I What Are the Empirical Challenges Facing Quantum Gravity Theories?

[ohwilleke]

TL;DR Summary

One of the leading researchers in Loop Quantum Gravity discusses existing observations that should guide future quantum gravity research in a three page paper.

Rovelli points to three pieces of existing observational evidence that should guide future quantum gravity research. Bottom line:

* abandon Lorentz invariance violating quantum gravity theories,
* abandon supergravity and string theory,
and
* stop working on the anti-deSitter/conformal field theory relationship in gravity and cosmology work (leave it to the condensed matter physicists who have legitimate uses for it).

I tend to agree but would welcome other views.

What other experimental data discredit large volumes of research papers?

Stop working on models of dark matter that predict NFW dark matter particle distributions would be a good start as they have repeatedly and convincingly been shown to be at odds with the observational data. If your theory predicts this, it is wrong. See, e.g., Pengfei Li, Federico Lelli, Stacy McGaugh, James Schombert, "A comprehensive catalog of dark matter halo models for SPARC galaxies" (January 28, 2020). arXiv 2001.10538; Marie Korsaga, et al., "GHASP: an Hα kinematics survey of spiral galaxies - XII. Distribution of luminous and dark matter in spiral and irregular nearby galaxies using Rc-band photometry" (September 17, 2018) Kyriakos Grammatikos, Vasiliki Pavlidou, "Getting the tiger by the tail: Probing the turnaround radius of structures with outer halo density profiles" (September 17, 2018); Antonino Del Popolo et al., "Correlations between the Dark Matter and Baryonic Properties of CLASH Galaxy Clusters" (August 6, 2018), Lin Wang, Da-Ming Chen, Ran Li "The total density profile of DM halos fitted from strong lensing" (July 31, 2017); here (2017), and here (2016), and here (2011). Since an NFW flows analytically (almost trivially) from a simple one component collisionless dark matter particle model, this means that reality is not well described by a simple one component collisionless dark matter particle model.

A. Lorentz Invariance

The breaking of Lorentz invariance at the Planck scale may simplify the construction of a quantum theory of gravity. This observation sparked a large theoretical enthusiasm for Lorentz-breaking theories some time ago, and rightly so. But that bubble of enthusiasm has been deflated by empirical observations. A large campaign of astrophysical observations has failed to reveal the Planck-scale breaking of the Lorentz invariance in situations where it would have been expected if this track for understanding quantum gravity had been the good one.

A methodological consideration is important at this point. Popperian falsifiability is an important demarcation criterium for scientific theories (that is, if a theory is not falsifiable, we better not call it “science”); however, Popperian falsification is rarely the way theories gain or lose credibility in science.

The way scientific theories gain or loose credibility in real science is rather through a Bayesian gradual increase or decrease of the positive or negative confirmation from empirical data. That is, when a theory predicts a novel phenomenon and we this to be right, our confidence in the theory grows; when it predicts a novel phenomenon and we do not find it, our confidence in the theory decreases. Failed predictions rarely definitely kill a theory, because theoreticians are very good at patching up and adjusting. But failed predictions do make the success of a research program far less probable: we loose confidence in it.

Hence, this has been the effect of not finding Lorentz violations in astrophysics: tentative quantum gravity theories that break Lorentz invariance might perhaps still be viable in principle, but in practice, far fewer people bet on them.

B. Supersymmetry

What I wrote above is particularly relevant to the spectacular non-discovery of supersymmetry at the LHC.

While in the Popperian sense, the non-appearance of supersymmetric particles at the TeV scale does not rule out all the theories based on supersymmetry, including string theory, in practice, the strong disappointment of not finding what was expected counts heavily as a strong dis-confirmation, in the Bayesian sense, of all those theories.

People have written that the non-discovery of supersymmetry is a crisis for theoretical physics. This is nonsense, of course. It is only a crisis for those who bet on supersymmetry and string theory. For all the alternative theoretical quantum gravity programs that were never convinced by the arguments for low-energy supersymmetry, the non-discovery of supersymmetry is not a crisis: it is a victory.

