Amazing bid by Thiemann to absorb string theory into LQG

[ranyart]

Originally posted by eigenguy
A rabbi and priest in a rowboat...

Indeed two 'Ed's' are better than one?

 

[Urs]

There is another simple example:

Look at the LQG-like quantization of the 1d non-relativistic point is
http://xxx.uni-augsburg.de/abs/gr-qc/0207106 . Then note equation (IV.5) and the one above it. In these equations care is taken that the usual quantum algebra is carried over to the LQG-like quantization scheme. If one were to remove the term \exp(-\alpha^2)/2 one would get instead the classical algebra, which corresponds to the way Thomas Thiemann 'quantizes' the LQG-string.

 

[selfAdjoint]

Urs, I was interested in Rehren's email that you posted on the Coffee Table, with its presentation of the various things that can happen, and the origin of the different levels of anomalies, central charges, and broken invariances. It was very educational. Do you have any comments on it? Does it have any bearing on these examples you give us from other quantizations?

 

[Urs]

selfAdjoint -

maybe we should carry this discussion to the Coffee Table where Rehren can see it.

Rehren argues that what Thiemann does is formally a form of 'quantization'. But let's not argue about words. The paper by Ashtekar, Fairhurst and Willis shows that with such a general notion of quantization one does not obtain the correct results even for the nonrelativistic 1d particle. The same holds true when applying this 'generalized' notion of quantization to the Maxwell field or the free relativistic particle. In each of these cases the results differ drastically from known physics (even if the large ambiguity in how to impose the exponentiated operator equations is used in a way that closely follows the usual quantization instead of following the classical theory).

This is not disputed by Ashtekar, Fairhurst and Willis. Their argument is the same as that by Thiemann: Maybe this drastically different notion of quantization is the correct one for gravity. Right, maybe it is. But maybe not. What I would like to see is some sort of motivation for why a radical departure from usual quantization is the right thing to do in quantum gravity. It is certainly not the physically right thing to do in the other systems that have been studied by this method.

 

[lethe]

Originally posted by Urs

Ok, here is a simple exercise that everybody who has followed our discussion should be able to solve:

lemme see..

1) Consider the Nambu-Goto action in 1+0 dimensions, which describes the free relativistic particle in Minkowski space (alternatively, for those who enjoy a bigger challenge: the charged particle in curved space with an electromagnetic field turned on)

start with Minkowski space, no background field:

S=m\int ds=m\int\sqrt{\eta_{\mu\nu}\dot{x}^\mu\dot{x}^\nu}d\tau

canonical momentum:
p_\mu=\frac{\delta L}{\delta\dot{x}^\mu}=\frac{m\dot{x}_\mu}{\sqrt{\dot{x}^\nu\dot{x}_\nu}}

2) Compute the single constraint of the theory.

from the expression for the canonical momentum, we see that

p^2=m^2

this is a first class constraint, since the equations of motion were not invoked.

3) Do a Dirac quantization by promoting this single constraint to an operator equation. Discuss the resulting quantum equation.

given a state |\psi>, look at its Fourier transform in Minkowski space

|\psi>=\int d^dk e^{ikx}|k>

since this is a generic Fourier transform over d-dimensional Minkowski space, the components of k are independent.

i guess i want to say something about enforcing only the expectation value of the constraint here, instead of the operator version of the constraint.

but i don't see why i have to do that here. let me see...

 

[Urs]

Hi lethe -

you can make a Fourier transformation, of course, but probably that's not necessary to make the point.

You have derived the classical constraint. Quantize it. Then impose Dirac quantization of constraints. Alternatively, impose LQG quantization of constraints. What do you get?

 

[Urs]

1st class constraints

By the way:

First-class constraints are those whose Poisson bracket closes on the set of constraints, i.e. is a linear combination of any of the constraints of the theory. Second class constraints are those whose Poisson brackte does not give another constraint.

In other words, the Poisson-bracket of 1st class constraints vanishes "weakly" or "on shell". The Poisson-bracket of second-class constraints does not.

Obviously only 1st class constraints have a chance to give a consistent set of operator constraint equations when quantized. Second class constraints must be eliminated by introducing "Dirac-brackets". This is a deformation of the usual Poisson bracket engineered in just such a way that all constraints become first class with respect to this new bracket. Dirac quantization really consists of replacing Dirac-brackets by commutators and sending the classical constraints to operator equations.

In any case, since for the relativistic particle there is only a single constraint it is trivially 1st class and we can quantize it immediately without worrying about 2nd class subtleties.

See for instance http://www.math.ias.edu/QFT/fall/faddeev4.ps

 

[eforgy]

Originally posted by Urs

You have derived the classical constraint. Quantize it. Then impose Dirac quantization of constraints. Alternatively, impose LQG quantization of constraints. What do you get?

Thanks lethe for starting to solve this exercise. It's been so long since I've looked at QM that I wouldn't have had a chance to get as far as you did.

But picking up where you left off, I think Urs simply wants you to look at the equation

<br /> p^2|\psi\rangle = m^2|\psi\rangle<br />

and note that this is the Klein-Gordon equation.


Eric

PS: By the way, why do you call this a "constraint"? Remember, it's been eons :)

PPS: I'd have no idea how to loop quantize this and I HAVE been following this thread :)

 

[eforgy]

Originally posted by eforgy

PPS: I'd have no idea how to loop quantize this and I HAVE been following this thread :)

I can do a little better than this. Looking back at the earlier posts here, I'd write down

<br /> e^{p^2-m^2}|\psi\rangle = |\psi\rangle.<br />

Since p^2 commutes with m^2, I guess we could write this as

<br /> e^{p^2} |\psi\rangle = e^{m^2} |\psi\rangle,<br />

but I don't know if this buys us anything. Would it make sense to look at

<br /> e^{\epsilon(p^2 - m^2)}|\psi\rangle<br /> = |\psi\rangle<br />

and combine terms of the same order in \epsilon? If we did this, I think we'd end up with

<br /> (p^2 - m^2)^n |\psi\rangle = 0<br />

for all n&gt; 0. Otherwise, we'd end up with a mess.

Eric

 

[Urs]

Hi Eric -

thanks for chiming in.

Ok, let me give it away:

As you said, the operator version of the single constraint of the free relativistic particle is obtained by the usual correspondence rule
p^\mu \to \hat p^\mu = -i \hbar \frac{\partial}{\partial x^\mu} and yields nothing but the Klein-Gordon equation
<br /> \partial^\mu \partial_\mu \phi = -m^2 \phi<br />
(up to factors of c,\hbar). I am calling this a constraint because, as lethe has derived, it is the operator version of the classical constraint \varphi = p^2 + m^2 = 0 of the action of the free relativistic particle. The classical free relativistic particle cannot move in all of its phase space, but has to stick to the subspace given by this equation, which is incidentally called the mass shell.

Ordinary quantization rules, found by Dirac and others, tell us that the quantum theory is given by demanding that this holds as an operator equation, which is nothing but the Klein-Gordon equation. This equation, together with its fermionic cousin, the Dirac equation, is very well tested experimentally, since it is the very basis on which all of QFT that can be measured in any accelerator is built.

Now let's see how LQG tells us to quantize the free relativistic particle:

There we are told not to consider the constraint \varphi itself but the group which is generated by it by means of Poisson brackets. I.e. we are supposed to look at the group elements
<br /> U(\tau) = \exp\left([\phi,\cdot]_\mathrm{PB}\right)<br />
where [\cdot,\cdot]_\mathrm{PB}$ is the Poisson bracket and this guy is supposed to act on classical observables, i.e. functions on phase space.

