This is not a question pertaining to interpretations, after the last one I realized I should not open Pandora's Box ;)

For theories to be consistent, they must reduced to known laws in the classical domains. The classical domain can be summed up as:

$$\hbar=0 ; c=\infty$$

Which is OK. I need to know, however, is that if QM is an independent and fundamental theory, why does it rely so heavily on the classical formalism. Is it necessary for a classical formalism to exist in order to have a quantum formalism? From as far as I have read, it does not seem so, and I find this puzzling. Suppose you have a dissipative system, or an open system when you cannot write an autonomous Hamiltonian in the classical case, how then do we approach these quantum mechanically, when we cannot even write down the corresponding Hamiltonian.

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    $\begingroup$ QM is a fundamental and independent theory -- logically it doesn't rely on the classical formalism, and yet historically it was invented by imperfect humans who knew classical mechanics. Physics exists independent of the equations we create to describe it, but gives us no power to divine the "right" equations from the aether. Such is life. Aside from that open quantum systems are an active research topic -- your intuition that a Hamiltonian approach is insufficient is correct: you must write down a "Liouvillian" superoperator (oh no more classical intuition!) $\endgroup$ – wsc Mar 30 '11 at 3:36
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    $\begingroup$ For quantum systems there is an approach due to Lindblad which allows you to include disspation. The general theory for open quantum systems was developed by Feynman and Vernon in 1963 $\endgroup$ – user346 Mar 30 '11 at 3:39
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    $\begingroup$ I'll take up here something in Roy Simpson's Answer (which I commend). From my PoV (not Roy's) $\hbar$ plays two roles in quantum theory: (1) it fixes the scale of Lorentz invariant quantum fluctuations, which can be accommodated in classical models perfectly well; (2) it fixes the scale of noncommutativity of measurement operators, which is entirely different from classical measurement theory. Hence, $\hbar=0$ has to be considered the classical case only with care. This is separate from the issues discussed in comments on David bar Moshe's Answer. $\endgroup$ – Peter Morgan Mar 31 '11 at 16:44

Quantum mechanics and quantum mechanical theories are totally independent of the classical ones. The classical theories may appear and often do appear as limits of the quantum theories. This is the case of all "textbook" theories - because the classical limit was known before the full quantum theory, and the quantum theory was actually "guessed" by adding the hats to the classical one. In a category of cases, the full quantum theory may be "reverse engineered" from the classical limit.

However, one must realize that this situation is just an artifact of the history of physics on Earth and it is not generally true. There are classical theories that can't be quantized - e.g. field theories with gauge anomalies - and there are quantum theories that have no classical limits - e.g. the six-dimensional $(2,0)$ superconformal field theory in the unbroken phase. Moreover, it's typical that the quantum versions of classical theories lead to new ordering ambiguities (the identity of all $O(\hbar^k)$ terms in the Hamiltonian is undetermined by the classical limit in which all choices of this form vanish, anyway), divergences, and new parameters and renormalization of them that have to be applied.

Also, the predictions of quantum mechanics don't need any classical crutches. Quantum mechanics works independently of its classical limits, and the classical behavior may be deduced from quantum mechanics and nothing else in the required limit. Historically, people discussed quantum mechanics as a tool to describe the microscopic world only, assuming that the large objects followed the classical logic. The Copenhagen folks divided the world in these two subworlds, in an ad hoc way, and that simplified their reasoning because they didn't need to study quantum physics of the macroscopic measurement devices etc.

But these days, we fully understand the actual physical mechanism - decoherence - that is responsible for the emergence of the classical logic in the right limits. Because of decoherence, which is a mechanism that only depends on the rules of quantum mechqnics, we know that quantum mechanics applies to small as well as large objects, to all objects in the world, and the classical behavior is an approximate consequence, an emergent law.

To know the evolution in time, one needs to know the Hamiltonian - or something equivalent that determines the dynamics. The previous sentence is true both in classical physics and quantum mechanics, for similar reasons, but independently. If a classical theory is a limit of a quantum theory, it of course also means that its classical Hamiltonian may be derived as a limit of the quantum Hamiltonian. Of course, if you don't know the Hamiltonian operator, you won't be able to determine the dynamics and evolution with time. Guessing the quantum Hamiltonian from its classical limit is one frequent, but in no way "universally inevitable", way to find a quantum Hamiltonian of a quantum theory.

