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  Is there a nonperturbative point-particle formalism for QFT?

+ 6 like - 0 dislike
6665 views

The field formalism for quantum field theory is well known--- you write down a path integral over the fields. This formalism defines field operators, and gives a field picture, which is self-consistent.

The perturbation expansion of such a path integral gives a particle picture, through the Feynman diagrams and their Schwinger parameters. The particle propagation is in space-time, and the main identity is the trivial looking integral:

$\int_0^\infty e^{-\tau (k^2 + m^2)} d\tau = {1\over k^2+m^2}$

When you Fourier transform it, the left-hand-side becomes a superposition of Gaussians spreading in $\tau$, which has the interpretation of an internal proper time ticking on the particle's world-line, while the right-hand-side becomes the (Euclidean) scalar propagator.

This identity gives a natural particle interpretation to perturbation theory, and as is well known, this was Feynman's motivation for the diagrams--- to give a pure particle picture. It's used to reparametrize loops too, to give an intuitive picture.

But this particle formalism, while elegant, is only defined perturbatively, Can one give a formalism for quantum field theory which is manifestly non-perturbative and manifestly particle based?

asked Sep 12, 2014 in Theoretical Physics by Ron Maimon (7,720 points) [ revision history ]

The joy of worldline formalism! http://ncatlab.org/nlab/show/worldline+formalism (Just to add pointers to references.)

@UrsSchreiber: Agreed, but there is a problem with the references, in that they take it for granted, on physical grounds, that the mathematical transformation for perturbative expansions can be extended to a nonperturbative correspondence. This correspondence simply has never been properly worked out in any rigorous way.

This issue is glossed over in all the references you find. This Mellin transform is fundamentally purely a perturbative transform on the propagators. Although one has physical intuition that it should work in general, and this is reinforced by string theory, one doesn't have a clear mathematical idea how to express many field theoretic concepts in worldline formalism, because they intrinsically involve non-perturbative processes, or infinitely many worldlines. for example, how do you represent an instanton mediated proton decay process in such a worldline formalism? how do you write down a spontaneously broken vacuum in particle picture? How do you write down background massless fields using worldlines (you can do it using classical sources, but how do you resum infinitely many soft worldlines into variations of the source)? how do you take the nonrelativistic limit of quantum field theory nonperturbatively formally? Many other questions (these questions come up here all the time, for example here: http://physicsoverflow.org/219/theorem-reduces-suitable-theorem-similar-ehrenfests-theorem ). Stochastic quantization completes the worldline formalism, you get a causal picture which makes it clear how the worldlines are affected by instantons, and how to represent Euclidean spontaneous symmetry breaking situations with a density of particles. You can also take the nonrelativistic limit rather easily, by identifying the stochastic time with the nonrelativistic time in the limit. Stochastic quantization is also one of the keys to Hairer's recent breakthrough in renormalization.

You could say "Do the Mellin transform on the Hamiltonian evolution", but this gives mathematically minded physicists the heebie jeebies, because the Hamiltonian domain is a separable Hilbert space, and when you are doing the transformation intuitively, you include also intuitive states with infinite particle number, an infinite number of worldlines, for example, those that describe the Higgs vacuum as a superconductor. You need a formalism which is fully nonperturbative to make sense of that.

@RonMaimon,  you are kindly invited to add whatever reference seems missing. If you don't want to edit the page yourself, give me your preferred list by some other means and I'll add it. 

Regarding non-perturbative business: sure, the default meaning of "the worldline formalism" is a thing in perturbation theory. That's notably "why" the string perturbation theory is about perturbation theory, of course. 

Oh, it's ok as it is. The modern references supersede the old ones. I just meant that it gives the impression that the correspondence is understood completely.

2 Answers

+ 5 like - 1 dislike

Parisi-Wu stochastic quantization provides the proper nonperturbative particle formalism, although this is not at all what it was originally proposed for.

The main issue with considering the Feynman expansion in a particle interpretion is the fact that the particles go back and forth in physical time, but their interactions are when they are at the same point in space-time. So you can't really say that "this particle got formed, then travelled some time, then met another particle" and so on, because the internal proper times are integrated over, so the other particle's proper time parameter has no order relation with the first particle's proper time parameter. The interactions are not ordered causally in terms of particle proper time, each propagator has a completely separate proper time which is not put together into a global notion of one universal proper time for all the particles.

