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Say you cook up a model about a physical system. Such a model consists of, say, a system of differential equations. What criterion decides whether the model is classical or quantum-mechanical?

None of the following criteria are valid:

  • Partial differential equations: Both the Maxwell equations and the Schrödinger equation are PDE's, but the first model is clearly classical and the second one is not. Conversely, finite-dimensional quantum systems have as equations of motion ordinary differential equations, so the latter are not restricted to classical systems only.

  • Complex numbers: You can use those to analyse electric circuits, so that's not enough. Conversely, you don't need complex numbers to formulate standard QM (cf. this PSE post).

  • Operators and Hilbert spaces: You can formulate classical mechanics à la Koopman-von Neumann. In the same vein:

  • Dirac-von Neumann axioms: These are too restrictive (e.g., they do not accommodate topological quantum field theories). Also, a certain model may be formulated in such a way that it's very hard to tell whether it satisfies these axioms or not. For example, the Schrödinger equation corresponds to a model that does not explicitly satisfy these axioms; and only when formulated in abstract terms this becomes obvious. It's not clear whether the same thing could be done with e.g. the Maxwell equations. In fact, one can formulate these equations as a Dirac-like equation $(\Gamma^\mu\partial_\mu+\Gamma^0)\Psi=0$ (see e.g. 1804.00556), which can be recast in abstract terms as $i\dot\Psi=H\Psi$ for a certain $H$.

  • Probabilities: Classical statistical mechanics does also deal with probabilistic concepts. Also, one could argue that standard QM is not inherently probabilistic, but that probabilities are an emergent property due to the measurement process and our choice of observable degrees of freedom.

  • Planck's constant: It's just a matter of units. You can eliminate this constant by means of the redefinition $t\to \hbar t$. One could even argue that this would be a natural definition from an experimental point of view, if we agree to measure frequencies instead of energies. Conversely, you may introduce this constant in classical mechanics by a similar change of variables (say, $F=\hbar\tilde F$ in the Newton equation). Needless to say, such a change of variables would be unnatural, but naturalness is not a well-defined criterion for classical vs. quantum.

  • Realism/determinism: This seems to depend on interpretations. But whether a theory is classical or quantum mechanical should not depend on how we interpret the theory; it should be intrinsic to the formalism.

People are after a quantum theory of gravity. What prevents me from saying that General Relativity is already quantum mechanical? It seems intuitively obvious that it is a classical theory, but I'm not sure how to put that intuition into words. None of the criteria above is conclusive.

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    $\begingroup$ I've removed some comments which didn't seem to be intended to request clarifications or suggest improvements. $\endgroup$ – David Z Apr 4 '18 at 21:44
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    $\begingroup$ Note that the appropriate answer to this question depends quite heavily on whether you mean "what distinguishes quantum theories from classical theories specifically", or "what distinguishes quantum theories from other theories in general" - for example, the class of what are often referred to as generalized probabilistic theories, which include classical, quantum and many other theories besides. In this latter class, classical theories are distinguished by many properties, and so the lack of any of these tells us we are dealing with a non-classical theory - but not necessarily a quantum one $\endgroup$ – Robin Saunders Apr 5 '18 at 0:19
  • $\begingroup$ @RobinSaunders Hmm that's actually a very good point, I like the way you put it. If you ever have some free time, please consider making that comment into an answer. Cheers! $\endgroup$ – AccidentalFourierTransform Apr 8 '18 at 2:11
  • $\begingroup$ I'm interested in why you say TQFT does not fit within the Dirac-von Neumann axioms. It's true that those axioms don't tell you much about the structure of the theory, but it's not really different for any QFT, for which there is a Hilbert space associated to any spatial manifold. I'd say those axioms are insufficiently strong, rather than being too restrictive. $\endgroup$ – Holographer Apr 8 '18 at 17:07

