A Structure of Matter in Quantum Field Theory

[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.

 

[DarMM]

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.

Fascinating. And what would you think of the fact that fields don't have well defined values at a point, i.e. $\phi(x)$ is undefined? That the fundamental things are the smeared fields, $\phi(f) = \int_{\mathcal{M}}{\phi(x)f(x)d^{4}x}$?

 

[PeterDonis]

what would you think of the fact that fields don't have well defined values at a point, i.e. $\phi(x)$ is undefined? That the fundamental things are the smeared fields, $\phi(f) = \int_{\mathcal{M}}{\phi(x)f(x)d^{4}x}$?

Well, since you and I are not point particles, I don't see the problem.

Seriously, since we never make measurements of anything at an exact point, I don't see the problem. "Smeared fields" seems like a better description of what we actually measure anyway.

 

[DarMM]

Well, since you and I are not point particles, I don't see the problem.

Seriously, since we never make measurements of anything at an exact point, I don't see the problem. "Smeared fields" seems like a better description of what we actually measure anyway.

Yeah I agree. This would tend to suggest that the fundamental object is really the local observable algebra and that any choice of fields is a particular basis/decomposition of it. That genuinely is invariant and carries the representations of symmetries we expect and is detached from the observer dependent notion of a Hilbert space.

Of course one could ask is this really what I'm made of, or if it simply defines a set of admissible and complementary classical descriptions of what I'm made of (i.e. the old Copenhagen "QM doesn't tell you anything about reality" that you also see in the Consistent histories view). However that's going into interpretations so I won't bother.

Regardless the fundamental thing is the sheaf of local observables.

 

[PeterDonis]

This would tend to suggest that the fundamental object is really the local observable algebra and that any choice of fields is a particular basis/decomposition of it.

I would agree.

 

[A. Neumaier]

And what's that if not particles?

We are made of matter described by states of a quantum field theory. Just as in classical mechanics we were supposed to be made of particles described by phase space coordinates.

 

[Peter Morgan]

And what would you think of the fact that fields don't have well defined values at a point, i.e. $\phi(x)$ is undefined? That the fundamental things are the smeared fields, $\int_{\mathcal{M}}{\phi(x)f(x)d^{4}x}$?

Those $\phi$'s really need an indication that they're operators, indexed by a set of test functions, $\hat\phi_f$, which, taken independently of any other observables we can take to represent a random variable, with the vacuum state providing a probability density. We can't measure $\hat\phi_f$ at a point, for $f$ a delta function, insofar as the variance is infinite (or, better, undefined, as you say) even for the free field, but, thinking very loosely, for any finite region an infinite sum of infinite variance random fields can perfectly well be finite and finite variance.
I find it helpful to use signal analysis language, with the test functions performing a function very close to that performed by "window functions" (signal analysis), Chris Fewster calls them "sampling functions" (and perhaps others do, but I haven't seen it from others). In signal analysis terms, we can think of $\hat\phi_g|0\rangle$ as a multiplicative modulation of the vacuum state, so that when a test function is used in this way it's appropriate to call a test function a "modulation function" (so signal analysis again).
In signal analysis terms, there's no a priori reason to think that the (pre-)inner product $\langle 0|\hat\phi_f^\dagger\hat\phi_g|0\rangle$ has to be a linear functional of $f$ and $g$ at all length scales and at all amplitudes, provided it's complex-linear in $\langle 0|\hat\phi_f^\dagger$ and $\hat\phi_g|0\rangle$, indeed our usual experience of nonlinearity in signal analysis suggests that we should expect it not to be, and yet the Wightman axioms insists it must be (for no physically justified principle), and the Haag-Kastler axioms, to approximately the same effect, insist on Additivity. Loosening this axiom results in a plethora of (what I find) interesting nonlinear models, as a result of which we can naturally construct multi-point operators (by polarization) that can be used to represent bound states (the account I've given here is obviously much too fast: a development that is as good as I could manage a few years ago can be found in arXiv:1507.08299, still very early days yet). FWIW, I see a connection between this account and the discussion in this thread of bound systems, with apparently no resolution, whereas for me this kind of approach offers at the least some possibilities — which, moreover, are moderately principled and empirically grounded in signal analysis concerns.