^{(j)}_{n} (x) \, \beta^{(j)}_{m} (p) \biggr) \, g_{m} (x) + \beta_{m} (p) \biggl( \sum_{j=1}^{\infty} (-)^{j} \, (i \hbar)^{j} g^{(j)}_{m}(x) \, \alpha^{(j)}_{n}(p) \biggr) \, f_{n} (x) $$
specially using
$$ \sum_{n=0}^{\infty} \alpha_{n} (p) \, f_{n}^{(j)} (x) = \frac{\partial^{j}}{\partial x^{j}} \biggl( \sum_{n=0}^{\infty} \alpha_{n} (p) \, f_{n} (x) \biggr) = \frac{\partial^{j}}{\partial x^{j}} F(p,x) $$
you see that you get the desired result (after changing the summations):
$$ \Biggl[ F(p,x), G(p,x) \Biggr] = \sum_{j=1}^{\infty} (-)^{j} \frac{(i \hbar)^{j}}{j!} \Biggl( \frac{\partial^{j} G}{\partial x^{j}} \frac{\partial^{j} F}{\partial p^{j}} - \frac{\partial^{j} F}{\partial x^{j}} \frac{\partial^{j} G}{\partial p^{j}} \Biggr) $$
because you see the only term that survives after dividing by (i \hbar) is the first one. This gives you the Poisson bracket. I didn't do any involved computations because they are long. It is more or less convincing.
Best Answer
Dirac's argument is on pp.85-86, and it goes like this:
The classical Poisson bracket obeys the following rules:
$$\{A,B\} = -\{B,A\}$$ $$\{aA + bB ,C\} = a\{ A,C\} + b\{B,C\} $$ $$\{AB,C\} = A\{B,C\} + \{B,C\}A $$ $$\{\{A,B\},C\} = \{B,\{A,C\}\} - \{A,\{B,C\}\}$$
Where I rewrote the Jacobi identity in the way that makes sense. Now Dirac asks whether you can define such a thing for noncommuting quantum objects, and he notes that you can, if
$$ i\hbar [A,B] = (AB - BA) $$
Where $\hbar$ is a proportionality context, fixed by dimensional analysis, while $i$ is there to ensure that the Poisson Bracket-analog is Hermitian, as observables should be by convention (the anticommutator is anti-Hermitian).
He deduces this by expanding the commutator: $[AB,CD]$ using the formal rules above as axioms, in two different ways. From this, he finds that
$$ [A,C](BD - DB) = (AC-CA)[B,D]$$
In the classical theory, this gives $0=0$, but in QM, the observables don't commute, so you learn you should identify commutators as the quantum analogs of Poisson brackets. He then argues that $\hbar$ commutes with everything, and therefore that the commutation relations should hold, and from there deduces that the Schrödinger picture is available.
The argument is streamlined, and not historically accurate. For Heisenberg's original argument (or something very close to it) see Wikipedia's Matrix Mechanics page.