I've studied QM and QFT for a couple of years now, so I'm familiar with the tersm "quantize", "quantization" and so on. I'm obviously also familiar with the Lagrangian description of Classical Mechanics. But I'm pretty sure that I lost a piece somewhere in the middle, because every time a classical theory is presented and the book/article/professor goes on saying "and now we quantize the theory" I don't fully understand what is meant.
What is the rigorous definition of the "process of quantization"? Is it just finding the conjugate variables and imposing commutation/anticommutation rules? Or is there more to it?

Bonus question: how does one use the definition of quantization to explain the historical name of QFT, the "second quantization"?

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    $\begingroup$ There is no rigorous notion of quantization as an automatic process. Taking a classical system and producing a quantum system involves choices. In math language, quantization is not a functor. However, given a quantum one-body system then "second quantization" is essentially automatic: replace the Hilbert space by its corresponding Fock space. This process is (essentially) a functor. $\endgroup$
    – mike stone
    Commented Mar 8, 2021 at 0:09
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    $\begingroup$ Quantisation is just a recipe for obtaining a quantum theory from a classical theory. It just involves obtaining in some way a collection of operators over a Hilbert space from a classical theory. As you can see here there are a lot of ways to do this, it is not exactly a rigorous procedure in many cases. $\endgroup$
    – Charlie
    Commented Mar 8, 2021 at 0:18
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    $\begingroup$ You seem to be talking specifically about canonical quantization (see here: physics.stackexchange.com/q/544748/247642). For the relation between the first and the second quantization you may look here: physics.stackexchange.com/a/542483/247642 $\endgroup$
    – Roger V.
    Commented Mar 8, 2021 at 0:56
  • $\begingroup$ You might enjoy this. "Second quantization" is a functor: a repackaging of an infinity of quantum oscillators to describe quantum fields, not the invention of a new theory, as in the case of ordinary quantization; it has proven an enduringly confusing inappropriate term. $\endgroup$ Commented Mar 8, 2021 at 1:50
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    $\begingroup$ If you wished to probe how arbitrary quantization is, try this question. $\endgroup$ Commented Mar 9, 2021 at 14:14

3 Answers 3


There are quite a number of notions of quantisation. It is a theorem by van Hove & Groenewald that the naive notion of quantisation that is usually taught where position and momentum are mapped to the usual operators cannot hold in all generality, that is the classical Poisson bracket is not mapped perfectly to the commutation relations we would expect.

Other quantisation methods include (and is not limited to) deformation quantisation, geometric quantisation & path integral quantisation.

On a more philosophical view, one might view that the whole process of quantisation is wrong-headed and really we should be looking at classical limits of axiomatically defined quantum systems. That is, the quantum theory comes first, and the classical system is a limit of it.

One example of this view is causal set theory, another example is algebraic quantum theory where nets of observables, thought of algebras, are posited on spacetime. One outcome here is that the algebra required is theory independent, being a hyperfinite von Neumann factor of type III.

  • $\begingroup$ Thanks for the references, I'll look into them. I know that the idea of Classical-to-Quantum is imprecise, because the "real deal" is the quantum theory and the classical one is just an approximation, but the idea of "quantization" is widely used and therefore I was looking for some unifying idea behind all kinds of quantizations. Is there such a thing? $\endgroup$ Commented Mar 10, 2021 at 8:39
  • $\begingroup$ For example, I have always been presented the anomalies as "symmetries of the classical actions that don't survive after quantization", implying a Classical-to-Quantum flow of thought. Is this definition wrong-headed too? $\endgroup$ Commented Mar 10, 2021 at 8:45

I can share with you what my volume of Merzbacher, 2nd Ed., says on this topic, and this, in turn, may become part of an answer that ultimately satisfies you as you piece this together from what others have to say as well. We know Hamilton's equations from classical mechanics: $$\dot q_k=\frac{\partial H}{\partial p_k}$$ and $$\dot p_k=-\frac{\partial H}{\partial q_k}$$

In chapter 15, Quantum Dynamics, Merzbacher derives the time rate of change of an operator: $$i\hbar\frac{d\langle A \rangle }{dt}= \langle AH-HA\rangle$$ or $$\frac{d\langle A\rangle}{dt}= \frac {\langle AH-HA\rangle}{i\hbar}$$ He then proceeds to apply this to a system of a mass point particle. He states: "If a quantum system has a classical analog, expectation values of operators behave, in the limit $\hbar \rightarrow0$, like the corresponding classical quantities."