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    $\begingroup$ @Luboš Upvoting this, but I reserve approval of your didactic statements about decoherence. Decoherence is definitely very interesting mathematics, but is it really true that "we" fully understand decoherence? I don't want to set an infinitely high standard for fully, and I know that many Physicists express similar sentiments about decoherence, but what we can do with the mathematics isn't yet enough. $\endgroup$ – Peter Morgan Mar 30 '11 at 13:15
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    $\begingroup$ I like this passage: "Guessing the quantum Hamiltonian from its classical limit is one frequent, but in no way "universally inevitably", way to find a quantum Hamiltonian of a quantum theory." It is a very healthy point of view, seriously. Not all microscopic (true) things can be "derived" from an average, classical picture. $\endgroup$ – Vladimir Kalitvianski Mar 30 '11 at 13:41
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    $\begingroup$ Thanks, Vladimir - also for reproducing the typo in "inevitable" that I just fixed. Peter, what I meant is that we know how to define decoherence and in principle, for any Hamiltonian, we may quantitatively calculate the decrease of the off-diagonal elements etc. In all cases I know, this ability is not just in principle, it's also in practice up to the accuracy that is enough to qualitatively divide the quantum and classical realms. Also, we know that decoherence is fully responsible for the classical-quantum boundary - nothing else has to occur. So we fully know it, don't we? If not, why? $\endgroup$ – Luboš Motl Mar 30 '11 at 16:59
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    $\begingroup$ I second @Peter's objection to @Luboš's apodictic (rather than didactic) statements about decoherence. As part of the mathematical formalism, decoherence is in no position to solve the persistent issue of interpretation — at least not all by itself. Decoherence shows that superpositions of differently localized states of macroscopic objects, well, decohere, but just like Copenhagen, it fails to give a rigorous definition of the term "macroscopic." $\endgroup$ – Koantum Mar 31 '11 at 2:42
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    $\begingroup$ @Koantum: the comments I wrote have nothing to do with MWI which is just a psychological crutch to "visualize" something. I don't think that MWI is a part of physics, and I don't believe MWI in any sense. My preferred interpretation is in terms of the Consistent Histories, but I view it just as a modern formalism that doesn't really change anything substantial about the canonical/Copenhagen interpretation of quantum mechanics. Probabilities of outcomes are everything that physics may predict and QM gives totally clear rules how to do it; everything else is philosophical flapdoodle. $\endgroup$ – Luboš Motl Apr 1 '11 at 10:05

Note that all the experiments we did prior to discovering quantum weirdness are still out there. Your complete theory has to be able to explain all the demos we do in a introductory freshman mechanics class.

However, if you can show that a new theory evolves to the old one in the realm where the old one was correct, you don't have to re-prove everything you already knew.

If you can build the new theory on the footing of the old one, your halfway to that point already.


I want to quote for you F.A. Berezin's point of view regarding this question as given in his seminal papers in "General concept of quantization" Commun. math. Phys. 40, 153-174(1975) and "Quantization" Math. USSR Izv. 8 (1974) 1109-1165.

In these papers, Berezin presented his theory of quantization having the correspondence principle as one of its postulates (at $\hbar \rightarrow 0$). The quantum and classical theories describe the same mechanical reality; the quantum description of reality is more detailed than the classical one, but once we have this detail, we have the freedom of giving up on some of it. Then we should reach the experimentally well established classical limit. This is like the case of one having a telescope but intentionally defocusing its lens.

It should also be mentioned that this interpretation is consistent with the nonfunctoriality point of view of (first) quantization, because (as emphasized in Berezin's articles) a lot of different detailed pictures of the reality may correspond to the same "defocused" limit of the classical theory.


Referring to Peter's comment. I should clarify that the quantum theory converges at the limit ($\hbar \rightarrow 0$) to a classical theory only for a subset of the states and observables of the system. For example Weyl operators on coherent states have a well defined limit. A detailed discussion of the classical limit (and why other states do not have a well defined limit) may be found in the following article by N.P. Landsman: "Between classical and quantum".

However, I think that Berezin's view as a general principle is still valid as a basis of our understanding the quantum phenomena.

  • $\begingroup$ The limit $\hbar\rightarrow 0$ is not a well-behaved limit, because there is violation of the Bell inequalities --there are states and observables for which one can construct the $2\sqrt{2}$ violation of the CHSH inequality-- for all non-zero $\hbar$. The classical case is equality, as Yayu puts it, $\hbar=0$, not the limit. A succinct, purely mathematical reference is L.J.Landau, Phys.Lett.A 120, 54(1987), the content of which is available at mth.kcl.ac.uk/~llandau/Homepage/Math/bell.html. $\endgroup$ – Peter Morgan Mar 30 '11 at 12:59
  • $\begingroup$ Peter - Thank you for your remark, I have added a clarification of this point in an update of my answer. $\endgroup$ – David Bar Moshe Mar 30 '11 at 16:13
  • $\begingroup$ Ah! Landsman! Not quite as succinct, but gold! $\endgroup$ – Peter Morgan Mar 30 '11 at 16:53

I would like to question some of the assumptions of the question itself here. Firstly the title:

Why is a Classical Formalism necessary for Quantum Mechanics?

