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  What do free equations describe and what is the interaction supposed to do?

+ 2 like - 3 dislike

From another thread:

3) Of course, a quadratic Poincare-invariant ''interaction'' changes a free theory into another free theory, so the renormalization can only have the effect of a change of mass. It is not really an interaction. 

Let us assume that we found (invented, advanced) an interaction that does not lead to renormalizations. What we expect from such an interaction to describe with respect to the solutions of our free equations? In my opinion, an interaction changes the occupation numbers of "free states", which otherwise would be constant in time. What do you think of it?

asked May 7, 2015 in Theoretical Physics by Vladimir Kalitvianski (102 points) [ revision history ]

What is this question? Of course interactions lead to changes in occupation numbers, and if you want an interaction with no renormalization effects, use a nonrelativistic field theory. This seems to be a rhetorical question.

@RonMaimon @dimension10: This is not a rhetorical question but a quest for clarification of the role of interaction terms in QFT.

1 Answer

+ 4 like - 0 dislike

Any quantum system is governed by a Hamiltonian $H$ that generates the dynamics via the Schroedinger equation. If the quantum system can be manipulated by an experimenter, the possible choices by the experimenter can be described by parameters in the Hamiltonian. Thus it is more appropriate to write $H=H(b)$, where $b$ is a vector of external parameters that can be varied. (Usually just a few parameters are considered, but in case of a fully controllable external magnetic field, say, $b$ would be this external field - considered as an infinite-dimensional parameter vector.) 

This gives a complete account of physical reality - reality knows not of interactions, only of dynamics and its generator $H$, and how the latter depends on experimental settings. Thus $H(b)$ tells, in principle, the full story. 

The interpretation in terms of interactions is introduced by the physicist in order to be able to compute the dynamics. Typically, it is assumed that $H$ is close to a Hamiltonian $H_0$ for which the dynamics can be solved explicitly. In this case one defines the interaction to be $V:=(H-H_0)/g$, where $g$ is of the order of the deviation. Then $H=H_0+gV$, which is the standard starting point for perturbation theory. In the context of perturbation theory, $H_0$ is called the unperturbed system and $H$ the perturbed system. Typically, $H_0$ has an interpretation as a simplified, idealized physical model, and the terms in $V$ can be given an interpretation in terms of this idealized model. Thus one talks about interaction terms, and names the interactions according to the intuition coming from the idealized model.

Simple systems can be prepared in a way that $g$ is actually a controllable parameter; in this case, the unperturbed system is physically real, and $g$ can be expressed as a function of the control parameter $b$ in $H(b)$.

In many other cases - actually in the majority of real applications -, $g$ is treated as if it were a controllable parameter, though it cannot be varied in practice. (For example, this is the case for anharmonic oscillators arising in quantum chemistry of simple molecules.) In all these cases, the unperturbed, idealized system is fictitious - physically nonexistent. But being mathematically realizable and simple to solve, perturbation theory can be carried out successfully. Hence the same perturbation terminology is used.

However, in all cases where the idealized system is fictitious, the system itself doesn't determine the interactions - it is the choice of the fictitious idealization defining $H_0$ that determines it. Even in the case where the idealized system is physically realizable, and realized for a particular value $k_0$ of $k$, the system doesn't determine the interactions in case that several values of $k_0$ lead to an exactly solvable system and hence are eligible to define $H_0$.

For example, we may consider a harmonic oscillator with Hamiltonian $H=1/2(p^2/m+k q^2)$ with $m,k>0$. (A system of corresponding relativistic oscillators was the context where the OP's quote was taken from.) We may think of $b=(m,k)$ to be the controllable parameters. This system is exactly solvable, so we could take $H_0:=H$, and the interaction is zero. On the other hand, we might think of it as a perturbation of the free particle with Hamiltonian $H_0:=p^2/2m$ by an interaction $V=kq^2/2$; the system itself is the same, though what is considered an interaction is different.