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I think this is a subtle question and I think it depends somewhat on how you choose to represent quantum mechanics. To see one extreme of this, consider the viewpoint put forth by Kibble in [1]. For simplicity I will be thinking of finite-dimensional quantum systems here; there are some subtleties in infinite dimensions but as far as I know the basic picture still holds. In this, he shows that if we describe the theory in terms of physical states (rays in the Hilbert space), then the dynamics of Schrödinger evolution correspond exactly to Hamiltonian evolution via the symplectic form from the Kähler structure on the projective Hilbert space (which is to say, the evolution is that of a classical system). However there are two distinctions which make quantum mechanics different from classical mechanics:

  • The phase space must be a projective Hilbert space (as opposed to just a symplectic manifold), and the Hamiltonian is restricted to being a quadratic form in the homogeneous coordinates on projective space. In classical mechanics any (sufficiently smooth) function is admissible as a Hamiltonian.
  • Composite systems are described differently. In classical mechanics the phase space of a composite system is the Cartesian product of the phase spaces. In quantum mechanics, it is the Segre embedding (which descends from the tensor product of Hilbert spaces). This is parametrically different; if the phase spaces of the two subsystems are $2m$ and $2n$, then in classical mechanics the composite system has dimension $2m+2n$, whereas in quantum mechanics it has dimension $2(n+1)(m+1)-2$. The extra states are the entangled states. Virtually all the observable consequences of QM come here, e.g. Bell inequalities. Of course if we consider identical particles things get even a bit more complicated.

If you ignore the second point, and focus only on a single quantum system, the surprising conclusion is that every quantum mechanical system is a special case of classical mechanics (with the provision that again I haven't checked the details in infinite dimensions but it is at least morally true). However, part of the structure of quantum mechanics is how it describes composite systems so you can't just ignore this second point. A mathematician would say that this gives an injective functor from the category of quantum mechanical theories to the category of classical theories which is not compatible with the symmetric monoidal structures on the two.

I want to point out that this is emphatically not how we typically think of the correspondence principle in quantum mechanics. That is, it is a mapping from a finite-dimensional quantum mechanical system to a finite-dimensional classical system (of the same dimension). Normally, if we think about e.g. a free particle in one dimension, the Hilbert space for that quantum system is infinite dimensional, yet it corresponds to a 2-dimensional classical phase space. But the point is that, at least in this question, we can't restrict to the ordinary notion of correspondence since we don't have a physical interpretation for the system of equations describing the theory.

Additionally, despite the above example, whether a theory is classical or quantum has essentially nothing to do with where the states live. Indeed, if we just want to consider a free particle in one dimension again, we would typically describe its state as a self-adjoint trace class unit trace operator $\hat \rho$ on the Hilbert space $L^2(\mathbb R)$. In contrast, in classical mechanics we would describe a state as a probability distribution $\rho$ on the phase space $\mathbb R^2$ (note that in the above example we had only pure classical states i.e. only those described by a $\delta$ function on the phase space whereas now we have mixed states). However we could just as easily describe the quantum state by its Wigner function, in which case it lives in exactly the same affine space as the classical distribution. However the Wigner function satisfies slightly different inequalities than the classical probability distribution; in particular it can be slightly negative and cannot be too positive. The details of this were first worked out in [2]. In this case, it is the dynamics that give away the quantum nature. Specifically, to go from classical to quantum mechanics, we must replace the Poisson bracket by the Moyal bracket (which has $O(\hbar^2)$ corrections), indicating the failure of Liouville's theorem in the phase space formulation of quantum mechanics: (quasi)probability density is not conserved along trajectories of the system.

All of this is to say that it seems difficult (and maybe impossible) to try to find a single distinguishing feature between classical and quantum mechanics without considering composite systems, so if that is what you want, I'm not sure I have an answer. If you do allow for composite systems though, it is a pretty unambiguous distinction. Given this, it is perhaps not surprising that all the experimental tests we have which demonstrate that the world is quantum and not classical are based on entanglement.

References:

[1]: Kibble, T. W. B. "Geometrization of quantum mechanics". Comm. Math. Phys. 65 (1979), no. 2, 189--201.

[2]: H.J. Groenewold (1946), "On the Principles of elementary quantum mechanics", Physica 12, pp. 405-460.