But because there is just a single constraint this group is nothing but the group of real numbers under addition:
<br /> U(\tau_1)\circ U(\tau_2) = U(\tau_1 + \tau_2)<br /> \,.<br />

So LQG-like quantization consists of finding an operator representation $\hat U(\tau)$ of the group of real numbers on some Hilbert space so that the operator product satisfies
<br /> \hat U(\tau_1) \hat U(\tau_2) = \hatU(\tau_1 + \tau_2)<br /> \,.<br />

Of course, due to the simpliciy of this example, we could just choose the usual Hilbert space of the free relativistic particle and set
<br /> \hat U(\tau) := \exp(i \tau (\hat p^\mu \hat p_\mu + m^2))<br /> \,.<br />
But we could also choose something very different. This is the great ambiguity that I was referring to. For instance, if we followed the tretament by Ashtekar, Fairhurst and Willis of the LQG-like quantization of the 1d nonrelativistic particle, than we'd want to use a nonseparable Hilbert space on which the momentum operator $\hat p$ is not representable. In this case, which is the precise analog of what Thomas Thiemann does in the 'LQG-string' the above choice for \hat U is not an option.

But of course, due to the great simplicity of this toy example, if you pick any (hermitian) operator \hat O on any Hilbert space, you can set
<br /> \hat U(\tau) := \exp(\tau i \hat O)<br />
and get a representation of the group of real numbers. You can furthermore choose operators that don't come from exponentiating other operators, of course.

LQG tells us that physical states are those invariant under the action of this group. Clearly, by choosing \hat U appropriately you can find an enormous number of states that satisfy the 'equation of motion'
<br /> \hat U(\tau)|\psi\rangle = |\psi\rangle<br /> \,.<br />
Almost anything goes.

For instance I could choose \hat O to be any projector. Than every state that is projected out by \hat O is physical, according to the LQG-like quantization prescription.

Or I could use strange Hilbert spaces, like, say, the 2-dimensional Hilbert space C^2.

The weirdest things are possible. The reason is, that by ignoring the form of the constraint \phi and just looking at the abstract classical group that it generates, we are loosing a lot of crucial physical information.

Thomas Thiemann similarly ignores the form of the Virasoro constraints and just cares about the classical group they generate. The result is no less strange than the above toy example. In fact, the above toy example is a subset of the full string quantization, namely that corresponding to strings that have no wiggly excitations.

I have pointed out the place at which this huge ambiguity appears in the Ashtekar, Fairhurst and Willis papers. There, too, one could choose almost anything else and get weird results. AF&W do choose to use a representation of the group operators that is very close to the usual one, but instead they choose a very strange Hilbert space. Other choices are possible. Nowehere in their paper is it explained why one choice should be preferred over the other.

 

[Urs]

Hi Eric -

yes, in your second post you demonstrate that exponentiating the KG constraint and demanding invariance yields the same thing as usual. The point is that LQG-like quantization allows you to exponentiate anything else, on any other Hilbert space and call it a quantization of the relativistic particle.

 

[eforgy]

Originally posted by Urs
Hi Eric -

yes, in your second post you demonstrate that exponentiating the KG constraint and demanding invariance yields the same thing as usual. The point is that LQG-like quantization allows you to exponentiate anything else, on any other Hilbert space and call it a quantization of the relativistic particle.

Hi Urs,

Let me state in my words how I am beginning to understand this. I am sure I am just repeating what you've been saying all along (assuming what I say is correct that is).

In the "usual" approach, you take your constraints and "quantize" them directly as operators on some hilbert space

<br /> C_I|\psi\rangle = 0.<br />

Now if you take this and exponentiate it, you get something like

<br /> e^{C_I} |\psi\rangle = |\psi\rangle,<br />

but this is still just an operator on the same Hilbert space we started with. These exponentiated operators form a group I suppose? An algebra?

Then, are you saying that the loop quantization procedure then looks for any group or algebra of operators that satisfy the same rules on any old Hilbert space and call this a quantization?

I'm sure I don't have that right.

Eric

 

[Urs]

Hi Eric -

yes, that's the idea - almost! :-)

There is a crucial subtlety:

Let me write C_I for the classical constraints, i.e. these are functions on phase space and the system is restricted to be at points in phase space on which C_I = 0.

The C_I are, already classically, generators of the gauge transformations of the system and physical observables must be gauge invariant and hence Poisson-commute with the constraints. We can write either
<br /> [C_I,A]_\mathrm{PB} = 0<br />
or
<br /> \exp([r^I C_I,\cdot]_\mathrm{PB})A = A<br /> \,.<br />
Classically this is both well defined and equivalent. The latter form makes us notice that the exponentiated constraints form a group (since they are 1st class)
<br /> \exp([r^I C_I,\cdot]_\mathrm{PB})<br /> \exp([s^I C_I,\cdot]_\mathrm{PB})<br /> =<br /> \exp([(r \times s)^I C_I,\cdot]_\mathrm{PB}) <br /> \,.<br />
This group is the gauge group of the system with elements
<br /> U(r^I) := \exp([r^I C_I,\cdot]_\mathrm{PB})<br /> \,.<br />

In standard Dirac quantization the classical constraints C_I are promoted to operators \hat C_I and the equations of motion become
<br /> \langle \psi | \hat C_I |\psi \rangle = 0<br /> \,.<br />
In simple cases this is equivalent to
<br /> \hat C_I |\psi \rangle = 0<br /> \,.<br />

In LQG-like quantization people like to choose a Hilbert space on which the C_I are not representable as operators. Then they claim that they can still 'quantize' the equation
<br /> \exp([r^I C_I,\cdot]_\mathrm{PB})A = A<br />
by doing the following:

- Find operators
<br /> \hat U(r^I)<br />
such that
<br /> \hat U(r^I)<br /> \hat U(s^I)<br /> =<br /> \hat U((r \times s)^I)<br /> \,.<br />

- Demand that physical states satisfy
<br /> \hat U(r^I)|\psi \rangle = 0<br /> \,.<br />

In previous posts I have mentioned that this presription introduces a large arbitrariness in the choice of the \hat U(r^I).
The example of the KG particle show that the physical information contained only in the abstract structure of the group generated by the classical constraints is way too little to reconstruct the physical system that one started with

 

[selfAdjoint]

And do the LQG people ever exhibit such a pretend quantisation of the Klein Gordon particle? I've never seen any but I'm a long way from completely famliar with their literature. It seems to me that you have to ask in each case, as Reheren does in his email, what are the degrees of freedom of the various things you are working with. Proving that the particle is too, umm, sparse to be quantized their way doesn't prove that their way is false in other cases.

And has anyone ever seen a KG particle, quantized or otherwise?

 

[lethe]

Originally posted by selfAdjoint

And has anyone ever seen a KG particle, quantized or otherwise?

yeah, i think they saw the pion already in the 30s

 

[Urs]

selfAdjoint -

I haven't seen the KG particle discussed by LQG-like methods. But whart I said about this quantization is exactly what Thomas Thiemann told me to do in general. Also note that, as I have said before, the 'LQG-string' contains the KG particle as a subcase. Nameley the 0-mode of the Virasoro constraints is nothing but the KG equation for the string, where the mass is given by internal oscialltions. So Thiemann dos not get the KG equation for the string.

Please note, as lethe has said, that all bosons that are found in nature are described by the KG equation and all fermions by the Dirac equation (in first quantized form). We could do exactly the same discussion for the Dirac particle and get exactly the same conclusions.

Finally note the example provided by LQG-people themselves: There are LQG-like papers on the 1d nonrelativistic particle as well as on the quantized EM field. In neither case is are the usual results obtained.

 

[marcus]

Larsson on SPR

the discussion of Thiemann's "Loop-String" paper continues at
SPR (sci.physics.research). Today Thomas Larsson posted the following. Comment? Any explications would be most welcome!

