## January 27, 2004

### Thiemann’s quantization of the Nambu-Goto action

#### Posted by urs

Last year there was a symposium called Strings meet Loops at the AEI in Potsdam at which researchers in the fields of String Theory and Loop Quantum Gravity were supposed to learn about each other’s approaches. In his introductory remarks H. Nicolai (being a string theorist) urged the LQG theorists to try to better understand how their quantization approach compares to known results.

Since the worldsheet theory of the (super)string is nothing but (super)gravity in 1+1 dimensions coupled to other fields it would be an ideal laboratory to compare the results of LQG in this setting to the usual lore, which in particular features the central extension of the Virasoro algebra as well as consistency conditions on the number of target space dimensions.

How does this model fit into the framework of canonical and loop quantum gravity?

A search on the arXive showed that so far only one paper had appeared which did address aspects of this simple and yet somewhat decisive question:

Starodubtsev concluded:

The suggested [LGQ-like] version of the Hamiltonian constraint leaves us with a theory which is considerably different from ordinary string theory. There are several indications that string theory in its usual form can probably not be recovered from the model obtained. […] the first version of Hamiltonian constraint is anomaly-free and the same is true of the diffeomorphism constraint.

When, after the symposium, I mentioned this reference to A. Ashtekar, a leading figure in LQG, he told me that he meanwhile was aware of this result and planning to analyze the problem in more detail.

Apparently this has borne fruit by now, since yesterday a paper by Th. Thiemann appeared on the arXive

which gives a detailed analysis of an LQG inspired canonical quantization of the 1+1 dimensional Nambu-Goto action for flat target space. The approach is a little different from that by Starodubtsev, but the results are similar in their unorthodoxy: Thiemann finds

- no sign of a critical dimension

- no ghost states

- no anomaly, no central charge

- no tachyon (and, indeed, not the rest of the usual string spectrum).

The claim is that all this is possible due to a quantization ambiguity that has not been noticed or not been investigated before: Instead of using the usual Fock/CFT representation and imposing the constraints as operator equations, Thiemann uses families of abstract representations of the operator algebra obtained by the GNS construction and solves the quantum constraints by a method called group averaging, or its more sophisticated cousin, the so-called Direct Integral Method.

Since these are the same methods used in LQG for quantizing the gravitational field in 3+1 dimensions it is somewhat interesting to see how vastly different the results obtained this way are from the standard lore. One might hence take this as a sign that the LQG approach to quantization is odd. But in some circles this is interpreted in just the opposite way, dreaming of the possibility that the new quantization method might improve on the standard approach to quantization in string theory. Indeed Thiemann himself speculates in his conclusions that his quantization prescription might

- solve the cosmological constant problem

- clarify tachyon condensation [?]

- solve the vaccum degeneracy puzzle

- help finding a working phenomenological model

- help proving perturbative finiteness beyond two loops .

To my mind these are surprisingly bold speculations.

I would much rather like to understand conceptually the nature of the apparent quantization ambiguity (if it really is one) that is the basis for all this. Do we really have this much freedom in quantizing the NG action? Why then do several different quantization schemes (BRST, path integral, lightcone quantization) all yield the standard result which strongly disagrees with the one obtained by Thiemann? What is the crucial assumption in Thiemann’s quantization that makes it different from the ordinary one?

I believe that these questions are what originally motivated H. Nicolai to initiate this investigation and their answer should teach us something.

In the remainder of this entry I shall try to look at some of the technical details of Thiemann’s paper, trying to understand what exactly it is that is going on.

We all know from Edward Nelson that

First quantization is a mystery.

But it should be possible to understand how precisely it is mysterious and how it is not.

After an intensive discussion and some false attempts to explain what is going on inThomas Thiemann’s paper, he finally chimed in himself and we could clarify the issue at the technical level. The crucial point is the following:

Thomas Thiemann does not perform a canonical quantization of the Virasoro constraints if we want to understand under canonical quantization that a theory with classical first-class constraints ${C}_{I}$ is quantized by demanding

(1)$〈\text{phys}\mid {\stackrel{̂}{C}}_{I}\mid \text{phys}〉=0\phantom{\rule{thinmathspace}{0ex}}.$

What Thomas Thiemann instead does (by his own account) is the following:

1) Find a representation ${\stackrel{̂}{U}}_{\phi }$ of the classical symmetry group elements $\phi$ on some Hilbert space. (Here the ${\stackrel{̂}{U}}_{\phi }$ need not have anything to do with the quantized ${\stackrel{̂}{C}}_{I}$, and in the case of the ‘LQG-string they don’t have anything to do with them.)

2) Demand that physical states are invariant under the action of the ${\stackrel{̂}{U}}_{\phi }$.

It is clear that this method explicitly translates the classical symmetry group to the ‘quantum’ theory and hence cannot, by its very construction, ever find any anomalies and related quantum effects.

An interesting aspect of this is that exactly the same method is used with respect to the spatial diffeomorphism constraints in Loop Quantum Gravity (while the Hamiltonian constraint is quantized more in the usual way). It must therefore be emphasized that LQG is not canonical quantization in the sense that the classical first-class constraints are not promoted to hold as expectation value equations in the quantum theory.

For me, this is the crucial insight of this discussion, and it shows that Hermann Nicolai’s question did address precisely the right problem. In the toy example laboratory of the Nambu-Goto string it is much easier for non-experts (like me) to follow the details and implications of what is being done, than in full fledged LQG. And it turns out, to my surprise, that what is being done is a speculative proposal for an alternative to standard quantum theory. This is not only my interpretation, but Thomas Thiemann himself says that the procedure, sketched above, for dealing with the constraints, should be compared to experiment to see if nature favors it over standard Dirac/Gupta-Bleuler quantization.

I am open-minded and can accept this in principle, but this has not been obvious to me at all, before. It means that, in the strinct sense of the word ‘canonical’, LQG is not canonical at all but rather similar in spirit to other proposed modifications of quantum theory, like for instance those proposed to explain away the black hole information loss problem by modifying Schroedinger’s equation.

I have tried to discuss some of these insights here.

So let me try to recapitulate the key idea in Thiemann’s quantization of the Nambu-Goto action, as far as I understand it.

Let ${\pi }_{\mu }$ be the canonical momentum to the embedding variable ${X}^{\mu }$. The usual left and right-moving bosonic fields are (pointwise)

(1)${Y}_{±}^{\mu }:={\eta }^{\mu \nu }{\pi }_{\mu }±{X}^{\prime \mu }\phantom{\rule{thinmathspace}{0ex}}.$

Smearing them over an interval $I$ of the circle and contracting with some reak ${k}_{\mu }$ yields

(2)${Y}_{±}^{k}\left(I\right):={\int }_{I}d\sigma \phantom{\rule{thinmathspace}{0ex}}{k}_{\mu }{Y}_{±}^{\mu }\phantom{\rule{thinmathspace}{0ex}}.$

This are the fields that we want to represent as operators on some Hilbert space with commutation relation given by

(3)$\left[{\stackrel{̂}{Y}}_{±}^{\mu }\left(\sigma \right),{\stackrel{̂}{Y}}_{±}^{\nu }\left({\sigma }^{\prime }\right)\right]=2i{\delta }^{\prime }\left(\sigma ,{\sigma }^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}}.$

From these one obtains bounded operators by exponentiation

(4)${\stackrel{̂}{W}}_{±}^{k}\left(I\right):=\mathrm{exp}\left(i{\stackrel{̂}{Y}}_{±}^{k}\left(I\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

The point is that for these bounded operators the GNS construction applies which tells us how to represent any unital *-algebra by bounded operators on some Hilbert space $ℋ$, which will be called the kinematical Hilbert space (up to some details).

Now, the crucial difference to the usual Dirac quantization ,where the constraints ${C}_{I}$ are imposed as

(5)${\stackrel{̂}{C}}_{I}\mid \psi 〉=0$

seems to be that instead the technique of group averaging imposes the exponentiation of this, namely

(6)$\mathrm{exp}\left({\stackrel{̂}{C}}_{I}\right)\mid \psi 〉=\mid \psi 〉$

(in the weak sense discussed between eqs. (5.4) and (5.5) of Thiemann’s paper). Naively this might appear to be the same thing, but it is not at all!

As an example, consider the commutator of one of the Virasoro constraints ${\stackrel{̂}{V}}_{±}\left(\xi \right)$ with ${\stackrel{̂}{W}}_{±}^{k}\left(I\right)$. There is an operator ordering issue and dealing with that yields the usual result that the conformal dimension of these ${\stackrel{̂}{W}}_{±}^{k}$ depends on $k$. But now instead look at the exponentiated expression

(7)${e}^{{\stackrel{̂}{V}}_{±}}{\stackrel{̂}{W}}_{±}^{k}\left(I\right){e}^{-{\stackrel{̂}{V}}_{±}}={e}^{{\stackrel{̂}{V}}_{±}}\mathrm{exp}\left(i{\stackrel{̂}{Y}}_{±}^{k}\left(I\right)\right){e}^{-{\stackrel{̂}{V}}_{±}}=\mathrm{exp}\left(i{e}^{{\stackrel{̂}{V}}_{±}}{\stackrel{̂}{Y}}_{±}^{k}\left(I\right){e}^{-{\stackrel{̂}{V}}_{±}}\right)=\mathrm{exp}\left(i{\stackrel{̂}{Y}}_{±}^{k}\left(\varphi \left(I\right)\right)\right)={\stackrel{̂}{W}}_{±}^{k}\left({\varphi }_{±}\left(I\right)\right)\phantom{\rule{thinmathspace}{0ex}},$

where ${\varphi }_{±}$ here denotes the group element of $\mathrm{Diff}\left({\mathrm{S}}^{1}\right)$ associated with ${V}_{±}={V}_{±}\left(\xi \right)$ ($\xi$ is some smearing function).

The exponentiation in a sense removes all operator ordering ambiguities, since the conjugation operation (the similarity transform) ${e}^{{\stackrel{̂}{V}}_{±}}\phantom{\rule{thinmathspace}{0ex}}\cdot \phantom{\rule{thinmathspace}{0ex}}{e}^{-{\stackrel{̂}{V}}_{±}}$ acts on every ${\stackrel{̂}{Y}}_{±}^{k}$ seperately and there is no operator ordering issue in the commutator $\left[{\stackrel{̂}{V}}_{±},{\stackrel{̂}{Y}}_{±}^{k}\left(I\right)\right]$.

Without this operator ordering issue there is no anomaly, hence no critical dimension, no tachyon, etc.

I therefore believe that the quantum ambiguity between the two sides of

(8)${\stackrel{̂}{C}}_{I}\mid \psi 〉=0↔\mathrm{exp}\left({\stackrel{̂}{C}}_{I}\right)\mid \psi 〉=\mid \psi 〉$

is what is at the heart of the difference between Thiemann’s quantization and the usual OCQ/BRST quantization.

Am I wrong?

Even if this is about right, there is something related which I don’t quite understand yet. Somehow the center-of-mass degree of freedom of the string is missing from Thiemann’s original Hilbert space. In section 6.4 he re-incorporates it by using a D-parameter familiy of his original Hilbert space, which hence clearly was just that of string oscillations. What I am puzzled about is that the 0-mode of the momentum operator does not seem to be the same thing as ${\pi }_{\mu }\left({p}_{\nu }\right)$ above equation (6.36). It seems to me that the two should be identified, somehow, and that then the question whether there is a tachyon or not should be addressed by actually constructing group-averaged and hence physical states.

Posted at January 27, 2004 3:34 PM UTC

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### Re: Thiemann’s quantization of the Nambu-Goto action

Here is a copy of Luboš’ answer to a related post of mine on sci.physics.research:

On 27 Jan 2004, Urs Schreiber wrote:

> I was trying to figure out what exactly it is in Th. Thiemanns
> quantization hep-th/0401172 of what he calls the ‘LQG-string’ that
> makes it so different from the usual quantization. I now believe that
> the crucial issue is how to impose the constraints.

Exactly. If physics is done properly, the (Virasoro) constraints are not
arbitrary constraints that are added by hand. They are really Einstein’s
equations, derived as the equations of motion from the action if it is
varied with respect to the metric - in this case the worldsheet metric.
The term R_{ab}-R.g_{ab}/2 vanishes identically in two dimensions, and
T_{ab}=0 is the only term in the equation that imposes the constraint. The
constraints are really Einstein’s equations, once again.

Moreover, because the (correct) theory is conformal, the trace
T_{ab}g^{ab} vanishes indentically, too, and therefore the three
components of the symmetric tensor T_{ab} actually reduce to two
components, and those two components impose the so-called Virasoro
constraints (which are easiest to be parameterized in the conformal gauge
where the metric is the standard flat metric rescaled by a
spacetime-dependent factor). For closed strings, there are independent
holomorphic and independent antiholomorphic generators - and they become
left-moving and right-moving observables on the Minkowski worldsheet
after we Wick-rotate.

Thomas Thiemann does not appreciate the logic behind all these things, and
he wants to work directly with the (obsolete) Nambu-Goto action to avoid
conformal field theory that he finds too difficult. Of course, the
Nambu-Goto action has no worldsheet metric, and therefore one is not
allowed to impose any further constraints. They simply don’t follow and
can’t follow from anything such as the equations of motion.

Thiemann does not give up, and imposes “the two” constraints by hand. It
is obvious from his paper that he thinks that one can add any constraints
he likes. Of course, there are no “the two” constraints. If he has no
worldsheet metric, the stress energy tensor has three components, and
there is no way to reduce them to two. Regardless of the effort one makes,
two tensor constraints in a general covariant nonconformal theory can
never transform properly as a tensor - because a symmetric tensor simply
has three components - and therefore his constraints won’t close upon an
algebra. His equations are manifestly general non-covariant, in contrast
with his claims.

Equivalently, because he obtained these constraints by artificially
imposing them, they won’t behave as conserved currents. (In a general
covariant theory without the worldsheet metric, we can’t even say what
does it mean for a current to be conserved, because the conservation law
nabla_a T^{ab} requires a metric to define the covariant derivative.) If
they don’t behave as conserved currents, they don’t commute with the
Hamiltonian, and imposing these constraints at t=0 will violate them at
nonzero “t” anyway (the constraint is not conserved).