Precisely for the same reason that the discovery of supersymmetry would have been a confirmation of the ideas supporting the string supersymmetry research direction, the non-discovery of supersymmetry at the LHC is a strong empirical indication against the search for quantum gravity in the direction of supersymmetric theories and strings.

Nature talks, and we better listen.

C. Cosmological Constant

A case similar to the one above but even stronger concerns the sign of the cosmological constant. The cosmological constant is a fundamental constant of nature, part of the Einstein equations (since 1917), whose value had not been measured until recently. An entire research community has long worked, and is still working, under general hypotheses that lead to the expectation for the sign of the cosmological constant to be negative. Even today, the vast majority of the theoretical work in that community assume it to be so.

Except that the sign of the cosmological constant is not negative. It is positive, as observation has convincingly shown.

Once again, this counts as a strong dis-confirmation of the hypotheses on which a large community has worked in the past, and is still working on today.

So far, we lack any direct evidence of a quantum gravitational phenomenon; however, the non-detection of Lorentz violations around the Planck scale, the nondiscovery of super symmetric particles at the LHC, and the measurement of a positive cosmological constant are strong indications from Nature that disfavor the tentative quantum gravity theories that naturally imply these phenomena.

[Submitted on 15 Nov 2021]

Considerations on Quantum Gravity Phenomenology​

Carlo Rovelli

I describe two phenomenological windows on quantum gravity that seem promising to me. I argue that we already have important empirical inputs that should orient research in quantum gravity.

Comments:	3 Pages
Subjects:	General Relativity and Quantum Cosmology (gr-qc); High Energy Physics - Theory (hep-th)
Journal reference:	Universe 2021, 7(11), 439
Cite as:	arXiv:2111.07828 [gr-qc]

 

[MathematicalPhysicist]

*abandon LQG...

 

[Fra]

I suspect the take on the strategy and guidance here, depends if one tries to see the unification of gravity and QM, as topic concerning trying to understand the "quantisation procedure" for gravity separate from the general unification of forces or if they are deeply entangled.

If one takes the latter view, dualities between very different theories (which need not be specifically AdS/CFT, which is just a major know explict case, it doesn't represent the general case even in conceptual sense) seems still like a highly relevant area, that is much more than just about the sign of the cosmological constant. Thinking in this direction does not necessarily mean advocating ST.

/Fredrik

 

[MathematicalPhysicist]

As far as I can see you cannot refute Superstring theory, LQG or Susy or Supergravity.
You can always say we need a bigger particle collider, larger energies.
If the models or theories don't have in them a built-in prediction of how much energies are needed, then they cannot be refuted.

 

[martinbn]

As far as I can see you cannot refute Superstring theory, LQG or Susy or Supergravity.
You can always say we need a bigger particle collider, larger energies.
If the models or theories don't have in them a built-in prediction of how much energies are needed, then they cannot be refuted.

That is in the article. His point is that even though you cannot disprove you can decrease/increase your confidence in a research approach. And that experiments and observations already show which programs are less promising.

 

[MathematicalPhysicist]

That is in the article. His point is that even though you cannot disprove you can decrease/increase your confidence in a research approach. And that experiments and observations already show which programs are less promising.

As I see it there is no difference between LQG, Superstring, etc.
Research in these will cease when the funding will dry out.

 

[martinbn]

As I see it there is no difference between LQG, Superstring, etc.
Research in these will cease when the funding will dry out.

Rovelli has a different view.

How can the funding dry out?! If you are hired at a university/institute you have your salary and you can do research. What other funding do you need?

 

[MathematicalPhysicist]

Rovelli has a different view.

How can the funding dry out?! If you are hired at a university/institute you have your salary and you can do research. What other funding do you need?

Your salary is composed of grants and teaching.
The committees that decide where the money goes might decide not to invest on these avenues.

 

[weirdoguy]

Your salary is composed of grants and teaching.

That is not true, there are scientists that don't teach and don't have grants but just normal salary for their scientific work. You tend to have a lot of strong but not really founded opinions...