---------Larsson's post---------

... expanded version of a post to the string coffee table,
http://golem.ph.utexas.edu/string/archives/000300.html . It is in response to a post by K-H Rehren, ...

<EDIT: I DECIDED TO MOVE LARSSON'S POST TO A SEPARATE THREAD,
SO FOR BREVITY THE MAIN PART IS SNIPPED OUT HERE>

This is the main algebraic difference between the LQG "lowest-A-number" reps and LE reps. This is an essential difference; in infinite
dimension, there is no Stone-von Neumann theorem that makes different
vacua equivalent.
--------------end of post-----------------

 

[ranyart]

Originally posted by marcus


This is the main algebraic difference between the LQG "lowest-A-number"
reps and LE reps. This is an essential difference; in infinite
dimension, there is no Stone-von Neumann theorem that makes different
vacua equivalent.
--------------end of post-----------------

There are a number of problems as I see it with commuters with respect to inter-dimensional transports, commutations.

The starting point is where one makes the initial 'start-point' for the equations in question. If one is within a 3-Dimensional Space, where Geometric Structures are whole, then reduce the structures down through compacted Dimensions of 3-D>> 2-D>> 1D, you are 'breaking' Geometric Transportation Laws, by going from 'whole' to 'bits'.

Now as far as I see it change the Dimension and you automatically change the effects one imposes on Space, in simplistic terms The Laws Of Physics will differ from frame to frame. You cannot be sure of a number of important factors, one major factor is that measure and what we are used to in a 3-D frame becomes something completely different. Distance and Geometric paramiters observed from a 3-D background are not Equivilent emmbedded into a 2-D background, a 'distance' in 3-Dimensions is not the same as a 'distance' in 2-Dimensions.

One can argue that 'quanta' and its effects are different from dimensional frame to frame. This I believe was taken up by Smolin and Magueijo and also Stepanov amongst others. In Geometric framing, a 3-D 'somthing' has a defined boundery, that's what shape/structure in Geometry is! In measuring structure, the devise you use MUST be in contact with what is being measured, for instance a tape-measure has increments that enable you to measure a three dimensional piece of wood form end to end, it is a precise 3-D object itself, as precise as one can be within 3-D.

The tape measure we are used to in our ordered 3-D world suits us well, the dimensionality of the tape measure itself does not come into question, it has length-width-depth in sufficient quantity as to be an ideal tool, but it has a dimensional limit that it constrains to.

If one was to use the piece of wood to measure the tape-measure? how would this effect measure? (try wrapping a piece of wood around a single increment of a tape measure!

The limits of measure from a 3-D frame to a 2-D frame relies upon what is certain ( 3-D structures) and what is not (2-D background). The HUP principle adheres to measure, for HUP to be effective then everything that is being measured must be connected, in our 3-D world this enables us to see matter in small enough quantities to pick-up things and move them around in 3-D Space. A speck of dust between your fingers is a sure thing if one can observe it and move it from A to B.

All well and good if A and B are dimensionally equivalent, but we know that a 3-D frame is not all there is. A good example is energy that commutes from a lower 2-D dimensional frame , up into our 3-D world that is Particle Creation!, 2-d >> 3-D. The converse of this is the reduction of Particles back down into a reduced dimensional field 3-D >> 2-d.

It is by no cincedence that HUP has to have continuous motion as a 'fixer' for Uncertainty, the more you constrain P the less position X exists for you to measure. The whole of HUP is inter dimensional transactions. As you measure something in a 3-D frame, you push part of it away from you and into another dimension (hidden variable). The continuation of measure ends from the perspective of the object in 3-D, it has a limit, this limit is in fact a Dimensional Geometry Bound.

Geometry = observation Measure within Three Dimensions, Quantum Measure is unobservable from this domain. In order to measure something in the Quantum Realm, you need an infinite amount of precision, you will always be taking the initial measure from a 3-D world, thus you cannot quantify a measure going from a 3-D world into the 2-d Quantum world, PRECISION IS FIXED INTO 3-Ds at all Times.

Let me state this again, the Precision for measure favours our 3-D world, because that's where we exist and that is where we always make our measurments from. If you cannot measure something precisely, then it cannot be quantizised, or discontinued. Put another way something that is in continuation cannot be measured in 'bits'.

There are inroads to the re-formulation of Einsteins SR and GR, the Laws of Relativity can be seen as the Laws of Dimensional Space(Geometry) while the SR can be seen as having a Varying Speed of Light,(because there are varying Dimensions of Space-Times) where there are domains of no Observers, less than 3-D space's.

Taken further, when we look out into the Cosmos from our 3-D constant frame, we observe Light as unchanging by default, it will always have a precision as long as we are within a 3-D frame(Galaxy), and will always be seen as 'coming' to us along a 3-D observation, but via a Background field that is 2-D, so it comes along a route from another Galaxy that is 3-D (other galaxy) >>2-D (EM-Vacuum-CMB-) >> (Our Galaxy) 3-D) Where it is Relative to our Dimensional Frame.

When we look inwards from our 3-D world down into the Quantum Realm 2-D, we expect a continuation of the same Laws, but this is not so purely by the fact that we are measuring and observing from 3-D to 2-D, and not as stated earlier with Cosmic Relative observations, there is a missing factor, the Cosmic Fact that there is other Galaxies and other 3-Dimensional worlds ,(us) 3-D >> 2-D >> 1-D (SPACE-SINGULARITY)2-D>> 3-D (them-other galaxies), while observing from inside our Galaxy to the Quantum Realm there is only (us) 3-D >>2-D>>1-D.

We do not observe other Galaxies from wihtin our 3-D world going in the direction of Space Reduction Quantumly, one just gets to a 2-D field of Particle Production, which technically replace's the Galaxies as you reduce down to microscopic Black Hole Singularity.

The problem I see it is String Theorists neglect the fact that their world-lines are continuous, and therby have no discreteness about them, plus the major fact they always expect their initial backgrounds to be Universal! they can not move along 3-dimensional spaces and down into the Quantum Realm, without an infinite amount of dimensional Explinations. There are no more than 3-Dimensional Space's in existence in Our Universe, this is bourne out by the fact that Structure(geometry) does not need excess 'Dimensions' for energy to transport from one frame to another, its 'where' you measure from that is fundemental.

 

[lethe]

Originally posted by Urs
By the way:

First-class constraints are those whose Poisson bracket closes on the set of constraints, i.e. is a linear combination of any of the constraints of the theory. Second class constraints are those whose Poisson brackte does not give another constraint.

hmm... my knowledge of classical mechanics is a little insufficient here. the definition of "First class constraint" that i mentioned above is that it is a constraint which holds even if the equaiton of motion is not satisfied.

so what do you mean that the Poisson bracket closes on the set of constraints? i calculated this:

[p^2-m^2,f]_\mathrm{P.B.}=-2p^\mu\partial_\mu f
where f is any function on phase space. is this supposed to vanish? or give some linear combination of the constraint p^2-m^2?

 

[lethe]

Originally posted by Urs

As you said, the operator version of the single constraint of the free relativistic particle is obtained by the usual correspondence rule
p^\mu \to \hat p^\mu = -i \hbar \frac{\partial}{\partial x^\mu} and yields nothing but the Klein-Gordon equation
<br /> \partial^\mu \partial_\mu \phi = -m^2 \phi<br />
(up to factors of c,\hbar).

