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Solitons in the paper "Quantum Meaning of Classical Field Theory"

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I'm reading the review paper Quantum Meaning of Classical Field Theory by R. Jackiw, suggested by Arnold, and it gives me a lot of confusions on solitons. I'm going to list a few in this post, but if someone thinks it's more reasonable to split them into multiple questions, then I'll do that. I'm speaking in the context outlined by the paper, e.g. scalar field theories ($\phi^4$, Sine-Gordon) in 1+1 dimension.

(1) How to understand the existence of soliton in the context of Wigner's classification (of elementary particles as irreducible unitary representations of the Poincare group)?

It seems single-soliton states do furnish an (or more?) irreducible representation of Poincare group. But there are also the "meson" states (i.e., the more mundane states that we can get even in free field theories), which also furnish an irreducible representation. And since soliton states are never superpositions of meson states, so we indeed have two copies of representations. Am I right to say when nonlinearities are present, one cannot tell how many types of elementary particles a field theory describes directly from the transformation property of the field(scalar, spinor, vector etc.)? Or is there a way to somehow view these solitons as emergent ones from the meson states? 

(2) How to view the wave profile of a soliton?

Although Jackiw explicitly wrote down solitary solutions of the classical field equation, he interpreted them as the Fourier transforms of $\langle P'|\Phi(0)|P\rangle$(equation 2.33), where $|P\rangle$ is a single soliton state with momentum P. So I cannot see how it is related the more familiar(to me) expression of the profile $\langle 0|\Phi(x)|P\rangle$(or arguably one might use Newton-Wigner basis), in fact, by the absolute stability of soliton assumed in the paper(page 684, below equation 2.25), $\langle 0|\Phi(x)|P\rangle$ is constantly 0, or did I just misinterpret the soliton stability?   

(3) Evaluation of the topological charge (This might be tightly related to confusion (2))

The current  $\epsilon^{\mu\nu}\partial_\nu\Phi$ (equation 2.49) is conserved because of the antisymmetry of $\epsilon^{\mu\nu}$, and the conserved charge is 

$$N=\int dx J^0=\int dx \Phi'=\Phi|^{x=+\infty}_{x=-\infty}$$

And indeed if we plug in the classical solitary solutions we'll get nonzero numbers. However, quantum mechanically isn't it more appropriate to evaluate $\langle P|\Phi(x)|P\rangle|^{x=+\infty}_{x=-\infty}$? But this is zero by translational invariance.

Partial or whole answers are both welcomed.

asked Aug 16, 2014 in Theoretical Physics by Jia Yiyang (2,465 points) [ revision history ]
edited Aug 17, 2014 by Arnold Neumaier

1 Answer

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A QFT is a highly reducible representation of the Poincare group, and lots of different irreps occur. (In a free massless spin 1/2 theory you even have all physical irreps present!) Thus seeing different representations in a construction is no cause for concern.

The defining representation of a QFT specifies the representation of the Heisenberg fields figuring in the action. The S-matrix, in contrast, employs asymptotic fields, one corresponding to each bound state. This is since it relates the asymptotic behavior at $t=-\infty$ and $t=+\infty$. The asymptotic limit is based on the interaction picture; see Chapter 3 of Vol III of Thirring's math physics series for a rigorous asymptotic limit and the definition of asymptotic observables in QM. 

If the two kinds of fields are essentially the same, the Fock space in which the free Heisenberg fields are defined can be identified with the Fock space of the asymptotic particles. This is the situation where perturbation theory is valid and gives a good S-matrix, and this is the situation discussed in the textbooks (usually without warning the reader that it works only under these circumstances).

In QFT, the discrepancy between the two kinds of fields is precisely the notorious infrared problem. It arises since if the field structure is different, the asymptotic Hilbert space of the S-matrix is structurally very different from the Heisenberg Hilbert space. (Note that all separable Hilbert spaces are isomorphic. Thus there exists some unitary isomorphism between the two, mediated by the so-called Moeller operator. But this operator is no longer perturbatively accessible.) Thus a match is impossible. The conclusion is that one cannot get a meaningful perturbative S-matrix if the bound state field content is different from the Heisenberg content of a field theory. It is also well-known from QM that the Born series diverges if there is a bound state - it must, because the T-matrix develops a pole. Path integral perturbation theory is just a covariant version of the Born series, hence has the same problem. See also http://www.physicsoverflow.org/808/different-kinds-of-s-matrices?show=927#a927 for a discussion of how the textbook S-matrix treatment may break down.

