Structure of Matter in Quantum Field Theory

[DarMM]

Of course in any rigorous construction, there is for every coupling constant some renormalized equivalent. But I don't know of any rigorous definition. How would you define a coupling rigorously, just given some Wightman quantum field theory - i.e., without reference to a particular method of constructing it?

Perhaps this is not what you are after, but the renormalized Hamiltonian would take the form:
$$H = H_{\phi} + H_{\psi} + g\colon\bar{\psi}\phi\psi\colon$$
where wick ordering is with respect to the vacuum of the interacting Hilbert space.
The coupling constant is then just $g$.

 

[A. Neumaier]

The coupling constants (...) are determined phenomenologically

... within a particular (and nonrigorous) construction from a given Lagrangian. But the Wightman axioms do not mention a Lagrangian, and it is not clear how one can associate to a given Wightman theory involving a family of basic fields a Lagrangian and interactions in a sensible and rigorous way.

 

[A. Neumaier]

Perhaps this is not what you are after, but the renormalized Hamiltonian would take the form:
$$H = H_{\phi} + H_{\psi} + g\colon\bar{\psi}\phi\psi\colon$$
where wick ordering is with respect to the vacuum of the interacting Hilbert space.
The coupling constant is then just $g$.

Is such an equation guaranteed to hold? No higher order terms introduced? Is $H_\phi$ even well-defined in general as an operator on the interacting Hilbert space? I haven't seen these things discussed anywhere.

 

[DarMM]

Is such an equation guaranteed to hold? No higher order terms introduced? Is $H_\phi$ even well-defined in general as an operator on the interacting Hilbert space? I haven't seen these things discussed anywhere.

Ah, I see what you mean. Indeed for $Yukawa_2$ one would have a family of Hamiltonians on Fock Space:
$$H(g,\kappa) = H_{\phi,\kappa} + H_{\psi,\kappa} + g\colon\bar{\psi}\phi\psi\colon - \delta m^2(g,\kappa)\colon\phi^{2}\colon - E(g,\kappa)$$
And then show that the vacuum of this Hamiltonian $\omega_{\kappa,g}$ has a limit, $\omega_{g}$, as a state on the algebra of local observables, though not of course a limit as a ket in Fock space.

The GNS representation of $\omega_{g}$ would be the interacting Hilbert Space of the Yukawa model. However no, the decomposition of the Hamiltonian would be invalid. The coupling constant would only be retained as a label of the GNS representation, with Wightman three-point $\phi\phi\psi$ correlators different in each rep (others as well obviously but that's the crucial one).

 

[A. Neumaier]

The coupling constant would only be retained as a label of the GNS representation

Yes, and in the $k$-parameter case one would just have a $k$-dimensional manifold of theories, and nothing definite to tell what the couplings of a particular theory from this family should be. The true parameters are the values of distinguished invariant observables such as asymptotic particle masses and charges.

 

[vanhees71]

... within a particular (and nonrigorous) construction from a given Lagrangian. But the Wightman axioms do not mention a Lagrangian, and it is not clear how one can associate to a given Wightman theory involving a family of basic fields a Lagrangian and interactions in a sensible and rigorous way.

It doesn't matter whether there's a Lagrangian. If you have a QFT somehow defined and Wightman functions those imply, via the standard interpretation of time-ordered N-point functions, measurable quantities like S-matrix elements and cross sections. These can be used to tune the physical parameters like coupling constants to some measured "standard cross section". Then you can predict cross sections for other scattering reactions and so test the validity of the theory. That's also how the standard model works.

 

[A. Neumaier]

These can be used to tune the physical parameters like coupling constants to some measured "standard cross section".

Yes, but If you have Wightman functions without a Lagrangian (as, e.g., in conformal field theory), they are parameterized arbitrarily, since couplings are not well-defined. Whatever parameters there are can be determined by fitting them to measured cross sections. So the physical parameters are not coupling constants but some of the measured stuff - masses, charges, decay probabilities, lifetimes, etc..

 

[vanhees71]

But the physically observable quantities, i.e., in the case of "vacuum QFT" cross sections are defined by the S-matrix elements which in turn are constructed from the N-point functions of the theory. What we call "coupling constants" is finally determined by cross sections, e.g., the electromagnetic coupling constant, usually quoted as the finestructure Sommerfeld constant, $\alpha=e^2/(4 \pi)$ (in the usually used natural HL units of the HEP community), can be defined, e.g., by $e^+ + e^- \rightarrow \mu^+ + \mu^-$ ("the" standard process of QED so-to-say) in the limit of small momentum transfer, leading to the value of about $1/137$. This value is entirely given by a measurable cross section and not necessarily tied to a Lagrangian, i.e., what you call "physical parameters" and the physicist usually calls "couplig constants" (and of course (pole) masses of particles (or resnonaces) which are defined by corresponding poles of N-point functions).