OK, so you just impose the operator version of the constraint.

for some reason i thought there was some issue about imposing only the expectation value of the constraint, as in Gupta-Bleuler. but i guess not.

incidentally, i have always been a little confused about in what sense i can call this quantization "the usual correspondence rule". in nonrelativistic quantum mechanics, we can perform the substitution x\rightarrow\hat{X} and p\rightarrow\hat{P}, subject to the canonical commutation relation [\hat{X},\hat{P}]=i[x,p]_\mathrm{P.B.}=i

it's easy enough to show that this is equivalent to \hat{X}\psi(x)=x\psi(x) and \hat{P}\psi(x)=-i\partial\phi/\partial x when working in the coordinate basis.

the canonical commutation relations give the basis independent quantization procedure. but in the relativistic theory, we don't have these canonical commutation relations, since there is no time operator, and it seems like we are forced to work in the position basis.

can we impose a more basis independent quantization procedure here?

Now let's see how LQG tells us to quantize the free relativistic particle:

There we are told not to consider the constraint \varphi itself but the group which is generated by it by means of Poisson brackets. I.e. we are supposed to look at the group elements
<br /> U(\tau) = \exp\left([\phi,\cdot]_\mathrm{PB}\right)<br />
where [\cdot,\cdot]_\mathrm{PB} is the Poisson bracket and this guy is supposed to act on classical observables, i.e. functions on phase space.

hmm... why is this thing a group now? i guess the Poisson algebra of observables is a Lie algebra, so we might expect that exponentiating it would yield a group, but i believe that infinite dimensional Lie algebras do not always exponentiate to Lie groups, only with finite dimensional Lie algebras do we have this guarantee.

the fact that you are exponentiating [\phi,\cdot]_\mathrm{P.B.} instead pf just \phi means that you are using the adjoint representation of this Lie algebra? this way we have a group of operators on the classical algebra of observables?

But we could also choose something very different. This is the great ambiguity that I was referring to. For instance, if we followed the tretament by Ashtekar, Fairhurst and Willis of the LQG-like quantization of the 1d nonrelativistic particle, than we'd want to use a nonseparable Hilbert space on which the momentum operator \hat p is not representable. In this case, which is the precise analog of what Thomas Thiemann does in the 'LQG-string' the above choice for \hat U is not an option.

OK, so they are considering representations of the group generated by the constraint, instead of the constraint itself.

this is the same way the Stone-von Neumann theorem goes, right? it says that there is only one theory that satisfies the exponetiation of the canonical commutation relations (the Weyl relation, i think this is called?). but i have read on s.p.r that there can be inequivalent (and perhaps even physically relevant) representations of the commutation relations themselves

so it seems like the choice to only look at the group version loses you generality?

so this choice is what allows Thiemann to get rid of the anomoly?

 

[Urs]

for some reason i thought there was some issue about imposing only the expectation value of the constraint, as in Gupta-Bleuler. but i guess not.

This is an additional subtlety but not the issue wrt LQG/standard quantization. Since the single constraint of the KG particle is self-adjoint it should make no difference. So if you want consider the Gupt-Bleueler quantization method. It doesn't alter the point about the LQG-quantization at all.

but in the relativistic theory, we don't have these canonical commutation relations, since there is no time operator,

I am not sure why you think so. We have
<br /> [\hat x^\mu , \hat p^\nu] \sim \eta^{\mu\nu}<br />
which translates to
<br /> [\partial_\mu , x^\nu] = \delta_\mu^\nu<br /> \,.<br />
There is however a subtlety with defining the Hilbert space of physical states, since these do not live in L^2(M^4), obviously (since they don't decay in the time direction). There are many ways to handle this, the most elegant and advanced being gauige fixing by means of BRST methods. But for our discussion all this does not really matter.

hmm... why is this thing a group now? i guess the Poisson algebra of observables is a Lie algebra, so we might expect that exponentiating it would yield a group, but i believe that infinite dimensional Lie algebras do not always exponentiate to Lie groups, only with finite dimensional Lie algebras do we have this guarantee.

I am not aware of the problems that you are hinting at, do you have a reference? Note that in the case of the Virasoro algebra, which is infinite dimensional of course, the classical group does exist all right. In any case, this would not affect the KG particle, which clearly has a finite constraint algebra.

OK, so they are considering representations of the group generated by the constraint, instead of the constraint itself.

Yes! That's the point. But note that 'representing the constraints themselves' is usually accompanied by much more structure. We are not just looking for any set of operators which has the same algebra as the constraints. We want these operators to be built from the canonical data of the classical system, i.e. canonical coordinates and momenta, by some sort of 'correspondence rule'. All this information about the physical system is lost in the 'represent the group without the rest'-approach.

this is the same way the Stone-von Neumann theorem goes, right? it says that there is only one theory that satisfies the exponetiation of the canonical commutation relations (the Weyl relation, i think this is called?). but i have read on s.p.r that there can be inequivalent (and perhaps even physically relevant) representations of the commutation relations themselves

Stone-von Neumann says that iff the Weyl algebra is represented weakly continuously, then the canonical coordinates and momenta \hat x,\hat p do exist as operators, too, otherwise they do not. And if they exist the Weyl algebra elements are the exponetiations of the Heisenberg algebra elements. See http://citeseer.nj.nec.com/355097.html

so it seems like the choice to only look at the group version loses you generality?

No, it gives you too much generality. Using these strange reps it is possible to built strange theories.

so this choice is what allows Thiemann to get rid of the anomoly?

Yes. There is no technical subtlety hidden in this 'getting rid of the anomaly'. There is the classical conformal group and Thomas Thiemann points out that one can built a Hilbert space on which operators exist which represent this classical group. That's nothing deep. On large enough Hilbert spaces there exist operators which represent almost everything.

 

[Urs]

so what do you mean that the Poisson bracket closes on the set of constraints?

That the Poisson-bracket of two constraints is again a linear combination of constraints. In your example f is not a constraint, so the bracket you compute need not be a constraint, either. Since there is only a single constraint for the KG particle the only bracket to test is the Poisson-bracket of the single constraint with itself, which vanishes.

But compare the Virasoro generators. The Poisson bracket of two Virasoro generators is again a Virasoro generator, up to a factor. Hence their algebra closes and they are 1st class.

 

[lethe]

Originally posted by Urs

I am not aware of the problems that you are hinting at, do you have a reference? Note that in the case of the Virasoro algebra, which is infinite dimensional of course, the classical group does exist all right. In any case, this would not affect the KG particle, which clearly has a finite constraint algebra.

i will check for a reference shortly, i am just talking out of my memory from class, so take it with a grain of salt, but i believe that the virasoro algebra does not generate a group. it generates conformal transformations in 2D, right? and something about there being local conformal transformations that do not have an inverse globally, and hence to not form a group.


hey, by the way, when is the voting for sci.physics.strings going to begin?

 

[jeff]

Hey lethe,

Originally posted by lethe
can we impose a more basis independent quantization procedure here?

I thought maybe you'd be interested in knowing that we can avoid having to choose specific canonical position and conjugate momentum variables entirely by formulating the classical theory in terms of a symplectic form &Omega;_&mu;&nu;, the non-degenerate closed 2-form) on phase space that serves as the fundamental structure needed to define hamiltonian dynamics. One then quantizes by replacing poisson brackets for &Omega;_&mu;&nu; with commutators etc.

 

[Urs]

lethe -

right, exponentiating a Lie algebra always only gives you the group locally.

Regarding s.p.s.: We are currently waiting for one of the 'Volunteer Votetakers' to volunteer taking votes. We are being told that this should happen in the days/weeks. I am hoping it will happen soon, but currently we cannot do anything to speed up the process.

 

[Urs]

Sci.Physics.Strings: Call For Votes

The Call For Votes for the proposed USENET newsgroup sci.physics.strings, supposed to be concerned with discussion of string theory, has now been published at

http://groups.google.de/groups?selm=1077593588.15146@isc.org .

Everybody may vote. Detailed instructions for how to vote are given at the above link.

