To put it a little better:
Is there more than one quantum system, which ends up in the classical harmonic oscillator in the classial limit?
I'm specifically, but not only, interested in an elaboration in terms of deformation quantization.
Quantum Mechanics – Is the Quantization of the Harmonic Oscillator Unique?
classical-mechanicsharmonic-oscillatorquantizationquantum mechanics
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Let me continue on from BeastRaban's exposition: In 1866, Mehler figured out how to carry out the sum for the $u_n(x)$ eigenfunctions of the quantum harmonic oscillator, Hermite polynomials, adding them all to an elegant and compact eponymous Mehler kernel, or equivalently.
The Green's function (propagator) for the quantum harmonic oscillator is then: $$ K(x,x';t)=\left(\frac{m\omega}{2\pi i\hbar \sin \omega t}\right)^{\frac{1}{2}}\exp\left(-\frac{m\omega((x^2+x'^2)\cos\omega t-2xx')}{2i\hbar \sin\omega t}\right) ~. $$ Consequently, given a configuration at t=0, in general a superposition of lots of components with characteristic frequency each and every one of them, you now know how the quantum oscillator hamiltonian will propagate it to arbitrary time t: $$ \psi(x,t) = \int_{-\infty}^\infty K(x,x';t) \psi(x',0) ~ dx'.$$
So the takeaway lesson of all such elaborate eigenfunction expansions is the Mehler kernel and one integral to perform, which accounts for everything.
I am not sure about the $ \frac{1}{\sqrt{A^2-x^2}} $ part in you approximation. In the asymptotic limit $n \rightarrow \infty$,the Hermite polynomials behave as follows:
The cosine part relates to the oscillations present in wavefunction which are visible even in fig 2.7b in Griffiths.The $(1-\frac{x^2}{2n})^{\frac{-1}{4}}$ part is the classical behaviour and in this case the graphs seems to match.
References:
Hermite Polynomials on Wikipedia
See the asympotic behaviour part for the above expression.
Best Answer
For every classical Hamiltonian $H(p,q)$ there are infinitely many quantized systems that reduce to it in the classical limit.
The reason is that adding to $H(p,q)$ an arbitrary number of expressions in $p$ and $q$ where at least one factor is a commutator doesn't change the classical syatem but changes the quantum version. To be specific, take, for example, $H'=H(p,q)+A(p,q)^*[p,q]^2A(p,q)$ for an arbitrary expression $A(p,q)$.
This holds for systems in which $p$ and $q$ are canonically conjugate variables. It is easy to generalize the argument to arbitrary quantum systems.
In short:
Quantization of individual systems is an ill-defined process.
Geometric quantization is more well-defined, as it effectively quantizes not a particular system but a particular group of symmetries. In the above case, it shows how to go from the classical $p$ and $q$ (i.e., from the classical Heisenberg Lie algebra with the Poisson bracket as Lie product) to the quantum version, but not how to go from a particular classical Hamiltonian (and hence a particular classical system) to its quantization.