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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.
Tfovid
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    You won't get the same quantum system if you use another set of canonical coordinates that mix $x,p$ non-trivially. – Void Jul 28 '20 at 06:58
  • @Void How does one ensure that $x$ and $p$ are converted unambiguously? – Tfovid Jul 28 '20 at 07:16
  • Generally $x$ has to be a truly spatial coordinate and $p$ a true momentum coordinate, no mixing. I have to admit I find quantization through symmetry considerations a better recipe (see Ballentine's QM book). Either way, quantization is just a formal heuristic, from today's perspective you would go more ask about the emergence of classical behaviour from the quantum dynamics. – Void Jul 28 '20 at 08:06
  • (a) In addition to ordering ambiguities mentioned by others, the presence of constraints will also complicate the prescription you have written down -- the book by Dirac explains how to handle this in great detail. (b) The names first and second quantization are historical artifacts and one shouldn't read too much into it. The "first quantization" refers to "quantizing a particle to get a wave function." The "second quantization" refers to "quantizing a wave function to get a wave functional (for a quantum field)." But, again, these terms are not very physically insightful. – Andrew Jul 28 '20 at 10:41
  • @Andrew Are first and second quantization as I referred to them in step #4 (more or less) accurate? – Tfovid Jul 28 '20 at 10:53
  • I've never heard of them described that way, I'd need to think about it. Normally I have heard "first quantization" meaning "quantize particle position and momentum to get a wavefunction" and "second quantization" meaning "quantize the wavefunction to get a wavefunctional." However, in modern understanding this is not an accurate description of what is really going on. In practice, "first quantization" refers to single-particle quantum mechanics, and "second quantization" to quantum field theory -- I don't think there is really more content that that. – Andrew Jul 28 '20 at 10:57
  • Please, please, please, do not associate second quantization with quantization, as a reverse classical limit. Second quantization is a deeply unfortunate established term for packaging an infinity of quantum oscillators (so already "1st"-quantized) to represent quantum fields. It is a functor, an unambiguous technical recipe, and has nothing, nothing, nothing to do with the rest of your question. – Cosmas Zachos Jul 28 '20 at 13:11
  • @CosmasZachos I'm not sure which part of first and second quantization, as I attempted to describe in #4, you find inaccurate. (And yes, that was a digression from the main question.) – Tfovid Jul 28 '20 at 13:47
  • More on the ambiguity of order. – J.G. Jul 28 '20 at 13:50
  • I don't understand its implications to declare inaccurate. My sense is fewer hackles would be raised if you simply dropped the sentence " and the number of excitations ...". It multiplexes the problem to an infinity of Schroedinger equations and an infinity of wave functions, etc... It really only serves to confuse the confusable. – Cosmas Zachos Jul 28 '20 at 13:52

3 Answers3

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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.

d_b
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    I didn't think of the ordering issue, thanks. I'll look at the posts that @ZeroTheHero posted below on the matter. Incidentally, do you agree about my understanding of first and second quantizations in step #4? – Tfovid Jul 28 '20 at 06:51
  • I don't really know what "first quantization" means. "Second quantization" just means writing states in the occupation number basis, and yes, these basis states have definite particle numbers. But a general state written in the occupation number basis need not have definite particle number. And anyway, these things have nothing to do with "quantization" as the process of constructing a quantum system from a classical system. – d_b Jul 30 '20 at 01:07
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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.

ZeroTheHero
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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.

doetoe
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  • It's that "something additional to support the physical validity" that seems quite ad hoc in my head. P.S.: Do you agree with my understanding of first and second quantization in my step #4? – Tfovid Jul 28 '20 at 09:58
  • @Tfovid In the end, the physical validity of a theory can only be tested by experiment. Your recipe comes with no guarantee that experiments will agree with it. – John Doty Jul 28 '20 at 13:32
  • @Tfovid it certainly is ad hoc, and inevitably so. There is no more general principle than "has to obey the rules of quantum mechanics" and "has to reproduce classical mechanics at everyday scales" (within this context). As John Doty says, it ultimately it's experiment/observation that determines the validity. To have good heuristics can guide you in designing good experiments. – doetoe Jul 28 '20 at 15:37
  • As for the second part of your question, when speaking of quantization, usually we mean starting out with a classical theory, and constructing the quantum theory of which it is a limit, which is what you do in the first 3 steps. What you do in the 4th step refers to the fact that in QM some quantities take only discrete values, which is a different kind of quantization. First and second quantization how I know it it are recipes for quantization in the context of quantum field theory (and not the observations you are referring to) – doetoe Jul 28 '20 at 15:45