 

[lethe]

Originally posted by Urs
I am not sure why you think so. We have
<br /> [\hat x^\mu , \hat p^\nu] \sim \eta^{\mu\nu}<br />
which translates to
<br /> [\partial_\mu , x^\nu] = \delta_\mu^\nu<br />

i infer from this expression that you have an operator on your Hilbert space \hat x^0 that acts on states like multiplication by t. this is the straightforward application of the nonrelativistic quantization to the relativistic particle.

but something that i have learned from reading s.p.r is that there is no such operator. since learning that fact, it has been a big question mark in my mind as to whether there actually exists a theory that could really be called relativistic quantum mechanics of a particle.

i have wanted to understand what is going on with that for a while. since we were doing quantization of the relativistic particle, i thought i would toss in a question about that for you, but i can certainly appreciate that it is a bit off topic for the current discussion

There is however a subtlety with defining the Hilbert space of physical states, since these do not live in L^2(M^4), obviously (since they don't decay in the time direction). There are many ways to handle this, the most elegant and advanced being gauige fixing by means of BRST methods. But for our discussion all this does not really matter.

perhaps this issue about decays is the reason for this thing that i have read about the nonexistence of the time operator, and perhaps this BRST process is the way to resolve it?

Originally posted by Urs
Note that in the case of the Virasoro algebra, which is infinite dimensional of course, the classical group does exist all right. In any case, this would not affect the KG particle, which clearly has a finite constraint algebra.

ahh... this is related to my other confusion. i was going to exponentiate the Poisson algebra of classical observables (which is infinite dimensional), not the subalgebra of constraints (which is closed under the Poisson bracket since the contraints are first-class). i suppose this is the reason i also screwed up and tried to take the Poisson bracket before with some generic classical observable instead of another constraint.

um... i guess i need to learn a bit more about this notion of the Poisson bracket closing on the constraints. and here, i thought i already knew all the classical mechanics i would ever need to know.

Originally posted by Urs
lethe -

right, exponentiating a Lie algebra always only gives you the group locally.

OK, i guess i am not so familiar with spaces which are only locally a group. but this sounds reasonable.

 

[lethe]

Originally posted by Urs

Yes! That's the point. But note that 'representing the constraints themselves' is usually accompanied by much more structure. We are not just looking for any set of operators which has the same algebra as the constraints. We want these operators to be built from the canonical data of the classical system, i.e. canonical coordinates and momenta, by some sort of 'correspondence rule'. All this information about the physical system is lost in the 'represent the group without the rest'-approach.




Stone-von Neumann says that iff the Weyl algebra is represented weakly continuously, then the canonical coordinates and momenta \hat x,\hat p do exist as operators, too, otherwise they do not. And if they exist the Weyl algebra elements are the exponetiations of the Heisenberg algebra elements. See http://citeseer.nj.nec.com/355097.html

yes, actually, i have read the paper you reference here, and i think that paper is exactly what i had in mind with my above comments.

in Theorem 1 of that paper, they state that any pair of family of operators that satisfies the Weyl form of the CCRs, is unitarily equivalent to the Schr&ouml;dinger representation (Weyl exponentiated form)

in theorem 2, they give some conditions that imply that any pair of operators P and Q that satisfy the canonical commutation relation (Heisenberg form [q,p]=i, not Weyl form) are equivalent to the Schr&ouml;dinger representation (Heisenberg/non-exponentiated form).

but then in the text, the authors makes reference to physically relevant systems for which those conditions are not met, and therefore may have inequivalent represntations to the Schr&ouml;dinger.

so it seems to me like finding a rep of P and Q is more general than finding a representation of their exponentiations. there are many inequivalent representations for the former, and only one representation for the latter.

this is what i took from the paper you referenced, so what am i missing?

No, it gives you too much generality. Using these strange reps it is possible to built strange theories.

see my above complaint. in short, there is only one rep of the exponentiated unitary form, and uncountably many reps of the self-adjoint non-exponentiated form of the CCR. so i conclude that the latter is more general, it allows for more systems. should i not conlcude that there is some ambiguity in choosing such a rep?

 

[Haelfix]

'OK, i guess i am not so familiar with spaces which are only locally a group. but this sounds reasonable.'


Exponentiating an algebra gives you a group, but it need not be the unique group, even in the finite dimensional case. There could be global topological features that have to be checked for. In fact, its a big pain, in practise you have to go through many tables to check for consistency.

In the infinite dimensional case, its even worse. For instance, there exists examples, where you can take an infinite dimensional group element arbitrarily close to the identity, but they are not exponentials of the lie algebra.

 

[Urs]

Lethe -

I bet that when you have heard that there is no time operator this was referring to an operator conjugate to a Hamiltonian. This is something different that we had been discussing.

Take a non-relativistic QM system with Hamiltonian H. Can there be an operator T such that [H,T] \sim 1?

As far as I remember the argument is that there cannot, because two operators satisfying a CCR as will act like multiplication/differentiation with respect to each other's eigenvalues and hence be unbounded from below and from above. But the spectrum of a decent Hamiltonian is supposed to be bounded from below (have a ground state), so it cannot satisfy any CCR.

But this argument doe not apply to systems which do not have an ordinary Hamiltonian. For instance the KG particle that we were discussing is governed by a constraint, not a Hamiltonian evolution. Here time is on par with the spatial dimenions.

If you wish, you can regard the constraint of the KG particle as the Hamiltonian with respect to parameter evolution, where the parameter is an auxiliary variable along the worldline of the particle. This plays formally the role of time in non-relativistic QM and the above argument would show that there is not operator associated with the worldline parameter which has the CCR with the constraint.

For more details on the quantization of the KG particle and its relations to non-relativistic QM you might want to have a look at http://www-stud.uni-essen.de/~sb0264/TimeInQM.html .

Regarding your summary of the Stone-vonNeumann theorem I do not quite agree. I think the message is that there are many reps of the Weyl algebra and that if and only if these reps are weakly continuous does the Heisenberg algebra exist and then the Weyl rep is the exponentiation of the Heisenberg algebra.

LQG like approaches play with the possibility that even if the Heisenberg algebra does not have a rep still a rep of the Weyl algebra exists.

 

[eforgy]

Originally posted by Urs


For more details on the quantization of the KG particle and its relations to non-relativistic QM you might want to have a look at http://www-stud.uni-essen.de/~sb0264/TimeInQM.html

Nice! :)

I just read this and highly recommend it.

Eric

PS: Urs, this constraint business is something I haven't thought much about. It gives me a new interpretation of the subspace of paths for which \partial^2 = 0. This is like the "physical" space.

 

[ranyart]

Originally posted by lethe
i infer from this expression that you have an operator on your Hilbert space \hat x^0 that acts on states like multiplication by t. this is the straightforward application of the nonrelativistic quantization to the relativistic particle.

but something that i have learned from reading s.p.r is that there is no such operator. since learning that fact, it has been a big question mark in my mind as to whether there actually exists a theory that could really be called relativistic quantum mechanics of a particle.

i have wanted to understand what is going on with that for a while. since we were doing quantization of the relativistic particle, i thought i would toss in a question about that for you, but i can certainly appreciate that it is a bit off topic for the current discussion.

When one moves from a Three Dimensional Relativistic network, down into the Quantum Mechanical 'Hidden-Variable-Network', the equations themselves have to be able to transform one set of data (with observations) back-and-forth. This is where the problem lay, and you clearly have touched upon a deeper meaning in your post quoted above?

What 'single' type of formula's can trancend both QM and GR?..will there be a single formula that can both describe an event in GR and QM without changing the Mathematical formula?..the answer is no, both theories are by their very nature uncompatable, you cannot Unify events in 3+1 dimensions with events in 1+1 dimensional reduced fields.