This gives the Feynman diagram picture an a-causal character, the integrals over the internal proper time aren't "this particle emitted that particle, and then later it was absorbed", because the notion of "later" is not there, the proper times don't fit together coherently to make a global proper time which is ticking along.

This interpretive problem is overcome (trivially!) in stochastic quantization. Recall that stochastic quantization involves a fictitious internal stochastic time, and one is only interested in the real space-time properties, so one considers states which are stationary in the stochastic time. I'll call the stochastic time $\tau$, and the stochastic quantization evolution equation is derived from varying the action:

$$\partial_\tau \phi = - {\delta S \over \delta \phi} + \eta $$

which gives, say for scalar $\phi^4$ theory, the equation

$$\partial_\tau \phi = \nabla^2 \phi - m^2 \phi - {\lambda \over 6} \phi^3 + \eta$$

The kinetic stochastic equation by construction makes a stationary Boltzmann distribution $e^{-S}$, so that it reproduces the original field theory in steady state.

The stochastic equation has a perturbation expansion also, which involves a purely classical (tree level) relation between $\phi$ and $\eta$, and then, if you are interested in stationary correlation functions, you contract the $\eta$s using their trivial correlations:

$$\langle \eta(x,\tau)\eta(x',\tau')\rangle = \delta(x-x')\delta(\tau-\tau')$$

This has the effect of sewing together the tree diagrams into loops, and in the case that one is only calculating expectations which are integrated over all $\tau$, you sew together exactly the Feynman diagrams of the original field theory.

But surprise! You've sewn the original Feynman diagrams together from a completely causal proper-time picture! The particle proper time is nothing other than the stochastic time, and the Parisi-Wu evolution is completely causal in proper time.

The particle propagation in stochastic time is only forward in time--- the propagator for $\tau$ is like a nonrelativistic propagator, not like a relativistic propagator. There is no backward in $\tau$ propagation for anything. At any given perturbative order in the coupling, you produce the same field theory diagram from several different chopped up stochastic versions of the diagram into two trees (this is explained in several places, including Parisi-Wu and reviews), but each tree only has causal in time particle interactions, the splitting of particles is according to the derivative of the original interaction, and the sewing together is once only at the joints.

What happened here is that the Schwinger parameter is promoted to a universal time, particles only interact at the same instant of stochastic time, but the physical space-time correlation functions are found by making an average over all stochastic times in steady state, and this removes the center of the $tau$ integrals, and turns tree-chopped but causal integrals into acausal but still particle based integrals.

Stochastic quantization is completely nonperturbative. The particle diffusion is due to the $\nabla^2$ term, the mass term is interpreted directly as a rate of production or removal of particles, and the cubic term as a 1-3 splitting of the particles causally in $\tau$.

Further, one can in principle analytically continue $\tau$ to a time variable to produce a nonrelativistic quantum mechanics (nonrelativistic in $\tau$) whose ground state is the field theory vacuum, and separately or together continue one of the x variables to be a physical time variable, so that one produces a nonperturbative nonrelativistic particle formalism.

answered Sep 12, 2014 by Ron Maimon (7,720 points) [ revision history ]

I don't know if this essentially trivial observation is original. If someone knows a reference, let me know.

In a old paper of Matt Strassler, one sees that there is a relation between color-trace ordering and proper-time ordering (formulae \((2.11) \to (2.13)\) ). How does Parisi-Wu stochastic quantization is managing non-abelian  YM QFT ?

This only gives a picture of Euclidean quantum field theory. When you try to analytically continue the imaginary time to real values you lose the concept of particle tracks, it seems to me.

@trimok: I don't know exactly, I really didn't work it out, I just noticed this property a week ago incidentally, while sketching out some stochastic computer simulations to clarify the gauge theory and renormalization discussions here. But to formulate gauge theory in stochastic quantization superficially is extremely easy, you don't need to fix a gauge (this is Parisi and Wu's motivation). The gauge unfixed stochastic equation is $\partial_\tau A^\nu = D_\mu F^{\mu\nu} + \eta$, where $\eta$ is vector white noise. It gives the right answer in numerical simulation for long times, but you don't reproduce gauge theory path-integral in the steady state if you don't fix gauge, because the diffusion of the gauge part of the evolution never settles down. When you fix gauge it certainly works the same way as for scalars above (I didn't do it in detail, but it's clear that the Feynman gauge propagator will do the same diffusion for the vectors, but you need to consider the ghosts too, and I didn't think about fermions yet. If you choose background field gauge you should reproduce Strassler's calculations in steady state, except embedded in a non-perturbative general framework). By the way, I read the Strassler paper just now, it's excellent and it should be reviewed (here:http://physicsoverflow.org/23578/field-theory-without-feynman-diagrams-loop-effective-actions). But ultimately Strassler is doing the same perturbative particle picture that Feynman and Schwinger did, the particle picture has just never been knowingly embedded in a nonperturbative framework. I say "knowingly" because Parisi and Wu did it, but I think unknowingly.