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As far as I know, the commutator relations make a theory quantum. If all observables commute, the theory is classical. If some observables have non-zero commutators (no matter if they are proportional to $\hbar$ or not), the theory is quantum.

Intuitively, what makes a theory quantum is the fact that observations affect the state of the system. In some sense, this is encoded in the commutator relations: The order of the measurements affects their outcome, the first measurement affects the result of the second one.

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    $\begingroup$ I think this answer is on the right track. In quantum mechanics, the transfer of information is intrinsically tied to the dynamics of the system, whereas in classical physics that is not the case. $\endgroup$ – DanielSank Apr 4 '18 at 20:16
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    $\begingroup$ I would agree with this. It was my answer also but I came too late. So, in any situation, what exactly is quantum is best shown in experiments such as Stern-Gerlach type. If you measure for x dirrection you get + and - or spin up or down, but if you measure in y, you get spins in that direction. If you measure first in x, thaen in y, you get as a rersult a y direction, but if you measure in x, then again in x, you get only x..... $\endgroup$ – Žarko Tomičić Apr 4 '18 at 20:17
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    $\begingroup$ I would say on the contrary that observations affect the state of a classical systems where everything is physical. $\endgroup$ – Bill Alsept Apr 4 '18 at 20:55
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    $\begingroup$ In MWI, observations don't affect the state of the system in some mysterious way. Rather, you should consider the composite Hilbert space describing both the system and the measuring device (large-dimensional Hilbert space). A measurement is a time-dependent interaction and in the measurement limit you produce a fully entangled state between the two. If you compute the reduced density matrix for the system of interest, you get a diagonal matrix of the probabilities. The point being that "observations affect the state of the system" is arguably really a statement about composite systems. $\endgroup$ – Logan M Apr 4 '18 at 21:07
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    $\begingroup$ @AccidentalFourierTransform I think that the existence of Poisson brackets is a semantic point. As long as you take the specific definition of a commutator as $[A, B] = A B - B A$ then this answer holds, IMO. After all, non-commuting operators, under this specific definition, are what lead to all the "quantum weirdness" such as Bell's theorem and the Uncertainty principles. All operators in KvN commute under this definition. $\endgroup$ – Bridgeburners Apr 5 '18 at 20:06
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Frame challenge: I think the question is based on a misleading premise.

While there are a number of characteristics typical of quantum theories as opposed to classical theories - some you've already listed in the question, and others have been suggested in the existing answers - there's no particular reason to expect there to be a single unambiguous rule that categorizes any arbitrary theory as either quantum or classical.

Nor is there any particular need for such a rule. You give the example of quantum gravity. However, the reason we want a quantum theory of gravity is not because it has the tag "quantum" attached to it, as if it were a handbag that would not be adequately fashionable without the correct label, but because we want it to be able to answer certain questions about reality which we already know General Relativity can't answer.

In short, don't worry about whether the theory is "quantum" or not - worry about whether it answers the questions you want answered or not.

Also relevant.


Addendum: the same goes for the existing theories, of course. We don't like the Standard Model because it is quantum. We like it because it works.

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    $\begingroup$ @JerrySchirmer, that's not really what this question asks, though. $\endgroup$ – Harry Johnston Apr 5 '18 at 19:37
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    $\begingroup$ It asks "what is it about a theory that makes it 'quantum'". And the answer would be "we apply quantization to some classical theory" $\endgroup$ – Jerry Schirmer Apr 5 '18 at 20:43
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    $\begingroup$ @JerrySchirmer, that's one possible answer, certainly. But I think the OP is asking for criteria that are based directly on the mathematical characteristics of a particular model, rather than on how the model was developed. (And I think in practice that, if presented with a theory with characteristics similar to other quantum theories, most physicists would call it a quantum theory regardless of whether it was derived from a classical model or not.) $\endgroup$ – Harry Johnston Apr 5 '18 at 21:20
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    $\begingroup$ ... incidentally, unless I've overlooked something, none of the existing answers mention quantization as a possible criteria so you might want to post that as an answer @JerrySchirmer $\endgroup$ – Harry Johnston Apr 5 '18 at 21:23
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    $\begingroup$ All that said, if I had to choose one feature that was the most important characteristic of quantum theories, I'd have to endorse Photon's answer. $\endgroup$ – Harry Johnston Apr 6 '18 at 23:06
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TL;DR: Correlations.