If one summarizes the situation, these constraints simply contradict the
equations of motion. It is not surprising. We are only allowed to derive
*one* equation of motion for each degree of freedom i.e. each component of
X, and this equation was derived from the action. Any further constraint
is inconsistent with such equations unless we add new degrees of freedom.

I hope that this point is absolutely clear. The equations of motion don’t
allow any new arbitrarily added constraints unless it is possible to
derive them from extra terms in the action (that can contain Lagrange
multipliers). The Lagrange multipliers for the Virasoro constraints *are*
the components of worldsheet metric, and omitting one component of g_{ab}
makes his theory explicitly non-covariant (even if Thiemann tries to
obscure the situation by using the letters C,D for the two components of
the metric in eqn. (3.1)).

The conformal symmetry is absolutely paramount in the process of solving
the theory and identifying the Virasoro algebra - isolating the two
generators T_{zz} and T_{zBAR zBAR} per point from the general symmetric
tensor. Conformal/Virasoro transformations are those that fix the
conformal gauge - i.e. the requirement that the metric is given by the
unit matrix up to an overall rescaling. Conformal theories give us T_{z
zBAR} (the trace) equal to zero, and this is necessary to decouple T_{zz}
and T_{z zBAR}. In two dimensions, the conformal transformations -
equivalently the maps preserving the angles - are the holomorphic maps
(with possible poles), and the holomorphic automorphisms of a closed
string’s worldsheet are generated by two sets of the Virasoro generators.

This material - why it is necessary to go from the Nambu-Goto action to
the Polyakov action and to conformal field theory in order to solve the
relativistic string and quantize it - is a basic material of chapter 1 or
chapter 2 of all elementary books about string theory and conformal field
theory. I think that a careful student should first try to understand this
basic stuff, before he or she decides to write “bombastic” papers boldly
claiming the discovery of new string theories and invalidity of all the
constraints (such as the critical dimension) that we have ever found.

In fact, I think that a careful student should first try to go through the
whole textbook first, before he publishes a paper on a related topic.
Thomas Thiemann is extremely far from being able to understand the chapter
3 about the BRST quantization, for example.

Thiemann’s theory has very little to do with string theory, and very
little to do with real physics, and unlike string theory, it is
inconsistent and misled. String theory is a very robust and unique theory
and there is no way to “deform it” from its stringiness, certainly not in
these naive ways.

> This may seem like essentially the same thing, but the crucial issue is
> apparently that the latter form allows to deal quite differently with
> operator ordering, which completely changes the quantization. In particular,
> it seems to allow Thiemann, in this case, to have no operator re-ordering at
> all, which is the basis for him not finding an anomaly, hence no tachyon and
> no critical dimension.

A problem is that you don’t know what you’re averaging over because his
“group” is not a real symmetry of the dynamics.

By the way, if you want to define physical spectrum by a
Gupta-Bleuler-like method, you must have a rule for a state itself that
decides whether the state is physical or not. In Gupta-Bleuler old
quantization of the string, “L_0 - a” and “L_m” for m>0 are required
to annihilate the physical states. This implies that the matrix element of
any L_n is zero (or “a” for n=0) because the negative ones annihilate the
bra-vector.

It is important that we could have defined the physical spectrum using a
condition that involves the single state only. If you decided to define
the physical spectrum by saying that all matrix elements of an operator
(or many operators) between the physical states must vanish, you might
obtain many solutions of this self-contained condition. For example, you
could switch the roles of L_7 and L_{-7}. However all consistent solutions
would give you an equivalent Hilbert space to the standard one.

The modern BRST quantization allows us to impose the conditions in a
stronger way. All these subtle things - such as the b,c system carrying
the central charge c=-26 - are extremely important for a correct
treatment of the strings, and they can be derived unambiguously.

> If this is true and Group averaging on the one hand and Gupta-Bleuler
> quantization on the other hand are two inequivalent consistent quantizations
> for the same constrained classical system I would like to understand if they
> are related in any sense.

No, they are not. What is called here the “group averaging” is a naive
classical operation that does not allow one any sort of quantization. You
can simply look that at his statements - such as one below eqn. (5.2) -
that in his treatment, the “anomaly” (central charge) in the commutation
relations (of the Virasoro algebra, for example) vanishes, are never
justified by anything. They are only justified by their simple intuition
that things should be simple. This incorrect result is then spread
everywhere, much like many other incorrect results. It is equally wrong as
simply saying that we have constructed a different representation of
quantum mechanics where the operators “x” and “p” commute with one
another.

The central charge - the c-number that appears on the right hand side of
the Virasoro algebra - is absolutely real and unique determined by the
type of field theory that we study (and the theory must be conformal,
otherwise it is not possible to talk about the Virasoro algebra). It can
be calculated in many ways and any treatment that claims that the Virasoro
generators constructed out of X don’t carry any central charge is simply
wrong.

There is absolutely no ambiguity in quantization of the perturbative
string. Knowing the background is equivalent to knowing the full theory,
its spectrum, and its interactions. There is no doubt that Thiemann’s
paper - one with the big claims about the “ambiguities” of the
quantization of the string - is plain wrong, and exhibits not one, but a
plenty of elementary misunderstanding by the author about the role of
constraints, symmetries, anomalies, and commutators in physics.

Let me summarize a small part of his fundamental errors again. He believes
many very incorrect ideas, for example that

* artificially chosen constraints can be freely imposed on your Hilbert
space, without ruining the theory and contradicting the equations of motion
* two constraints in 2 dimensions can transform as a general symmetric
tensor, and having a tensor with a wrong number of components does not
spoil the general covariance
* he also thinks that the Virasoro generators have nothing to do with the
conformal symmetry and they have the same form in any 2D theory
* in other words, he believes that you can isolate the Virasoro generators
without going to a conformal gauge
* classical Poisson brackets and classical reasoning is enough to
determine the commutators in the corresponding quantum theory
* anomalies in symmetries, carried by various degrees of freedom,
can be ignored or hand-waved away
* there is an ambiguity in defining a representation of the algebra of
creation and annihilation operators
* the calculation of the conformal anomaly does not have to be treated
seriously
* the tools of the so-called axiomatic quantum field theory are useful
in treating two-dimensional field theories related to
perturbative string theory
* if a set of formulae looks well enough to him, it must be OK and the
consistent stringy interactions and everything else must follow

Once again, all these things are wrong, much like nearly all of his
conclusions (and completely all “new” conclusions).

Thiemann himself admits that this is the same type of “methods” that they
have also applied to four-dimensional gravity. Well, probably. My research
of the papers on loop quantum gravity confirms it with a high degree of
reliability. Every time one can calculate something that gives them an
interesting but inconvenient result, they claim that in fact we don’t need
to calculate it, and it might be ambiguous, and so on. No, this is not
what we can call science. In science, including string theory, we have
pretty well-defined rules how to calculate some class of observables, and
all things calculated according to these rules must be treated seriously.
If a single thing disagrees, the theory must be rejected.

The inevitability of conformal symmetry for a controlled quantization of
the relativistic string - and for isolation (in fact, the definition) of
the Virasoro generators - is real. The theorems of CFT about its being
uniquely determined by certain data are also real. The conformal anomalies
of certain fields are also real. The two-loop divergent diagrams in
ordinary GR are also real. We know how to compute and prove all these
things, and propagating fog and mist can only obscure these
well-established facts from those who don’t want to see the truth.

I guess that this paper will demonstrate to most theoretical physicists -
even those who have not been interested in these “alternative” fields -
how bad the situation in the loop quantum gravity community has become.
There are hundreds of people who understand the quantization of a free
string very well, and they can judge whether Thiemann’s paper is
reasonable or not and whether funding of this “new kind of science”
should continue.

All the best
Lubos

Posted by: Urs Schreiber on January 28, 2004 3:01 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

Hi Luboš,

I see your general point, but would like to look at some of the issues you raised in more detail.

You say that the Nambu-Goto action is ‘obsolete’. But of course the NG action is classically equivalent to the Polyakov action and I think that in the critical number of dimensions the equivalence extends to the quantum theory. Furthermore, the Nambu-Goto action for the string is essentially the Dirac-Born-Infeld action (up to the worldsheet gauge field) of the D-string.

As far as I can see the constraints that Thiemann arrives at in equation (2.4) of his paper follow from standard canonical reasoning. One finds that the canonical momenta ${\pi }_{\mu }$ of the Nambu-Goto action as well as of the DBI action classically satisfy two identities which can be identified as constraints. At the classical level these constraints are precisely the (classical) Virasoro constraints that one also obtains by varying the worldsheet metric in the Polyakov action. Since the two actions are classically equivalent this is no surprise.

My point is that there should be a priori nothing wrong with looking at the Nambu-Goto action when studying the string. Indeed this is frequently done for instance when F-strings and D-strings are considered at the same time, as for instance in

Y. Igarashi, K. Itoh, K. Kamimura, R. Kuriki, Canonical equivalence between super D-string and type IIB superstring.

In equations (2.3) and (2.4) of this paper the authors in particular give the same two bosonic constraints of the Nambu-Goto action that Thiemann arrives at. Their action also involves superfields and the worldsheet gauge field, but this does not affect the general result that the Virasoro constraints follow from a canonical analysis of the Nambu-Goto action. I have spelled out the derivation (for the bosonic DBI action) in a recent entry. (By setting the worldsheet gauge field and the $C$ fields to zero this derivation directly restricts to that for the ordinary Nambu-Goto action).

My point is that it is maybe not fair to say that Thiemann artificially or freely chooses the constraints - at least not at the classical level. The constraints that he uses are, classically, the Virasoro constraints of the closed bosonic string.

My suspicion is rather that Thiemann devitates from standard reasoning when he defines what he wants to understand under quantizing the Virasoro constraints. Would you agree with this?

Let’s ignore the way on which we arrived at the classical Virasoro constraints (by starting from one of various classically equivalent actions) and concentrate on the question what it means to quantize them.

The standard procedure is to make Gupta-Bleuler quantization and use either creation/annihilation operator normal ordering or CFT techniques to make sense of the quantum representation of the classical Virasoro generators. This leads in the usual way to the anomaly, the shift a in (L_0 - a) and so on.

Thiemann claims (based on a large literature on quantization of constrained systems that is also the basis for loop quantum gravity) that there is an at least superficially different technique that can also be addressed as quantization of the Virasoro constraints. In the simple case at hand this is imposing the constraint the way mentioned right below equation (5.4), which essentially says that

(1)$〈\psi \mid \mathrm{exp}\left(\mathrm{constraints}\right)\mid {\psi }^{\prime }〉=〈\psi \mid {\psi }^{\prime }〉\phantom{\rule{thinmathspace}{0ex}},$

where the Hilbert space and the representation of the operators is not necessarily the usual Fock representation.

This is not equivalent to and not even implied by saying that

(2)$〈\psi \mid \mathrm{constraints}\mid {\psi }^{\prime }〉=0\phantom{\rule{thinmathspace}{0ex}}.$

Of course when I write this I am ignoring issues of what we really mean by writing $\mathrm{exp}\left(\mathrm{some}\mathrm{operator}\right)$, i.e. whether this is supposed to be normal ordered or regulated or what. I am trusting that this is taken care of by Thiemann’s rigorous construction of Hilbert spaces and operators on them, but I guess that Luboš disagrees with this. :-)

Posted by: Urs Schreiber on January 28, 2004 4:14 PM | Permalink | Reply to this

### Huh?

Thomas Thiemann does not appreciate the logic behind all these things, and he wants to work directly with the (obsolete) Nambu-Goto action to avoid conformal field theory that he finds too difficult. Of course, the Nambu-Goto action has no worldsheet metric, and therefore one is not allowed to impose any further constraints. They simply don’t follow and can’t follow from anything such as the equations of motion.

As I teach my students in the first days of my String Theory class, the Virasoro constraints follow straightforwardly from a canonical treatment of the Nambu-Goto string.

That is hardly the issue.

Posted by: Jacques Distler on January 29, 2004 2:58 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

Here is another reply by Luboš:

Dear Urs,

Concerning your comments that you can get rid of all ordering constants by
exponentiating something, I hope that you don’t really believe it because
this would be a complete misunderstanding of the singularities in quantum
field theory. The exponentials of something always store the same
information as “something”, and if one of them has some ordering constant
contribution, you see it in the other as well.

For example, X(z) X(0) have logarithmic OPEs. This implies that
exp(i.K.X(z)) has a power law OPE with exp(-i.K.X(z)). It’s totally
nonsensical at quantum level to imagine that exp(-i.K.X(z)) is an inverse
operator to exp(i.K.X(z)). Do you understand why? This is a very
important point.

While for the Virasoro group without the central charge you would be able
to write the explicit “exponentiated” elements of the reparameterization
group and - because they have a clear geometric interpretatino, you could
invert them without anomalies, it is simply not true for the Virasoro
operators generating the reparameterization of X’s. Because of the term
c/z^4 in the OPE of two stress energy tensors, you must know very well
that exp(-V) can’t be treated as the inverse of exp(+V). You can only
imagine that exp(V) is an honest element of a group if the OPEs of V with
itself - and all other “V“‘s that you want to use - only have the 1/z
term, corresponding to the commutator. Recall that

O1(z) O2(0) ~ [O1,O2] (z) / z

the coefficient of 1/z is schematically the commutator of the two
operators. If you integrate a stress energy tensor etc., it is also OK to
have the 1/z^2 term in the OPEs of the stress energy tensor because it
reflects the worldsheet dimension of the stress energy tensor and tells
you how should you integrate it to get scalars etc.

But the OPE of the stress energy tensor (of the X^mu CFT) with itself
contains an extra 1/z^4 term. This is just a fact that you can calculate
in many ways, and this simply means that exp(V) where V is a Virasoro
generator, or some integrated combination of the stress energy tensor,
does not behave as an honest element of some group, and exp(-V) is not in
any naive sense inverse to exp(V) because these two *operators* have
singularities.