Is this premise actually true? One could imagine setting up a quantum structure (Hilbert Spaces, $\Psi$, Operators) without any reference to anything classical. The closest one comes is in formulating the experimental meanings of the eigenvalues.

For theories to be consistent, they must reduced to known laws in the classical domains.

I dont agree that this is a requirement for theory consistency. Of course it depends what "reduced" means here, but some quantum theories live quite happily without an (apparent) classical counterpart (spin?). Anyway theory consistency is about whether if P(x) is deduced then not P(x) is not also deducible ie a logical requirement.

The point that I suspect is intended here is that there is expected to be "theory correspondence" from Quantum to Classical: this is weaker than requiring "reduction".

The classical domain can be summed up as $\hbar=0$

Well not quite, it is now known that h was discovered in Classical Statistical Mechanics by Gibbs as an area proportionality term.

As far as writing Hamiltonians is concerned the Classical theory could be said to simply provide a heuristic guide - in a sense it is not necessary since it may not exist. That is if we have the classical Hamiltonian H(p,q), then we know that the quantum one is unordered as $H(-i\hbar\partial /{\partial x},x)$, but we still dont know what the ordered (correct) form is.

The real question that is being asked here is how to write a quantum Hamiltonian when there is no (apparent) classical theory. Perhaps the answer lies in Feynman's saying "Guess, then try it out" or perhaps there is some greater theory of Quantum Hamiltonians yet to be found...

  • $\begingroup$ @Roy, Useful. It prodded me to discuss whether $\hbar=0$ is necessary for classicality in the Question's comments stream, qv. I like the "Guess, then try it out" quote. Good Engineering, I suppose, is educated guessing, knowing how to get it right first time. Does your first comment refer to presenting a quantum theory by presenting an algebra of observables and presenting or constructing a Hilbert space representation of it, from which we can construct the Hamiltonian as the generator of time-like translations? Are we also thinking Mackey? $\endgroup$ – Peter Morgan Mar 31 '11 at 16:57
  • $\begingroup$ @Peter, thanks for the vote(s). I did not mean to be as specific as Mackey there - anything even more productive would do! I will have to study your top comments about $\hbar$ more as well: maybe they are related to something in one of your papers? $\endgroup$ – Roy Simpson Apr 1 '11 at 15:45
  • $\begingroup$ @Roy I wouldn't expect you to be channeling Mackey specifically. Your Answer here just suggested to me some affinity with that type of tradition. The comments on $\hbar$ are associated with the math in Phys.Lett.A338,8-12(2005), quant-ph/0411156. One can construct a classical field that has Lorentz invariant fluctuations with a scale that is determined by $\hbar$; as a classical field, for the commutator we have $[\hat\phi(x),\hat\phi(y)]=0$ for all points $x,y$. One could also mention Stochastic Electrodynamics. $\endgroup$ – Peter Morgan Apr 2 '11 at 13:34
  • $\begingroup$ OK. My reference there to productive was because you are using it in comments on other theories, so I thought I'd mention the term, and because I have studied the idea of trying to formalise the idea of a "productive theory". But this latter topic isnt physics (as yet), but a branch of formal logic. $\endgroup$ – Roy Simpson Apr 2 '11 at 15:00
  • $\begingroup$ @Roy, I noted the reference to productive. It seems hard to include new approaches in a formalized "productivity", insofar as they're new. To evade that, one class of measures of productivity would be any monotonic function of the number of people who think there's enough promise to spend time on an approach. A more nuanced measure would somehow include the status in a wider network of people in the network of people working on an approach. Defining an "approach" is a problem because a tight definition makes any approach the work of only one person. Have you something on "productive"? $\endgroup$ – Peter Morgan Apr 2 '11 at 15:49

This is an excellent question and the answer is it follows from the crucial postulates of quantum mechanics. Namely, the operator corresponding to a dynamical variable is obtained by replacing the classical canonical variables by corresponding quantum mechanical operators and any pair of canonically conjugate operators will satisfy the Heisenberg commutation rules. Physically speaking, it has definite relevance to these remarks of the great Neils Bohr, "However far the phenomena transcend the scope of classical physical explanation, the account of all the evidence must be expressed in classical terms". This is because "by the word 'experiment' we refer to a situation where we can tell others what we have done and what we have learned" so that "the account of the experimental arrangement and of the results of the observations must be expressed in unambiguous language with suitable application of the terminology of the classical physics". (Note that for some observables like spin there is no classical analog and we take different approach.)