First things first: since the OP asks for a criterion to tell whether a model is quantum mechanical, the answer has to involve observables. After all if you could rewrite your "quantum" model as a "classical" model, those labels would not be worth much after all.

Furthermore all quantum theories (that I know of) are probabilistic, therefore this answer focuses on probabilistic observables, i.e. correlation functions.

The fundamental difference between a quantum theory and a classical theory is their correlation structure. That is, quantum theories can show correlations that classical theories cannot.

The historically first and simplest example of this is Bell's inequality. By now there are many such inequalities for all kinds of observables, a frequently used one being the CHSH inequality. In general these inequalities set bounds on correlation functions that cannot be violated by a classical probability theory, where the latter can be made precisely (see below). Quantum probability theories can violate some of these inequalities, which makes them intrinsically different.

Interestingly, there are also theories that have correlations that are even stronger than in quantum theory. These are known as Popescu-Rohrlich boxes and they have been shown to allow maximal violation of the so called Tsirelson bound, another inequality which is however fulfilled by quantum theory.

Making these statements (which all work on the level of probability distributions on a space of observables) is a whole field. Some references (I'll try to put some more tomorrow, too tired now):

  1. One can try to uniquely single out quantum theory as a 'special' probability theory by starting from certain information theoretic postulates: https://arxiv.org/abs/1203.4516
  2. So called 'loophole free' Bell tests have shown that we live in a world that violates classical probability theory (even though some people will argue against that): https://www.nature.com/articles/nature15759
  3. A nice presentation about the ideas mentioned above of a guy who (unlike me) actually knows what he is talking about: http://www.math.umd.edu/~diom/RIT/QI-Spring10/ClassvsQuantInfo.pdf
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Here is an experimentalist's answer:

A mathematical system, either algebraic or differential equations, has axioms and theorems and is self contained and self consistent.

A physics theory is a subset of a mathematical system that is defined by imposing extra axioms, called laws or postulates, which are necessary by construction, in order to pickup from the overall mathematical set, those solutions which fit data, i.e. measurements and observations.

Classical theories are those that use classical laws, such as: Newton's laws for mechanics, the set of laws of electricity and magnetism unified in Maxwell's equations, the thermodynamic laws (and maybe etc).

Quantum theories are the ones obeying quantum mechanical laws, i.e. the postulates of quantum mechanics, no matter the mathematical formulation.

In order to fit the data and observations, quantum mechanics postulates were necessary, and this is what distinguishes classical from quantum, IMO.

Edit after comments:

In your list:

Dirac-von Neumann axioms: These are too restrictive (e.g., they do not accommodate topological quantum field theories).

This was the first time I met Topological Quantum Field Theories (TQFT). (Such introductions are one of the reasons I follow this site - to get whiffs of new-to-me physics.)

The gauge is, if this set of theories fit data and predict measurements.

In axiomatic mathematical theories, theorems can be set up as axioms, and then the former axioms have to be proven as theorems, for a self consistent theory. Usually the axioms are chosen as the simplest expression from a set of consistent theorems.

Since TQFTs fit data and are predictive of quantum states, it is necessary that from the axiomatic postulates for TQFT one should be able to derive the postulates of quantum mechanics (possibly in a very complicated mathematical method). The wikipedia article on TQFT seems to indicate this. This is necessary for a theory to be quantum IMO.

I.e. it is the postulates that connect measurements to the mathematical formulas, by construction.