Note that his naive operation, involving the (wrong) application of the
formula

exp(C.D.C^{-1}) = C exp(D) C^{-1}

which is OK for matrices, is incorrect in our “usual” representation of
CFT, because of singularities between C and C itself. You can’t imagine
that C^{-1} is inverse to C - there are just no meaningful operators on
the Hilbert space that would look like C=exp(V) and were inverse to one
another. Because C^{-1}.C is not really one, you can’t derive the formula
you derived either, unless c=0. Note that it even requires you, for
C=exp(V), to consider exp(exp(V)…). These are heavily singular
operators, and all these confusions simply come from his/their wrong
intuition that you can work with the operators in CFT as with ordinary
classical numbers. They don’t understand where the normal ordering terms
come from, they don’t understand singularities of operators in quantum
field theories, they don’t understand the difference between classical and
quantum field theory.

It’s just totally pathetic, and every student in theoretical physics
should be able to identify all these errors.

All the best
Lubos

Posted by: Urs Schreiber on January 28, 2004 4:17 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

Hi again, Luboš!

Yes, I understand everything that you say here. I know that $:\mathrm{exp}\left(-V\right):$ is not the inverse to $:\mathrm{exp}\left(V\right):$ in CFT and I do understand where the $1/{z}^{4}$ terms come from. When you go back to my original entry you’ll see that I address precisely this phenomenon by mentioning that things like $:\mathrm{exp}\left(k\cdot X\right):$ have conformal dimension depending on $k$ in CFT, which is another aspect of this phenomenon.

But, yes, I was taking for granted that Thiemann is using a rep of his operators that allows him to ignore all normal ordering issues and work with them as with matrices and hence not as in CFT. He is referring to lot’s of mathematical theorems, using the GNS construction etc. (that I obviously haven’t checked myself and I am trusting that he applies them correctly) and even though he does not say so explicitly I deduced from his paper, in particular from the the third paragraph on p. 20, that he does use

(1)$\mathrm{exp}\left(C\cdot D\cdot {C}^{-1}\right)=C\mathrm{exp}\left(D\right){C}^{-1}\phantom{\rule{thinmathspace}{0ex}}.$

I do understand that this does not make sense in CFT (or even any other quantum field theory in the usual sense) but I also believe that a large number of mathematically versed people in the LQG camp do think that this can be given good meaning by using all these mathematical constructions that Thiemann alludes to. Unfortunately I am not an expert on this stuff.

I think the key ingredient is the GNS construction, which tells you that a unital *-algebra can be represented faithfully i.e. without normal ordering issues just like matrices on some Hilbert space. That’s the content of the relation in the 9th line from below on p.15:

(2)$\left[a\right]\left[b\right]=\left[\mathrm{ab}\right]\phantom{\rule{thinmathspace}{0ex}}.$

On the right hand side is the classical multiplication of the algebra, on the left hand side we have operator multiplication. Whenever this is true we do have

(3)${\left(\mathrm{exp}\left({\pi }_{\omega }\left(a\right)\right)\right)}^{-1}=\mathrm{exp}\left(-{\pi }_{\omega }\left(a\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

There is some fine print to this construction which I am maybe not fully aware of. In particular things need to be bounded for this to make sense. That’s why Thiemann uses the operators $\stackrel{̂}{W}=\mathrm{exp}\left(i\stackrel{̂}{Y}\right)$ instead of the $\stackrel{̂}{Y}$ themselves, because these would be unbounded.

Posted by: Urs Schreiber on January 28, 2004 4:41 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

For the general discussion of Thiemann’s paper I think it is important to realize that much of the usual lore of QFT is not supposed to apply. In particular, there is, as far as I understand, nothing like a double Wick contraction in the commutator of two Virasoro generators.

Let me spell this out in detail:

Assume that we have operators $Y\left(\sigma \right)$, $\sigma \in {S}^{1}$ which have the commutator

(1)$\left[Y\left(\sigma \right),Y\left({\sigma }^{\prime }\right)\right]=-{\delta }^{\prime }\left(\sigma ,{\sigma }^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}},$

as in equation (6.4) of Thiemanns paper. Next assume that one can make sense of products of these operators $Y\left(\sigma \right)Y\left(\sigma \right)$ at equal points, without introducing any notion of normal ordering. This can be either thought of as pertaining to the classical Poisson algebra or, according to Thiemann et. al (if I understand correctly), by using a special representation on a special Hilbert space ${ℋ}_{\omega }$ obtained by the GNS construction. Anyway, assume that the following expression makes sense:

(2)${L}_{\xi }:=\frac{1}{2}{\int }_{{S}^{1}}d\sigma \phantom{\rule{thinmathspace}{0ex}}\xi \left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}Y\left(\sigma \right)Y\left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}.$

The point is not to worry, for the moment, how this object is supposed to act on some state, but merely to regard its algebraic relations.

(3)$\left[{L}_{\xi },Y\left(\sigma \right)\right]={\left(\xi \left(\sigma \right)Y\left(\sigma \right)\right)}^{\prime }\phantom{\rule{thinmathspace}{0ex}}.$

This is nothing but what one also gets by using classical Poisson brackets, too.

For convenience, let me introduce some notation: For a general field $A\left(\sigma \right)$ let $w\left(A\right)$ be the classical conformal weight of $A\left(\sigma \right)$ iff

(4)$\left[{L}_{\xi },A\left(\sigma \right)\right]=\xi \left(\sigma \right){A}^{\prime }\left(\sigma \right)+w\left(A\right){\xi }^{\prime }\left(\sigma \right)A\left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}.$

It is easy to check that

(5)$w\left(A\left(\sigma \right)B\left(\sigma \right)\right)=w\left(A\left(\sigma \right)\right)+w\left(B\left(\sigma \right)\right)$

so that

(6)$w\left(Y\left(\sigma \right)\right)=1$

and

(7)$w\left(Y\left(\sigma \right)Y\left(\sigma \right)\right)=2$

and so on.

Now, denote for any field $A\left(\sigma \right)$ and any complex-valued function $\xi$ on ${S}^{1}$ the $\xi$-mode of $A$ by ${A}_{\xi }$, i.e.

(8)${A}_{\xi }:=\int d\sigma \phantom{\rule{thinmathspace}{0ex}}\xi \left(\sigma \right)A\left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}.$

Using again the naive quantum mechanical commutation relations (or Poisson brackets) one finds the following transformation of such modes

(9)$\left[{L}_{{\xi }_{1}},\phantom{\rule{thinmathspace}{0ex}}{A}_{{\xi }_{2}}\right]={A}_{\left(w-1\right){\xi }_{1}^{\prime }{\xi }_{2}-{\xi }_{1}{\xi }_{2}^{\prime }}\phantom{\rule{thinmathspace}{0ex}}.$

This implies in particular that

(10)$\left[{L}_{{\xi }_{1}},{L}_{{\xi }_{2}}\right]={L}_{{\xi }_{1}^{\prime }{\xi }_{2}-{\xi }_{1}{\xi }_{2}^{\prime }}\phantom{\rule{thinmathspace}{0ex}}.$

This is of course nothing but the usual relation known from classical Poisson brackets of the classical Virasoro constraints, as reviewed for instance by Thiemann in his equation (3.3). There is no anomaly because one assumed to have no need to consider normal ordering as in $:{L}_{\xi }:$ or the like and all operator products are assumed to behave like classical products. But the important point seems to be that Thiemann claims that secion 6.2 of his paper gives us a way to make sense of the above algebraic expressions as relations between operators that are well defined on some Hilbert space ${ℋ}_{\omega }$. This is how he gets a representation of the conformal group on his Hilbert space without having a conformal anomaly.

Posted by: Urs Schreiber on January 28, 2004 6:48 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

I do not understand Thiemann’s paper at all so I’m going to ask totally naive questions. It seems clear that if Thiemann’s construction is consistent, this new theory is nothing like a 1+1 dimensional field theory, so it might be that my questions will not even make sense in this framework.

First, how does he calculate the spectrum? He says in some place that the graviton state is gauge-dependent, which just boggles me out. What are the observables?

Secondly, can he write down the operator corresponding to X, and see what its commutation relations are?

Let me say I also share Lubos’ view about such grandiose claims. It doesn’t improve my confidence in the paper when he blithely disregards all the previous literature about quantizing the string.

Posted by: Arvind on January 28, 2004 11:17 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

Just a very quick and brief comment for the moment: Thiemann claims to be able to construct an operator representation $\stackrel{̂}{W}$ of the classical observables $W\left(\xi \right)=\mathrm{exp}\left(\int d\sigma \phantom{\rule{thinmathspace}{0ex}}\xi \left(\sigma \right)\left(i\frac{\delta }{\delta X\left(\sigma \right)}±{X}^{\prime }\left(\sigma \right)\right)$ that essentially behaves just as the classical $W$. This way the Pohlmeyer charges, uncontroversial classical invariants of the string, become quantum ‘charges’ for him.

I believe that if instead of Pohlmeyer charges we use classical DDF states (in the hopefully obvious sense) this construction would even give something similar to the usual string spectrum (up to the offset $a$ and the existence of null states).

We should (at least I should) try to understand if and why the claim about $\stackrel{̂}{W}$ can be correct. This is where the mystery lies, I believe.

Posted by: Urs Schreiber on January 29, 2004 2:01 AM | Permalink | Reply to this

### There’s no place like home

Why don’t I just close my eyes, click my heels and wish away all anomalies?

What are the rules here?

It is well known that it is impossible to preserve all of the relations of the classical Poisson-bracket algebra as operator relations in the quantum theory.

What principle allows Thiemann to decide which relations will be carried over into the quantum theory?

Where does he discuss which relations fail to carry over?

Posted by: Jacques Distler on January 29, 2004 3:20 AM | Permalink | Reply to this

### Re: There’s no place like home

Maybe the issue is seperability of the Hilbert space.

Is the Hilbert space Thiemann constructs in his paper seperable? Unless I am missing something it is apparently not. This might explain why things work very different on this Hilbert space than on the ordinary seperable one.

Compare the situation in the what is called “loop quantum cosmology”. There, after the dust has settled, what is done is essentially an ordinary quantization of the Wheelder-DeWitt equation but on a nonseperable Hilbert space, where, if $a$ is the scale factor of the universe, all states of the form $\mathrm{exp}\left(\mathrm{ip}a\right)$ for real $p$ are orthonormal simply by postulating a non-standard scalar product $〈\cdot \mid \cdot 〉$ with respect to which

(1)$〈{e}^{i{p}_{1}a}\mid {e}^{i{p}_{2}a}〉:=1\mathrm{if}{p}_{1}={p}_{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{and}=0\mathrm{otherwise}\phantom{\rule{thinmathspace}{0ex}}.$

This is what makes the operator $\partial /\partial a$ technically have a discrete spectrum (where eigenstates are normalizable) even though its eigenvalues have a continuous range. (I am doing this from memory. Probably I mix up some details. Maybe the role of $\stackrel{̂}{a}$ and $\partial /\partial a$ in the above has to be exchanged.) This is the basis on which loop quantum cosmology obtains a discrete evolution of the scale factor, somehow (unfortunately I didn’t understand how precisely this follows when hearing a talk about this once).

So quantization on a non-seperable Hilbert space leads to radically different quantum theories. Maybe that’s what happens in Thiemann’s paper on the quantization of the string?

Posted by: Urs Schreiber on January 29, 2004 7:17 AM | Permalink | Reply to this

### Re: There’s no place like home

Hi all,

I can’t follow any of this in detail at the moment, but my impression from talking to some of Ashtekar’s student is that non-separable Hilbert space is generic in their quatization. So, Urs may be on the correct reasoning here. Now , then the issue is how to recover the ordinary classical world from it. I know Josh Willis did some work (and still working on it) on the issue.

Demian Cho

Posted by: Demian on January 29, 2004 2:16 PM | Permalink | Reply to this

### Re: There’s no place like home

Hi Demian,

I have just finished posting a lengthy message to s.p.r. explaining why I think Thiemann’s Hilbert space is indeed non-seperable - when your comment comes in! :-)

Good, so my memory was correct that LQG usually deals with non-seperable Hilbert spaces. I didn’t fully realize this until I heard a talk by Bojowald at ‘Strings meet Loops’ where he mentioned that this is a crucial issue in his ‘loop quantum cosmology’.

You write:

Now , then the issue is how to recover the ordinary classical world from it. I know Josh Willis did some work (and still working on it) on the issue.

Hm. Maybe I am confused, but right now it seems that there is way too much classicality in Thiemann’s paper and that we’d rather like to understand how to recover the ordinary quantum world in his approach! :-)

For instance, he emphasizes that his choice of inner product = choice of $\omega$ is just meant to be a simple example (although he also mentions that other examples may be hard to come by). Would other $\omega$ maybe yield seperable Hilbert spaces?

As far as I understand this could be possible, but it does not appear to be likely. Maybe Josh Willis could comment on this point?

Posted by: Urs Schreiber on January 29, 2004 2:45 PM | Permalink | Reply to this

### Re: There’s no place like home

On p. 115 of

it says indeed

We remark that the spin-network basis is not countable because the set of graphs in $\sigma$ is not countable, whence ${ℋ}^{0}$ is not separable. We will see that this is even the case after moding out by spatial diffeomorphisms although one can argue that after moding out by diffeomorphisms the remaining space is an orthogonal, uncountably infinite sum of superselected, mutually isomorphic, separable Hilbert spaces.

Posted by: Urs Schreiber on January 29, 2004 3:01 PM | Permalink | Reply to this

### Re: There’s no place like home

Too many forums! :)

I just posted some of my thoughts at the Physics Forum

http://www.physicsforums.com/

Which is the preferred place to discuss this: there, here, spr? :)

The point is, Urs, I think our work may be relevent to “fixing” the problem of Hilbert spaces in LQG.

You and I haven’t talked much about topology in our approach, but of course, any topology we have will be non-Hausdorff. However, it does have a property that I sometimes think of as being “weakly Hausdorff”. I hope I am not clashing with standard terminology there. Points in our space are not separable in general, but they are weakly separable. What I mean by that is that two points are separable if they are not contained in the same D-diamond. This gives a kind of “blurriness” down at the level of individual cells, but carries the usual notion of separability as long as you back away from the cells a bit.

Very exciting stuff :)

Eric

Posted by: Eric on January 29, 2004 3:15 PM | Permalink | Reply to this

### Re: There’s no place like home

Hi Eric!