Now perturbation theory works well only if the spectra are close, which is not the case for a harmonic oscillator and a free particle. In the latter case, perturbation theory is completely meaningless (just as naive perturbation theory in QFT).

But we could also take $H_0:=1/2(p^2/m+k_0 q^2)$ with $k_0\approx k$, which have a similar spectrum, and the interaction term would be $V=gq^2/2$ with small $g=k-k_0$. Perturbation theory now works (at least for the small eigenvalues), so this is a meaningful use of the term interaction. In this particular case it is possible to sum the perturbation series to infinite order, and one gets the correct results. In particular, the ground state frequency $\omega_0$ of the unperturbed system changes into the ground state frequency $\omega$ of the true system. One says that the frequency has been renormalized by resumming the series.  This is sort of an academic exercise for the exactly solvable harmonic oscillator; this is why in my remark that you had quoted I referred to this situation as ''It is not really an interaction''.

But the same freedom of choosing what deserves to be called the interaction becomes numerically very relevant already for the anharmonic oscillator with Hamiltonian$H=1/2(p^2/m+k q^2)+cq^4$. Here the textbook choice $H_0:=1/2(p^2/m+k q^2)$ that leads to the interaction $V=cq^4$ is much inferior compared to an adaptive choice of a harmonic oscillator Hamiltonian $H_0$ with an improved, renormalized frequency, typically determined by variational perturbation theory. (You can find plenty of references by entering "renormalization anharmonic oscillator" - without the quotation marks - into scholar.google.com.) 

If the same is done in quantum field theory, mass takes the place of frequency, and one speaks of mass renormalization (and for other constants, charge renormalization and field renormalization). Note that in QED we are always in the situation that only the total Hamiltonian is realized in Nature, so $H_0$ has to be chosen by the physicist doing the perturbative analysis. Depending on what you declare to be $H_0$ you get different interaction terms. If you take an arbitrary free Hamiltonian with the same symmetries as in $H$ you get as interaction terms the same kinds of terms as if you simply choose the quadratic part of $H$ as $H_0$, but with subtracted coefficients. You can see that the origin of the subtraction terms in the QFT renormalization procedure lies in the fact that $H_0$ and hence the interactions are not determined by Nature but by how to make perturbation theory successful. 

In short: If you choose the right $H_0$ (i.e., in QM, the one with the true ground state frequency, and in QFT, the one with the no-cutoff limit taken after having determined the optimal coefficients by low order loop calculations at fixed cutoff) then no further renormalization is needed. But the conventional $H_0$ that simply truncates the normally ordered Hamiltonian at second order is poorly adapted to the real physics and needs renormalization.

If a quantum system is described not directly by a Hamiltonian but instead in terms of a  Lagrangian $L$, the problem and its handling is essentially the same.

answered May 8, 2015 by Arnold Neumaier (15,787 points) [ revision history ]
edited May 10, 2015 by Arnold Neumaier

Thank you, Arnold, for your answer. I have to run; I will read it later.

Arnold, surely there's no guarantee that the Hamiltonian (the component of the generator of translations in a time-like direction) is an element of a given algebra of observables. We can define it in terms of active translations, something like $$P_\mu\hat\phi(x_1)\cdots\hat\phi(x_n)|0\rangle=\left.\frac{1}{\mathrm{i}}\frac{\partial}{\partial z^\mu}\hat\phi(x_1+z)\cdots\hat\phi(x_n+z)|0\rangle\right|_{z=0},$$ but in any case it's nonlocal. Specifically in the interacting QFT case, Haag's theorem has it that the interaction picture does not exist, there's no unitary transformation from initial to final, if we work with the free field.

@PeterMorgan: I do not understand what your Hamiltonian describes. Is it $H$ or $H_0$? Also, $H_0$ is not to be literally "free" Hamiltonian, but that whose solutions can be easily found. It may include some part of interaction.