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  • $\begingroup$ +1 Thank you for the answer, but I'm not convinced. As I said in the OP, the postulates of QM are too restrictive. There are systems that we deem quantum-mechanical, yet they fail to satisfy these axioms. For example, topological quantum field theories (which have their own set of axioms). $\endgroup$ – AccidentalFourierTransform Apr 7 '18 at 19:47
  • $\begingroup$ These topological theories, do they fit any data? ? If they fit the data, then this just means that some of the postulates (linked above) of quantum mechanics can be relaxed/ignored. Otherwise , as when theorems in axiomatic mathematics can be turned into axioms, they become theorems.Or are they just a science fiction game with mathematics $\endgroup$ – anna v Apr 8 '18 at 3:12
  • $\begingroup$ Wow, that's a very condescending comment. Just because you don't find them useful does not make them "science fiction games". Wow, just wow. I really didn't expect that attitude from you... $\endgroup$ – AccidentalFourierTransform Apr 8 '18 at 3:13
  • $\begingroup$ And of course they fit data; TQFT's are essential to study the low-energy behaviour of some condensed matter systems. $\endgroup$ – AccidentalFourierTransform Apr 8 '18 at 3:16
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    $\begingroup$ +1 for a very good point: “Quantum theories are the ones obeying quantum mechanical laws, i.e. the postulates of quantum mechanics, no matter the mathematical formulation.” $\endgroup$ – AlQuemist Apr 9 '18 at 8:24
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I would say that something intrinsically quantum is the way in which probabilities and the function which obeys the partial differential equation are related.

As you note, both interference and probabilities are present in classical theories. What's new are probability amplitudes where interference leads to a supression of probabilities which is not possible in classical theories.

For the finite-dimensional case, there's also Lucien Hardy's proposal "Quantum Theory From Five Reasonable Axioms" (https://arxiv.org/abs/quant-ph/0101012). There, the distinguishing factor between quantum theory and classical probability theory is that "there exists a continuous reversible transformation on a system between any two pure states of that system."

Another reference along similar lines is Chapter 9 of Scott Aaronson's book "Quantum Computing since Democritus".

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  • $\begingroup$ Isn't interference of probabilities basically how we express wave-particle duality mathematically? $\endgroup$ – asmaier Apr 24 '18 at 13:06
  • $\begingroup$ I am not sure what you are getting at. Frist, there is no interference of probabilities but only probability amplitudes and second, sure, the physical phenomenon of wave-particle duality is related to this mathematical mechanism. $\endgroup$ – Marc Apr 25 '18 at 14:15
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tl; dr

Erm... You do.

Say you cook up a model about a physical system ...

Equations do not exists by themselves, they always have a surrounding. The head is assumptions and the tail usually describes limitations of said mathematical model. So really, it is up to your interpretation of question at hand OR the data available to you, that can consistently (deterministically?) predict if a theory is "Quantum".

Conversely, if you do not have a head and tail, you can make a lot of cases about what an equations is talking about but can't say anything concretely.

All the answers here are inspiring, and frankly sexy, but take time to consider my rudimentary examples below


This way of thinking "what characteristic of equation predicts its applicability in <name of physics branch>" is a misuse of mathematics.
Maths is, perhaps, the ultimate but we must remember that in physics we use it as a tool. My illustration below might seem childish but please consider the following equations

Equation 1:

$$ x^2 + x - 6 = 0 $$

Equation 2:

$$ 2x + 5y = 20 $$

Just looking at these, a mathematician can happily say that

  • Equation 1
    • has two solutions +2 and -3, and
    • the curve is upward facing, with maxima at x = -0.5
  • Equation 2
    • has a slope of -0.4
    • has intercepts 4 and 10
    • has infinite ordered pairs (x, y) satisfying the equation
    • describes a curve that encloses the origin

And we would all agree with the above points.
But the wise physicist stays mum, because s/he knows that these equations aren't just scribblings of some dyslexic Vulcan but are models of something, they represent something or some phenomena. So a physicist agrees with the mathematician but doesn't come to a conclusion.

Let us look at the questions which lead us to these equations

Question 1:

The product of a quantity and one more than itself is 6, find the value of this quantity if
a. the quantity is money lent
b. the quantity is time

Question 2:

Two times the number of my sons and five times the number of my daughters always equals two times the number of appendages a normal person has on his hands. How many sons and daughters do I have?

