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.