For example, in QED one has an infrared problem since an asymptotic electron, being charged, is dressed with an intrinsic electromagnetic field (which means that, formally - before the IR limit - it is a superposition of a Heisenberg electron and infinitely many virtual photons and virtual electron-positron pairs), while the Heisenberg electron has charge but no associated electromagnetic field.

In QCD (and in the standard model) one has a much more severe infrared problem since asymptotic bound states (mesons, baryons, glueballs - and in the standard model also nuclei, atoms, and molcules) are white, while the Heisenberg fields (quarks and gluons) have color. 

The asymptotic states emerge naturally (as in QM) from the interactions by taking an asymptotic limit. Before the limit is taken, they are (formally) superpositions of arbitrarily many Heisenberg particles, but in the limit, the formal contact with the Heisenberg representation gets lost. (This is why in QM, bound states are never discussed in the Heisenberg picture - it is useful only for finite times!)

In a theory where the Heisenberg fields are bosons it is possible because of topological subtleties that an asymptotic field is a fermion field (there are explicitly solvable rigorous examples in 1+1D). In the classical limit, the latter are seen as solitons. This is inverse to the phenomenon in 1+1D  called bosonization. The simplest rigorous example is that of a free fermion theory, where the current field satisfies the commutation rule of a free boson field. Conversely, one can reconstruct (by an explicit but messier construction) inside the algebra of operators on a boson Fock space a (soliton) field with the commutation relations of a fermion field, in such a way that its current field equals the original boson field. The result is that in 2D there is no theorem relating spin and statistics. In particular, the bosonic Sine-Gordon QFT and the fermionic Thirring model are equivalent QFTs, since one can express the operators of each in terms of the other.

In 4D, similar things can apparently happen (because solitons actually exist in some cases), but only for interacting theories. The spin-statistics theorem - a theorem about representations of  free fields - forbids this in the free case (only).

Soliton states are orthogonal to all boson states (which includes the vacuum), for essentially the same reason that bound states of different energy are orthogonal. When approximated at finite IR and UV cutoff, solitons are bound states of infinitely many bosons and therefore only approximately orthogonal to boson states. But when the IR cutoff is removed they move into a different superselection sector of the theory. (Due to renormalization effects, the Hilbert space structure changes radically in this limit; in fact, in 4D, it seems that no sensible Hilbert space is left, which is the cause for the hardness of the YM millennium problem.)  Note that the superpositions of the bosons that make approximate fermions cease to be superpositions only in the infrared limit (where the compactness of space is removed) where the approximation becomes exact and solitons appear. (Solitons cannot exist in a compact space, as they are defined by different boundary conditions at spatial infinity when $t\to \pm\infty$.)

Therefore, it doesn't make sense to regard the solitons as bound states in the final, covariant theory. Indeed, at the level of the S-matrix, one has democracy of particles - all bound states appear in the same way. (This inspired Heisenberg's idea about using the S-matrix as basic principle to organize the in his time ill-understood structure of the particle zoo.) What is elementary and what is composite is not determined by the S-matrix but by the action that generates it. If - as in 1+1 D in the example mentioned - two different actions produce isomorphic QFTs (in the sense that their Borchers class of local fields are isomorphic) and hence equivalent S-matrices, the elementary particle content is not unique, and depends on the action preferred by the user. This is the same kind of user preferences as in a classical theory written in different coordinate systems. The physics is independent of the coordinate system, but not the resulting interpretation. In the same way, a QFT is in an intrinsic sense action-independent. (There are even 1+1D QFTs satisfying the Wightman axioms but where no associated action is known.)

To see the wave functions of a solitonic bound state one cannot use the Fock space formulation but needs to work in the functional Schroedinger picture. There $\psi$ is a functional of the classical field $\phi$, and a semiclassical interpretation in terms of density and phase is possible, which gives some limited intuitive understanding. See http://www.physicsoverflow.org/22012/functionals-of-quantum-states-in-qft?show=22145#a22145

I cannot answer your query (3) since all Jackiw does is just plausible reasoning, picking the arguments that were found working in practice. Most of QFT is not yet a theory where logic can be applied everywhere. (Actually, it seems that your argument is wrong since a soliton state $|P\rangle$ is not translation invariant. - Will check this.)

answered Aug 16, 2014 by Arnold Neumaier (12,365 points) [ revision history ]
edited Aug 17, 2014 by Arnold Neumaier
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In 1+1D  there is the phenomenon called bosonization. The simplest example is that of a free fermion theory, where the current field satisfies the commutation rule of a free boson field. Conversely, one can reconstruct (by an explicit but messier construction that I'd have to look up) inside the algebra of operators on a boson Fock space a field with the commutation relations of a fermion field, in such a way that its current field equals the original boson field.