The transformation has to occur with the Mathematical interpretations, for instance a 3+1 Network has collisions in space, dimensionally 'whole'?.. Particles move around and collide in the 3+1 network that allows them this freedom. In reduced QM Dimensional Networks, this cannot happen as there are no 3 Dimensional 'Whole-Particles' that can exist similtainiously in 3-D and 2-D!

If one uses a formula that traces a Particle wherever it goes (which is what Einstien formulated in GR!), then there comes a point where not only does the formula cease to exist, the Particle itself has been removed from the 3+1 network, as the point in Spacetime(3+1)is reduced to a point in just Space..no TIME = no Observation = Hidden Vaiables = no Collisions = just space/fields = dimensional backgrounds = different (Special) formula's!

Now the bigger picture can be viewed by many 'theorists' into whatever formula's takes their fancy, for instance a simplyfied String Theorist would create 'extra' formulas to exist in 'extra' dimensions, all of which are technically sound, as they have no need of verifacation, and cannot-be verified by observations, the removal of 'SpaceTime' is a natural consequence of the removal of observations. The extra dimensions that some Mathematicians 'create' just simply do not exist.

From a dimensional perspective within GR and SR, one can go from 3+1 (4-D), to 2+1 (3-D) to 1+1 (2-D) TO A SINGULARITY NETWORK that is 0+1...1+0 . Of course the energies that are replacing Particles in 'Identity' terms as one reduces the Particles into Fields, also end up as Creationary Energies when one reaches the simplistic 1+1 AREA NETWORK! around a Blackhole, which happens to reside at the Core of every single Galaxy. Some would offer an explination that Science needs a Dimensional perspective alteration to the Existence of our place within a Spacetime Galaxy (3-D+t), surrounded by Fields of QM Networks that is Electro-Magnetic-Vacuum Space (2+1) with no further need of Mathematical Extensions to Reality.

 

[lethe]

Originally posted by ranyart
When one moves from a Three Dimensional Relativistic network, down into the Quantum Mechanical 'Hidden-Variable-Network',

hidden variable network? wtf?

 

[Urs]

hidden variable network? wtf?

Maybe, maybe, maybe he is thinking of Smolin's latest attempt at merging Nelson's stochastic QM with quantum gravity

http://xxx.uni-augsburg.de/abs/gr-qc/0311059

where spin networks are indeed used as 'hidden variables' to produce QM dynamics from classical statistics.

Last time Lee Smolin tried the same with BFSS Matrix Theory

http://xxx.uni-augsburg.de/abs/hep-th/0201031 .

I used to consider this interesting,

http://groups.google.de/groups?selm=ahe52s$1a2q$1@rs04.hrz.uni-essen.de

though I am not so sure anymore.

Anyway, this is what ranyart's avant-garde poetry reminded me of. As with every piece of modern art, you have to search the answer within yourself. ;-)

 

[lethe]

Originally posted by Urs

As far as I remember the argument is that there cannot, because two operators satisfying a CCR as will act like multiplication/differentiation with respect to each other's eigenvalues and hence be unbounded from below and from above. But the spectrum of a decent Hamiltonian is supposed to be bounded from below (have a ground state), so it cannot satisfy any CCR.

right, this is what i had in mind.

But this argument doe not apply to systems which do not have an ordinary Hamiltonian. For instance the KG particle that we were discussing is governed by a constraint, not a Hamiltonian evolution. Here time is on par with the spatial dimenions.

ok, this is interesting

If you wish, you can regard the constraint of the KG particle as the Hamiltonian with respect to parameter evolution, where the parameter is an auxiliary variable along the worldline of the particle. This plays formally the role of time in non-relativistic QM and the above argument would show that there is not operator associated with the worldline parameter which has the CCR with the constraint.

i think i can see that now. thank you, that was very helpful for me.

Regarding your summary of the Stone-vonNeumann theorem I do not quite agree. I think the message is that there are many reps of the Weyl algebra and that if and only if these reps are weakly continuous does the Heisenberg algebra exist and then the Weyl rep is the exponentiation of the Heisenberg algebra.

LQG like approaches play with the possibility that even if the Heisenberg algebra does not have a rep still a rep of the Weyl algebra exists.

before i think about your point here, I am confused as to what you are referring to when you say "Weyl algebra". i am thinking it should be the set of operators you get after exponentiation, but do these things form an algebra? i expect them to form a group, but i wouldn't expect the sum of two of these guys to be another one of these guys.

in short, the operators in the Weyl relation are the Lie group corresponding to the Lie algebra spanned by the operators in the canonical commutation relations (the Heisenberg algebra)

 

[Urs]

Hi lethe -

in short, the operators in the Weyl relation are the Lie group corresponding to the Lie algebra spanned by the operators in the canonical commutation relations (the Heisenberg algebra)

Wait, we have to get out nomenclature in sync.

What I am calling a Weyl algebra are operators U(a),V(a) which satisfy
U(a)V(b) = \exp(i 2\pi ab/\hbar)V(b)U(a). These
need not come from exponentiating elements of a Heisenberg algebra. But the Stone-vonNeumann theorem tells us that iff U and V are weakly-continuous, then they do come from an exponentiated Heisenberg algebra. Otherwise they don't. If they are weakly continuous, then you are right that U and V give the Lie group of the Heisenberg algebra, namley the Heisenberg group.

LQG is based on throwing away the Heisenberg algebra and concentrating on reps of the Weyl algebra U and V which are not weakly continuous.

 

[ranyart]

Originally posted by Urs
Maybe, maybe, maybe he is thinking of Smolin's latest attempt at merging Nelson's stochastic QM with quantum gravity

http://xxx.uni-augsburg.de/abs/gr-qc/0311059

where spin networks are indeed used as 'hidden variables' to produce QM dynamics from classical statistics.

Last time Lee Smolin tried the same with BFSS Matrix Theory

http://xxx.uni-augsburg.de/abs/hep-th/0201031 .

I used to consider this interesting,

http://groups.google.de/groups?selm=ahe52s$1a2q$1@rs04.hrz.uni-essen.de

though I am not so sure anymore.

Anyway, this is what ranyart's avant-garde poetry reminded me of. As with every piece of modern art, you have to search the answer within yourself. ;-)

Ah!..with a little detective work I see what you mean (which is not a literal reference to me seeing into your mind!)

 

[arivero]

Hmm yes, "QM from QG".

 

[Urs]

QM from QG

Right, that's what the Smolin paper is called. Unfortunately the statement implied by the title is either a tautology or circular.

 

[arivero]

No, if the idea is that QG is a new theory above QM, so that any classical field or particle defined inside the QG theory will magically be a quantum field or particle. Still, the paper does not go as fas as his title, because Nelson stochasticity is imposed, not deduced.

 

[ranyart]

This paper has some relevence to this post.

http://arxiv.org/PS_cache/hep-th/pdf/0403/0403108.pdf

 

[arivero]

Originally posted by ranyart
This paper has some relevence to this post.

http://arxiv.org/PS_cache/hep-th/pdf/0403/0403108.pdf

But quantisation of the strings is a very different matter, isn't it? To begin with, the string is already by itself a many-particle entity, so first quantisation should be enough.

 

[pelastration]

Originally posted by arivero
... the string is already by itself a many-particle entity

many-particle ... can you explain more?

 

[selfAdjoint]

Thomas Larson's SPR post

Thomas Larson has today posted a possible way forward for LQG of sci.physics research, here . Recall that Urs had said LQG required a factor in the commutator that he called V to obtain the Virasoro algebra, and noted that LQG theorists set V = 1.

Now Larson points us to a paper on the math-ph arxiv which discusses a great many (all?) the possible candidates for V.