Here, there, everywhere (imagine the respective Beatles tune :-)

I would vote for discussion here at the Coffee Table. You can be sure that I read this, while I will not regularly check the Physics Forum, in general.

Regarding your point on separability: Is the notion of separability of a Hilbert space really related to separability of points in the sense of Hausdorff/non-Hausdorff? I think with respect to Hilbert spaces separability simplye means ‘has a countable basis’. Is this related to a Hausdorff property, somehow?

Posted by: Urs Schreiber on January 29, 2004 3:31 PM | Permalink | Reply to this

### Re: There’s no place like home

Regarding your point on separability: Is the notion of separability of a Hilbert space really related to separability of points in the sense of Hausdorff/non-Hausdorff? I think with respect to Hilbert spaces separability simplye means ‘has a countable basis’. Is this related to a Hausdorff property, somehow?

Hi Urs,

Don’t forget that my math sucks :) I don’t know of any theorem that relates the two ideas, but it feels right. What you wrote (somewhere between here, there, and everywhere :)) made me think that the two illnesses were related. For example, you said

the W(I) are not sensitive to ‘neighbouring’ W(J): The Hilbert space is by construction so large that W({x}) and W({x+epsilon}) can sit right next to each other without noticing each other.

I am probably totally off and should just shut up :) At least I can say I’m having fun :)

The real point is that their inner product seems to be sick, and I think that our work (or maybe Harrison’s) could be step in the direction of trying to fix it. Maybe not.

Eric

Posted by: Eric on January 29, 2004 3:45 PM | Permalink | Reply to this

### Re: There’s no place like home

What’s a separable Hilbert space, and why does it matter?

Posted by: Arvind on January 29, 2004 4:45 PM | Permalink | Reply to this

### Re: There’s no place like home

A Separable Hilbert Space is one with a countable basis.

I don’t think it matters a whit.

Posted by: Jacques Distler on January 29, 2004 5:14 PM | Permalink | Reply to this

### Re: There’s no place like home

Hm, don’t you think that the reason that Thiemann can work with his operators essentially as if dealing with a classical Poisson algebra is due to the peculiar nature of the Hilbert space that he chooses?

Posted by: Urs Schreiber on January 29, 2004 5:19 PM | Permalink | Reply to this

### Re: There’s no place like home

No, I don’t believe there’s any quantization scheme that takes the full Poisson-bracket algebra of the classical theory and carries it over — unaltered — into the operator algebra of the quantum theory.

Depending on the quantization scheme, you may be able to carry over some subalgebra (the prototypical example, being the CCRs).

Posted by: Jacques Distler on January 29, 2004 7:34 PM | Permalink | Reply to this

### Re: There’s no place like home

I don’t believe either that the full Poisson algebra carries over. (There is a theorem showing that this cannot work in general.) That’s why I added the qualifier ‘essentially’. My point is, which I have been discussing with Luboš here, that in Thiemann’s quantization many more properties of the classical algebra carry over than just the CCR. That, of course, not the entire classical algebra is reproduced is the content of section 6.6 of Thiemann’s paper, where he discusses the quantum deformations of the classical invariant algebra of the Pohlmeyer charges.

But what is crucial for Thiemann’s removal of the anomaly is that things like

(1)${\left(\mathrm{exp}\left(\left({\pi }^{\mu }+i{X}^{\prime \mu }\right)\right)\right)}^{-1}=\mathrm{exp}\left(-\left({\pi }^{\mu }+i{X}^{\prime \mu }\right)\right)$

do hold true in his quantization, as oppsosed to the analogous normal-ordered relations in CFT. You can see this explicitly in his equation (6.7) and implicitly in the absolutely crucial relation

(2)$\alpha \left(W\left({Y}_{±}\right)\right)=W\left(\alpha \left({Y}_{±}\right)\right)$

in the third paragraph on p. 20. (Here $\alpha$ is the action of the exponentiated Virasoro generators.)

As far as I can see Thiemann’s quantization is technically correct (no mathematical errors). So there must be some physical assumption which makes him part company with the usual lore.

I think that it is crucial that he allows himself to work on non-separable Hilbert spaces. His construction of a Hilbert space by applying the GNS theorem to the Weyl algebra of $W$ operators is what allows the above-mentioned non-standard quantum relations, but it also leads to a non-separability of the Hilbert space.

Of the kinematical Hilbert space that is. It is not too surprising that the physical Hilbert space is separable again, because it is obviously much ‘smaller’ in general. But, when comparing his quantization with the OCQ or BRST quantization (instead of the LCQ, where only the physical Hilbert space appears because the constraints are solved before quantization) we have to look at the kinematical Hilbert space, because the Hilbert space on which the CFT operators are represented in the usual approach is also kinematical (contains non-physical states). The physical Hilbert space in the usual approach is that generated by the DDF operators acting on physical massless/tachyonic states.

Of course non-separable Hilbert spaces do appear in practice from time to time, but then we are always dealing with uncountably many superselection secors, each of which is separable. Thiemann’s non-countable Hilbert space (and, by the way, I have just received email by him confirming that the kinematical Hilbert space in his paper is non-separable) does however not separate into superselction sectors each of which would carry a representation of the constraints.

I think there are two alternatives:

1) Either there is a technical, mathematical error in Thiemann’s paper and hence his conclusions are wrong. If you believe that this is the case, that his quantization in particular is flawed, then please point out where you think the mistake lies.

2) Or the math is correct (which I am pretty convinced that it is). In this case we need to talk about if the assumptions that are made before the crank of the formalism is turned are viable. I am suggesting that the assumption of a non-separable kinematical Hilbert space may be a physically non-viable assumption.

Posted by: Urs Schreiber on January 30, 2004 12:16 PM | Permalink | Reply to this

### Re: There’s no place like home

OK, so you (he) claim(s) that there is a quantization in which the commutation relations of $X\prime \left(\sigma \right)$, $\Pi \left(\sigma \right)$, ${T}_{++}\left(\sigma \right)$ and ${T}_{--}\left(\sigma \right)$ are carried over from the classical Poisson-bracket algebra, unaltered (i.e., the commutators of the $T$’s do not pick up a central term)?

Certainly, that’s not true if the $T$’s lie in the universal enveloping algebra generated by $X\prime \left(\sigma \right)$, $\Pi \left(\sigma \right)$ — as is conventionally the case.

Posted by: Jacques Distler on January 30, 2004 3:17 PM | Permalink | Reply to this

### Re: There’s no place like home

So what is wrong with this?

Posted by: Urs Schreiber on January 30, 2004 3:26 PM | Permalink | Reply to this

### Re: There’s no place like home

You mean aside from the fact that none of the symbols are well-defined?

Look, this is elementary stuff.

We can expand everything in Fourier modes. If ${T}_{++}$ is in the universal enveloping algebra of the Fourier modes of $X\prime$ and $\Pi$, then its Fourier modes (conventionally called ${L}_{n}$) are some expressions quadratic in those modes.

Since the Fourier modes of $X\prime$ and $\Pi$ (the “oscillators”) don’t commute, you need to specify an ordering. I don’t care what ordering you choose, but I insist that you choose one.

Now compute the commutator of two ${L}_{n}$’s. Again, you will obtain something which is at most quadratic in oscillators (there will, in general, also be a piece ${0}^{\mathrm{th}}$-order in oscillators). And it must be re-ordered to agree with your original definition of the ${L}_{n}$s.

Carrying out this computation, you obtain the central term in the Virasoro algebra, and I believe that it is a theorem that the result is independent of what ordering you chose for the ${L}_{n}$s.

Note that I never mentioned what Hilbert space I hope to represent these operators on. So I don’t see where its separability (or lack thereof) enters into the considerations.

Posted by: Jacques Distler on January 30, 2004 5:02 PM | Permalink | Reply to this

### Re: There’s no place like home

You mean aside from the fact that none of the symbols are well-defined?

I don’t see why as an algebra these symbols should not be well defined. All the caveats that I included pertained only to the representation of these things as operators.

[…] you need to specify an ordering. I don’t care what ordering you choose, but I insist that you choose one.

Ok, let me choose the ordering the way it drops out from the Fourier decomposition without reordering:

(1)${L}_{m}=\frac{1}{2}\sum _{k=-\infty }^{\infty }{\alpha }_{m-k}{\alpha }_{k}\phantom{\rule{thinmathspace}{0ex}}.$

I could open Green, Schwarz & Witten on p. 73, where they derive the classical algebra of this object and check that in going from their (2.1.83) to (2.1.84) there is no re-ordering involved. But let me write it out here in a different way:

Using

(2)$\left[{L}_{m},{\alpha }_{k}\right]=-k{\alpha }_{k+m}$

one gets

(3)$\left[{L}_{m},{L}_{n}\right]=\frac{1}{2}\sum _{k}\left[{L}_{m},{\alpha }_{n-k}{\alpha }_{k}\right]$
(4)$=\frac{1}{2}\sum _{k}\left(\left[{L}_{m},{\alpha }_{n-k}\right]{\alpha }_{k}+{\alpha }_{n-k}\left[{L}_{m},{\alpha }_{k}\right]\right)$
(5)$=\frac{1}{2}\sum _{k}\left(\left(k-n\right){\alpha }_{n+m-k}{\alpha }_{k}-k{\alpha }_{n-k}{\alpha }_{m+k}\right)$
(6)$=\frac{1}{2}\sum _{k}\left(\left(k-n\right){\alpha }_{n+m-k}{\alpha }_{k}+\left(m-k\right){\alpha }_{n+m-k}{\alpha }_{k}\right)$
(7)$=\left(m-n\right)\frac{1}{2}\sum _{k}{\alpha }_{n+m-k}{\alpha }_{k}$
(8)$=\left(m-n\right){L}_{m+n}\phantom{\rule{thinmathspace}{0ex}}.$

There is no reordering involved in this.

I believe that it is a theorem that the result is independent of what ordering you chose for the ${L}_{n}$’s.

Do you have a reference to this theorem?

Posted by: Urs Schreiber on January 30, 2004 6:40 PM | Permalink | Reply to this

### Re: There’s no place like home

Good God! If you’re going to be that sloppy manipulating divergent quantities, we had better quit discussing this now.

Cut off those infinite sums (i.e., rather than ${\sum }_{k=-\infty }^{\infty }$, consider ${\sum }_{k=-N}^{N}$) and try again.

The only ${L}_{n}$ with an ordering ambiguity is ${L}_{0}$, so it suffices define an ordering for it. To compute central term, it suffices to compute the commutator of $\left[{L}_{n},{L}_{-n}\right]$.

Posted by: Jacques Distler on January 30, 2004 7:15 PM | Permalink | Reply to this

### Re: There’s no place like home

Dear Jacques -

you write:

Good God! If you’re going to be that sloppy manipulating divergent quantities, we had better quit discussing this now.

I hope I am not annoying you. I very much appreciate that you take the time to discuss these things with me.

You seem to be very convinced that Thiemann (and myself, for that matter) are confused about a very elementary point. As for myself I don’t see my mistake yet, but it is of course well possible that I am subject to misapprehensions. Certainly you don’t have infinite time to waste on this - but please be at least assured that your contributions are very valuable to me and probably to others, who are interested in Thomas Thiemann’s work.

I believe that if you, or other string theorists, can point out technical mistakes in Thiemann’s paper that this will have considerable effect on the Loop Quantum Gravity people in general. Thiemann’s quantization can be regarded as a testing ground, a laboratory, for the techniques used in LQG. The LQG camp is well known for its high esteem of mathematical rigour and it would be very important to them to be made aware of a technical mistake. Physical viability of their approach is another matter, but I do expect that they care about the consistent definition of the objects that they are dealing with.

Thomas Thiemann, as you know, is one of the more prominent people working on LQG, and he has a record of papers with a rather high technical level. I have heard string theorists criticising his papers as being games of math instead of physics. But in any case the claim is that this math is well done. So I bet that he and many others in the LQG field would highly appreciate if string theorists can spot technical mathematical mistakes in their work.

Because of this I would kindly ask you not to give up on me and my attempts to answer to your charges. I may not be the most suitable person for that task and am indeed hoping that somebody more knowledgeable will chime in to help me out. I have indeed contacted Thomas Thiemann by email and he says that next week he’ll be back from a conference and willing to discuss his paper. Surely he’ll be a better advocate of his work than I am.

That said, let me try to answer your latest comments. Unfortunately, as you will see below, I will still not be able to completely understand your criticism. Please bear with me. Thanks!

You write:

Cut off those infinite sums […] and try again.

Ok. At least I can reassure you that I do understand that cutting off these sums does modify the algebraic relation $\left[{L}_{n},{L}_{m}\right]=\left(n-m\right){L}_{n+m}$.

But, alas, I don’t see what the cutting off of these sums has to do with the question whether there is a non-commutative algebra such that it has commutators which reproduce the Poisson-brackets of the oscillators ${a}_{n}$ and the generators ${L}_{n}$.

You are saying that I should be more careful with manipulating divergent terms. This puzzles me a little. All I wrote down are infinite sums of products of elements of an infinite-dimensional non-commutative algebra generated by elements ${a}_{n}$. Until I talk about representing these as operators on some space there are no numbers which could diverge, I think.

Of course I do understand that when I acted with the ${L}_{0}$ generator with the ordering as given in my previous comment on a Fock vaccum state which is annihilated by the ${a}_{n}$ for positive $n$, that the result would be ill defined because it would formally contain an infinite real number multiplying the Fock vacuum.

But this leads precisely to the idea that I tried to discuss before: The claim by Thiemann is essentially that the noncommutative algebra that I indicated in my previous comment, with the ordering as given there (which is equivalent to the definitions in that other comment) can be represented on a non-separable Hilbert space in such a way that the objects that I manipulated in my last comment have a perfectly well defined action on this Hilbert space, without any divergencies.