 

[MathematicalPhysicist]

That is not true, there are scientists that don't teach and don't have grants but just normal salary for their scientific work. You tend to have a lot of strong but not really founded opinions...

So they can work on whatever they want to work on, without any need for writing proposals for their research?
I am skeptical...

 

[Demystifier]

So they can work on whatever they want to work on, without any need for writing proposals for their research?

Yes. For example, me.

 

[Demystifier]

As I see it there is no difference between LQG, Superstring, etc.
Research in these will cease when the funding will dry out.

By that criterion, do you see a difference between LQG and history of Roman Empire?

 

[weirdoguy]

I am skeptical...

What's funny is that you are skeptical on something that is a standard practice. Grants are relatively rare and really hard to get. Do you really think that each and every arXiv preprint is connected with some grant?

 

[MathematicalPhysicist]

Yes. For example, me.

Who pays for your work?

 

[MathematicalPhysicist]

By that criterion, do you see a difference between LQG and history of Roman Empire?

I guess both don't get funded.

 

[MathematicalPhysicist]

What's funny is that you are skeptical on something that is a standard practice. Grants are relatively rare and really hard to get. Do you really think that each and every arXiv preprint is connected with some grant?

Grant or some scholarship.

I know that for postdoc and phd you need to apply for some scholarships to fund your research.
So do you say that most professor's research isn't funded?
So what makes up their salary?

 

[Fra]

Having noting todo with the research politics but:

Rovelli wrote:

"As always in science, a priori everything is possible,but there is a profound difference between an implausible wild speculation and the predictions of a plausible, coherent framework. This is a distinction a bit too much disregarded in today’s fundamental physics, in my opinion."

This is true, but a problem is still that what IS a "plausible, coherent framework", is subjective so it seems still like am empty statement in the context. What appears to be a wild speculation from one perspective, may be the good inference from another perspective, because one has chose different fundamental starting points. There is no conflict in this, as there is no unique way to extrapolate known facts to guesses of the unknown. Only the future will tell.

As a lot of theoretical research on physics is not about producing explicit phenomenological predictions, but about playing around in a theoryspace, which is defined and constrained in different ways depending on what paradigm one is using. The hope is that one day something nice will come out. So as I see, focus is more on which "theory toolbox" is likely to be the most efficient and successful one, in eventually either making explicit phenomenolgical predictions (that are DOABLE) or making process by increaseing explanatory power (which I see as the prime mission) in the sense of for example, reudcing the number of free parameters, and providing a conceptual framework for navigating in theory space. There will never be an consensus agreement on this among scienticst, and there need not be. This is just the way it is, it is not a foundational problem per see.

Lets say we one does not like Strings, not LQG, then what is the alternative? Shall we just stop wasting time on anything that is not explicit phenomenology?

From a pragmatic perspective, and short time scale resource planning, it may seen like an option. But there is also a problem with such an approch, that "science" risks beeing just a big "statistics", with less and less explanatory power. I think what drives some of us, is not just about collecting statistics about the past, but to gain deeper insight in how things are causally related. This is for me what foundations of physics is about, not just "applying" the mature theory to experiments.

/Fredrik

 

[Demystifier]

Who pays for your work?

Government. (Which gets money from taxpayers.)

 

[Demystifier]

I know that for postdoc and phd you need to apply for some scholarships to fund your research.

It depends on the country (among other things). Where do you live? I suspect in US, because it's typical for US citizens to assume that the whole world works the same way as their country does. (Of course, not all US citizens think that way, but in my experience many do.)

 

[Demystifier]

So do you say that most professor's research isn't funded?
So what makes up their salary?

The primary job of a professor is teaching. But leaving that aside, getting a salary for research doesn't imply that you have to tell in advance what will you study and what are your expected results. In fact, in my opinion (with which not everybody will agree), it is contrary to the spirit, the idea and the purpose of fundamental research to make a specific research proposal before actually doing the research. If you know in advance what your results will be, then maybe you are doing straightforward stuff which will not produce a true progress. As Einstein said, if we knew what it was we were doing, it wouldn't be called research, would it?