An example where a decay probability is used to define a coupling is the "pion decay constant", which determines the overlap of a asymptotic free pion state with the corresponding axial-vector current. This is also a good example where one has no Lagrangian or only a Lagrangian of a low-energy effective theory like chiral perturbation theory.

 

[A. Neumaier]

since quark and gluon fields are an unphysical expansion of the physical (hadron and glueball) fields on a Hilbert-Krein space

Could you please check the papers by Kugo and Ojima in the context of my doubts here? They ignore infrared problems (which are at the heart of confinement) and then get - if I understand them correctly - for pure Yang-Mills a physical asymptotic 1-gluon operator. But I think this is impossible since it contradicts confinement.

 

[DarMM]

Could you please check the papers by Kugo and Ojima in the context of my doubts here? They ignore infrared problems (which are at the heart of confinement) and then get - if I understand them correctly - for pure Yang-Mills a physical asymptotic 1-gluon operator. But I think this is impossible since it contradicts confinement.

In essence Section IV of their paper is a sort of formal (in the sense of non-rigorous) demonstration that the typical perturbative BRST treatment of Yang-Mills leads to a unitary S-matrix (again, term-by-term perturbatively) between the states which seem physical at the perturbative level.

Of course there is non-perturbative confinement, so this entire approach is really nothing more than showing perturbation theory isn't pathological and self-contradictory. They in fact discuss these issues in Sections V-VI, including statements that the Gluon fields are not physical later in Section VI.

 

[DarMM]

I'm still looking for rigorous work on the flavor fields. Most references still seem to say the hadron fields are fundamental, but I don't see why, as A. Neumaier's reasoning makes more sense to me.

So let's just tentatively assume that. The basic objects are flavour and glueball fields.

First question:
Since particles are only an asymptotic notion, can we then say that QCD is "about" the non-commutative statistics of localized flavor and glueball field observations?

Second question:
SU(3) symmetry is usually phrased in terms of quarks and gluons. What does it reflect about the colorless physical states?

 

[vanhees71]

Which SU(3) symmetry are you talking about? If you mean Gell-Mann et al's "eight-fold way", it's usually phrased in terms of hadrons. When it was formulated, QCD still wasn't known and "quarks" were seen by Gell-Mann et al just as "mathematical constructions". This quickly changed thanks to Feynman's famous visit to SLAC and the discovery of Bjorken scaling, which gave rise to the "parton model" and finally the identification of (constituent) quarks with partons.

 

[DarMM]

Which SU(3) symmetry are you talking about?

Color gauge SU(3)

 

[vanhees71]

Of course, this is almost a synonym for QCD. Of course, there by definition you deal with quark and gluon fields in the Lagrangian. That these nonetheless give not rise to particles, is what's called "confinement", and it's in my opinion not completely understood. That QCD is correct beyond its applicability where it can be evaluated by perturbative methods (like in deep-inelastic scattering experiments) also in the low-energy sector, is mainly known from lattice calculations, reproducing, e.g., the hadron-mass spectrum quite accurately. This, however, still does not provide a true understanding of how the (mathematical) "mechanism" of confinement "really works".

 

[A. Neumaier]

let's just tentatively assume that. The basic objects are flavour and glueball fields.

First question:
Since particles are only an asymptotic notion, can we then say that QCD is "about" the non-commutative statistics of localized flavor and glueball field observations?

It is primarily about the flow represented by flavor and glueball fields, as embodied in their expectation values and correlation functions, and as computed numerically by means of QCD hydrodynamics (1PI theory) or QCD kinetics (2PI theory). In addition, it is about extracting asymptotic information about the fate of colliding rays of such fields at large times, i.e., after particle freeze-out, where the asymptotic hadron descriptions is needed to classify the resulting distribution of particle measurements. The latter can be interpreted in terms of the Haag-Ruelle S-matrix for hadrons.

The perturbative S-matrix is unphysical since it is in terms of quarks and gluons. It can describe only intermediate time situations before the hadron freeze-out.

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Second question:
[Color gauge] SU(3) symmetry is usually phrased in terms of quarks and gluons. What does it reflect about the colorless physical states?

