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  Is there a theorem that says that QFT reduces to QM in a suitable limit? A theorem similar to Ehrenfest's theorem?

+ 11 like - 0 dislike

Is there a theorem that says that QFT reduces to QM in a suitable limit? Of course, it should be, as QFT is relativisitc quantum mechanics. But, is there a more manifest one? such as Ehrenfest's theorem which dictates the transition from QM to Classical Mechanics.

I am curious as the basic object in QM is wave function whereas the basic object in QFT is quantized operator, which looks very different, and wondering how the comparison works.

(Of course, I am aware that QFT can calculate corrections for Coulomb potential energy, which may seem to be a transition from QFT to QM, but I want one more manifest than that.)

This post has been migrated from (A51.SE)
asked Oct 11, 2011 in Theoretical Physics by Youngsub (55 points) [ no revision ]

5 Answers

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Yes there is. In fact there's not much difference between QFT and QM (as far as the formalism is concerned.) The point is,

the basic object in QM is wave function whereas the basic object in QFT is quantized operator, which looks very different,

is not quite true. In QM you can use Hamiltonian formalism bringing the operators to the fore, and in QFT you can use wave function of a state to the fore, too.

Note that in QFT, a wave function is associated to a constant time surface. If there are two particles on that constant time surface, the wave function of the QFT is the wave function of these two particles. If you write down the Hamiltonian time evolution of this two-particle wave function using relativistic QFT, it manifestly has the form of

standard non-relativistic two-particle Hamiltonian with interaction terms + relativistic corrections.

Maybe you've read too many QFT textbooks with only an emphasis of correlation functions; try reading a QFT textbook for, say, condensed matter theory. Then you'll clearly see that QFT can also be formulated just as a quantum mechanics of variable-number particle systems (which in particular contains a fixed-number subsystem in the non-relativistic limit. )

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answered Oct 11, 2011 by Yuji (1,395 points) [ no revision ]
The OP asked for a theorem, and there is a central difficulty, in that for the case of a reduction to a field theory of bound-states, as is required for protons and nuclei, the theory is lacking, as far as I know.

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The OP asked whether QFT reduces to QM in a suitable limit, under the right set of circumstances, for which the answer is given nicely here (no theorem exists because the result is way too simple to merit the effort). It is true that QFT is not reduced to QM in unsuitable circumstances, many of which do exist including the one you point out, but in my mind that is besides the point.

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+ 9 like - 0 dislike

The answer to this depends on what degree of subtlety you want to take the question to. The move from finite to infinite DoFs is always delicate in parts.

If, for example, you've read literature on the thermodynamic limit of statistical mechanics, and it all seems straightforward to you, then yes, QFT reduces to QM, as declared by Yuji and lots of upvoters, although I consider the lack of any citation in Yuji's Answer somewhat telling. If you think the thermodynamic limit is fraught, then you will likely think the same of the reduction from QFT to QM. AFAIK, however, an effective and simple statement of what the problem(s) might be, if any, does not exist, as exemplified by Vladimir's problematic Answer.

Renormalization, as a way to handle the move to infinite DoFs on a Lorentzian manifold with nontrivial evolution, certainly plays a part in any reservation one might have, but many Physicists now take the view that the renormalization group is an adequate way to deal with the mathematics. Most of the books on condensed matter theory that Yuji recommends, whether QFT or otherwise, largely gloss the harder mathematical worries one might have about renormalization.