 

[weirdoguy]

So what makes up their salary?

University just pays them for doing research and writing papers on that. That's the way it is in Poland. Grants are relatively rare. The same goes with PhD, one doesn't need to apply for any funding.

 

[MathematicalPhysicist]

It depends on the country (among other things). Where do you live? I suspect in US, because it's typical for US citizens to assume that the whole world works the same way as their country does. (Of course, not all US citizens think that way, but in my experience many do.)

No, I don't live in the USA.
I got this wrong impression from google search.
I did pursue two MScs one in maths and the other in physics but didn't complete them (the average grade in the courses for maths was 95 and for the physics was 75, I didn't complete my thesis writing component in the required alleged time).
I didn't need to apply for scholarships, but I did work as a grader (a work which I still have).

But for PhD I guess from what I had seen, you need to write research proposals, and to explain your progress in detailed reports that's at least how it looks in the University which I did my two Msc above.

 

[MathematicalPhysicist]

But I agree, if you know what your results will be then it cannot be called research.
But nowadays, in the capitalist's eyes do people still conduct such research as Albert Einstein is quoted?

 

[Fra]

But nowadays, in the capitalist's eyes do people still conduct such research as Albert Einstein is quoted?

To a certain extent, there would be a compromise. Either one have to accept some bias over the research direction, in order to get payed (for example if your goal is to become a professional researcher), or one has to accept to not get payed and do it on your free time (if the goal is to ansewr your own questions). But in the latter case the compromise is still that one has less time to spend. I suspect that those people that backed up financially and are free to do whatever they want are rare.

I made the latter choice long time ago. Had I aligned and followed advice of supervisors at the time I should have pursued string theory as that is where the "opportunities" were.

/Fredrik

 

[Demystifier]

But nowadays, in the capitalist's eyes do people still conduct such research as Albert Einstein is quoted?

Those who pay for the research must have a lot of money, and those who have a lot of money tend to see everything from the capitalist point of view. Fundamental research is just one of the victims of that. The result is that grant proposals for fundamental research look like business proposals. To get a grant, a scientist must write a lot of bull...t that makes little sense from a scientific point of view. It is supposed to make sense to bureaucrats who make decisions about grants, but I doubt that it makes sense to them either. The way how scientists actually conduct research has little to do with what they write in the grant proposals.

 

[mitchell porter]

I just went and looked at the five most recent papers on hep-th. All of them acknowledged support from some kind of grant, in countries as different as US, Japan, China, various EU, and Chile.

 

[maline]

Summary:: One of the leading researchers in Loop Quantum Gravity discusses existing observations that should guide future quantum gravity research in a three page paper.

stop working on the anti-deSitter/conformal field theory relationship in gravity and cosmology work

I don't see that he's saying that. AdS/CFT never made any claim that our universe should be AdS. Their claim is that the duality defines a quantum gravity theory, which happens to live in AdS. Hopefully, gravity in our world works more-or-less similarly to how it works in AdS, at least as far as the high-energy stuff is concerned; so finding a full quantum gravity for AdS would be a great and very relevant achievement.

I understood Rovelli's remark to be referring to the "swampland" problem, that string theories with Calabi-Yau compactifications seem to predict a negative cosmological constant.

 

[Demystifier]

Hopefully, gravity in our world works more-or-less similarly to how it works in AdS

Yes. AdS is somewhat like a harmonic oscillator; our world is not exactly a harmonic oscillator, but understanding of the harmonic oscillator helps a lot to understand the real world as well.

 

[martinbn]

Yes. AdS is somewhat like a harmonic oscillator; our world is not exactly a harmonic oscillator, but understanding of the harmonic oscillator helps a lot to understand the real world as well.

Is it like the harmonic oscillator or is it hoped to be?

 

[Demystifier]

Is it like the harmonic oscillator or is it hoped to be?

Good point, it's more like a hope.

 

[Nullstein]

So is Rovelli arguing against LQG now? As far as I'm concerned, it isn't locally Lorentz-invariant either, due to the inevitable singular excitations of geometry, at least at sufficiently small scales.