@VladimirKalitvianski, $P_\mu$ as I've defined it here doesn't "describe" a dynamics as much as it is defined by the infinitesimal action of active translations. I would take the dynamics to be defined in some different way. $P_\mu$ as I've defined it here would apply equally well for an interacting field $\hat\xi(x)$ (for any commutation relations and vacuum state, assuming we can somehow construct a $\hat\xi(x)$) as it does for a free field $\hat\phi(x)$. For the free field, $P_\mu$ has the same expectation value as $\int k_\mu a^\dagger(k)a(k)\mathrm{d}^4k$ (up to a scale, depending on the definition of the field relative to the creation and annihilation operators), but for an interacting field $P_\mu$ might not be expressible as a function of the field and/or creation and annihilation operators (indeed, Haag's theorem says it will not be expressible within the free field algebra).

@PeterMorgan: The observable algebra contains many nonlocal operators - e.g., all exponentials of smeared fields, their products, and their linear combinations. It also has to contain the exponentials $e^{itH/\hbar}$ of the Hamiltonian $H$ (and the momentum operators), since translations must be unitarily implemented. Your formula is correct if $\hbar=1$, the fields are the interacting fields, and the $x_i$ are distinct.

Since nowhere a free field enters (unless you want to do perturbation theory), this does not contradict Haag's theorem. It is also independent of the S-matrix, as the latter is traditionally defined by a limit whose existence is difficult to establish, and exists at best if the right choice of $H_0$ is made. (Haag-Ruelle theory which guarantees the existence of an S-matrix without reference to an $H_0$ applies only in the case of massive theories where the IR problem is absent.)

Haag's theorem says not only that $H$ will not be expressible within the corresponding free field algebra, but that the interacting field operators won't be either. It says nothing about how $H$ is composed of interacting field operators.

@ArnoldNeumaier, thanks. I suppose I distinguish between observables and energy-momentum in the sense that Haag's §II.1.2 for the Wightman axioms and §III.1 for the Haag-Kastler axioms distinguish between operators $\hat\phi_{\!f}$, smeared by test functions taken from something like the function space of Laurent Schwartz, or algebras of local observables $\mathcal{A}\hspace{-0.07em}(\!\mathcal{O}\!)$, in contrast to the "representors $U\!(\!a,\alpha\!)$ of the symmetry transformations". Energy-momentum is different from local measurement insofar as we can make a change to the state at arbitrary space-like separation with the same effect on the energy-momentum as making the same change right next to an observer, whereas for local observables we require cluster separation.

@PeterMorgan: The algebra of observables in the Wightman sense is bigger than the union of the $A({\cal O})$, which contains only observables with compactly localized support. Indeed, the observables of primary interest in physics (that gave name to the concept of the observable algebra) are the various currents (which are local) and the associated conserved quantities (the spatial integral of the time component of  currents).

 +1. All these(and recent discussions)  make me more favor the constructive approach, which in some sense is the way physicists have chosen, except that physicists do it non-rigorously, while the term "constructive field theory" seems to be reserved for those mathematically rigorous endeavors.   

@JiaYiyang: "make me more favor the constructive approach" - over what? What do you mean by ''the constructive approach" in a nonrigorous setting?

@ArnoldNeumaier, it's too late in my time zone, I'll write a reply tomorrow.

 By "make me more favor the constructive approach" I meant to compare axiomatic approach, algebraic approach and constructive approach, and I'm feeling constructive approach is the most natural and probably the most practical approach. 