Your apparent acceptance of Ehrenfest as a good enough reduction from quantum to classical mechanics suggests that for you the answer to your Question is yes. However, Ehrenfest's theorem is by no means entirely acceptable to all Physicists. It's a digression from the QFT/QM topic of your Question, but, for example,

Phys. Rev. A 50, 2854–2859 (1994)
Inadequacy of Ehrenfest’s theorem to characterize the classical regime
L. E. Ballentine, Yumin Yang, and J. P. Zibin

Abstract: The classical limit of quantum mechanics is usually discussed in terms of Ehrenfest’s theorem, which states that, for a sufficiently narrow wave packet, the mean position in the quantum state will follow a classical trajectory. We show, however, that that criterion is neither necessary nor sufficient to identify the classical regime. Generally speaking, the classical limit of a quantum state is not a single classical orbit, but an ensemble of orbits. The failure of the mean position in the quantum state to follow a classical orbit often merely reflects the fact that the centroid of a classical ensemble need not follow a classical orbit. A quantum state may behave essentially classically, even when Ehrenfest’s theorem does not apply, if it yields agreement with the results calculated from the Liouville equation for a classical ensemble. We illustrate this fact with examples that include both regular and chaotic classical motions.

PRA link: http://link.aps.org/doi/10.1103/PhysRevA.50.2854
DOI: 10.1103/PhysRevA.50.2854

For someone who moves very well between classical, QM, and QFT, from whom I got the link above, I suggest

Between classical and quantum
N.P. Landsman
Abstract: The relationship between classical and quantum theory is of central importance to the philosophy of physics, and any interpretation of quantum mechanics has to clarify it. Our discussion of this relationship is partly historical and conceptual, but mostly technical and mathematically rigorous, including over 500 references. On the assumption that quantum mechanics is universal and complete, we discuss three ways in which classical physics has so far been believed to emerge from quantum physics, namely in the limit h -> 0 of small Planck's constant (in a finite system), in the limit of a large system, and through decoherence and consistent histores. The first limit is closely related to modern quantization theory and microlocal analysis, whereas the second involves methods of C*-algebras and the concepts of superselection sectors and macroscopic observables. In these limits, the classical world does not emerge as a sharply defined objective reality, but rather as an approximate appearance relative to certain "classical" states and observables. Decoherence subsequently clarifies the role of such states, in that they are "einselected", i.e. robust against coupling to the environment. Furthermore, the nature of classical observables is elucidated by the fact that they typically define (approximately) consistent sets of histories. We make the point that classicality results from the elimination of certain states and observables from quantum theory. Thus the classical world is not created by observation (as Heisenberg once claimed), but rather by the lack of it.
Comments: 100 pages, to appear in Elsevier's Handbook of the Philosophy of Physics [which it has]

This post has been migrated from (A51.SE)
answered Oct 11, 2011 by Peter Morgan (1,230 points) [ no revision ]
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QFT results are not always reducible to QM. But let us consider a simple case of charge motion in an external filed. Let us suppose the energy is smaller than necessary for pair creation. In the first Born approximation the QFT result looks like a QM one: the charge is scattered elastically and that's it. However in QFT there are higher (radiative) corrections. Taking them into account changes radically the picture. Briefly, the probability of elastic scattering is zero. You cannot scatter a charge without loosing some energy into radiation. But if you calculate an inclusive cross section (sum of all inelastic ones), it coincides practically with the elastic one calculated in the first Born approximation. Thus, it is the inclusive picture that can be compared reasonably to QM results. Inclusive and average are akin - they imply summation.

The inclusive picture cannot be presently obtained from QFT equations, unlike Ehrenfest relationships. It is attained with summation of all orders of the perturbation theory and with performing renormalizations. It is so because the charge-field interaction is written in a wrong way. In other words, it is so because the QFT equations are wrong. My toy model (electronium) shows how it could be reformulated.

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answered Oct 11, 2011 by Vladimir Kalitvianski (102 points) [ no revision ]
reshown Aug 23, 2014 by dimension10
Most voted comments show all comments
@Yuji: I completely agree with your general statement, but may be you will be so kind to be more specific? I know I am always guilty in eyes of some, so I am not surprised with downvotes. Still, apart from general statements, please be specific, keep to the point.