As far as I can see, this is an interpretational issue. The testable part of quantum mechanics is a probability calculus. It serves to assign probabilities to possible outcomes of measurements on the basis of actual outcomes of other measurements. In other words, it concerns correlations between measurement outcomes. The rest (for instance, claims about what happens between measurements) is "not even wrong" in Wolf­gang Pauli’s felic­i­tous phrase. In other words, it is a matter of interpretation.

The piv­otal role played by mea­sure­ments in all stan­dard for­mu­la­tions of quantum mechanics is an embar­rass­ment known as the “mea­sure­ment problem.” As an anony­mous ref­eree for a phi­los­ophy of sci­ence journal once put it to me, “to solve this problem means to design an inter­pre­ta­tion in which mea­sure­ment processes are not dif­ferent in prin­ciple from ordi­nary phys­ical inter­ac­tions.” Yet how can reducing mea­sure­ments to “ordi­nary phys­ical inter­ac­tions” solve this problem, con­sid­ering that quantum mechanics describes “ordi­nary phys­ical inter­ac­tions” in terms of cor­re­la­tions between the pos­sible out­comes of mea­sure­ments per­formed on the inter­acting sys­tems? This kind of “solu­tion” merely sweeps the problem under the rug. The way I see it, to solve the mea­sure­ment problem means to design an inter­pre­ta­tion in which the cen­tral role played by mea­sure­ments in all stan­dard axiom­a­ti­za­tions of quantum mechanics is under­stood. (For my two cents, see the links on this page at my personal website.

  • $\begingroup$ Seeing sb1's answer, posted almost simultaneously with mine, I should add that Niels Bohr did indeed more than anyone else to explain the central role played by measurements in the axiomatics of quantum mechanics. $\endgroup$ – Koantum Mar 30 '11 at 4:01
  • $\begingroup$ I know what you mean, and I'm sure you do, but "cor­re­la­tions between the prob­a­bil­i­ties" is not a happy turn of phrase. Something like "correlations between outcomes" or "joint probabilities of outcomes", perhaps. As an aside, I almost share your almost instrumentalist starting point, but I regret to say that I haven't found the details of Pondicherry productive. I'm thinking of asking a Question that would give you some scope, but I haven't yet thought of one that wouldn't be [closed]. I felt that I had to read between the lines to find this Answer useful. $\endgroup$ – Peter Morgan Mar 30 '11 at 12:56
  • $\begingroup$ @peter, I have corrected the objectionable phrase "cor­re­la­tions between the prob­a­bil­i­ties." (I got it right in the first paragraph.) Agreed, as an interpretation, what I used to call the "Pondicherry interpretation of quantum mechanics" is not productive — at least not in a narrow sense of the word. $\endgroup$ – Koantum Mar 31 '11 at 2:23

As an addition to the previous answers, let me give my own point of view on this subject. Briefly, the classical picture is an inclusive quantum picture - it includes many different quantum events in one classical "object" by definition. It is just like taking a photograph with photon by photon, point by point, pixel by pixel. Each point is not classical but their ensemble taken together and associated with an "average" or "inclusive" image ("object") is usually described with the classical physics (deterministic point-like or wavy classical behavior). The brightest example - an interference picture. It is a multi-photon stuff.

Formally one can see it from the Ehrenfest equations if one remembers that an "average" means "inclusive" and "normalized". There is no need to tend here $\hbar$ to zero.

It is funny but the very well known Coulomb potential is also an inclusive rather than "elastic" picture, see my opus here.

EDIT: Seeing downvotes, I conclude that too few understand or know the Ehrenfest theorems and ascribe the inclusive picture to my personal "speculations". Too bad!

  • $\begingroup$ Please stop pushing your own theory forward all the time. $\endgroup$ – Olaf Mar 30 '11 at 16:34
  • $\begingroup$ If my results are published, recognized and answer the questions, I answer to help people. $\endgroup$ – Vladimir Kalitvianski Mar 30 '11 at 16:39
  • $\begingroup$ @MisterO I think we have gone over this territory before. As the Beatles said, Let it be. $\endgroup$ – user346 Mar 31 '11 at 6:14

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