What Are the Empirical Challenges Facing Quantum Gravity Theories?

[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?

 

[Nullstein]

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

You're welcome! You can check out the reference above, but the case of the harmonic oscillator is also discussed here.

 

[Fra]

Your paper is also from 2004.

You seem right, I focused on the content first. I don't know however, why a paper that looks published 2004, has the date 2018 written inside it? A typo? Look at the first page.

/Fredrik

 

[gentzen]

You seem right, I focused on the content first. I don't know however, why a paper that looks published 2004, has the date 2018 written inside it? A typo? Look at the first page.

No, not a typo, just a typical artifact of how arXiv compiles and caches tex. You can find similar artifact in thousands of other papers on arXiv. Try not to make any unwarranted conclusions from it.

 

[Demystifier]

only countable sums of spin network states are allowed in the LQG Hilbert space

Why? In ordinary QM a wave packet such as a Gaussian is a superposition of uncountably many plane waves, so why isn't something similar allowed in LQG?

If your reply is that plane waves are not in a Hilbert space, then I note that they are in a rigged Hilbert space. In that case, can a rigged Hilbert space be constructed for LQG and can it solve the problem you are referring to?

 

[Nullstein]

Why? In ordinary QM a wave packet such as a Gaussian is a superposition of uncountably many plane waves, so why isn't something similar allowed in LQG?

If your reply is that plane waves are not in a Hilbert space, then I note that they are in a rigged Hilbert space. In that case, can a rigged Hilbert space be constructed for LQG and can it solve the problem you are referring to?

Riggend Hilbert spaces are used in LQG as well, but this is a different situation. You can use plane waves in QM and they are elements of the rigged Hilbert space, but in order to form a physical state, you have to integrate them over all $k$, rather than take an uncountable sum over all $k$ (that's not possible). On the other hand, you have, e.g., the harmonic oscillator eigenstates, labeled by natural numbers $n$, which form a basis of the actual Hilbert space and in order to form physical states, you take (countable) sums over all $n$. In the case of harmonic oscillator eigenstates, it doesn't make sense to integrate over $n$. Now if you define a wave packet as an integral over uncountably many plane waves from the rigged Hilbert space, then the resulting state can still be expanded into countably many harmonic oscillator eigenstates.

The spin network states in LQG are like the harmonic oscillator states in QM. They are members of the actual Hilbert space (not a rigged Hilbert space) and labeled by discrete numbers. One has to sum them, rather than integrate. However, they form an uncountable basis, because the LQG Hilbert space is not separable. It's then just a fact of mathematics, unrelated to any physical theory, that sums of uncountably many terms only exist if at most countably many terms are non-zero. So the expansion of any state is necessarly an at most countable sum of spin network states. You can't expand a state into something like an integral of spin network states. Now even if you introduced a rigged Hilbert space on top of the LQG Hilbert space and defined states as integrals over an uncountable number of generalized states from this rigged Hilbert space, the resulting state would be in the actual Hilbert space and could be expanded in the spin network basis with at most countably many non-zero coefficients.

Objects	How many are there?	Where do they live?	How do we obtain physical states?
Harmonic oscillator eigenstates	Countably many	(separable) Hilbert space of QM	Sum of countably many terms
Plane waves	Uncountably many	Rigged Hilbert space on top of QM Hilbert space	Integral over k
Spin network states	Uncountably many	(non-separable) Hilbert space of LQG	Sum over countably many terms
Elements of a potential rigged Hilbert space on top of the LQG Hilbert space	Uncountably many	Rigged Hilbert space on top of LQG Hilbert space	Integral over something

 

[Fra]

I think topic also calls for reflection on what is distinguishable and measurable from the perspective of an observer/agent. This is a weak point in as far as I recall, most theoretical camps. The avoidance of these question is clear even when you read Rovellis RQM. Already there one can smell a suspicious application of QM. At least it's far from conceptually clear. Yet one continues to build onto a questionable principles.

Where is the physical "hilbert space" or the corresponding information, "encoded" in LQG?

/Fredrik

 

[Demystifier]

They are members of the actual Hilbert space (not a rigged Hilbert space) and labeled by discrete numbers. One has to sum them, rather than integrate. However, they form an uncountable basis, because the LQG Hilbert space is not separable.

I don't understand that on the level of set theory. How can an uncountable set be labeled by discrete numbers?

 

[Nullstein]

I think topic also calls for reflection on what is distinguishable and measurable from the perspective of an observer/agent. This is a weak point in as far as I recall, most theoretical camps. The avoidance of these question is clear even when you read Rovellis RQM. Already there one can smell a suspicious application of QM. At least it's far from conceptually clear. Yet one continues to build onto a questionable principles.

Where is the physical "hilbert space" or the corresponding information, "encoded" in LQG?

LQG is just a regular quantum theory and you can interpret it according to your favorite interpretation, so new problems don't really arise regarding that in LQG. The new interpretational problem that arises in LQG (and in most other quantum gravity theories for that matter) is the problem of time.

I don't understand that on the level of set theory. How can an uncountable set be labeled by discrete numbers?

Well, in practice, it's really labeled in terms of a set of embedded graphs, equipped with the discrete topology (no two graphs are in any way similar) and angular momentum quantum numbers on the edges. That's a discrete, but uncountable set. If you want to stick to numbers, you could use the well-ordering theorem and instead label it by uncountable ordinal numbers (again equipped with the discrete topology). But anyway, the point is that the index set is discrete.

To put it completely in set theoretic terms: Any Hilbert space is isomorphic to a sequence Hilbert space $l^2(\kappa)$, where $\kappa$ is some cardinal number. A basis for this space is given by $e_\omega$, where $\omega <\kappa$ are the ordinal numbers smaller than $\kappa$. The most general vector in this space can be written as $v=\sum_{\omega<\kappa} c_\omega e_\omega$ with at most countably many non-zero $c_\omega$.

 

[jim mcnamara]

Fair warning. Sigh. We seem to be back on topic, which is Physics, not funding Physics. Please continue in that vein. Thank you.

 

[*now*]

..., because the LQG Hilbert space is not separable.

Was a reference provided specific to Rovelli when this was raised earlier in this thread?