I reach this disposition by thinking about the meaning of cutoffs. I've wondered why cutoffs are so not welcomed, because for example we surely shouldn't expect the Minkowskian QFT to literally describe the realistic physics in infinite volume since we know well the large distance spacetime structure is not Minkowskian, so why are we not satisfied with a QFT in a finite volume? To me the reasonable answer is that, though there's nothing physically wrong with a finite volume QFT, there is something physically wrong if your QFT is not well controlled as a function of volume, for example if the cross section of your theory depends on the volume at the order of $O(V)$, which means the physics sensitively depends on large distance, which clearly is not the case for our physical reality. This is why it would be nice if the theory is constructed so that the physics depends on, say, $O(V^{-1})$. But once you have this you basically can remove the cutoff. So in this sense what we should expect from a QFT is a theory with cutoffs but well controlled, rather than insisting from the start that there shall be no cutoffs. The two perspectives might be mathematically equivalent at the end of the day, but I find the former much more natural.     

It seems to me the constructive approach in its spirit is representing the first  perspective, while axiomatic and algebraic approach the latter. And with hindsight the former really is what physicists have been doing, for example the regularization and renormalization programs, but is most of the time only at the perturbative level, and this is what I meant by "non-rigorous constructive approach".

I still appreciate the clarification power that axiomatic and algebraic approach sometimes possess, but it seems hard to take them as starting point to make useful progress in physics. 

@YiaYiyang as I understand it, cutoffs should naturally occure where the effective theory valid for a certain range of scales stops being valid, and new physics kicks in (at the GUT, QG scale etc for example). Aslo, it is not true that renormalization can only be done perturbatively, the exact (or functional) renrmalization group (ERG) method works non-perturbatively too for example.

@Dilaton, I think my earlier argument applies to effective field theories too, that's why I didn't bother to mention it. Whether your QFT can be UV/IR completed or not, you would need some good control. If your theory has a completion, it means it's perfectly controlled;if your theory is a EFT valid only in a certain range, it means the control is limited, and it still may describe physics at the regime of good control but not at other regimes where the control is bad.  

Aslo, it is not true that renormalization can only be done perturbatively   

I never claimed that, I said "most of the time", and "non-perturbative" does not equal "rigorous" in the context I'm speaking of, for example almost all work in lattice gauge theory can be considered non-perturbative, but few are rigorous in the constructive sense.

@JiaYiyang all quantum field theories are only effective theories with a limited domain of validity, as the physics effects needed to describe the system change when going to larger or smaller scales. For example the SM is expected to break down at some point when going to higher energies. This is why the LHC is looking for new physics. The new physics does not need to kick in at LHC reachable scales though.

Axiomatic, algebrai, constructive, perturbative, and lattice QFT should not be thought of as being competing choices but as being complementary techniques that reveal different insights into QFT.

In finite volume, there is neither Lorentz symmetry nor equilibrium thermodynamics. But both are essential for realistic physics. Thus one must be able to take the limit.

Whether one actually takes the limit or assumes that the cutoff is astronomically large doesn't make a significant difference once one assumes that a good large distance behavior. But the problem (of constructive QFT) in removing the finite volume constraint is precisely the difficulty in proving (for 4D YM, say) estimates of the sort you and Ron assume without qualms.

The difficulties are due to the severe IR problems of gauge theories, manifesting themselves physically in confinement and instantonic (i.e.., O(e^{1/g})$ type) contributions. The traditional approaches give no clue at all about how these should be addressed in finite volume - gauge fields are invisible at large distances but glueballs and hadronic degrees of freedom are.

Here is the strength of axiomatic/algebraic QFT - it can study what has to be the case independent of any approximation scheme and passing to the limit. For example, that a good scattering theory exists in the massive case (Haag-Ruelle theory), or the structure of superselection sectors.

 Axiomatic/algebraic QFT was never intended to replace constructions, but it may inform constructions of things that must be met.


 all quantum field theories are only effective theories with a limited domain of validity

Realistically yes, but the same can probably be said to any theory, including string theories, and the boundary between QFT and string seems to be increasingly blurred. Mathematically no, Yang-Mills is widely believed to have a completion in both IR and UV. 

@Dilaton: Maybe I am wrong, but UV incompleteness does not necessary mean the cut-off is unavoidable. An incomplete theory may "work" even out of its region of validity, if properly formulated (I mean, no catastrophes).