@ArnoldNeumaier: I was sort of glib in the answer, the propagators in the tree expansion of the stochastic equation very easily analytically continue to sensible nonrelativistic particle propagators, so there is no problem with the particle picture. But there is an issue with the noise, in that the gluing together for the noise doesn't continue simply to ordinary nonrelativistic quantum mechanics. If you write down the action integrating out the noise, schematically, it's $(\partial_\tau \phi)^2 + (\nabla^2 \phi)^2 + V'(\phi)^2$ (schematically means up to derivative cross terms and Parisi-Sourlas determinant to enforce the supersymmetry in the formulation, but these are all well understood). Although this looks like a horrible higher derivative theory of some kind, it is nothing to be scared of, it just has no Lorentz symmetry in the extra stochastic time, and when you consider the ground state it turns into the field theory path integral. You can continue this action to imaginary $\tau$ and imaginary $t$ together or separately, it's a perfectly ordinary stochastic action. Although it's got complicated looking diagrams, they just mean "simulate the stochastic equation", but it's not the Schrodinger field action, because you integrated out the white noise.

Continuing $\tau$ to imaginary values in the perturbation series, while perturbatively possible, doesn't just involve forward in $\tau$ paths, I didn't say the right thing in the answer, but there is a joining together at the eta-eta joints of a stochastically split diagram of forward in time propagators on one half of the diagram, and backward in time propagators on the other half. But each half is perfectly causal in $\tau$ even in real time, and there is still a perfectly consistent particle interpretation, it's just that the diagrams are split in two along the joints in this crazy Parisi-Wu way. I didn't work it out in real space in detail, just the imaginary time scalar version (because that's what I needed now), and this is definitely nonperturbative and particle.

+ 1 like - 1 dislike

No, it is not possible.

I agree with Ron's answer only to the extent that stochastic dynamics (i.e,. Euclidean field theory) can be given a particle picture since the path integral has a measure supported by paths. But these paths have nothing to do with what one gets by analytically continuing the Euclidean time to Minkowski time.

In Minkowski time, a particle picture is intrinsically tied to Fock representations, which is needed to define particle number and states with a definite number of particles.  Without a Fock representation there is no notion of particles at finite time, and hence no notion of particle dynamics. But Haag's theorem implies that relativistic QFT cannot be represented in a Fock space.

In Minkowski space, one only has the notion of asymptotic particles at times $t\to\pm\infty$, defined by Haag-Ruelle scattering theory. These are the particles detected in high energy experiments.

But at finite times, these asymptotic particles do not have paths in any meaningful sense, except at length scales of the order of the Compton wavelength (or their nonelectronic analogues), where semiclassical methods (corresponding for photons to geometric optics) are applicable. These are the paths seen (through their ionization products) in bubble chambers.

answered Sep 14, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

This answer somehow reminds me also of the fact that in QM it is pointless (no pun intended) to talk about the exact state corresponding to an observable before a measurement has happend (?) and about the path integral formalism ...

In quantum mechanics one cannot ask about the precise path.

But the mean path and the root means square deviation from the mean path have a well-defined meaning and define a smeared version of the semiclassical path. This is the part that is objective about the motion of quantum mechanical particles. For an electron in a bubble chamber, reproduces the observed path; for an electron in the ground state of a hydrogen atom it gives a thick world line centered at the world line of the center of mass of the atom with an uncertainty of the order of the Bohr radius of the hydrogen atom - a very intuitive picture. 

“But Haag's theorem implies that relativistic QFT cannot be represented in a Fock space”

I wonder if Haag's theorem is strong enough to imply this. Surely it implies the Hilbert space of interacting theory is not unitarily equivalent to that of free theories, but how does it imply it must be non-Fock? In fact, two free scalar field theories with different masses should already be unitarily inequivalent, but both their Hilbert spaces are Fock.