 

[martinbn]

So is Rovelli arguing against LQG now? As far as I'm concerned, it isn't locally Lorentz-invariant either, due to the inevitable singular excitations of geometry, at least at sufficiently small scales.

How so!?

 

[Nullstein]

How so!?

Local Lorentz-invariance means that spacetime is locally Minkowski, i.e. given any point in spacetime and any desired accuracy, there must be a small neighborhood around that point such that this neighborhood is isometric to a region of Minkowski spacetime, at least up to the specified accuracy. However, in all LQG models, geometry is defined by spin networks or spin foams. These are lattice-like structures embedded into the spacetime manifold. If you choose a point directly on a spin network, you will find its geometry to be excited. But if you pick any neighborhood of that point and any point within that neighborhood (other than the ones that lie on the lattice), the geometry around this other point will not be excited. There is a discontinuous jump. So there is no neighborhood around the points that lie on a spin network/foam that are isometric to a region of Minkowski spacetime. Minkowski spacetime is translation-invariant. You would have to be able to shift one point onto each other and still have the same geometry. But that's impossible due to the discontinuity.

 

[Fra]

I see a conceptual difference in violating a continuous symmetry due to that it is in fact violated(and spacetime still is a justified continuum), or just because the continuum model may not be physically justified. Then it seems that, at before point where the symmetry is "violated", the set which is subject to the symmetry is no longer well defined. I think is may be more of a problem for the continuum model itself.

/Fredrik

 

[martinbn]

Local Lorentz-invariance means that spacetime is locally Minkowski, i.e. given any point in spacetime and any desired accuracy, there must be a small neighborhood around that point such that this neighborhood is isometric to a region of Minkowski spacetime, at least up to the specified accuracy.

This of course is not true. The neighborhood will be homeomerphic not isometric to a region in Minkowski.

However, in all LQG models, geometry is defined by spin networks or spin foams. These are lattice-like structures embedded into the spacetime manifold. If you choose a point directly on a spin network, you will find its geometry to be excited. But if you pick any neighborhood of that point and any point within that neighborhood (other than the ones that lie on the lattice), the geometry around this other point will not be excited. There is a discontinuous jump. So there is no neighborhood around the points that lie on a spin network/foam that are isometric to a region of Minkowski spacetime. Minkowski spacetime is translation-invariant. You would have to be able to shift one point onto each other and still have the same geometry. But that's impossible due to the discontinuity.

I know very little about this, so I cannot really respond, but I think that your description is not right (of course I am probably wrong). Can you point to a source about this?

 

[Nullstein]

This of course is not true. The neighborhood will be homeomerphic not isometric to a region in Minkowski.

No, it must be isometric, i.e. not only topologically isomorphic, but the metric must be locally Minkowski. You need a homeomorphism which maps the local metric onto a Minkowski metric (up the the desired accuracy), hence an isometry.

I know very little about this, so I cannot really respond, but I think that your description is not right (of course I am probably wrong). Can you point to a source about this?

You can check Thiemann's book for example. Any neighborhood of an at least trivalent vertex of a spin network has a finite volume, whereas any small enough neighborhood of a point that doesn't lie on the spin network, has zero volume. Hence, the whole volume is concentrated on the points that lie on the spin network.

Here's another way to look at it: Loop quantum gravity models are lattice models, just like lattice QCD. Sure, LQG gives to the freedom to choose the lattices as fine as you want to and you're not limited to regular lattices and can pick arbitrary graphs, but after you have made the choice, they remains lattices. Hence, some continuous symmetries such as local translation invariance are broken, just like in lattice QCD. The hope is that they are restored on macroscopic scales where the granularity of the lattice becomes irrelevant.

 

[Demystifier]

So is Rovelli arguing against LQG now? As far as I'm concerned, it isn't locally Lorentz-invariant either, due to the inevitable singular excitations of geometry, at least at sufficiently small scales.