Color SU3 is by definition no longer visible in the vacuum sector. It is a scaffolding needed for getting the representation right.

 

[vanhees71]

Hm, as a simple-minded partictioner, I'd say the observed asymptotic free states of QCD are hadrons, which are all "composite objects" or "bound states". If course, I'm well aware that we are not able to calculate those asymptotic free states from first principles except in an approximate sense as using numerical (Monte-Carlo) techniques (aka lattice-QCD calculations) and exploiting various "accidental symmetries" to build "effective hadronic models" (like (unitarized) chiral perturbation theory in the light-quark sector (up, down(, strange)) and heavy-quark effective theory and or NRQCD in the heavy-quark sector (heavy quarkonia, open-heavy flavor mesons).

This at least deals with entities that are really detected by HEP, nuclear physics and HIC experiments ;-)).

 

[A. Neumaier]

the observed asymptotic free states of QCD are hadrons, which are all "composite objects" or "bound states".

Composite only in terms of quark fields that defy an interpretation in terms of standard quantum mechanics since there is no Hilbert space on which they act, hence no probability interpretation.

The question we are discussing here is what of the standard (indefinite space) QCD picture remains if one insists on quantum field theory on a Hilbert space. Now only color invariant fields are permitted as constituents, just as in the effective hadronic models. But in these, the hadrons are no longer composite in a naive sense.

 

[DarMM]

Of course, this is almost a synonym for QCD. Of course, there by definition you deal with quark and gluon fields in the Lagrangian. That these nonetheless give not rise to particles, is what's called "confinement", and it's in my opinion not completely understood. That QCD is correct beyond its applicability where it can be evaluated by perturbative methods (like in deep-inelastic scattering experiments) also in the low-energy sector, is mainly known from lattice calculations, reproducing, e.g., the hadron-mass spectrum quite accurately. This, however, still does not provide a true understanding of how the (mathematical) "mechanism" of confinement "really works".

I actually think this isn't stated enough, that confinement isn't a real physical mechanism, but a mathematical question of how the elements of the larger Hilbert-Krein space of perturbative QCD are related to the physical states in the Hilbert space of nonperturbative QCD. Often confinement is presented as some kind of physical effect we don't understand.

The perturbative S-matrix is unphysical since it is in terms of quarks and gluons. It can describe only intermediate time situations before the hadron freeze-out.
Color SU3 is by definition no longer visible in the vacuum sector. It is a scaffolding needed for getting the representation right.

Do you mean in order to ensure that the hadrons come out as fermions? i.e. if you decompose the hadrons into quarks and only had those quarks carry the physical properties of spin and flavour, they would incorrectly produce the hardons as bosons. For that reason we must add a fictitious degree of freedom to the unphysical quarks.

 

[vanhees71]

There are fermionic (baryons) and bosonic (mesons) hadrons, i.e., in the usual physicists' language bound states of three valence quarks or a valence quark and a valence antiquark; described by parton-distribution functions in the perturbative treatment of deep-inelastic scattering. There may be more bound states like glue balls, tetra-quark states, which are not yet unanimously discovered in experiments.

 

[DarMM]

There are fermionic (baryons) and bosonic (mesons) hadrons

I'm just focusing on the fermionic ones, as they are often used as the reason one needs color.

The main question would still be what does SU(3) color charge reflect about the physical states.

 

[A. Neumaier]

The main question would still be what does SU(3) color charge reflect about the physical states.

Color SU3 is by definition no longer visible in the vacuum sector. It is a scaffolding needed for getting the representation right.

Do you mean in order to ensure that the hadrons come out as fermions? i.e. if you decompose the hadrons into quarks and only had those quarks carry the physical properties of spin and flavour, they would incorrectly produce the hadrons as bosons. For that reason we must add a fictitious degree of freedom to the unphysical quarks.

No; $qqq$ is always a fermion if $q$ is, independent of any gauge structure. The color SU(3) is needed among others to explain the observed collection of baryons (in particular $\Delta++=uuu$) in terms of quarks without violating the spin-statistics theorem. An alternative (today minority view) explanation in terms of representation theory was through parastatistics rather than color gauge theory - so maybe some remnant of that is visible in a pure Hilbert space theory. I don't understand enough about parastatistics, though.