 

[marcus]

More about Thiemann's LQG-string
(the project to realize string theory within the context of LQG)

Robert C. Helling, Giuseppe Policastro
String quantization: Fock vs. LQG Representations
19 pages
http://arxiv.org/abs/hep-th/0409182

---abstract---
We set up a unified framework to compare the quantization of the bosonic string in two approaches: One proposed by Thiemann, based on methods of loop quantum gravity, and the other using the usual Fock space quantization. Both yield a diffeomorphism invariant quantum theory. We discuss why there is no central charge in Thiemann's approach but a discontinuity characteristic for the loop approach to diffeomorphism invariant theories. Then we show the (un)physical consequences of this discontinuity in the example of the harmonic oscillators such as an unbounded energy spectrum. On the other hand, in the continuous Fock representation, the unitary operators for the diffeomorphisms have to be constructed using the method of Gupta and Bleuler representing the diffeomorphism group up to a phase given by the usual central charge.
---end quote---

 

[marcus]

Robert Helling's papers go back to 1998, he has co-authored with
Hermann Nicolai, has been much of the time at Albert Einstein Institute,
Potsdam MPI, has specialized in M-theory (from the looks of it)

It looks like Hermann Nicolai, a director at AEI Potsdam, who organized last year's String meet Loop conference (with Abhay Ashtekar), has perhaps encouraged Thiemann to try this merger of theories in the first place

this was what the StringMeetLoop conference last October was supposedly to lay the groundwork for. Nicolai does String/M and particle theory and he co-organized it with Ashtekar who does Loop and is a relativist.

At the conference they wanted to get the HEP people---the particle physicists---talking to the relativists---the General Relativity people. Both being concerned with quantizing gravity in their respective fashions.

But after, when Thiemann took the first step the reception was not so hopeful or encouraging, as I thought. More like the bluejays in the front yard when the neighbor cat comes to visit. All kinds of reasons offered why it could not possibly be right.

This Robert Helling article has a different tone of voice

here is the original Thiemann article, in case you have not already seen it:

The LQG -- String: Loop Quantum Gravity Quantization of String Theory I. Flat Target Space
Thomas Thiemann
46 pages
http://arxiv.org/hep-th/0401172

 

[selfAdjoint]

"...at least in the case of the quantum oscillator the polymer state is unphysical." That seems to be the death-knell fot the Thiemann approach. I wasn't aware of the fierce properties of the polymer state (no momenta), which as the authors say, is much used in LQG derivations. Back to the drawing board.

 

[marcus]

Helling's post without the LM comment

---quote from Robert Helling SPS post---
Thanks for noting our paper. Unfortunately, I am about to leave
Cambridge (my next postdoc is at IU Bremen, back in Germany) and all
my papers and notes are stored away in boxes and unaccesible to me at
the moment. So I cannot answer Urs' qustions about signs and
(anti)-commutativity. He might well be right and we screwed those up
but those would be just typos and wouldn't change anything substantial
in the conclusions. Furthermore, I don't have my laptop's network
connection currently running, thus I have to use google groups rather
than my regular news reader.

> Their description of the harmonic oscillator looks particularly strange
> because if we did not agree what is the physics of the quantum harmonic
> oscillator, we could probably agree about nothing in the world.[/color]

Mybe you missed that point but our philosophy was to say "this is what
you get when you apply LQG methods to the harmonic oscillator". Be
careful with them I am general because the _physical_ consequences (esp.
the spectrum) are not what you meassure in this easy example. It was
important to us not just to say that functions that jump have no place
in physics because you could never observe them. That would be too
easy (and in fact plain wrong: If you describe a D-brane (whose
physical existence I understand Lubos does not doubt) by a skyscraper
sheef then this is done exactly by a function (of the transverse
coordinates) that is zero everywhere except at one point where it has
a finite value).

[Moderator's note:...LM]

So the point of our discussion of the harmonic oscillator is that there
are measurable consequences of doingit 'the wrong way'.

[Moderator's note:... LM]

> I am sure that many of us tried to deal with divergent sums and divergent
> integrals, and various other singular objects in various ways that can be
> proved "wrong" on physical grounds.[/color]

Actually, this is a major reason for why this formalism is more
involved than the usual one: In the algebraic language (and this is
just mathemtically more careful language, no physical difference to
the usual approach) great care is taken to avoid divergent (and
similar) sums etc so no ambiguities (or ways to do it wrong) appear
from that. This is why careful people deal with the Weyl operators
e^{ix} and e^{ip} instead of the usual x and p: The Weyl operators are
bounded and thus problems with domains of definition etc do not arise.
For example, in the position representation p is the deriviative. But
not all functions in L^2(R) are differentiable. Only a dense subset is. But all are translateble. Thus one saves some complications (if
ones intend is to be careful) if one uses the better defined Weyl
operators instead. I am not saying that it cannot be done with x and
p, it's just you either close your eyes to mathematical subtleties
(which is what we physicists do most of the time and it works fine
most of the time) or you have to deal with limits and that stuff.

[Moderator's note: ... LM]

> All Hilbert spaces obtained from these wrong assumptions are
> non-separable, unphysical, and the only way how the non-separability can
> be cured is if the resulting theory is completely topological and all
> these values of "x" are eventually unphysical, perhaps except for their
> ordering.[/color]

The separability is not the issue.

[Moderator's note:... LM]

In fact, there are (accepted) physical systems with a non-separable
Hilbert space: One (as we remark in a footnote) are Bloch electrons.
That is electrons in a periodic potential.

[Moderator's note:...LM]

Then you know that the wave function is periodic as well. Ahem no, not
quite, only the physics is periodic. So the wave function is periodic up
to a phase. And by doing the intergral over the whole infinite crystal,
you find that two wave functions with different phases are orthogonal. So
for each point in the interval [0,2pi) of phases there is an orthogonal
sector in the Hilbert space. Thus the total Hilbert space is kind of the
L^2 of the unit cell to times the number of points in that interval, clearly a non-separable space.

[Moderator's note:... LM]

You could say that this happens only in the infinite crystal size
limit.

[Moderator's note:... LM]

But this idealization people usually are happy to make. Otherwise (with an
IR cut-off) there would for example be no phase transitions. But that is a
different matter.

> The states in this model represent unphysical mixtures of a
> hugely infinite number of superselection sectors - it is another
> description of Helling et al. comments about "discontinuity" of
> Thiemann's representation, I think. Each of these sectors is made of a
> single state. In physics, it is legitimate to study each single
> superselection sector separately - and if these superselection sectors are
> made of a single state, the theory is physically vacuous.[/color]

At least in the mathematical sense (and that is supposed to coincide
with the physical sense), a super-selection sector is a representation
of the quantum algebra. To states are in different sectors if they
are in inequivalent representations.

> > Everybody knows that first quantization is a mystery.[/color]
>
> "Why the world is quantum?" may be a mystery and the most counterintuitive
> insight about the world, but the mathematical operation behind the first
> quantization does not seem mysterious in any way, and it also does not
> seem ambiguous. Moreover, I don't know why you chose the "first
> quantization" because even it is a mystery, it is a smaller mystery than
> the "second quantization". ;-)[/color]

Was this just a joke? If not, here is why people say this (and usualy
this continues with "second quantization is a functor"): Of course if
your classical system has R^n as configuration space and its cotangent
bundle as the symplectic space then every child knows how to quntize:
Take L^2(R^n) as your Hilbert space and replace all x's by
multiplication operators and all p's by derivatives. Oh, and when
there's an ordering ambiguity, follow one or the other prescription
(but do that consistently).

However, what do you do, if I just give you some symplectic space and
don't tell you which are the simple preferred position and momentum
coordinates (and that's what x and p are, just coordinates). And as
the real world does not come with coordinates written on everything
one should have some recepy how to deal with this more general
situation. And then check that this reduces to the usual story (or an
equivalent one) in the simple situation. And this is what we have done
in the paper.