This is the crucial claim. It is about operator representations of the abstract non-commutative algebra of the ${a}_{n}$ and the ${L}_{n}$ (in the ordering indicated before) or equivalently the smeared $Y\left(\sigma \right)$ and $Y\left(\sigma \right)Y\left(\sigma \right)$, I believe. The claim is essentially that there is an operator representation where the ${L}_{n}$ (in the ordering that I have given, or equivalently, the ${\int }_{{S}^{1}}d\sigma \phantom{\rule{thinmathspace}{0ex}}\xi \left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}Y\left(\sigma \right)Y\left(\sigma \right)$) are well defined and act without producing divergencies. This applies to the ${L}_{n}$ with the sums going from $-\infty$ to $\infty$, because this is what one gets when Fourier-decomposing the $Y\left(\sigma \right)$ in ${L}_{n}\propto \int d\sigma \phantom{\rule{thinmathspace}{0ex}}{e}^{\mathrm{in}\sigma }Y\left(\sigma \right)Y\left(\sigma \right)$.

But there is some fine print. Maybe that’s what is at the heart of the matter:

Actually Thiemann does no explicitly construct a representation of the ${a}_{n}$ on a Hilbert space such that the above is true. What he does construct is a represenation of the exponentiated oscillators $\mathrm{exp}\left(i{a}_{n}\right)$. That’s because these give bounded operators, which is what he needs to apply the GNS construction, as far as I understand.

I am not sure that I fully understand what this implies for the representation of the ${a}_{n}$ themselves. At the beginning of section 6.5 it says that

Since the Pohlmeyer Charges ${Z}_{±}$ involve polynomials of the ${Y}_{±}$ [$\sim {a}_{n}$] rather than polynomials of the ${W}_{±}$ [$\sim \mathrm{exp}\left({\mathrm{ia}}_{n}\right)$] it seems that our representation does not support the Quantum Pohlmeyer charges.

I think that he then goes on to show how to resolve this apparent problem. But I am not sure that I fully understand his solution. It seems that he is claiming that by dealing carefully with the various terms the Pohlmeyer Charges, and hence the ${a}_{n}$ are represented on his Hilbert space. Right above his equation (6.41) it says

we write the regulated invariants as polynomials in the ${W}_{±}\left(s\right)$ and then remove the regulator and see whether the result is well-defined and meaningful.

This sounds like problems could be hidden here. Even more so since he does not address the question whether what holds true for the Pohlmeyer charges in this context also holds true for the $\mathrm{exp}\left(i{L}_{n}\right)$. Maybe this is where the problem lies? I would like to understand this better. If this is a problem then it is at least not a trivial, elementary and obvious problem, is it?

Posted by: Urs Schreiber on January 30, 2004 8:57 PM | Permalink | Reply to this

### Re: There’s no place like home

Ok. At least I can reassure you that I do understand that cutting off these sums does modify the algebraic relation $\left[{L}_{n},{L}_{m}\right]=\left(n-m\right){L}_{n+m}$.

I am saying more than that. I am telling you that you can derive the value of the central charge by doing this cutoff calculation, and then taking $N\to \infty$.

Do it! It’s a worthwhile exercise.

Actually Thiemann does no explicitly construct a representation of the ${a}_{n}$ on a Hilbert space such that the above is true. What he does construct is a represenation of the exponentiated oscillators ${e}^{i{a}_{n}}$ . That’s because these give bounded operators, which is what he needs to apply the GNS construction, as far as I understand.

I thought the claim was that he had found a quantization in which the Virasoro algebra is unextended in the quantum theory. If we are not able to represent the oscillators ${a}_{n}$ on the string Hilbert space, then I return to my previous position of having no idea what he is talking about.

I’m curious, though, what he means by the spacetime “graviton” in a theory where he does not know how to represent the ${a}_{n}$ on the string Hilbert space.

Posted by: Jacques Distler on January 30, 2004 9:46 PM | Permalink | Reply to this

### Re: There’s no place like home

Do it! It’s a worthwhile exercise.

Ok, I’ll do it tomorrow. My girlfriend just called and said I should come home (it’s almost midnight already). :-)

I would still like to understand why the comuptation without the cutoff should be inconsistent (as long as I don’t claim to apply these non-normally ordered objects to a Fock vacuum).

If we are not able to represent the oscillators a n on the string Hilbert space, then I return to my previous position of having no idea what he is talking about.

Yeah. I did not clearly realize that this might be a problem before I wrote my previous comment. Actually what Thiemann needs is that the $\mathrm{exp}\left(i{L}_{n}\right)$ are represented on his Hilbert space because he demands states to be invariant under the action of $\mathrm{exp}\left(i{L}_{n}\right)\phantom{\rule{thinmathspace}{0ex}},\forall n$. I’ve sent him an email asking him about it. Let’s see. Alternatively I could try to figure it out myself - but that might take longer. :-)

Posted by: Urs Schreiber on January 30, 2004 10:31 PM | Permalink | Reply to this

### Re: There’s no place like home

Hi Jacques -

this morning I did the calculation that you told me to do. Sorry for having been dense. I now see that when I introduce a regulator and remove it after calculating the commutator I find an anomaly even if I don’t normal order anything.

Thanks for healing me from this confusion. Before I had been under the impression that the anomaly is entirely due to the normal ordering.

I need to recheck my calculation, though, because the first attempt gave me a prefactor 1/8 in front of the ${m}^{3}$ term. Could it be that this factor depends on the chosen ordering?

Ok, so now I finally get your point about Kansas, etc. ;-). I admit that the problem in Thiemann’s paper is not related to the non-separability of his Hilbert space. Thanks for your help!

Posted by: Urs Schreiber on February 2, 2004 4:18 PM | Permalink | Reply to this

### And your little dog too …

I need to recheck my calculation, though, because the first attempt gave me a prefactor 1/8 in front of the m3 term. Could it be that this factor depends on the chosen ordering?

Nope. That coefficient should be universal. The choice of ordering affects the non-universal (m-independent) terms.

But, computational details aside, I think you can now see the point of my comment above.

If the Lns are quadratic expressions in the oscillators, then you get a central extension. If you think you don’t get one, you’ve made a mistake.

I will leave it to others to draw whatever conclusions seem warranted about the rest of Thiemann’s paper.

Posted by: Jacques Distler on February 2, 2004 5:21 PM | Permalink | Reply to this
Read the post Not in Kansas
Weblog: Musings
Excerpt: There's a lively discussion over at the String Coffee Table of a recent paper by Thomas Thiemann on a new,...
Tracked: January 30, 2004 7:17 AM

### Re: Thiemann’s quantization of the Nambu-Goto action

Hopefully Thomas will join this discussion himself. But for now let me say that he tells me that

1) he is aware of the fact (which took me a while to appreciate) that one cannot get a quantization of the ${L}_{m}$ without anomaly, no matter which ordering is chosen

2) this does not affect his approach because he defines the action of the operators representing the exponentiated constraints by

(1)${U}_{±}\left(\varphi \right){\pi }_{\omega }\left({W}_{±}\left(s\right)\right){\Omega }_{\omega }={\pi }_{\omega }\left({\alpha }_{\varphi }^{±}\left({W}_{±}\left(s\right)\right)\right){\Omega }_{\omega }\phantom{\rule{thinmathspace}{0ex}}.$

Here ${U}_{±}\left(\varphi \right)$ (in Thomas’ paper this is a \varphi) is the operator which represents the exponentiated Virasoro element $\mathrm{exp}\left({\sum }_{n}{c}^{n}{L}_{n}^{±}\right)$ which again implements the diffeomorphism $\varphi$ on one half ($+$ or $-$) of the algebra.

${\pi }_{\omega }\left({W}_{±}\left(s\right)\right)$ is the operator version of ${W}_{±}\left(s\right)$, which is essentially an exponentiated oscillator.

${\Omega }_{\omega }$ is sort of a vaccuum in the GNS-Hilbert space (all states are obtained by acting on ${\Omega }_{\omega }$ with the ${\pi }_{\omega }\left({W}_{±}\left(s\right)\right)$.

Thomas says that the oscillators themselves are not represented on his Hilbert space, only the exponentiated oscillators are. Also the Virasoro generators ${L}_{m}$ are not represented on the Hilbert space, only the exponentiated ${U}_{±}\left(\varphi \right)$ are, he says.

My question would be:

Is it ok to just define the action of the ${U}_{±}\left(\varphi \right)$ as above, without writing them out in terms of the canonical coordinates-momenta/oscillators?

If this were ok, could’t I just do the same in the usual Fock quantization of the string by simply declaring that

(2)$\mathrm{exp}\left(\sum _{n}{c}^{n}{\stackrel{̂}{L}}_{n}\right){\stackrel{̂}{\alpha }}_{-m}\mid 0〉={\left(\mathrm{exp}\left(\left\{\sum _{n}{c}^{n}{L}_{n},.\right\}\right){\alpha }_{-m}\right)}^{\stackrel{̂}{\cdot }}\mid 0〉$

(where now ${\alpha }_{-m}$ is a worldsheet oscillator and hats $\stackrel{̂}{\cdot }$ distinguish Poisson algebra elements from operators and $\left\{,\right\}$ is the Poisson bracket.

Posted by: Urs Schreiber on February 2, 2004 7:28 PM | Permalink | Reply to this

### Damn those pesky oscillators!

Is it ok to just define the action of the ${U}_{±}\left(\varphi \right)$ as above, without writing them out in terms of the canonical coordinates-momenta/oscillators?

You can define whatever the heck you want. Above, I made a very important stipulation that the ${L}_{n}$s should be in the universal enveloping algebra generated by the oscillators. If I relax that assumption, I can get the central charge of the Virasoro algebra to be anything I want (including 0) by adding to the ${L}_{n}$s a piece that commutes with the oscillators, and contributes (negatively) to the total central charge.

In the critical bosonic string, that’s what happens when you add in the ghost contribution to the Virasoro generators.

If Thiemann has only the exponentiated oscillators, rather than the oscillators themselves, then he’d better build the Virasoro generators (or their exponentiated versions) out of those, instead.

Otherwise, one has said nothing.

More prosaically, as I said above, if the oscillators are not represented on the String Hilbert Space, I wonder how the heck the graviton (or any other tensor field) is.

Posted by: Jacques Distler on February 2, 2004 8:03 PM | Permalink | Reply to this
Weblog: Musings
Excerpt: In the Sisyphean task of implementing the conversion of LaTex symbols to MathML Named Entities, here's another update to the itex2MML executable used by my plugin.
Tracked: February 3, 2004 4:39 AM

### Re: Thiemann’s quantization of the Nambu-Goto action

Thomas Thiemann has asked me to forward the following email message to the Coffee Table discussion.

In reply to an email by myself he answers (the quoted text is from my original mail):

[begin forwarded text]

let me try to rephrase the objections that have been raised in terms of the following question:

By the logic of your paper, what keeps you from constructing analogues of the operators ${U}_{±}\left(\phi \right)$ on the standard Fock space of string states?

nothing, however, they I am worried that they won’t act unitarily because they mix the standard annihilation and creation operators. in other words, the operators ${U}_{±}$ are defined in the standard Fock rep. of the string but I am not sure whether they define a unitary rep. of the reparameterization group. That’s a good question, see below.

There certainly exist these operators (defined by their very action on the states) which represent the conformal group without anomaly on this Fock space. But they are not expressible in terms of the ${L}_{n}$. So in which sense could one claim that these ${U}_{±}\left(\phi \right)$ are obtained from Dirac’s quantization scheme, if they are not expressible in terms of the quantized first class constraints?

Good question. Here is the simple answer: Suppose you have a classical phase space and a constraint function $C$ on it which generates infinitesimal gauge transformations as

(1)${\delta }_{t}F=t\left\{C,F\right\}$

where $F$ is any function on phase space and $\left\{.,.\right\}$ is the Poisson bracket. One can exponentiate this infinitesimal action to the Hamiltonian flow of $C$ given by the automorphisms

(2)${\alpha }_{t}\left(F\right)=\mathrm{exp}\left(\left\{C,.\right\}\right)\cdot F$

The Dirac observables of the system are those functions which are gauge invariant, that is,

(3)${\alpha }_{t}\left(F\right)=F\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall t$

In a quantization you want to find a representation $\pi$ of the functions $F$ as operators $\pi \left(F\right)$ on a Hilbert space $H$ with a cyclic “vacuum” $\mid 0〉$. We now define a representation of the one parameter group of automorphisms by

(4)$U\left(t\right)\pi \left(F\right)\Omega :=\pi \left({\alpha }_{t}\left(F\right)\right)\Omega$

It follows that $\Omega$ is invariant since $\pi \left(1\right)=1$, it is a physical state. More generally, physical states are defined by the condition

(5)$U\left(t\right)\mid \mathrm{phys}〉=\mid \mathrm{phys}〉\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall t.$

There is a beautiful interplay between physical states and Dirac observables because they obviously map $\Omega$ to physical states.

(6)$i\pi \left(C\right):={\left[\frac{d}{\mathrm{dt}}U\left(t\right)\right]}_{t=0}$

First of all this works at best only if $U\left(t\right)$ are unitary operators as otherwise $\pi \left(C\right)$ cannot be self-adjoint. If $\pi \left(C\right)$ is not self-adjoint you have violated a basic quantization principle, namely that real valued functions $C$ should be represented as self-adjoint operators. The necessary and sufficient criterion for $U\left(t\right)$ to be unitary is to check whether the functional

(7)$\omega \left(F\right):=〈0\mid \pi \left(F\right)\mid 0〉$

is ${\alpha }_{t}$ – invariant, that is

(8)$\omega \cdot {\alpha }_{t}=\omega .$

[Editor’s note: The original message here has a \circ instead of a \cdot $\cdot$.]

If that is the case, Stone’s theorem of functional analysis says that $\pi \left(C\right)$ exists if and only if the one parameter unitary group $t↦U\left(t\right)$ is weakly continuous, that is

(9)$\underset{t\to 0}{\mathrm{lim}}〈\psi ,U\left(t\right)\psi \prime 〉=〈\psi ,\psi \prime 〉\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall \psi ,\psi \prime \in H$

In my representation this condition is violated. However, this is unimportant because obviously $U\left(t\right)$ is a bona fide quantization of ${\alpha }_{t}$ and secondly one can use the $U\left(t\right)$ in order to define physical states, the $\pi \left(C\right)$ are not needed for that.