The fact that naive perturbation theory produces infinite corrections, no matter which finite counterterms are used, proves that no Fock space works. See divergence-perturbation-theory-interacting-hilbert-space.

Regarding the last comment--- the "infinite corrections" for Haag's theorem are infrared non-convergence of matrix elements in infinite volume, they have nothing to do with counterterms. Haag's theorem is true on a lattice just as well, it has nothing to do with ultraviolet issues.

I am not taking about a Fock-space representation at all. Fock space is a finite number of particles. I am talking about a path-integral identity, which includes infinite number of particles (in infinite volume).

You might say "there is no formalism for that", but yes there is. The path integral extends the Fock space picture naturally to include states of infinite particle number. You can only speak about such a non-perturbative correspondence in a formalism where you can deal comfortably with infinite particle number (for example when you shift  the vacuum).

In this case, I showed you exactly how it works, and it does work. There is nothing special about Euclidean space--- the kinematics of Euclidean space--- the state space-- is exactly the same as the kinematics of Minkowski space. You can talk about a 2 pion state in Lattice QCD. Only the dynamics is different, so the two pion state decays in Euclidean time instead of oscillating, and the theory is O(4) invariant.

The point of Stochastic quantization is that it automatically makes you write lattice QCD in terms of evolving quark/gluon worldlines nonperturbatively, so that you understand what instantons look like in the worldline picture (they describe certain collections of worldlines, a theta-term adds an asymptotic boundary term to the stochastic evolution which unfortunately is only stochastic (real) at $\theta=0$ and $\theta=\pi$) and quark-condensates (which are described by a density of worldlines). It is exactly what is required to understand world-line physics outside of perturbation theory, the worldlines interactions are causally evolving in stochastic time.

Haag's theorem holds in any non-free QFT where the Wightman axioms can be verified, and even more generally (L Streit, A generalization of Haag's theorem, Il Nuovo Cimento A62 (1969), 673-680).

Infrared problems are absent in massive $\phi^4$ theory, but Haag's theorem is still valid there (if the theory exists at all). Thus Haag's theorem is not an infrared problem.

On the other hand, Minkowski path integrals are ill-defined except in perturbation theory, and as long as this is the case I do not accept any argument using these as convincing evidence for nonperturbative properties. 

Nonperturbative means that one actually has a representation in a Hilbert space. And a particle picture means nothing if this Hilbert space contains no single-particle and few-particle states. 

"Haag's theorem holds ... generally." Yes I know. So what?

Infrared problems for vacuum matrix elements are not absent in massive $\phi^4$. The Haag theorem can be demonstrated even in ultralocal massive $\phi^4$ on a lattice with no derivative terms! Consider the two lattice actions on an infinite lattice of x's:

$$ S= \sum_x  m^2 \phi^2 $$

$$ S = \sum_x m^2 \phi^2 + \lambda \phi^4 $$

What is the inner product of the two vacua? It's zero. For any one lattice site, it's nonzero, it's some number $0<I<1$, but on N lattice sites it's $I^N$, and in the limit of infinite volume, when N goes to infinity, it gives zero.

Haag's theorem is the same thing turned into a general proof using translation invariance, and it's obvious. There was a long chat on this  (here: http://www.physicsoverflow.org/22462/discussion-regarding-connection-renormalisation-physical )  It holds all the time precisely because it comes from infinite volume considerations. It has no relevance to this question.

"Nonperturbative" means you can calculate it exactly on a computer, and derive the Hilbert space in the limit of a few excitations on top of a vacuum. One way of doing this is to talk about the Hilbert space, but a better way of doing this is to include path integral configurations in the description which are not described by any reasonable Hilbert space, but which appear in the theory.

Such states are, for example, the different vacua of a Mexican hat potential, each of which is different from any other by infinitely many particles. The Hilbert space for the different vacua are completely disjoint (all vectors are orthogonal) in infinite volume, you can't describe a different vacuum using Fock space excitations over one vacuum (although you can approximate it in finite volume). But these different vacua are clearly part of the same path integral, and in a rigorous stochastic quantization they are all part of the description, because the stochastic description includes infinitely many particles also, not just finite excitations over the vacuum. This is why I am not talking about Hilbert spaces, only path integrals.