Now, I hope you have an aha! moment. The answer of Q1 b is just +2 because time can not be negative (we've all solved such questions as kids) and the answer to Q2 can be quite surprising - 5 sons and 2 daughters - because physicists are good people and don't make fractional children or negative children.

Did you see that -- one equation, two variables, and we still get a unique answer - constraints.

So mathematician (the equation) and physicist (the big picture) are both correct where they stand. But the physicists wins, because

  • we are at physics.stackexchange.com
  • math in itself is very strong, pure, almost unpalatable; we need both the background information and the constraints to understand what this wonderful tool is trying to tell us through equations.

On a serious note, I'd like to point out that there's probably no (respectable) book on classical physics which teaches F = ma without first explicitly-and-clearly stating the following:

  • Assumptions required e.g. frictionless surfaces and perfectly rigid bodies
  • Newton's Three Laws of Motion (word-by-word)
  • That dF = d(m.v), which can be simplified if mass is (almost) constant
  • and most importantly, the fact that objects we are dealing with are not of super-tiny scale, i.e. larger than 10-9m in diameter.

Authors don't do this for pedagogy, most 9th grade students wouldn't give a damn about rigidity, but in fact they do it because these statements are necessary for the equation/theory to work.

Trying to predict if an equation describes a Quantum thingy is a discussion-based question at best, or meta-math.


To the OP specifically,

If you are an inventor, working on something like GUT (why else would you have a equation whose origin you do not know) and you are curious if it applies equally well to big and small bodies - apply constraints. I do not have the mathematical foresight but logically I can say that variations in constraints will define the way system behaves for Quantum and Classical bodies.

In Thinking Fast and Slow there's a chapter which illustrates that we have a tendency to support what is popular/fancy rather than what is correct/plausible. I think the question is primarily opinion based.

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  • $\begingroup$ Apropos of equation 1, a mathematician would perhaps say minimum rather than maxima (sic). $\endgroup$ – Deepak Apr 9 '18 at 15:50
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Physical models are determined by their lattice of events. The set of physical events form an algebraic lattice with the two binary operators that serve as the OR and AND between events. We assume the lattice of events to be sigma-additive and orthomodular. We call this lattice the logic of the model. In this sense events are the elements of logic. System states are probability measures over this algebra. Physical quantities are mappings between statements on measurements of a quantity (think of Borel-sets of the reals) and the logic.

The logic of a classical model is isomorphic to a set algebra so it is distributive ( a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and vice versa) and fully atomic.

The logic of a quantum model isomorphic to the lattice of the subspaces of a Hilbert space and therefore it is not distributive but also fully atomic.

The above alone is sufficient to explain many features associated with quantum models, including

  • real valued physical quantities can be represented as self-adjoint operators
  • commutation relations
  • superposition of states
  • the Schrödinger equation
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    $\begingroup$ Can you please add some references? I think the answer could benefit from that. $\endgroup$ – Kiro Apr 7 '18 at 7:35
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TLDR: Wave-Particle duality

I want to answer this question from a historical perspective:

According to our current understanding a quantum theory shows features of both classical mechanics and electrodynamics (e.g. light) at the same time. The first person to notice such a connection between mechanics and the theory of light was Hamilton. He developed Hamiltonian Optics, which described light as a particle (aka corpuscle). Theorists soon recognized that Hamiltonian Optics cannot account for light phenomena like interference, diffraction, and polarisation. They realized that Hamiltonian Optics is only an approximation, which works well as long as the wavelength of light is much smaller than the measurement apparatus (e.g. for geometrical optics based on light rays and lenses). Nevertheless, the language of Hamiltonian Optics worked perfectly to describe classical mechanics, which is now commonly known as Hamiltonian mechanics.

Maxwell's field theory of Electrodynamics was a more correct description of light, but then came Planck and Einstein. They showed that to describe Black Body Radiation and the photoelectric effect it was necessary to assume that light cannot be a field with infinite divisibility (i.e. continuity) as assumed in Maxwell's Wave theory of light. Rather, light must consist of countable entities they called "quanta". But, this theory was ad hoc and not consistent with special relativity. (Note: the consistent version is Quantum Electrodynamics.) Although immature, the Planck and Einstein explanation of these phenomena was the first quantum theory because it showed (or better, assumed) wave-particle duality. (Note: Quantisation doesn't mean going from a wave theory of light back to a corpuscle theory like Hamiltonian Optics. Rather it combines features of waves and particles.)