@ArnoldNeumaier: Thank you for your answer. +1! (I still cannot vote up directly.)

@JiaYiyang what are these completions in case of Yang-Mills? And at what scales is YM to be expected to break down, both in the UV and in the IR?

@VladimirKalitvianski no, (effective) theories can not be made valid beyond their domain of validity, as new effects have to be considered that kick in the original theory can not describe.


I'm not objecting usefulness of the study of taking infinite volume limit, but rather discussing which perspective is more natural: Is infinite volume a fundamental setting and finite volume an convenient artifact, or the other way around? I think the latter. The earlier discussion on perturbation theory and Haag's theorem can be naturally resolved by adopting this perspective and Ron's answer there: at any stage of the construction we are doing scattering theory in finite volume, and we throw out the vacuum bubbles at this finite volume, which is a well defined procedure. Then we take the infinite volume limit, and through this procedure we get a good control of the S-matrix, all this just means our constructed S-matrix won't depend sensitively on large distance stuff, which is all we really wanted. This is a legitimate way to get a limitedly controlled theory, and this control is good enough to study much interesting physics, though not all of it. However from axiomatic/algebraic philosophy the success of perturbation theory looks still mysterious, though not in direct contradiction.

 But the problem (of constructive QFT) in removing the finite volume constraint is precisely the difficulty in proving (for 4D YM, say) estimates of the sort you and Ron assume without qualms.

I'm not on Ron's side regarding this issue, and my whole comment was promoting the importance of the work in constructive field theory direction!

@Dilaton: I meant specifically the absence of catastrophes, not predictions of physics not yet implemented in EFT. I would like you all to read and think over my opus on this subject, and, maybe, leave your reviews after that.

@Dilaton, Mathematically YM is valid at all scales, UV is completed by asymptotic freedom, IR is not yet done but clearly it still has to be a YM, given all the evidences provided from the empirical success of lattice QCD. Realistically quantum gravity has to pop up at planck scale, and since gravity couples to everything, Yang-Mills at this scale would almost surely fail empirically if we don't take gravity into the picture.  

@JiaYiyang: For me a natural perspective must be one in which the fundamental symmetries of the universe are taken serious. Lattices break all continuous symmetries and introduce physically irrelevant parameters (size and shape of the fundamental cell), hence are never natural. They are just convenient for the numerical analysis, just as finite element grids in solving partial differential equations.

at any stage of the construction we are doing scattering theory in finite volume,

On the lattice one cannot do scattering theory, as all lattice calculations are done in the Euclidean analytic continuation. One calculates correlations functions, continues these analytically back to Minkowski spacetime, and indirectly infers scattering information. See, e.g., http://arxiv.org/abs/1208.4059, http://arxiv.org/abs/0909.0200, or http://arxiv.org/abs/1102.4299.

@VladimirKalitvianski if you apply a theory only inside its domain of validity, no "catastrophies" occur. The bad things happen only due to inapproriate extrapolations.


For me a natural perspective must be one in which the fundamental symmetries of the universe are taken serious. 

That might be true in UV, but in large volume we all know Poincare symmetry is not exact anyway.

On the lattice one cannot do scattering theory, as all lattice calculations are done in the Euclidean analytic continuation. 

I'm a using finite volume setting but not Euclidean, it's just a finite volume cutoff of Minkowskian QFT, and I'm arguing finite volume in fact is very physical, but a sensitive dependence on it is not physical. 

The bad things happen only due to inapproriate extrapolations.

@Dilaton: I think you are "excessively certain" in that.


in large volume we all know Poincare symmetry is not exact 

In quantum gravity, Poincare symmetry is an exact symmetry, part of the even bigger diffeomorphism symmetry. (Indeed, the latter is induced by a massless spin 2 representation of the Poincare group. See http://journals.aps.org/pr/abstract/10.1103/PhysRev.138.B988 .)

a finite volume cutoff of Minkowskian QFT

Could you please give a reference where this is actually being done and gives (in 4D spacetime) useful physical results? 