The single particle and few-particle states are derived in confining field theories at long distances once you know the vacuum structure and operators that excite the vacuum. The rigorous Hilbert space for gauge theory is a glueball Hilbert space. This is not the appropriate language for making the theory make sense, this is why stochastic quantization beats Hilbert space quantization, it's more general, it includes states of infinite particle number, which you need in infinite volume, not for renormalization reasons, but for reasons of infinite volume.

Our philosophies are extremely different.

Haags theorem is even valid on a spherical (but continuous) space, where the volume is finite and no infinite volume limit must be taken. 

That Haag theorem also holds on a lattice doesn't imply that my argument with counterterms is invalid: If Haag's theorem were invalid, i.e., if the interacting theory would happen in a fixed Fock space, ordinary perturbation theory around the correct particle mass would have to give finite perturbation results. 

Your argument is circular if you regard Haag's theorem as a consequence of infrared problems and take the disjointness of the noninteracting and the interacting vacuum as proof of infrared problems. For me, an infrared problem is solely something causes by an IR divergence in naive perturbative procedures, and this is invariably associated with long range forces due to massless particles. 

What you call exact computations on a computer are not really exact since they have an error scaling at best like $O(N^{-1/2})$ for a lattice with $N$ sites. This is far from being exact. I regard lattice calculations as approximately nonperturbative only. You cannot perform the limit $N\to\infty$ exactly on a computer; even the existence of the limit is highly nontrivial and requires an extremely careful renormalization analysis if it is to give the exact QFT theory. 

Multiple vacua can be described by some Hilbert space, just not by a Fock space. One takes as starting point the direct sum of a family of Fock spaces parameterized by the different stable vacua. For example, this is one of the standard ways one may proceed in exactly solvable 1+1D QFTs.

The glueball Hilbert space is only the asymptotic Hilbert space of bound states. In the finite time dynamics there are no glueballs, as these only form asymptotically. Thus the particle picture is present only at $t\to\pm\infty$, and there are no particle paths.

Haag's theorem is only valid on a finite box in a continuum limit in certain cases, and mostly conjecturally (although the conjecture is certainly true). Those cases are when you are adding infinitely many bare particles at high k to the vacuum when you add the interaction. This happens, sure, but it is not what Haag's proof does. What Haag proved is zero inner product at infinite volume, because this is easy to prove without considering details of the theory.

It is true that you add infinitely many bare particles ot the vacuum when you turn on a $\phi^4$ interaction, or change a coupling constant in a normal theory, but it is not true for the separation of mexican hat vacua (these vacua are not disjoint in finite volume, they mix together into one vacuum), or the continuum of vacua of supersymmetric theories. Inasmuch as the low-energy theory of N=2 SUSY models involves different particles and couplings, any finite-volume drift between these is a violation of Haag's theorem (the finite volume breaks the vacuum degeneracy though).

The analog of Haag's theorem is also not true in finite volume in pure stochastic theories, like the statistical theory of a 3d membrane with elastic forces pulled by noise along a  fourth dimension, or other stochastic theories without Lorentz invariance. In this cases, which are supersymmetric stochastic, you can move from theory to theory at finite volume in the same Hilbert space.

The Haag theorem is a limitation of Fock space descriptions, and I am not restricting to Fock space descriptions or separable Hilbert spaces, rather to anything stochastic you can simulate on a computer, and provably take a stochastic limit on.

"Provably taking a stochastic limit" means that you have a sequence of algorithms $A_k$ which pick out a mathematical object from a set at random at running time infinity, and provably converge (in measure) to a random pick from the space. This defines integration in general, modulo some technical issues due to mathematician sillyness (which are overcome by Hairer, because he is working without defining measures on distribution spaces, rather starting from Wiener processes and showing the algorithmic convergence by hand explcitly) and it allows you to speak about states with infinitely many particles with no special problems.

The particle descriptions of stochastic quantization surpass the limitations of Fock space, they allow you to easily and naturally talk about states with an infinite number of particles, whether the infinity comes from infinitely many particles at high k (renormalization Haag theorem, which only applies in finite volume when there is an infinite renormalization of the vacuum), or just from infinitely many particles at large V which is how the general theorem is proved, and has nothing to do with the former thing, which requires more detailed analysis to prove (it was Glimm Jaffe proved the Haag theorem for interacting 3d $\phi^4$ in finite volume, not Haag).

Part of the disagreement here seems to be due to different meaning of the word "particle". In the worldline formalism there is clearly in some sense a particle on some worldline, whence the name of the formalism in the first place. Whether this particle may be intrinsically recovered by looking for Fock space presentations of the resulting field theory is a different question, and seems to not be the question that Ron is after here.