Thus if the operator A above is x (position) or p (momentum), the quantum expression must reduce to the appropriate Hamilton equation above in the limit that $\hbar \rightarrow 0$.

He then writes:

All these conditions can be satisfied if we do the following:

(1) Let H be a Hermitian operator identical in form with $H_c$ [the classical Hamiltonian], but replacing all coordinates and momenta by their corresponding operators.

(2) Postulate the fundamental commutation relations between the Hermitian operators representing coordinates and momenta.

He then proceeds to show that for any two general functions of the coordinates and momenta, F and G, the expression on the RHS of the last equation above is just the Poisson bracket of these two functions from classical mechanics, {F,G} (Note: the quantum expression will have an $\hbar$ factor in it that must be taken to 0 as the correspondence principle requires to complete the equality between the two expressions).

Finally, to come full circle he writes:

In order to test the quantization procedure just outlined, we must show that the quantum mechanics obtained from it are identical with the equations of wave mechanics which we know to give an accurate description of many physical phenomena.

From the title he gives the section, Wave Mechanics Regained, you won't be surprised to learn that he rederives the Schrodinger equation.

As I say, I am just regurgitating what I knew to exist in Merzbacher's QM book. I hope it contains some piece of the puzzle you are trying to put together!

  • $\begingroup$ "the expression on the RHS of the last equation above is just the Poisson bracket" is it literally the Poisson bracket or just structurally very similar to the Poisson bracket? I've read that it is similar to it algebraically but it's still a commutator and a distinct thing. $\endgroup$
    – user288901
    Commented Mar 8, 2021 at 1:19
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    $\begingroup$ Hi @ExpertNonexpert . Thanks for your question! I have added a parenthetical statement to my answer above to make it more complete. Merzbacher does a calculation where $G=x^2$ and $F=p_x^2$. He computes both the Poisson bracket for these two functions and the $x^2p_x^2 - p_x^2x^2$ using the commutation relation for $x$ and $p_x$ and shows he gets the exact same result after taking $\hbar$ to zero in the latter. $\endgroup$
    – CGS
    Commented Mar 8, 2021 at 3:10
  • $\begingroup$ That's interesting. $\endgroup$
    – user288901
    Commented Mar 8, 2021 at 3:38

There is a section of Leonard Susskind's Quantum Mechanics titled "Quantization."

...We started with a well known and well trusted classical system -the free particle- and quantized it. We codify this procedure as follows:

  1. Start with a classical system. This means a set of coordinates $x$ and momenta $p$... The coordinates and momenta come in pairs, $x_i$ and $p_i$. The classical system also has a Hamiltonian, which is a function of the $x$'s and $p$'s.
  1. Replace the classical phase space with a linear vector space. In the position representation, the space of states is represented by a wave function $\psi(x)$ that depends on the coordinates -in general, all of them.
  1. Replace the $x$'s and $p$'s with operators $X_i$ and $P_i$. Each $X_i$ acts on the wavefunction to multiply it by $x_i$. Each $P_i$ acts according to the rule $P_i \rightarrow -i\hbar\frac{\partial}{\partial x_i}.$
  1. When these replacements are made, the Hamiltonian becomes an operator that can be used in either the time-dependent or time-independent Schrodinger equation. The time-dependent equation tells us how the wave function changes with time. The time-independent form allows us to find the eigenvectors and eigenvalues of the Hamiltonian.
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    $\begingroup$ The problem is that this approach is not rigorous. It does nog work in general. $\endgroup$
    – NDewolf
    Commented Mar 8, 2021 at 10:17
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    $\begingroup$ @NDewolf The problem is that quantization can't be rigorous, since there are quantum phenomena without classical analogues. $\endgroup$
    – John Doty
    Commented Mar 8, 2021 at 17:17
  • $\begingroup$ @JohnDoty I can sort of see what you mean, but even the simplest quantisation procedure introduces effects without classical analogue so that's not a very clear statement. There is also nothing wrong with a procedure that only gives an injection from classical to quantum systems (nobody expects anything more than that as far as I know). $\endgroup$
    – NDewolf
    Commented Mar 8, 2021 at 17:51
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    $\begingroup$ An example of a quantum phenomenon that fails to emerge from naive quantization is the Pauli Exclusion Principle. But I agree with you that quantization is a valuable technique that often yields workable models. However, this is physics, not math. Rigor is neither necessary nor sufficient as a path to a good model. Experiment is the final decider, not mathematics. $\endgroup$
    – John Doty
    Commented Mar 8, 2021 at 18:58

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