In quantum gravity, Poincare symmetry is an exact symmetry, part of the even bigger diffeomorphism symmetry.

I haven't read Weinberg's paper, but it seems irrelevant to my argument. I see nothing wrong of constructing an infinite-volume Minkowskian QFT to describe local physics even if our universe if compact, you would have to acknowledge in this case infinite volume limit is necessary only in the sense that our QFT needs to be controlled, rather than respecting some nonexistent exact symmetry of the universe.

Could you please give a reference where this is actually being done and gives (in 4D spacetime) useful physical results? 

I don't have a reference, I'm referring to the textbook treatment of the irrelevance of vacuum bubbles which are manifestations of Haag's theorem. We throw away vacuum bubbles by a division to get S-matrix. If you assume a infinite volume setting from the start, this is literally dividing infinity by infinity which doesn't make sense, the way it makes sense is to consider S-matrix in finite volume where such procedure makes sense, and then discover that the so-constructed S-matrix is well-controlled in large volume, so that the IR cutoff can seemingly be removed. But there's no guarantee the IR cutoff cutoff is safely removed if you start looking at other aspects of your QFT. This is a case where we have gained a limitedly controlled construction, both in the sense that we do it perturbatively, and in the sense that we have only required a controlled S-matrix instead of demanding more. The example was provided by Ron in my first Haag's theorem post and I'm just repeating it, but the "control" perspective is my own, Ron may or may not subscribe to it.  


1. Your claim ''in large volume we all know Poincare symmetry is not exact'' is not correct. This is independent of any limits. Poincare covariance is an exact symmetry of our universe, as far as we can tell.

2. If our universe is compact, there is no S-matrix, as it requires that particles can separate at infinite distance. 


We throw away vacuum bubbles by a division to get S-matrix.

We don't get an S-matrix in finite volume, as the interpretation of a time-ordered expectation value as S-matrix element presupposes the possibility of infinite separation of asymptotic states. 


1. Your claim ''in large volume we all know Poincare symmetry is not exact'' is not correct. This is independent of any limits. Poincare covariance is an exact symmetry of our universe, as far as we can tell.

Alright, if you say so I'll admit the possibility I'm wrong about Poincare symmetry, though I don't even know what Poincare symmetry means in a curved space time(translations seem problematic, and there seems no guarantee that diffeomorphism invariance transformations give the correct Poincare group structure together with the local Lorentz transformations). But it still is beside the point, would you at least acknowledge we'll still be very interested in infinite volume limit of QFT even if the universe is compact but astronomically large? It should be clear in such case we're trying to achieve control.

To your 2,3, all I can say is that you seem to be holding a too stringent view of S-matrix.  I'd say even in a compact universe S-matrix will be a very good tool to describe reality, and in reality we never scatter truly asymptotic particles anyway.

@JiaYiyang: The S-matrix is of practical interest only for microscopic scattering processes; on this scale, gravitation can be neglected and spacetime can be regarded as flat. This is why we are interested in QFT on flat spacetime. Of course the infinite volume limit is here of interest since we get something covariant only in this limit. The macroscopic distances of scattering experiments are indeed essentially infinite compared to the tiny distances where the interactions happen. Thus having control over the infinite volume limit is relevant, but only in the case of a flat spacetime. (In a compact spacetime it is impossible to control it, for the reasons told.)

I have the standard view about S-matrices, as can be found in every QM book. A Hamiltonian defined on a compact space necessarily has a discrete spectrum, but the S-matrix is associated with the continuous spectrum.

The QFT textbook of Weinberg introduces the S-matrix in Vol. I, Chapter 3. One has to begin with free asymptotic in- and out-states to define the S-matrix. Already the term asymptotic is meaningless if space is compact as there will never be a limit in which the particles can be considered free, and the Moeller operator $\Omega(\mp\infty)$ used in (3.1.13) will not exist. One can still make the formal manipulations but they are devoid of a physical meaning. 