The crazy genius of deBroglie and Schrödinger was needed to apply this theory in the opposite direction - to particles. They noticed that if Maxwell's wave theory of light must be extended to contain quanta/particles, classical theory (which consists only of particles) must be extended to produce the features of waves. They saw classical theory could be an approximation like Hamiltonian Optics, which is valid only for short wavelengths. Thus, Schrödinger developed wave mechanics not by postulating quanta, but by reversing the approximations necessary to go from Maxwell's theory of light to Hamiltonian Optics. In opposition to Electrodynamics, Classical Mechanics needed to be "wavized" to become a complete theory showing wave-particle duality. (Note: here again, quantisation is not going from a particle theory to a complete wave theory of infinite divisibility, rather, it combines features of both worlds.)

So, a theory is "Quantum" when it integrates/combines the features of both waves and particles. A classical theory is either only waves/fields or only particles.

Regarding the quantisation of General Relativity, it is instructive to compare this classical field theory with another classical field theory, namely fluid dynamics. What both theories have in common is their high non-linearity. Both can only be quantised if they get linearized first. If one linearizes fluid dynamics, one gets the equation for sound waves. If one linearizes the equations of GR, one gets the equations of gravitational waves. If one quantizes the equation of sound waves, one gets phonons. If one quantizes gravitational waves, one gets gravitons. Again, both Gravitons and Phonons show wave-particle duality. But in both cases, we need to linearize our theory first to be able to quantize it. (Note: Phonons only exist in solids. Gravitons might also only exist in "solid" space-time.)

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I'm astonished that nobody appears to mention that a quantum theory describes quantities which have discrete values. All quantities which appear continuous on the macroscopic level can only take on discrete values in a quantum theory. The differences are "communicated" by "particles" (photons etc.). That's the heart of a quantum theory.

Describing the states and interacting particles has not been achieved, or has only been tentatively achieved, for gravitation.

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    $\begingroup$ -1 This answer is basically incorrect; in particular, “All quantities which appear continuous on the macroscopic level can only take on discrete values in a quantum theory”. $\endgroup$ – AlQuemist Apr 9 '18 at 14:32
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    $\begingroup$ @PeterA.Schneider No, that's a very simplistic view of classical mechanics (and physics in general): a single system always has an infinite number of different descriptions, some of which are typically more accurate than others. It's turtles all the way down: you can always add more levels of sophistication to a certain model. In this sense, speaking of a "coin" is not meaningful: you have to decide which degrees of freedom you want to study (only heads/tails? or also it's final temperature? what about any possible deformation due to the impact?) (1/2) $\endgroup$ – AccidentalFourierTransform Apr 9 '18 at 16:52
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    $\begingroup$ (2/2) At some point you truncate the problem, and pick a certain finite set of degrees of freedom. Once you do this, you should be able to decide whether the model is classical or quantum-mechanical independently of other "more sophisticated" models. The binary model is consistent in and of itself, independently of more accurate descriptions. It is a valid model, and complete as far as the degrees of freedom we chose to describe is concerned. Whether there is a Newtonian description that is more accurate is completely irrelevant. FWIW, I appreciate your answer anyway, and I upvoted it. $\endgroup$ – AccidentalFourierTransform Apr 9 '18 at 16:54
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    $\begingroup$ @PeterA.Schneider Take guitar string or some other resonating system - you get discrete results. $\endgroup$ – Arvo Apr 10 '18 at 12:03
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    $\begingroup$ As with @Arvo I immediately glanced at classical standing waves. As with quantum systems they discreteness comes from the application of boundary conditions. As with quantum systems they are a steady-state effect and you can observe results that don't meet the quantization condition in the immediate aftermath of disturbing the system. $\endgroup$ – dmckee Apr 11 '18 at 19:01

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