Notice that all of this is standard knowledge in constraint quantization, it is not my invention. The beauty of the construction is that you get everything for free once you have a positive linear functional $\omega$. The functional of standard string theory is positive only in $D=26$ and $a=1$, so you need the tachyon there. It is a good exercise to check whether that functional is invariant under the Virasoro group or only under the algebra. I have not done that calculation yet.

A related question comes to mind: In the LQG quantization of 1+3d gravity, is the representation of the constraints there similar to those in the ‘LQG-string’?

yes and no. the spatial diffeomorphism group is represented very much in analogy to the diffeomorphism group of the circle for the lqg string. The Hamiltonian constraint is represented in the form $\pi \left(C\right)$.

Hope that helps. Best,

Thomas

PS: Maybe you can upload this to the coffee table, I somehow can’t make this work.

end forwarded text

Posted by: Urs Schreiber on February 3, 2004 5:06 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

Hi Thomas -

you write:

$U\left(t\right)$ is a bona fide quantization of ${\alpha }_{t}$ and secondly one can use the $U\left(t\right)$ in order to define physical states, the $\pi \left(C\right)$ are not needed for that.

Apparently this is the crucial point that is controversial.

I’d think that Dirac quantization forces us to impose

(1)$〈\mathrm{phys}\mid \pi \left(C\right)\mid \mathrm{phys}〉=0\phantom{\rule{thinmathspace}{0ex}}.$

Among other reasons, this is what you get from path-integral approaches.

With respect to the LQG-constraints of 1+3d gravity you write

the spatial diffeomorphism group is represented very much in analogy to the diffeomorphism group of the circle for the lqg string. The Hamiltonian constraint is represented in the form $\pi \left(C\right)$.

What is the rationale behind this difference? Is it that for the spatial diffeos you don’t have the weak continuity condition that you were referring to, while for the temporal diffeos you do?

Posted by: Urs Schreiber on February 3, 2004 5:21 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

Dear Urs,

I understand that the question whether constraint quantization can be done with
the group or the algebra is controversial.
The uneasy feeling may come from your
experience with Fock spaces of which perturbative path integral quantization
is just another version. In those representations one usually deals with
the algebra, however, notice that one can
work as well with the group. So you question my procedure by using an example where both approaches work. I would say
that there is no evidence for concern.
For instance in LQG we have a similar
phenomenon with respect to the spatial
diffeomorphism group. We can only quantize the group, not its algebra. Yet the solution space consists of states which are supported on generalized knot classes
which sounds completely right. There are
other examples where the group treatment,
also known as group averaging or refined algebraic quantization
See for instance [23] and references
therein.

Coming to your second question, the reason for the unsymmetrical treatment of Hamiltonian and spatial diffeomorphism
constraint in Loop Quantum Gravity
is that the constraint algebra
of GR is much more complicated than for the string. The temporal diffeomorphisms
are treated differently than the spatial ones because the Hamiltonian constraint,
also known as Wheeler-DeWitt constraint,
in contrast to the spatial diffeomorphism
constraint, is not a quadratic expression
in the basic variables.
One can show that in the LQG representation that we use, also called
Ashtekar – Isham – Lewandowski representation, we
indeed get a unitary rep. of the spatial
diffeo group which like for the particular
rep. of the LQG string that I studied is
not weakly continuous. On the other hand, the Hamiltonian constraint can be defined
as a self-adjoint operator. So we use a
mixture of both procedures to quantize
GR.

Hope that helps,
bests,

Thomas

Posted by: Thomas Thiemann on February 3, 2004 7:22 PM | Permalink | Reply to this

### Out, out, damned anomaly!

We now define a representation of the one parameter group of automorphisms by $U\left(t\right)\pi \left(F\right)\Omega :=\pi \left({\alpha }_{t}\left(F\right)\right)\Omega$

In other words, there is no such thing as an anomalous symmetry? The gauge group of the classical theory is by definition promoted to a gauge symmetry of the quantum theory.

Umh, sorry, but things don’t work that way in quantum field theory.

Posted by: Jacques Distler on February 4, 2004 4:00 AM | Permalink | Reply to this

### Rather than re-inventing the wheel…

I’d recommend the classic paper,

Alvarez-Gaumé and Nelson, “Hamiltonian Interpretation Of Anomalies,” Commun. Math. Phys. 99 (1985) 103.

for how to understand anomalies from this point of view.

Posted by: Jacques Distler on February 4, 2004 4:21 AM | Permalink | Reply to this

### Re: Out, out, damned anomaly!

So may we conclude that the single most problematic technical step in LQG is that the spatial diffeo constraints are not imposed as operator constraints but that instead the classical spatial diffeo group is, as Jacques says,
by definition promoted to a gauge symmetry of the quantum theory?

Posted by: Urs Schreiber on February 4, 2004 4:11 PM | Permalink | Reply to this

### LQG

I was making no comment about LQG, merely about the matter at hand — the quantization of the Nambu-Goto string.

I would hope that the LQG formalism would be “smart” enough to know that it is supposed to crash and burn if you attempt use it to quantize a theory with gravitational anomalies.

But this example (in which the un-centrally-extended $\mathrm{Diff}\left({S}^{1}\right)$ is simply postulated to hold in the quantum theory) does not make one sanguine.

In this particular case, the central extension does not prevent one from carrying out the quantization (one merely has to split the constraints, in the conventional fashion). But a formalism that doesn’t even notice the existence of the central extension is too naive to be of any use, either here or, presumably, elsewhere.

Posted by: Jacques Distler on February 4, 2004 5:07 PM | Permalink | Reply to this

### Re: LQG

Ok, so let’s assume we’d take the classical ADM constraints of 1+3d gravity, perform canonical quantization, introduce a cutoff the way you have taught me to do, compute the quantum commutators and remove the regulator. Would we find an anomaly?

Posted by: Urs Schreiber on February 4, 2004 5:29 PM | Permalink | Reply to this

### Re: LQG

Such a procedure has not a prayer of working because you will find yourself unable to remove the regulator.

Surely that’s well-known to anyone who’s thought seriously about the subject.

Posted by: Jacques Distler on February 4, 2004 5:40 PM | Permalink | Reply to this

### Re: LQG

I see. So what do you mean when you say:

I would hope that the LQG formalism would be ‘smart’ enough to know that it is supposed to crash and burn if you attempt use it to quantize a theory with gravitational anomalies.

If we demand that canonical quantization has to have $〈\mathrm{phys}\mid \pi \left(C\right)\mid \mathrm{phys}〉=0$ and if, according to your last comment, we cannot even in principle make sense of this, then we have to conclude that ‘canonical quantum gravity’ is an oxymoron. So what is it that you hope the LQG formalism to know?

Posted by: Urs Schreiber on February 4, 2004 5:57 PM | Permalink | Reply to this

### Just Say No!

I’m really not interested in getting into a discussion of LQG.

If we want to discuss the quantization of the Nambu-Goto string, where everything can be made rigorous and well-defined, then I will be happy to lend any insight I might have.

If this discussion of Thiemann’s paper leads you to some new understanding about what the LQG people are attempting to do, that’s great.

But I, personally, would like to stay on a subject where I know what I’m talking about.

Posted by: Jacques Distler on February 4, 2004 7:44 PM | Permalink | Reply to this

### Re: Just Say No!

Hi,

it seems you guys are getting into a religious discussion about whether it is
allowed to quantize the constraints the
way I did.

I am a disbeliever of any religion and I have therefore nothing useful to add to this
discussion. However, maybe you may find it
useful to remember that the way I
treated the constraints is EXACTLY the same as one quantizes the Poincare group of ordinary QFT.

Cheers,

Thomas

Posted by: Thomas Thiemann on February 5, 2004 12:34 PM | Permalink | Reply to this

### Faith-based calculation

it seems you guys are getting into a religious discussion about whether it is allowed to quantize the constraints the way I did.

The existence or nonexistence of anomalies is not a matter of religious faith.

In the absence of anomalies, any old slapdash, illegitimate set of manipulations will obtain the “right answer.” In the presence of anomalies (as Urs discovered above), one needs to be more careful.

From your description, your method can be used to promote any gauge symmetry of the classical theory to a gauge symmetry of the quantum theory. The symmetry in the quantum theory can never be spoiled by anomalies.

I don’t believe that. And our difference of opinion is not merely a religious one.

Posted by: Jacques Distler on February 5, 2004 1:50 PM | Permalink | Reply to this

### Alternative quantization procedure?

Hi Thomas -

you wrote:

it seems you guys are getting into a religious discussion about whether it is allowed to quantize the constraints the way I did.

Let me emphasize that I, for one, are just sitting here trying to understand what is going on.

I am already quite fond of the fact that we apparently managed to pinpoint the very spot at which your approach parts company with the standard quantization of the string. As you can see by looking through the discussion here, I first thought the difference lies somewhere else.

I believe that I was a little bit mislead by your discussion of group averaging. From equation (5.2) and (6.25) of your paper I got the impression that you had found a way to quantize the Virasoro constraints without getting an anomaly. The absence of anomalies is of course the necessary condition for group averaging to be applicable (which is pretty obvious but also confirmed for instance by Giulini and Marolf in their gr-qc/9902045).

I am glad that we could clarify that no group averaging in the sense of using exponentiated constraints is used in your ‘LQG-string’ and that in fact the constraints are not representable on your Hilbert space.

If I understand you correctly, then you are proposing (and using in LQG) a procedure for quantizing a classical constrained theory which can be summarized as

Find an operator representation of the classical symmetry group (which need not be constructible from the quantized first class constraints) and define physical quantum states to be those that are invariant under the action of these operators.

I am absolutely no expert on LQG, but I have heard talks by yourself and other people working in LQG and have also looked at parts of your introductory papers on LQG. Unfortunately this, apparently crucial, fact has so far escaped my attention. I appreciate that we could isolate it as the key difference between your treatment of the string and the standard quantization.

Let me assure you that I have no religious prejudices about what quantization is supposed to be. I am just trying to understand physics. As I wrote in the introduction to this discussion here, there is a well known saying due to E. Nelson, I believe, who said that “first quantization is a mystery” (of course his point was mainly to emphasize that second quantization is not). As you know, this means that it is an empirical fact that the world is quantum and that we sometimes have to guess the correct quantum rules from knowledge of the classical limit that we observe.

There are however some prescription for how to obtain a quantum theory from an action functional, among them are path integral quantization and Dirac quantization of first class constraints. For simple enough systems, at least, like the Nambu-Goto string, these can be shown to lead to the same result.

Now you are saying that nature might be described by a quantum theory which follows neither from the path integral nor from Dirac/Gupta-Bleuler quantization but is radically different from these.

Since so far this alternative approach has been used only to tackle non-perturbative quantum gravity, where nobody has much of a clue what the result is supposed to look like, it seems that the radical departure of the LQG quantization procedure from standard quantization procedures has not been widely noticed.

Do you agree with this assessment?

There are several proposals for alternative quantization procedures out there. Some people try to modify Schroedinger’s equation, other invent things like ‘Event Enhance Quantum Mechanics’ or the like. Modifications of the quantum principle have been proposed to explain away the information loss problem in black holes, for instance. So in a sense, when it comes to quantum gravity, one might take the viewpoint that everything we know about the world far above the Planck scale is subject to scrutinization again.

Still, all modifications of the standard procedure of quantization should reduce to the well known physics in an appropriate limit, far from the Planck scale. You mention that your quantization of the constraints is precisely that of the Poincare group in QFT. But the quantized Poincare generators are not anomalous, so that no difference is to be expected here.
But do you have an argument why and how your alternative quantization could reproduce the standard quantization in some limit?

Posted by: Urs Schreiber on February 5, 2004 5:04 PM | Permalink | Reply to this

### Re: Alternative quantization procedure?

Dear Urs,

I appreciate very much your interest, thanks
a lot.

I agree that weakly discontinuous representations of gauge symmetries are
quite unfamiliar in standard QFT. I strongly
believe that this is related to the fact that in ordinary QFT we have a background
metric to build on while in LQG or for the
string, which is a worlsheet metric independent 2d QFT, there is no such structure available. My intuition comes from the fact that without a background metric it is meaningless to say whether
two objects are close or far apart. Close or far wrt which metric if there is none?
This leads to very discontinuous behaviour. For instance for the LQG string
two states are orthogonal if the intervals
on which they are supported are not identical, no matter how “little” they
differ in one coordinate system, because in another they maybe drastically different.

Notice, however, that I did not show that
this happens for all reps., there might be
more continuous ones.

It is true that the Poincar'e group is
represented w/o anomaly in standard QFT
as required by the axioms but maybe one
can view the recently discussed defomations of the Poincar'e group as
a “anomalous” realization.

In any case, look at sections 6.6, 6.7, 6.8 where I show that one construct
coherent states for the LQG string such that the W_\pm and hence the Z_\pm have expectation values
as close to the classical one as you want
so that semiclassically you get the
correct limit.

Hope that helps,
cheers,

Thomas

Posted by: Thomas Thiemann on February 5, 2004 11:25 PM | Permalink | Reply to this

### The Virasoro anomaly from regulated generators

Since I have been asked by others about it and generally for the record, I’d like to present the calculation, recommended by Jacques Distler as a worthwhile exercise and of some importance for the present discssion, of the Virasoro anomaly by means of using regulated generators. In retrospect all this is pretty obvious and I am sufficiently ashamed to have been confused about it, but that’s life.

Everybody knows the various derivations of the anomaly as done in in GSW and Polchinski. Here I am going to discuss what could be called the canonical, functional perspective, because this is a perspective that might confuse one into missing the anomaly - as I have unfortunately demonstrated.

So let there be a canonical coordinate field $X\left(\sigma \right)$ on the circle, with canonical momentum $\pi \left(\sigma \right)=-i\frac{\delta }{\delta X\left(\sigma \right)}$ such that

(1)$\left[X\left(\sigma \right),\pi \left({\sigma }^{\prime }\right)\right]=i\delta \left(\sigma ,{\sigma }^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}}.$

From these the ‘chiral’ field

(2)$Y\left(\sigma \right):=\frac{1}{\sqrt{2}}\left(i\frac{\delta }{\delta X\left(\sigma \right)}+{X}^{\prime }\left(\sigma \right)\right)$

is constructed, which has the commutator

(3)$\left[Y\left(\sigma \right),Y\left({\sigma }^{\prime }\right)\right]=-i{\delta }^{\prime }\left(\sigma ,{\sigma }^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}}.$

The task is to make sense of the commutator algebra of squared $Y$.