It is known, that there is no unique way to do this map from a
symplectic space to a Hilbert space with operators in general. There
are further choices involved.

[Moderator's note:... LM]

> I did not quite like their comments describing the algebras with different
> central charges as different "representations" (with an exclamation mark).[/color]

When we say algebra, we mean the C*-algebra of the observables. And
those are indeed the same (and only the represenations differ).

[Moderator's note:... LM]

These algebras are the algebras of the X's (or the W(f) after some
massaging). Then this Weyl algebra has representations. And on those
representations there is a symmetry (Lie-)algebra acting by unitary
operators. And this symmetry algebra is some Virasoro alegbra in both
cases. But these symmetry algebras have different central charges in
the two representations of the Weyl algebra. So: There are two kinds
of algebras, don't confuse them.

Furthermore, even if we didn't talk about it, in the Virasoro algebra
the central charge is just an abstract element usally called c. It
commutes with everything, so in an irreducible representation it is
represented by a number. And again, this number depends on the
representation. This number, together with h, the eigenvalue of L_0 in that representation, label a highest weight representation of the
algebra.

[Moderator's note:... LM]

> The only way how can one understand this sentence is that they claim that
> the Virasoro algebras with different central charges are isomorphic to
> each other.[/color]

Nobody claimed that. As I just said: In the algbra, c is an abstract
element, it becomes a number only in a representation. And nobody
claims that representations with different c are equivalent.

[Moderator's note:... LM]

> There is a common theme in Thiemann's papers which, I'm afraid, may
> unfortunately be shared by Helling and Policastro - which is that they
> often look at the "classical limit" of an algebra, and treat all the
> modifications implied by quantum mechanics as unimportant - and perhaps
> annoying? - details whose relevance for any of their conclusions goes to
> zero.[/color]

Could you be more specific with this claim?

[Moderator's note:... LM]

> By the way, this purely quantum viewpoint will be even more important if
> we want to get more insight into the (2,0) theory or even M-theory at the
> generic point of the moduli space - because these theories (at least in
> some backgrounds) clearly indicate that they cannot be fully obtained from
> a classical theory by quantization - and they almost certainly cannot be
> obtained from a *unique* classical theory.[/color]

Quantization is a game that always involves a classical system.
However nobody claimed that every qunatum system arises from the
quantization of a classical theory.

[Moderator's note:... LM]
-----end quote---

 

[marcus]

Part 2 of Helling's post

there was so much mod comment inserted into Helling's text that it was hard to see the overall intent of his post, I have elided the mod comment,
as in the preceding, to get a sense of the original.

---quote from Helling's SPS post---
> There are other examples in which Thiemann et al. try to make this sort of
> "quantization without quantization". They want the commutators to be
> always equal to the Poisson brackets;[/color]

I hope you don't include us in "et al". We impose the Poisson goes to
commutator rule only for linear combinations of what would be x and p,
not for higher powers. And I doubt Thomas would do commit that crime
either.

[Moderator's note:... LM]


> One may be trying to obtain a completely different framework of
> "quantization" - but there are several but's.[/color]

We tried hard to spell out the general framework of the quantization
procedure used in the two approaches. We say, that you can include the
polymer quantization if you do not impose the at first rather
technical condition of weak continuity. But then this has huge
observable consequences. So don't confuse framework and consequences.

[Moderator's note:... LM]

Could you spell out the rules for your framework that clearly rules
out the LQG one? It should be a machine that turns a symplectic space
with its observables into a Hilbertspace with its operators.

[Moderator's note:... LM]

> First of all, this procedure
> is not really quantization because it tries to preserve those properties
> of the classical theory that *cannot* hold in what is normally called a
> "quantum theory" - such as the exact equality between the commutators and
> the Poisson brackets.[/color]

Nobody imposes that. We only ask for a unitary representation of the
diffeomorphism symmetry. And those might obey the group law of the
diffeo group or not (because of an anomaly).

[Moderator's note:...LM]

> Second of all, it is not physics because no one has
> certainly seen a Thiemannian harmonic oscillator[/color]

Right. That we meant by "(un)physical" in the abstract.

- and no one ever will,
> simply because non-separable Hilbert spaces cannot be "seen".[/color]

See above.

> I find it mildly entertaining that the normal procedures of quantization -
> including quantization of the harmonic oscillator - are themselves
> pictured as an alternative approach.[/color]

Where?

[Moderator's note:... LM]

We describe both quantizations in a single framework. There is
one choice to be made. And that has physical consequences. In the
mechanics example, those consequnces are unphysical, so the choice was
wrong. Everybody is free to deduce something about the choice in the
string case.

> Well, of course that we do not need
> nonsensical non-separable spaces to describe the harmonic oscillator. Not
> only that: non-separable spaces do *not* describe the harmonic oscillator
> and they never did. Moreover, the standard procedure has been known since
> the mid 1920s, and it is the only one that can give physical predictions
> that reduce to the classical oscillator in the appropriate limit. It's
> great to rediscover this cool method of quantization in 2004, but it
> should not be viewed as something new.[/color]

We didn't say there is anything wrong with the standard harmonic
oscillator. Rather we used it as a test bed for the quantization
procedure. This was to counter arguments along the lines of 'nobody
has yet seen a string in nature".

> That's nice to hear because Robert Helling was the person who patiently
> required (in "Re: Background Independence", 2004-09-14 04:30:48 PST) that
>
> > RH: 4) The ground state of (quantum) GR should be diffeomorphism invariant
> > as otherwise diffeomorphisms would be spontaneously broken.[/color][/color]

Let's face it: In that thread you didn't get it that I was playing the
devil's advocate.

[Moderator's note: Well, sometimes I am confused who is your devil and who
is your God, or whoever is your devil's "alternative". ;-)

> > it is discussed that the singular GNS state can be interpreted as a thermal
> > state of infinite temperature![/color][/color]

One should be carefule as this is world sheet temperature and not
target space.

> > I didn't know this before and like that insight, because it points at a way
> > to understand a larger framework in which various "different" quantizations
> > (of the string for instance) appear as different aspects of the same thing.[/color]
>
> I am not getting the purpose of these attempts. Is the goal to be nice and
> to prove that no one can ever be completely silly?[/color]

If you like to express it that way...

Sorry, right now, I do not have more time to reply to the more polemic
parts of your post.

Robert
---end quote---

 

[marcus]

"...at least in the case of the quantum oscillator the polymer state is unphysical." That seems to be the death-knell fot the Thiemann approach. I wasn't aware of the fierce properties of the polymer state (no momenta), which as the authors say, is much used in LQG derivations. Back to the drawing board.

Your take on this was shared by Thomas Larsson today
https://www.physicsforums.com/showthread.php?t=44495
He says the way (according to Helling/Policastro) that LQG handles
the harmonic oscillator is fatal for LQG.

In effect, he tells the Loop Gravitists to go "back to the drawing board".

the first thing I notice is I am pleased with our reflexes
I posted notice of H/P on 19 September
then it appeared (I think the next day) on SPS
and now discussion has started (24 September) on SPR

You were sadly shaking your head 5 days before Thomas Larsson.

By now I am accustomed to surprises so I am waiting to see how this turns out and cannot really give a reaction.

I recall that Rovelli (and Daniele Colosi) had a paper last year about the Harmonic Oscillator. Their paper was called:
A simple background-independent hamiltonian quantum model
"...Our main tool is the kernel of the projector on the solutions of Wheeler-de Witt equation, which we analyze in detail..."
http://arxiv.org/abs/gr-qc/0306059

this is not to say that their paper has any bearing on the fatal harmonic oscillator disease discovered by Helling/Policastro (wouldnt seem to but I suppose it might)