$\Omega(\mp\infty)$ does not even exist for flat space with long-range interactions (massless particles). This is already the case in QED and is the source of its IR troubles. The textbook treatments avoid these (and the associated nonexistence of a proper S-matrix) by giving the photon a tiny mass, though this is inconsistent as it spoils gauge invariance. 

I don't even know what Poincare symmetry means in a curved space time(translations seem problematic

It survives as a local symmetry, i.e., on the Lie algebra level. Infinitesimal translations are unproblematic; they are just a special case of vector fields = infinitesimal diffeomorphisms. The symmetry under local translations is promoted to the symmetry under all local deformations. Local Lorentz transformations behave correctly when treating them separately  in the tetrad formalism. Thus the full symmetry group of the universe on a spacetime manifold $M$ is essentially $Diff(M) \times SO(1,3) \times SU(3) \times SU(2) \times U(1)$, whose Lie algebra contains that of the standard model, whose group is essentially $R^{1,3} \times SO(1,3) \times SU(3) \times SU(2) \times U(1)$.

Look, guys, when we take a simplified QED, no problems exist:

1) QED with a classical current and quantized EMF. We have S-matrix solutions with coherent EMF states. Such a theory is incomplete (no pair production is described), but no Volume and IR difficulties arise.

2) When we have quantized fermion field, but practically classical EMF, again, no those problems arise (Interaction of atoms at long distances, for example). Such a theory is also incomplete (no photons are radiated).

The problems arise when we couple the quantized EMF with quantized fermion filed, and I think, they arise due to bad way of coupling. So it is not a property of QED/ QFT, but problems of a particular QFT formulation. No matter how hard you try to put it on a "mathematically solid base", you still deal with a badly formulated theory.

@VladimirKalitvianski: Things are not so simple. A quantum electron field in an external classical field has already the problem that the textbook S-matrix does not exist. 

See Derezinski's lecture notes. In the introduction (p.8) he writes:

The infrared problem means that in the naive theory some integrals are divergent for small momenta. This problem appears already in non-relativistic quantum mechanics – in scattering theory with Coulomb forces. These forces are long-range, which makes the usual definition of the scattering operator impossible. Its another manifestation is the appearance of inequivalent representations of canonical commutation relations, when we consider scattering of photons against a classical 4-current that has a different direction in the past and in the future. Thus, even in these toy non-relativistic situations the usual scattering operator is ill-defined. 


I don't see why scattering is impossible in finite volume, even in finite volume I can easily picture two  wavepackets coming near to each other, collide and then go away. 

Arnold, I meant literally a "static" interaction with no photon radiation. I read it in Akhiezer-Berestetski QED. Namely, page 302, formula (24.26). It is used later to calculate the long-distance potential of interacting atoms.

@JiaYiyang: Interaction of a charge with IR part of the EMF spectrum makes the corresponding wave packet infinite (larger than any finite volume). Se my electronium model. This fact is known from T. Welton's paper.

@JiaYiyang: Of course, there is nothing uncommon about two  wavepackets coming near to each other, collide and then go away. But this happens in finite time. If you look at what happens for $t\to\pm\infty$ (which is intrinsic to any discussion of the S-matrix), you'll see that the wavepackets spread and ultimately (after sufficiently many revolutions if necessary) recombine in a more and more delocalized manner, and asymptotically one cannot say anything. One is and remains in a superposition of bound states. Nothing like an S-matrix appears. 

