Is this four-step recipe for quantization always valid?

Question

I know that there is more than one way to go about quantization, but operationally, I find it useful to have a go-to set of steps that can convert a classical system to its quantum analog. Is there any step in this four-step recipe that isn't valid?

1. Formulate the classical Hamiltonian which shall represent the mode to be quantized,
2. identify the pair of canonically conjugate variables $\left(x, p\right)$ that satisfy Hamilton's equations $\frac{dx}{dt} = \frac{\partial H}{\partial p}$ and $\frac{dp}{dt} = -\frac{\partial H}{\partial x}$,
3. convert the dynamical variables in the Hamiltonian into their quantum counterparts $x \rightarrow \hat{x} = x\times$ and $p \rightarrow \hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}$, and finally
4. solve Schrödinger's equation $\hat{H}\phi_n = E_n \phi_n$ for the eigenfunctions $\phi_n$ and eigenenergies $E_n$. If the potential is bounded, one will then see that $\phi_n$ exhibits discrete nodes, hence the first quantization, and the number of excitations (i.e., particles) can only increase by discrete energy increments $n$, hence the second quantization.

Answer 1

Suppose your classical hamiltonian is $H(x,p) = x^2 p^2$. What quantum hamiltonian operator will your recipe produce? You might say it's $\hat{H} = \hat{x}^2 \hat{p}^2$. However classically $x$ and $p$ are just real-valued functions on phase space, so they commute and we could just as well write $H(x, p) = p^2 x^2$, $H(x,p) = xpxp$, etc. and make the same naive replacement. Since the quantum operators don't commute, we end with different quantum hamiltonians depending on the ordering we choose.

Another problem is that the classical hamiltonian might be $H=0$. This is the case in, for example, pure Chern-Simons theory in 2+1d (with no Maxwell term or matter fields). In spite of this, the theory can be quantized canonically, leading to interesting kinematic structure. But it is not clear how one could quantize such a theory following your recipe.

Answer 2

No. There remains an ambiguity of ordering. See for instance this post for an example where there could be different outcomes of quantization depending on the ordering, and where your 4-step approach would be ambiguous.

Also relevant is this post.

Answer 3

Very generally speaking, classical mechanics is a limiting case of quantum mechanics.

A quantization recipe gives you an informed guess as to how to get a valid quantum system whose large scale behavior is in accord with the classical system.

It might be the correct system, but like always when taking limits, there are many inequivalent systems having the same limit, however many additional (classical) constraints you add to the system.

As void said, you get a heuristic, but you will need something additional to support the physical validity.