Naively one might write “$\left[\frac{1}{2}Y\left(\sigma \right)Y\left(\sigma \right),\frac{1}{2}Y\left({\sigma }^{\prime }\right)Y\left({\sigma }^{\prime }\right)\right]=-i{\delta }^{\prime }\left(\sigma ,{\sigma }^{\prime }\right)\left(Y\left(\sigma \right)Y\left({\sigma }^{\prime }\right)+i\frac{1}{2}{\delta }^{\prime }\left(\sigma ,{\sigma }^{\prime }\right)\right)$”, where the second term on the right is due to reordering, is hence classically absent and essentially the quantum anomaly - except for the fact that all this is not well defined since it involves products of distributions.

To deal with these, the fields have to be smeared appropriately or, equivalently, their sum over modes have to be truncated. There are many possible smearings and truncations. After some experimenting the most convenient one I found is obtained by using

(4)${Y}^{\left(N\right)}\left(\sigma \right):=\frac{1}{\sqrt{2}}\left(i\frac{\delta }{\delta X\left(\sigma \right)}+\int d{\sigma }^{\prime }\phantom{\rule{thinmathspace}{0ex}}f\left(\sigma -{\sigma }^{\prime }\right){X}^{\prime }\left({\sigma }^{\prime }\right)\right)\phantom{\rule{thinmathspace}{0ex}},$

where $f\left(\sigma -{\sigma }^{\prime }\right)$ is an approximation for $\delta \left(\sigma -{\sigma }^{\prime }\right)$:

(5)$f\left(\sigma -{\sigma }^{\prime }\right):=\frac{1}{2\pi }\sum _{k=-N}^{N}{e}^{\mathrm{ik}\left(\sigma -{\sigma }^{\prime }\right)}\phantom{\rule{thinmathspace}{0ex}}.$

This one turns the $Y$-$Y$ commutator into

(6)$\left[{Y}^{\left(N\right)}\left(\sigma \right),{Y}^{\left(N\right)}\left({\sigma }^{\prime }\right)\right]=-i{f}^{\prime }\left(\sigma -{\sigma }^{\prime }\right)$

and the regulated modes of the Virasoro generators ${L}_{m}^{\left(N\right)}$ are simply

(7)${L}_{m}^{\left(N\right)}=\int d\sigma \phantom{\rule{thinmathspace}{0ex}}{e}^{-\mathrm{im}\sigma }\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}{Y}^{\left(N\right)}\left(\sigma \right){Y}^{\left(N\right)}\left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}.$

The commutators in question are now computed formally just as for the ill-defined expression mentioned above:

(8)$\left[{L}_{m}^{\left(N\right)},{L}_{-m}^{\left(N\right)}\right]=\int d\sigma \phantom{\rule{thinmathspace}{0ex}}d\kappa \phantom{\rule{thinmathspace}{0ex}}\left(-i{e}^{-\mathrm{im}\left(\sigma -\kappa \right)}\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(\sigma -\kappa \right){Y}^{\left(N\right)}\left(\sigma \right){Y}^{\left(N\right)}\left(\kappa \right)+\frac{1}{2}{e}^{-\mathrm{im}\left(\sigma -\kappa \right)}\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(\sigma -\kappa \right){f}^{\prime }\left(\sigma -\kappa \right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

The second term in the integral, which is again due to the reordering, should essentially give the sought-after anomaly. Indeed, it can easily be evaluated explicitly, which yields

(9)$\frac{1}{2}\int d\sigma \phantom{\rule{thinmathspace}{0ex}}d\kappa \phantom{\rule{thinmathspace}{0ex}}{e}^{-\mathrm{im}\left(\sigma -\kappa \right)}\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(\sigma -\kappa \right){f}^{\prime }\left(\sigma -\kappa \right)=\frac{1}{12}\left({m}^{3}-m\right)-m\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}N\left(N+1\right)+F\left(N\right)\phantom{\rule{thinmathspace}{0ex}},$

where $F\left(N\right)$ is a polynomial in $N$ which is independent of $m$. The first term is the standard anomaly

(10)$A\left(m\right)=\frac{1}{12}\left({m}^{3}-m\right)\phantom{\rule{thinmathspace}{0ex}}.$

The second is a shift that can be re-absorbed into the definition of ${L}_{0}^{\left(N\right)}$:

(11)${\stackrel{˜}{L}}_{m}^{\left(N\right)}:={L}_{m}^{\left(N\right)}-{\delta }_{m,0}\frac{1}{2}\frac{N\left(N+1\right)}{2}$

so that the non-normal-ordered ${\stackrel{˜}{L}}_{0}^{\left(N\right)}$ has finite expectation value in the Fock vacuum.

The term $F\left(N\right)$ would diverge when the regulator is removed (when $N$ is sent to infinity). It should somehow cancel. To see that we need to look at the first term of the above $\left[{L}_{m}^{\left(N\right)},{L}_{-m}^{\left(N\right)}\right]$. From the Jacobi identity it follows that the anomaly can contain only ${m}^{3}$ and ${m}^{1}$ terms. Therefore it makes sense to look at terms containing different powers of $m$:

(12)$-i\int d\sigma \phantom{\rule{thinmathspace}{0ex}}d\kappa \phantom{\rule{thinmathspace}{0ex}}{e}^{-\mathrm{im}\left(\sigma -\kappa \right)}\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(\sigma -\kappa \right){Y}^{\left(N\right)}\left(\sigma \right){Y}^{\left(N\right)}\left(\kappa \right)=-i\int d\sigma \phantom{\rule{thinmathspace}{0ex}}d\kappa \phantom{\rule{thinmathspace}{0ex}}\left(1-\mathrm{im}\left(\sigma -\kappa \right)+\cdots \right)\phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(\sigma -\kappa \right){Y}^{\left(N\right)}\left(\sigma \right){Y}^{\left(N\right)}\left(\kappa \right)\phantom{\rule{thinmathspace}{0ex}}.$

For even powers of $m$ the coefficient in front of the $Y$s is an odd function of $\sigma -\kappa$. This means that for even powers of $m$ we may replace ${Y}^{\left(N\right)}\left(\sigma \right){Y}^{\left(N\right)}\left(\kappa \right)$ in the integrand with $\frac{1}{2}\left[{Y}^{\left(N\right)}\left(\sigma \right),{Y}^{\left(N\right)}\left(\kappa \right)\right]=-{\mathrm{if}}^{\prime }\left(\sigma -\kappa \right)$.

For ${m}^{0}$ this yields

(13)$-\frac{1}{2}\int d\sigma \phantom{\rule{thinmathspace}{0ex}}d\kappa \phantom{\rule{thinmathspace}{0ex}}{f}^{\prime }\left(\sigma -\kappa \right){f}^{\prime }\left(\sigma -\kappa \right)=-F\left(N\right)$

which indeed precisely cancels the previously found potentially diverging term $F\left(N\right)$. From the Jacobi identity it now follows that all other even powers of $m$, which could give c-numbers, will disappear. Since we are implicitly using symmetric ordering (no (normal-)reordering in the Fourier decomposition) the odd powers of $m$ don’t give rise to further c-number terms, either.

This means that the regulator can now be removed, which finally yields

(14)$\underset{N\to \infty }{\mathrm{lim}}\left[{\stackrel{˜}{L}}_{m}^{\left(N\right)},{\stackrel{˜}{L}}_{-m}^{\left(N\right)}\right]=2m{\stackrel{˜}{L}}_{0}^{\left(\infty \right)}+\frac{1}{12}\left({m}^{3}-m\right)\phantom{\rule{thinmathspace}{0ex}}.$
Posted by: Urs Schreiber on February 5, 2004 7:16 PM | Permalink | Reply to this

### Re: The Virasoro anomaly from regulated generators

Dear Urs,

thanks for the calculation.

I guess this is to show that the Virasoro
anomaly is unavoidable and unique. I insist
that this conclusion is wrong since it
is representation dependent.

Namely to represent the momentum as
\pi(x)=i\delta/\delta X(x)
the Hilbert space is L_2([dX]). Leaving
aside the fact that the Lebesgue measure [dX] does not exist in infinite dimensions
suppose that I choose a different rep.
L_2(d\mu) where \mu is a different measure. Unless the two measures are
mutually absolutely continuous, the two
reps. are not unitarily equivalent and
the value of the anomaly will change.

Maybe there is an additional argument that
one must use, but the above calculation
is not sufficient to show uniqueness of the anomaly.

Maybe I am missing something. Looking forward to your answwer,

Thomas

Posted by: Thomas Thiemann on February 5, 2004 11:40 PM | Permalink | Reply to this

### Wishful thinking

Urs never used the expression $i\frac{\delta }{\delta X\left(\sigma \right)}$. He was just being unnecessarily fancy. What he did use was that $\pi \left(\sigma \right)$ obeys the canonical commutation relations

(1)$\left[X\left(\sigma \right),\pi \left(\sigma \prime \right)\right]=i\delta \left(\sigma ,\sigma \prime \right)$

From that, he constructed properly-regularized versions of the Virasoro generators, and computed their commutators.

Never was the Hilbert space mentioned, nor were any Hermiticity assumptions made.

This is really important to understand.

You can experiment with other smearing functions all you want. You will never succeed in making the anomaly go away.

More generally, do you still contend that all anomalies (including this one) can be defined away by royal fiat?

Posted by: Jacques Distler on February 6, 2004 1:13 AM | Permalink | Reply to this

### Re: Wishful thinking

Dear Jaques,

thanks for clarifying this. If really only
the CCR’s are used then I believe the result
and this is probably the no-go theorem that

However, I was really never doubting this.
What I am saying is that it is not necessary
to define the L_m, you can live with the
finite transformations in order to construct the physical Hilbert space. The finite transformations can be quantized w/o
anomalies, basically because one can
quantize w/o going back to the L_m.
In other
words I am quantizing the group rather than the Lie algebra. There are standard
techniques in QFT for doing that and this
is what I have done.

You may have an intuition against such a procedure but please notice that this is
just how one quantizes the Poincar’e group
in QFT. I would like to know what your
intuition is so that I can better understand where the discussion is going.

Cheers,

Thomas

Posted by: Thomas Thiemann on February 6, 2004 1:23 PM | Permalink | Reply to this

### Here, there, … everywhere

I can assure you that the group, too, receives a central extension (look up the phrase, “Schwarzian derivative”).

However, the computation is bound to be more involved, because defining the properly-regulated group elements is more subtle.

In any case, the result is a foregone conclusion, because if you take group elements infinitesimally-close to the identity, you will just reproduce the Lie-algebra computation.

Earlier, you claimed refuge in the thought that the Lie-algebra elements that would be obtained would not be self-adjoint. But, as you can see, no assumptions about self-adjointness were made in this derivation.

The Poincaré group (and, indeed, its Lie algebra too) do not receive any corrections, because there are no ordering-ambiguities in defining the generators, so the computation can proceed “naively”.

Not so, for instance, if you go to Light-cone gauge. In Light-cone gauge, the Lorentz-generators, ${M}^{i-}$, (here, I label the coordinates, ${X}^{±}=\frac{1}{\sqrt{2}}\left({X}^{0}±{X}^{1}\right)$ and ${X}^{i}$, $i=2,...,d-1$) have ordering ambiguities and need to be smeared. When you compute the commutator $\left[{M}^{i-},{M}^{j-}\right]$, you get an additional term, showing that for general $d$, the Poincaré algebra does not survive quantization in Light-cone gauge.

The computation is very similar to the one Urs just did for you. I commend it highly as a very illuminating exercise to do.

Posted by: Jacques Distler on February 6, 2004 2:06 PM | Permalink | Reply to this

### Rules of the game

I must say, I am not 100% convinced by Urs argument. What I think one needs to show would be the following: Start with the classical Poisson algebra generated by the ${a}_{n}$’s, that is all functions (polynoms or some power series or something like that) of the ${a}_{n}$. This is the classical algebra $A$. Let’s pretend that the ${L}_{n}$ live in this algebra (there might be a problem with the infinite sums. I am willing to ignore that but if one introduces some regulator one has to show that the result does not depend on that choice).

Now we quantize $A$, that is we find a linear map

(1)$q:A\to \mathrm{Op}\left(H\right)$

where $\mathrm{Op}\left(H\right)$ are operators on a Hilbert space (not neccesarily bounded) with the property that for $f,g\in A$ we have

(2)$q\left(\left\{f,g\right\}\right)=i\hslash \left[q\left(f\right),q\left(g\right)\right]+O\left({\hslash }^{2}\right).$

We further require the representation to be irreducible, that is it shouldn’t have any invariant subspaces (this should say that we are quantizing only the ${a}_{n}$’s and nothing else). And then I would like to see that this implies

(3)$\left[q\left({L}_{n}\right),q\left({L}_{m}\right)\right]=...$

Note that I don’t want to assume a highest weight representation! One example of the above is that $q$ is normal ordering but the claim would be that the central charge is independent of the choice of $q$. Can you show that while being careful with the $q$’s and the terms of higher order in $\mathrm{hbar}$?

Posted by: Robert on February 10, 2004 6:03 PM | Permalink | Reply to this

### Re: Rules of the game

Hi Robert,

assume I introduce a regulator and work with finite sums only, the way I have sketched above. Fixing Weyl-ordering I get

(1)$\left[{L}_{m}^{\left(N\right)},{L}_{-m}^{\left(N\right)}\right]=2m\left({L}_{0}^{\left(N\right)}-\frac{1}{2}\frac{N\left(N+1\right)}{2}\right)+\frac{1}{12}\left({m}^{3}+m\right)+\mathrm{stuff}\mathrm{that}\mathrm{disappears}\mathrm{as}N\to \infty \phantom{\rule{thinmathspace}{0ex}}.$

But since this is a result for finite sums I can without any problems (I think) reorder the ${a}_{n}$ in this expression. This is for free in the ${L}_{m}$ (since we can assume that $m\ne 0$) and gives a term proportional to $m$ for ${L}_{0}$. The ${m}^{3}$-term is in any case unaffected.