On the other hand, chemist compute S-matrices (and certain information derived from them -  chemical reaction rates) from computations in finite volumes. But this requires a lot of attention to details. In particular, you cannot use a manifold without boundary but need to impose artificial absorbing boundary conditions, to avoid that what I describes happens. This changes the Hamiltonian to a nonhermitian operator; hence the dynamics becomes nonunitary. How to get nevertheless a unitary S-matrix (or rather a unitary approximation to the intended one) is a nontrivial story that involves the true S-matrix (on the unbounded flat space) and a significant amount of theory.

@VladimirKalitvianski: But what I said shows that the problems appear far earlier than you locate them. In particular, IR problems already appear when a single charged particle scatters in the presence of an external electromagnetic field. Scattering is no longer simply described by an S-matrix anymore but they can still be solved using coherent states. In fact, coherent states are sufficient to solve all IR problems of QED. 

@ArnoldNeumaier: You are nearly right. I say "nearly" because the coherent states have uncertain energy and the energy is not really conserved. Electronium excitations preserve the energy conservation law and are nearly coherent. Nearly - because they do not include too high frequencies, which are always present in the coherent states. Too high frequencies are not excited just because of lack of energy.


But this happens in finite time. If you look at what happens for t→±∞

I think we have very conflicting views on what shall be called "scattering". I'd say from a physical point of view the "two wavepackets go near then collide then fly away"  is really as much of a baggage as you should put into the term "scattering", this definition can be made more specific depending on you context, e.g. finite volume or infinite volume etc. On the other hand, you insist on a way too rigid definition, this is not how we use this piece of language. For example, if you are consistent throughout,  you may say to a QFT practitioner "actually there's no scattering in QED because the wave operator doesn't exist, you may call it scattering but it's not a real scattering...", and this gets completely pedantic.

@JiaYiyang: With your definition of scattering you have a nice intuitive picture but no formal apparatus to discuss it. Not a single formula can be derived from your view.

Everybody in physics (rigorous or not) who treats scattering in the continuum uses theory formally based on the limit $t\to\pm\infty$ and then manipulated to give integrals computable by Feynman diagrams.

And those doing QFT on a lattice never compute S-matrices - they compute correlation functions in Euclidean space, and use the theory for the continuum to deduce information about scattering amplitudes in flat space at $t\to\pm\infty$.

Thus to turn the intuitive idea of "two wavepackets go near then collide then fly away" into something that can be represented by the traditional scattering formalism you embed the region where this happens into flat space and pretend that flying away extends to infinity. Then you have a good framework for predicting what happens, and when the scattering objects are small enough the predictions will be appropriate. (This sort of idealizaion is done everywhere in physics.)

But if you embed it instead in a compact space you won't get anything useful since its behavior at infinity looks very different from the intention ''flying apart''.

Thus, effectively, you must treat scattering as if it happens in the tangent space of the space $M$ at the point where the scattering event happens, rather than in a compact space, even though the true universe might be compact. Nothing should depend on more than the region where your experiments take place, and this is effectively flat space.

In the application to experiment one uses of course that large positive and negative finite times are good approximations to $\pm\infty$ and silently applies the theoretical results to finite times that are neither too large nor too small for providing a sensible approximation. In particular, the sizes are always chosen so that space is flat in the measuring region.


Thus, effectively, you must treat scattering as if it happens in the tangent space of the space M at the point where the scattering event happens, rather than in a compact space, even though the true universe might be compact. Nothing should depend on more than the region where your experiments take place, and this is effectively flat space.

Exactly! This is why I think the constructive philosophy looks more natural at least on the surface. The stipulation of a sensible infinite volume limit just means our physics should be insensitive to large distance: when you do collision experiment in your lab, the shape of the universe shoudn't matter. To me this stipulation is not so much connected to the requirement of exact symmetry,  that's all I'm saying. This philosophy naturally encompasses both UV/IR completed QFT and effective field theory. 

@JiaYiyang: But still, if you want to describe your local scattering by an S-matrix you must make the idealization of a spacetime with a timelike Killing vector and noncompact space slices. Thus talking about compact space in the context of scattering amounts to choosing the wrong model for the analysis.

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