This post has been migrated from (A51.SE)
No, being East Asian, I just loosely quoted a word of Confucius. (The original parable was that he advised a person to eat more, and advised another person to eat less. A disciple who was seeing this was confused, and asked the master why he gave opposite advice. Confucious said that the first person was somewhat fat and the second person was somewhat thin, and the advise should be made according to the person.) To be a bit more specific, I guess your answer (about infrared divergence, etc.) was too high level for the original question, leading to the down votes. I didn't down vote you.

This post has been migrated from (A51.SE)
I think people are overreacting--- the infrared problem is real, and needs to be mentioned in this context, because it is similar to the problem of relativistic bound states with large numbers of soft gluons (although there, you have a nuclear scale cutoff). Both problems are insufficiently well understood, and this prevents a general theorem, even in the nonrelativistic case, in my opinion.

This post has been migrated from (A51.SE)
Even in classical mechanics, if you have a weak oscillator ($k\to 0)$, any sudden excitation leads to a very large amplitude $\vec{r}_{osc}\propto 1/\omega$. Only time averaging makes the result finite. Averaging implies inclusion of all positions within a period (inclusive picture).

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@VladimirKalitvianski: I didn't downvote either, but I can see why people might have. I've said this before to you, but I'll say it again: the problem is that on a first reading, the answer appears to entirely miss the point of the question. May I suggest that you prefix your answer with a paragraph either explaining the conventional view before going into why it falls apart in the details. I think it would help a reader ease into the style.

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For Vlad's sake: you're being downvoted because the question was specifically about when and where one can find the limit. For instance, it should be clear that a 1-particle model for the hydrogen atom is not grossly wrong; one can sensibly ask for a first principles "derivation" of it. Scare quotes used because it would probably be highly not rigorous.

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@genneth: I answered as I could, in particular, I pointed out to the inclusive picture which coincides with the "mechanical" one. My answer is correct, what else should I do?

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+ 2 like - 0 dislike

A textbook reference for his comment is N. Nagaosa QFT in Condensed Matter Physics. However, if you are looking to reverse the procedure of second quantization then look in the beginning of Zee's book QFT in a Nutshell. I'm sorry I can't point out the exact page number but I'll do the best I can to describe the example, as I don't own the book anymore.

I think the example starts with a classical field theory and you are asked to treat it like a mattress with a finite number of springs. After following along for a bit, if you have any mathematical background in analysis this technique should be troubling, because any limiting procedure used after linearization will not restore the continuum of states, ie there are not enough rational numbers to fill the entire real-line. So any explanation you are given of second quantization will always be a slight fudge in going from a countable number of degrees of freedom to an uncountable number of degrees of freedom, uncountable in the sense that set of suitable states will be as bad as the cardinality of the real numbers.

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answered Oct 11, 2011 by MA Perez (90 points) [ no revision ]
Physically speaking, you shouldn't care about those niceties. Nobody knows if the real word coordinates is really described by real numbers. They may just be rational numbers, or maybe of the form `(very, very small number) x integer`. Real numbers might be our approximation.

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It depends on your perspective. From the perspective of AQFT, whether or not the continuum is fundamentally necessary in the construction of your theory is something that has gained positive traction in operator theories of algebras of observables. People like C. Flori, [see arxiv: 1106.5660], C. Isham, A. Doering, among others are trying to develop a framework that does not rely on this rather innocuous assumption.

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I've spent a long time with my eyes open looking for this reduction and have not seen it. (The suitable limit is not c-->infinity. Consider a slow positron running into an electron.)

As for h-->0 limits: These are sometimes easy to obtain FORMALLY if you express the theory as a mapping from observables into an abstract C*-algebra. With h = 0, the algebra is commutative, and its irreducible representations correspond to classical solutions. For QM, you get classical mechanics. For QED, you get free electrodynamics. [The fermionic fields all die.] For an interacting QFT without fermions, I'm over my head.

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answered Apr 12, 2012 by Greg Weeks (50 points) [ no revision ]

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