There have been analogous disputes years back over whether string theory really contains strings. In string theory the "worldline theory" becomes a "worldsheet theory" and as such is clearly given by a string, but whether and how that string may be recovered in terms of any Focks states in the resulting effective QFT is a completely different question.

Yes, our concept of particles is quite different.

Ron is talking about virtual particles, which have (in real time) no trace of existence except for their occurrence in path integrals and Feynman diagrams. This is only a particle picture for visualization purposes. One cannot assign any time-dependent properties to these particles, as the formalism doesn't allow to distinguish between different states of the particles at different times. Given a state of the system, one cannot tell in principle even simple things like the mean number of these particles in a given space region at a given time.

I am talking about real particles whose existence is more solidly based, so that one can properly talk about (and in principle compute) properties such as the above mean number and many other observables. In particular, one can discuss on a formal level the conditions for the presence of a particle track close to a given trajectory.

But I see now that the original question was only about the first kind of particle. I don't care about their existence.

Please don't mangle what I am asking for: I don't claim that there is a Fock space description either of unregulated interacting field theories, or of regulated interacting field theories in infinite volume! It is not true, I know it is not true, and there are many ways to show it is not true. Your argument above regarding infinite dressing of operator products is a less obvious way, the Haag theorem things is a more obvious way.

It is manifestly obvious that changing the coupling g introduces infinitely many particles associated to the vacuum in infinite volume, and infinitely many particles per unit volume in the limit that the regulator is relaxed, or infinitely many bare particles attached to any one particle when there is no regulator and when you add an interaction (as you correctly say in equations above). That means you can't express the unregularized field theory or the infinite volume field theory inside a normal Fock space, because Fock space is always restricted to finitely many particles. It's not deep.

I was not asking about Fock space formalism. I was asking for a nonperturbative path-integral formalism which has a direct interpretation in terms of consistent interactions of particle worldlines, and corresponds directly to the perturbative particle-worldline picture known since Feynman and Schwinger and which is generalized in string theory.

Such a formalism never fits inside a Fock space formalism, because there is no restriction to finite numbers of particles. When you take the renormalization limit for interactions, by adding a noise-mollifier and removing the mollifier, you also produce infinitely many particles around any one particle, so it doesn't fit inside Fock space locally either. It doesn't matter.

If you have a regulator, and a finite volume, then everything does fit inside Fock space, because there are always finitely many particles introduced by the perturbations, because you can't have a divergence at short distances (because of the regulator) or at long distances (because of finite volume). Naive intuitive methods only work because they have an implicit ultraviolet and infrared regulator inside.

But it is just as easy conceptually to define a path integral with infinitely many particles as with finitely many, although the infinite number of particle case always lies outside the Fock space description. It's no more difficult than defining an infinite volume lattice gas Monte-Carlo simulation, although in this limit, technically speaking, the probability of any configuration is exactly zero (this is the analog of these theorems in statistics), and in the limit that you make the lattice teeny tiny, the probability of any precise configuration of given lattice-gas density in any finite volume also becomes zero. It's just because there are infinitey many particles in the limiting description. So what. It's still a particle description because we can calculate it in terms of particles in a regulator, and relax the regulator to get a limit.

The problem which you opaquely phrase as "states outside the Fock space" can always be rephrased more clearly as "states with infinitely many particles". When you do, you see that it is no obstacle to a nonperturbative path-integral formalism.

"Nonperturbative" does not mean "Hilbert space". It means that you can  simulate the theory on a computer with a parameter which, when you take to infinity, you reproduce the correlation functions. This is provided by noise-regulated stochastic quantization, which can be so simulated, and you can see the effect of instantons and so on. Further, you can explicitly do the same simulation using particles branching and statistically getting annihilated, in a limit of large density, infinite volume, and smooth (mollified) noise.

@dimension10 @ArnoldNeumaier, I'll create a question. [CREATED]

@ArnoldNeumaier I've deleted the comments and moved Ron's comments back to the old answer. I hope this was the intention. By the way, shouldn't @RonMaimon's comment be reproduced both here and on @JiaYiyang's new question, since it's a comment to both the answer there and the comment here?

@dimension10, I think Ron's comment is more relevant here, although it is clear that now this conversation is having a gap because Arnold's answer is moved.

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