On the other hand, the unphysical features of quarks are tied directly to the SU(3) gauge formulation with SU(3) as a local symmetry group. The Krein space (indefinite Hilbert space) formulation of gauge theories is a reflection of the fact that the classical Poisson bracket of a gauge theory (the $\hbar\to0$ limit of $i[.,.]/\hbar$ in the quantum theory of gauge invariant operators) is not symplectic, and needs to be embedded into a bigger, unphysical symplectic space that (in perturbation theory) can be canonically quantized, though only with an indefinite inner product. The gauge invariant subalgebra of the Poisson algebra is the physical field algebra, but no way is known to quantize it canonically without the detour through Krein spaces. An (apparently not yet developed) infinite-dimensional generalization of Kontsevich quantization of Poisson manifolds (also defined only in perturbation theory, through a deformation procedure) should give the physical quantum field algebra directly on a Hilbert space. For 2 spacetime dimensions see https://arxiv.org/abs/hep-th/9910133..

 

[vanhees71]

I'm just focusing on the fermionic ones, as they are often used as the reason one needs color.

The main question would still be what does SU(3) color charge reflect about the physical states.

Well, the most direct evidence is the ratio $R$ of the inclusive reaction $\mathrm{e}^+ + \mathrm{e}^- \rightarrow \text{hadrons}$, which let's you count the number of colours in a literal sense (given the -1/3 and 2/3 charges of each down-up-quark pair of the 3 families). The famous plot is in

http://pdg.lbl.gov/2018/hadronic-xsections/rpp2018-sigma_R_ee.pdf

 

[A. Neumaier]

Well, the most direct evidence is the ratio $R$ of the inclusive reaction $\mathrm{e}^+ + \mathrm{e}^- \rightarrow \text{hadrons}$, which let's you count the number of colours in a literal sense (given the -1/3 and 2/3 charges of each down-up-quark pair of the 3 families). The famous plot is in

http://pdg.lbl.gov/2018/hadronic-xsections/rpp2018-sigma_R_ee.pdf

Please explain the plot and/or point to an explanation. Just looking at the plot one cannot see the connection you describe.

 

[vanhees71]

The plot shows the total cross section for the annhilation of an electron and a positron to hadrons, diveded by the "elementary crossection" $\mathrm{e}^+ + \mathrm{e}^- \rightarrow \mu^+ + \mu^-$ as a function of center-momentum energy $\sqrt{s}$. The naive quarkmodel predicts
$$R=N_{\text{color}} \sum_{q=\text{active quarks}} Q_q^2.$$
Of course, from the conserved-quantum numbers of the entrance channel it's clear that you get hadrons the vector channel (and the corresponding isovector and isoscalar channels), i.e., vector meson excitations. So "active quarks" at a given $\sqrt{s}$ means the sum goes over all vector-meson channels with these quantum numbers and below this $\sqrt{s}$.

For $\sqrt{s} \leq 1.5 \; \mathrm{GeV}$ the naive expectation obviously doesn't work very well, but you see peaks of the corresponding vector mesons (around 770-780 MeV you see a broad peak of the $\rho$ meson, going down to the $2m_{\text{e}}$ threshold and a very narrow peak on top (the $\omega$ meson) and around $1 \; \text{GeV}$ the $\phi$ meson. Above this you see some $\rho'$ resonance(s) but otherwise the predicted plateau (below the $J/\psi$ peak). There the vector channels built by u, d, and strange quarks are "open" and thus the predicted value (also drawn as a horizontal line) is $N_{\text{col}} [(2/3)^2+(-1/3)^2 + (-1/3)^2]=N_{\text{col}} \cdot 2/3$. Since the value is close to 2 you can see already from this most naive quark model that $N_{\text{col}}=3$. Of course to get more accurate you need to include higher QCD corrections (indicated also in the plot). Above the very narrow $J/\psi$ peak (and below the also very narrow $\Upsilon$ peak) you see again a plateau. This adds the charm charm channel which opens at the $J/\psi$ channel. Then you've to add another $(2/3)^2$ to the bracket in the above bracket, and indeed in the plot you see the corresponding jump from the value 2 to $10/3=3.333$. The same happens when also the b-quark channel is open (indicate by the $\Upsilon$ peak), there you add another $(1/3)^2$ to the bracket, and even this jump you can see in the data enhancing $R$ to $11/3=3.667$. The next peak around $\sqrt{s}$ is of course from the (vector part of) the $\mathrm{Z}^0$ boson, one of the intermediate vector bosons of the weak interaction.

 

[mitchell porter]

confinement isn't a real physical mechanism

But you can have deconfinement at high temperatures, and then reconfinement.