Local Lorentz-invariance means that spacetime is locally Minkowski, i.e. given any point in spacetime and any desired accuracy, there must be a small neighborhood around that point such that this neighborhood is isometric to a region of Minkowski spacetime, at least up to the specified accuracy. However, in all LQG models, geometry is defined by spin networks or spin foams. These are lattice-like structures embedded into the spacetime manifold. If you choose a point directly on a spin network, you will find its geometry to be excited. But if you pick any neighborhood of that point and any point within that neighborhood (other than the ones that lie on the lattice), the geometry around this other point will not be excited. There is a discontinuous jump. So there is no neighborhood around the points that lie on a spin network/foam that are isometric to a region of Minkowski spacetime. Minkowski spacetime is translation-invariant. You would have to be able to shift one point onto each other and still have the same geometry. But that's impossible due to the discontinuity.

Here is what Rovelli says about this, in his book "Quantum Gravity":

 

[Nullstein]

It seems like he admits that Lorentz-violation effects might be present. In fact, I'm not arguing against his second argument, which I agree with. I'm arguing that "the short-scale structure of a macroscopically Lorentz-invariant weave might break Lorentz-invariance" as he puts it. He then argues that superpositions might help, but he doesn't provide a proof and it is easy to see that it isn't true: In the LQG Hilbert space, there is an uncountable number of orthogonal spin network states, but at most a countably infinite subset of them is allowed in a superposition. If the state is supposed to be Lorentz-invariant, it is necessary that each spin network in the superposition must be mapped onto another spin network that was already in the superposition). However, there is a continuum of Lorentz transformations and they generate a continuum of new spin network states when acted on a spin network state. Since only countable sums of spin network states are allowed in the LQG Hilbert space, not all of those transformed spin networks can be present in the sum, so there exist Lorentz transformations that don't leave the state invariant.

 

[martinbn]

No, it must be isometric, i.e. not only topologically isomorphic, but the metric must be locally Minkowski. You need a homeomorphism which maps the local metric onto a Minkowski metric (up the the desired accuracy), hence an isometry.

This is not the case. Otherwise the manifold will have zero curvature, and definitely not all have zero curvature. In fact "most" often it is not the case.

 

[martinbn]

It seems like he admits that Lorentz-violation effects might be present. In fact, I'm not arguing against his second argument, which I agree with. I'm arguing that "the short-scale structure of a macroscopically Lorentz-invariant weave might break Lorentz-invariance" as he puts it. He then argues that superpositions might help, but he doesn't provide a proof and it is easy to see that it isn't true: In the LQG Hilbert space, there is an uncountable number of orthogonal spin network states, but at most a countably infinite subset of them is allowed in a superposition. If the state is supposed to be Lorentz-invariant, it is necessary that each spin network in the superposition must be mapped onto another spin network that was already in the superposition). However, there is a continuum of Lorentz transformations and they generate a continuum of new spin network states when acted on a spin network state. Since only countable sums of spin network states are allowed in the LQG Hilbert space, not all of those transformed spin networks can be present in the sum, so there exist Lorentz transformations that don't leave the state invariant.

What happens in the case of QFT with a seperable Hilbert space? Are the states invariant? There are still a continuum of boosts.

 

[Nullstein]

This is not the case. Otherwise the manifold will have zero curvature, and definitely not all have zero curvature. In fact "most" often it is not the case.

Which is why I wrote "up to the desired accuracy." If you choose pick a desired $\epsilon$, there must be neighborhood such that there exists a homeomorphism that is an isometry up to the given accuracy $\epsilon$.

What happens in the case of QFT with a seperable Hilbert space? Are the states invariant? There are still a continuum of boosts.

The QFT Hilbert space is separable. All states can be expanded in a countable basis, so if you apply a boost to a sum of basis vectors, you can expand the resulting state in the same basis. That's no longer true in LQG. The set of spin network states forms a basis, but it is uncountable. A transformed spin network is orthogonal to the original spin network even if you change it only a tiny bit. There are no infinitesimal diffeomorphisms in LQG and the finite ones are discontinuous.

 

[Demystifier]

The set of spin network states forms a basis, but it is uncountable.