The result is that in 2D there is no theorem relating spin and statistics. In particular, the bosonic Sine-Gordon QFT and the fermionic Thirring model are equivalent QFTs, since one can express the operators of each in terms of the other.

In 4D, similar things can apparently happen (mainly because solitons actually exist in some cases) but not in the free case, where the spin-statistics theorem forbids this.

Thank you Arnold, a few questions:

(1)You said

The defining representation of a QFT specifies the representation of the Heisenberg fields figuring in the action. The S-matrix, in contrast, employs asymptotic fields, one corresponding to each bound state. This is since it relates the asymptotic behavior at t=−∞ and t=+∞.  

So is the asymptotic fields defined as $lim_{t \to \pm\infty}\Phi(x,t)$? It's hard for me to see how such a limit can take the Heisenberg operator $\Phi(x,t)$ out of the Hilbert space it originally lives in, I mean, I can understand a limit might not always exist, but how do you get a new Hilbert space by taking the limit of operator? And how to see it is related to infrared?

(2) You said 

Actually, I now see that your argument is wrong since a soliton state |P⟩ is not translation invariant

But |P> is assumed to be a momentum eigenstate, hence\(\langle P|\Phi(x)|P\rangle=\langle P|e^{-i\hat{P}x}\Phi(0)e^{i\hat{P}x}|P\rangle=\langle P|\Phi(0)|P\rangle\), which is translationally invariant.

(3)In your reply to Dilaton,

 but not in the free case, where the spin-statistics theorem forbids this

But I thought spin-statistics also applied to 4D interacting QFT? I can understand the violation if solitons are bound states of mesons, but it seems solitons are just as elementary. 

The asymptotic limit is more involved, as it involves the interaction picture. For the correct asymptotic limit and the definition of asymptotic observables see Chapter 3 of Vol III of Thirring's math physics series. It is also well-known from QM that the Born series diverges if there is a bound state - it must, because the T-matrix develops a pole.. The path integral perturbation theory is just a covariant version of the Born series, hence has the same problem. See also http://www.physicsoverflow.org/808/different-kinds-of-s-matrices?show=927#a927 for a discussion of how the textbook S-matrix treatment breaks down.

Spin statistics is a theorem about representations of  free fields only. Solitons are - when approximated at finite IR and UV cutoff - bound states of infinitely many mesons, but as the cutoff is removed they move into a different superselection sector of the theory. (Due to renormalization effects, the Hilbert space structure changes radically in this limit; in fact, in 4D, it seems that no sensible Hilbert space is left, which is the cause for the hardness of the YM millennium problem.)  It doesn't make sense to regard them still as bound states in the final, covariant theory.

At the level of the S-matrix one has democracy of particles - all bound states appear in the same way. What is elementary and what is composite is not determined by the S-matrix but by the action that generates it. If - as in 1+1 D in the example mentioned - two different actions produce isomorphic S-matrices, the elementary particle content is not unique, and depends on the preferences of the user. This is the same kind of admissible preferences as  in a classical theory written in different coordinate systems. The physics is independent of the coordinate system, but not the resulting interpretation. In the same way, a QFT is in an intrinsic sense action-independent. (There are even 1+1D QFTs satisfying the Wightman axioms but where no associated action is known.)

I need to check about P with what is in Jackiw; I wrote my answer without having the paper.at hand. In any case, the classical calculation is what is needed for a correct interpretation of QCD (theta angles).

Note that the superpositions of the bosons that make approximate fermions cease to be superpositions only in the infrared limit (where the compactness of space is removed) where the approximation becomes exact and solitons appear. (Solitons cannot exist in a compact space, as they are defined by different boundary conditions at spatial infinity when $t\to\pm\infty$.)

For a simple exposition of bosonization in 1+1D, see arXiv:1402.5061 . This contains the basics and illustrative examples but no proofs. For bosonization in 1+2D see, e.g., http://arxiv.org/pdf/hep-th/9306065.pdf . This contains also references to theory in the 1+1D case. For bosonization in 1+3D see, e.g., http://arxiv.org/pdf/hep-th/9408128 and http://arxiv.org/pdf/hep-th/9909100.pdf , where the connection to solitons is made.

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In a theory where the Heisenberg fields are bosons it is possible because of topological subtleties that an asymptotic field is a fermion field (there are explicitly solvable rigorous examples in 1+1D.

Hi Arnold, can you give some examples about this? I do not understand how how asymptotic fermion fields can arise from a theory of bosonic Heisenberg fields ...

thanks Arnold, that's a lot of reading on my side.

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