 

[DarMM]

But you can have deconfinement at high temperatures, and then reconfinement.

In another thread here:
https://www.physicsforums.com/threads/how-can-quarks-exist-if-they-are-confined.958432/

I mentioned that the finite density (and I should also say high temperture) phase of QCD is probably the answer.

However I don't think what these phases mean in quantum field theory is as simple as is often thought, for they lie in a different folio to the normal QCD vacuum. This means that high temperature and high density QCD isn't just a bunch of particles sitting in the normal QCD vacuum and tightly squeezed together and having large kinetic energy.

There's no unitary (or even Louville-VonNeumann) evolution from the normal vacuum sector states to these states.

 

[A. Neumaier]

This means that high temperature and high density QCD isn't just a bunch of particles sitting in the normal QCD vacuum and tightly squeezed together and having large kinetic energy.

But it means that there is a family of theories (foilia) parameterized by temperature and density, of which the vacuum sector is just the limit of zero density and temperature. (See also here.) This family of theories is what is called QCD, and since it has quarks and gluons in certain sectors, at least as a kind of quasiparticles, one must be able to account for them somehow even in a rigorous view of QCD.

 

[DarMM]

But it means that there is a family of theories (foilia) parameterized by temperature and density, of which the vacuum sector is just the limit of zero density and temperature. (See also here.) This family of theories is what is called QCD, and since it has quarks and gluons in certain sectors, at least as a kind of quasiparticles, one must be able to account for them somehow even in a rigorous view of QCD.

Thinking a bit about this I was just wondering about its relation to the Unruh effect, perhaps somebody could tell me where I'm wrong.

Let's take an inertial observer and some test object. The inertial observer uses QCD + QED to described the object and thus that it is composed of neutrons, protons and electrons.

However a highly accelerating observer would due to the Unruh effect view the object as being in a Thermal (KMS) state at temperature $T = \frac{\hbar a}{2\pi ck_B}$. If the temperature is high enough the accelerating observer would have his state be a deconfined quark-gluon plasma state.

So whether something is made of nuclear matter or quark-gluon plasma seems to depend on the observer.

 

[A. Neumaier]

So whether something is made of nuclear matter or quark-gluon plasma seems to depend on the observer.

Yes. Observer-independent is only the field. The particle interpretation of the field is frame- and hence observer-dependent - for an abstract observer.
In practice, one cannot observe a test object highly accelerated with respect to an observer - one observes only some mean values of the part of the field it generates that is close to the observer during the observation time. Thus the relevant frame is that of the matter field comoving with the observer.

 

[DarMM]

And even when they do, those notions can be observer-dependent, as illustrated by, for example, the Unruh effect.

This one is tough for me, that what I am made of is observer dependent to some degree. Hard to understand intuitively, at least for me!

 

[PeterDonis]

what I am made of is observer dependent to some degree

No, it isn't; what you are made of is invariant. A description of what you are made of in terms of "particles" might not be--but that's not a problem with what you are made of, it's a problem with thinking that a description in terms of "particles" has to be invariant, when in fact it doesn't.

 

[DarMM]

No, it isn't; what you are made of is invariant

And what's that if not particles?

[Moderator's note: rest of post removed as it has now been moved to this thread.]

 

[PeterDonis]

And what's that if not particles?

Short answer? Quantum fields.

 

[DarMM]

Just to say, this is more to tease this out, I do think (and hope for my intuitions sake!) that you are correct that the fundamental "stuff" things are made of is invariant and it is simply the particle description that is variant. This is more just a question of if that stuff is quantum fields or are quantum fields still just observables with no stronger a claim to be the constituents of matter than the particle observables.

Also of course QFT might be ultimately wrong and there might be another layer beneath it, but let's assume it's as correct as it seems for now.

Short answer? Quantum fields.

So to sketch this out. Would you say for example that the fundamental constituents of matter involve the pre-symmetry breaking electroweak fields or the post-symmetry breaking electromagnetic and $W^{\pm}$ and $Z^{0}$ fields?

 

[PeterDonis]

Would you say for example that the fundamental constituents of matter involve the pre-symmetry breaking electroweak fields or the post-symmetry breaking electromagnetic and $W^{\pm}$ and $Z^{0}$ fields?

Yes.

These are not different sets of fields. They are different descriptions in terms of fields. I look at it the same as a change of inertial frames in SR: you're describing the same underlying thing, just in different coordinates. Similarly, symmetry breaking doesn't change the underlying thing, but it does change which description of it is most useful.