Rovelli and Vidotto in the book "Covariant Loop Quantum Gravity" say the opposite:

 

[Nullstein]

Rovelli and Vidotto in the book "Covariant Loop Quantum Gravity" say the opposite:
View attachment 293072View attachment 293073

Rovelli and Vidotto is about a different model than the one discussed in Rovelli or Thiemann. The model in Rovelli or Thiemann definitely has an uncountable basis. An uncountable basis is necessary if all diffeomorphisms are to be implemented as unitary transformations.

 

[Demystifier]

The model in Rovelli or Thiemann definitely has an uncountable basis.

Can you support this claim by a reference?

An uncountable basis is necessary if all diffeomorphisms are to be implemented as unitary transformations.

Why would diffeomorfisms would be implemented as unitary transformations in the first place? Shouldn't diffeomorphism equivalent states be counted as the same state in the Hilbert space?

 

[Nullstein]

Can you support this claim by a reference?

See Thiemann, p. 241: "We remark that the spin-network basis is not countable because the set of graphs in $\sigma$ is not countable, whence $\mathcal H_0$ is not separable."

Why would diffeomorfisms would be implemented as unitary transformations in the first place? Shouldn't diffeomorphism equivalent states be counted as the same state in the Hilbert space?

You first need to construct the kinematical Hilbert space, where the constraints are represented. The constraint algebra of GR contains the infinitesimal diffeomorphisms as subalgebra, so there must be a representation of the diffeomorphism group on the kinematical Hilbert space. In fact, in LQG, there is only a representation of the group and not the algebra, because LQG uses a discontinuous representation. No infinitesimal generators exist. After solving the diffeomorphism constraints, you end up on a diffeomorphism invariant Hilbert space, but it is still non-separable (see Thiemann as well).

Also check out Ashtekar, Lewandowski, "Background Independent Quantum Gravity: A Status Report":
"Note that there are continuous families of 4 or higher valent graphs which can not be mapped to one another by C n diffeomorphisms with n > 0. Consequently, states in Hdiff based on two of these graphs are mutually orthogonal. Thus, even though we have ‘factored out’ by a very large group Diff, the Hilbert space Hdiff is still non-separable."

 

[Fra]

I'm not a dedicated fan of LQG anyway, but I noted this paper as well which seems far more recent than then Ashtekar/Lewandowski review from 2004? Any takes on this?

Separable Hilbert space in Loop Quantum Gravity
Winston Fairbairn, Carlo Rovelli, Oct 25, 2018

"...In the standard construction, the kinematical Hilbert space of the diffeomorphism-invariant states is nonseparable. This is a consequence of the fact that the knot-space of the equivalence classes of graphs under diffeomorphisms is noncountable. However, the continuous moduli labeling these classes do not appear to affect the physics of the theory. We investigate the possibility that these moduli could be only the consequence of a poor choice in the fine-tuning of the mathematical setting. We show that by simply choosing a minor extension of the functional class of the classical fields and coordinates, the moduli disappear, the knot classes become countable, and the kinematical Hilbert space of loop quantum gravity becomes separable."

-- https://arxiv.org/abs/gr-qc/0403047

/Fredrik

 

[Nullstein]

I'm not a dedicated fan of LQG anyway, but I noted this paper as well which seems far more recent than then Ashtekar/Lewandowski review from 2004? Any takes on this?

Your paper is also from 2004.

Separable Hilbert space in Loop Quantum Gravity
Winston Fairbairn, Carlo Rovelli, Oct 25, 2018

"...In the standard construction, the kinematical Hilbert space of the diffeomorphism-invariant states is nonseparable. This is a consequence of the fact that the knot-space of the equivalence classes of graphs under diffeomorphisms is noncountable. However, the continuous moduli labeling these classes do not appear to affect the physics of the theory. We investigate the possibility that these moduli could be only the consequence of a poor choice in the fine-tuning of the mathematical setting. We show that by simply choosing a minor extension of the functional class of the classical fields and coordinates, the moduli disappear, the knot classes become countable, and the kinematical Hilbert space of loop quantum gravity becomes separable."

In this papers, they enlarge the $Diff(\sigma)$ symmetry subgroup of GR (the full group also includes the symmetries generated by the Hamiltonian constraint) to include more symmetries. If you require invariance under a larger symmetry group, the space of solutions may become smaller and even separable. However, there is no justification for the enlarged symmetry subgroup. $Diff(\sigma)$ arises as the symmetry subgroup of classical GR during the Dirac constraint analysis, but the extended group mentioned in that paper does not. Sure, if you give yourself the freedom to extend the symmetry group during quantization, you can shrink to Hilbert space and even make it zero-dimensional if you want to. But there is no justification for enlarging the symmetry subgroup.

Moreover, making the Hilbert space separable doesn't even help. Separability is only necessary for local Lorentz-invariance, but not sufficient. The general argument still holds, because it just so happens that the geometry in LQG is concentrated on singular objects. Being able to expand the transformed state in the same basis as the original state doesn't automatically lead to the equality of the two states, it's just a necessary condition for equality. So even if one were able to restrict the state space to a separable subspace of $\mathcal H_{diff}$, one would then still need to prove local Lorentz-invariance.

To the best of my knowledge, the closest that LQG people have gotten to implementing Lorentz-invariance in a variant of LQG is to formulate a theory with $SL(2,\mathbb C)$ symmetry and implementing Lorentz-invariance at the vertices of a spin foam. However, it doesn't even make sense to apply Lorentz transformations to points that are not vertices, so the crucial feature, i.e. having a whole local neighborhood to resemble a region of Minkowski spacetime, has not been achieved.

 

[Demystifier]

@Nullstein Generally I find very interesting that LQG is quite different from QFT. I vaguely remember that somewhere I saw a statement that when one applies the LQG method of quantization to a simple harmonic oscillator, one obtains a result not equivalent to harmonic oscillator quantized by standard techniques. Do you perhaps know a more precise statement and the relevant reference?

 

[Nullstein]

@Nullstein Generally I find very interesting that LQG is quite different from QFT. I vaguely remember that somewhere I saw a statement that when one applies the LQG method of quantization to a simple harmonic oscillator, one obtains a result not equivalent to harmonic oscillator quantized by standard techniques. Do you perhaps know a more precise statement and the relevant reference?

Yes, quantization works very differently in LQG. The position eigenstates $\left|x\right>$ are normalizable and orthogonal ($\left<x|x'\right>=\delta_{xx'}$) as compared to ordinary quantum mechanics, where you have $\left<x|x'\right>=\delta(x-x')$ ($\delta_{xx'}$ is the Kronecker delta instead of the Dirac delta). While it is still possible to define the position operator ($\hat x\left|x\right>=x\left|x\right>$), it becomes impossible to define the momentum operator in such a way that the canonical commutation relations are obeyed. Instead, one quantizes the exponentiated momentum operators ($e^{i \epsilon p} \rightarrow \widehat{e^{i \epsilon p}}$) and they still acts as a translation operators: $\widehat{e^{i \epsilon p}}\left|x\right> = \left|x-\epsilon\right>$ However, in this Hilbert space, $\left|x-\epsilon\right>$ is not in any sense close to $\left|x\right>$. Hence, it becomes impossible to calculate the derivative of $\widehat{e^{i \epsilon p}}$ at $\epsilon=0$ and so the infinitesimal generator $\hat p$ doesn't exist. So how are you going to define the Hamiltonian of the free particle or of the harmonic oscillator? The LQG people just define $\hat p_{LQG} = \frac{1}{2i\epsilon}\left(\widehat{e^{i \epsilon p}}-\widehat{e^{-i \epsilon p}}\right)$ for some sufficiently small, but non-zero $\epsilon$, which thus becomes a free parameter of the theory. The Hamiltonian of the harmonic oscillator is then defined as $\hat H = \frac{1}{2}\hat p_{LQG}^2 + \frac{1}{2}x^2$ and naturally, you get deviations from the standard theory if you do this.

For a reference, you can check out Nicolai et al. "LQG: an outside view"

 

[Demystifier]

Thanks @Nullstein! Do you also have a reference for more details?