So dealing with finite sums removes all further subtleties related to different choices of ordering, I’d say. Not so?

Posted by: Urs Schreiber on February 10, 2004 6:46 PM | Permalink | Reply to this

### Re: Rules of the game

I must say, this does not satisfy me. Maybe I am not aware of some basic theorem of quantization but how do I know that all ambiguities in quantization of the U(1) algebra come from ordering problems?

Unfortunately, in my previous post the hbar’s came out as question marks (at least in my netscape that keeps warning me that it needs more fonts), but in general the rule “Poisson brackets go to commutators” is only true up to higher order terms in hbar!

Let me give an illustrating however not convincing example: Let’s quantize the usual Heisenberg algebra of $x$’s and $p$’s (this of course is unique by Stone-von-Neumann but I would like to show the higher order corrections). For definiteness, let us quantize by “normal ordering” that is by moving all $p$’s to the right. Then compute for example $\left\{{x}^{2}{p}^{4},{x}^{4}{p}^{2}\right\}=-12{x}^{5}{p}^{5}$ However in the quantized theory, compute $\left[{x}^{2}{p}^{4},{x}^{4}{p}^{2}\right]={x}^{2}{p}^{4}{x}^{4}{p}^{2}-{x}^{4}{p}^{2}{x}^{2}{p}^{4}$ and then restore the ordering on the RHS. To do this you have to do more than one reordering but the term you would get from the Poisson bracket captures only the term coming from one reordering, the others are higher order in hbar.

The upshot is: You cannot assume you know the commutators of your quantum operators (and you use them to compute the central charge above), you only know them up to higher order terms. What we usually do is we assume that there are some “elementary” ones, like $x$ and $p$ or ${a}_{n}$ andpostulate that they don’t have higher order corrections, but how do we justify this? We could use other coordinates in the Poisson manifold and then rrequire those not to have higher order corrections.

In the Heisenberg case we know that the quantization is unique up to unitary equivalence. But what about the $U\left(1\right)$ current algebra? That looks like many copies of Heisenberg algebras but there might be functional analysis issues in the infinite tensor product.

Posted by: Robert on February 11, 2004 9:43 AM | Permalink | Reply to this

### Re: Rules of the game

After this concrete question I also have a more philosophical one: What exactly do we mean by “Quantum theory Q is a quantization of a classical theory C”? They should have the same symmetries and the same field content (representation data of the symmetries). But that surely isn’t enough. In the Lagrangian setting this just determines the gauge groups and the fiels but not the action in any way. It might be that this specifies the action (maybe after getting rid of irrelevant operators) but for example there could be truely marginal ones that we cannot get rid of. But as we all know the moduli space they generate usually is very different between the quantum theory and the classical theory (in cases we believe we know how to quantize, like on the lattice or we can use enough susy to render some classical reasoning valid in the quantum theory). Just doing an expansion in hbar does not seem like a good idea, especially in strongly coupled theories.

We would believe that the classical theory at least specifies a perturbative theory and if there is a regime where the coupling is small (UV for QCD) we could require the quantum theory to converge to the pertubation theory but such a regime is not always available.

So what do we require ofsomebody that claims he has quantized the string? $U\left(1\right)$ current algebra +Vir? Any dynamics?

Posted by: Robert on February 11, 2004 9:59 AM | Permalink | Reply to this

### Re: Rules of the game

So what do we require of somebody that claims he has quantized the string?

I think this is precisely the question around which the discussion about Thiemann’s approach revolves. More concretely, in this particular case of a constrained system the question is:

What do we mean by the quantization of a theory with classical (1st class) constraints ${C}_{I}=0$?

There once was a little discussion of this point over at s.p.r.

There Aaron pointed out, that the form of the quantum constraints follows from the path integral. I am not sure how we would handle that in case of the Nambu-Goto action, but I think that if a proposed ‘quantization’ is not derivable by means of path integral techniques we will hesitate to accept this procedure.

Looking back at this old thread, I was reminded of very relevant comments by Marc Henneaux from pp. 156 of his string theory lecture notes, where he writes

Third question: Conversely, should one not try to use a different representation of the string operators so as to avoid the central charge? Again, it might very well be possible to construct such a representation and, if so, it is very likely that the resulting quantum theory would be very different from the one explained here. It could be that this yet-to-be-constructed theory would possess an intrinsic interest of its own […]. Moreover, because that theory would not be based on the use of oscillator variables, it might be more easily extendable to higher-dimensional objects, such as the membrane. However, to the author’s knowledge, this subject has not been investigated.

He even mentions the relation of this question to the canonical quantization of gravity:

Second question: Is it conceivable that one should somehow weaken the Wheeler-De Witt equations of quantum gravity, as it would be necessary if a (c- or q-number) “central charge” appears in the constraint “algebra”? Yes it is, but no work along these lines has been done.

Posted by: Urs Schreiber on February 11, 2004 4:11 PM | Permalink | Reply to this

### Re: Rules of the game

What we usually do is we assume that there are some ‘elementary’ ones, like $x$ and $p$ or ${a}_{n}$ and postulate that they don’t have higher order corrections, but how do we justify this?

Ok, now I get your point. Apparently we agree that my little calculation above demonstrates that the anomaly is independent of ordering if the commutator of the ${a}_{n}$ is taken to be

(1)$\left[{a}_{m},{a}_{n}\right]=m{\delta }_{n,-m}\phantom{\rule{thinmathspace}{0ex}}.$

But your point is that this is already an unnececarily restrictive assumption, since we might as well have

(2)$\left[{a}_{m},{a}_{n}\right]=m{\delta }_{n,-m}+𝒪\left(h\right)\phantom{\rule{thinmathspace}{0ex}}.$

In the Heisenberg case we know that the quantization is unique up to unitary equivalence. But what about the U (1) current algebra? That looks like many copies of Heisenberg algebras but there might be functional analysis issues in the infinite tensor product.

I see. Well, being lazy I would tend to answer this by again pointing to the fact that in my above calculation only finite sums appear, hence also only finitely many copies of the Heisenberg algebra, hence every possibility is unitarily equivalent to the one I have been discussing.

Probably that is not the answer that you want to see. If you have a solution to this problem which uses the way of reasoning that you are getting at, please let me know.

Posted by: Urs Schreiber on February 11, 2004 11:21 AM | Permalink | Reply to this

### Re: Rules of the game

To address Robert’s question, note that the canononical commutation relations for $X\prime \left(\sigma \right)$ and $\pi \left(\sigma \right)$ are just a copy of the Heisenberg algebra for each Fourier mode. One can, I think legitimately, take them to hold exactly (with no $O\left({\hslash }^{2}\right)$ term).

The anomaly that you compute in the commutator of two Virasoro generators is then a term which is, indeed $O\left({\hslash }^{2}\right)$. The $O\left(\hslash \right)$ term in the commutator is the same as in the classical theory. The leading ($\propto {m}^{3}$) piece of the anomaly is $O\left({\hslash }^{2}\right)$. But, given the canonical commutation relations, it is completely universal.

Posted by: Jacques Distler on February 11, 2004 1:57 PM | Permalink | Reply to this

### Choice of ordering

Just to make clear a point that Urs did not dwell on: the different choices of ordering of ${L}_{0}$ (Weyl-ordering, as used here, versus normal-ordering, say) differ by an $N$-dependent constant.

Thus the shift in ${L}_{0}$ that Urs talks about above will differ for different choices of ordering of ${L}_{0}$. What is completely universal is the coefficient of ${m}^{3}$ in the above computations.

Posted by: Jacques Distler on February 6, 2004 12:52 AM | Permalink | Reply to this

### Re: The Virasoro anomaly from regulated generators

I wrote:

Therefore it makes sense to look at terms containing different powers of $m$:

Unfortunately this is wrong and makes the entire argument wrong.

Naively one might think that it is enough to count powers of $m$. But there are subtle effects. Since $m$ is integer we cannot distinguish $m$ from $m\mathrm{cos}\left(2\pi m\right)$, for instance.

I realized that this is a problem when trying to run through the same regulator calculation for the superstring.

Now I am totally at a loss how to properly evaluate

(1)$\underset{N\to \infty }{\mathrm{lim}}{\int }_{0}^{2\pi }d\sigma \phantom{\rule{thinmathspace}{0ex}}d\kappa \phantom{\rule{thinmathspace}{0ex}}{e}^{-\mathrm{im}\left(\sigma -\kappa \right)}{f}^{\prime }\left(\sigma -\kappa \right)Y\left(\sigma \right)Y\left(\kappa \right)\phantom{\rule{thinmathspace}{0ex}}.$
Posted by: Urs Schreiber on March 4, 2004 7:28 PM | Permalink | PGP Sig | Reply to this

### Anomaly Monopoly

Since this topic is all-but-beaten to death now, I thought it might be fun apply Thomas’s methods to a theory people actually care about (nobody gives a crap about the bosonic string).

Classical Yang-Mills theory is dilatation-invariant.

Exercise 1: Construct the generator, $D$, of dilatations in classical Yang-Mills.

The action of dilatations (just like the action of Poincaré) doesn’t quite commute with the Hamiltonian. Rather, the Poisson-bracket of $D$ with the Hamiltonian is proportional to $H$. Equivalently, the action of a 1-parameter group of dilatations is to rescale the Hamiltonian, $H\to \lambda H$.

Exercise 2: Now, apply Thomas’s procedure to construct the action of this 1-parameter group of dilatations in the quantum theory (“just like Poincaré”, as Thomas would say).

Excellent! Since $H$ rescales under the action of the dilatation group, we have proven that the spectrum of $H$ in the quantum theory is continuous near zero.

In other words, the quantum theory “has no mass-gap.”

Bzzzt! Do not pass GO, do not collect \$1 million!

Posted by: Jacques Distler on February 6, 2004 2:59 PM | Permalink | Reply to this

### Re: Anomaly Monopoly

I assume that we are now talking about YM on a flat background. If ${T}_{\mu \nu }$ is the (symmetric) energy-momentum tensor of the action

(1)${ℒ}_{\mathrm{YM}}=-\frac{1}{4{g}^{2}}{F}^{2}$

the dilatation current is

(2)${D}_{\mu }={T}_{\mu \nu }{x}^{\nu }$

whose divergence is the trace of ${T}_{\mu \nu }$:

(3)${\partial }_{\mu }{D}^{\mu }={T}^{\mu }{}_{\mu }\phantom{\rule{thinmathspace}{0ex}}.$

Classically this trace vanishes, but quantumly scale transformations are associated with running of the coupling constants $g\to g+\beta \left(g\right)$ so that in the quantum theory

(4)${\partial }_{\mu }{D}^{\mu }=\beta \left(g\right)\frac{2}{{g}^{3}}{F}^{2}\phantom{\rule{thinmathspace}{0ex}}.$

This is the trace anomaly and should be related to what Jacques is getting at.

My question to Thomas is: It has been argued that because LQG has the exact EH action at the Planck-scale it cannot possibly have this as an effective action at low energies, due to renormalization. Could it be that this disagreement between what LQG proponents are expecting to happen and what some other people are expecting, is also due to the fact that there are no quantum anomalies present in the LQG approach?

Posted by: Urs Schreiber on February 6, 2004 7:06 PM | Permalink | Reply to this

### Re: Anomaly Monopoly

This is the trace anomaly and should be related to what Jacques is getting at.

Of course they are related. If Thomas can quantize this particular 2D theory without picking up the Virasoro anomaly, he can presumably quantize 4D Yang-Mills without picking up the trace anomaly.

But, to make things easy for Thomas, you really should have written out the stress tensor in canonical form,

(1)${T}_{00}=\frac{1}{2}\mathrm{Tr}\left({E}^{2}+{B}^{2}\right),\phantom{\rule{1em}{0ex}}{T}_{0i}=\mathrm{Tr}\left(E×B\right)$

where ${B}^{k}={ϵ}^{\mathrm{ijk}}\left({\partial }_{i}{A}_{j}+\frac{i}{2}\left[{A}_{i},{A}_{j}\right]\right)$ is the chromomagnetic field, and ${E}_{i}$ is the chromoelectric field*, canonically-conjugate to ${A}_{i}$. The generator of the dilatation symmetry is, as you said,

(2)$D=\int {d}^{3}x{T}_{0\mu }{x}^{\mu }=tH+\int {d}^{3}x{T}_{0i}{x}^{i}$

and satisfies

(3)$\frac{dD}{dt}=\left\{D,H\right\}+\frac{\partial D}{\partial t}=0$

Promoting $U\left(\lambda \right)={e}^{-\lambda \left\{D,\cdot \right\}}$ to a symmetry of quantum Yang-Mills, one easily finds an example of Dirac’s dictum:

All theorists should first apply their theories to themselves.

* I should point out, just for completeness, that there is no quantum-mechanical obstruction to imposing the Gauss-law constraint $〈\text{phys}\mid {D}^{i}{E}_{i}\mid \text{phys}\prime 〉=0$.

Posted by: Jacques Distler on February 6, 2004 9:57 PM | Permalink | Reply to this

### Re: Anomaly Monopoly

You can have an anomaly free rep. of the
symmetry and still get drastic quantum corrections at the Planck scale. In LQG
geometrical operators corresponding to
length, area and volume of a curve, surface
and region respectively have discrete
spectrum which is a drastic departure from
the smooth classical structure. At scales
way above the Planck scale things look smooth again semiclassically.

BTW:
In all books about renormalization you will find the statement that violating
a continuous gauge symmetry (as e.g. the
local gauge symmetry of YM theory) leads to an inconsistent theory. As the gauge symmetry of the string is continuous as well, it fits to the pattern when trying to representing it w/o anomaly. You may
violate rigid symmetries such as the chiral symmetry corresponding to the ABJ
anomaly. Now the string is curious in the sense that in the usual rep. we do have an
anomaly and yet get a consistent theory

Finally, although I do not know of any rep. for non abelean ym theory by itself
which supports the hamiltonian, the coupled YM - EH action can be quantized
in the LQG representation
and supports the Hamiltonian. That Hamiltonian IS NOT dilatation invariant.

Posted by: Thomas Thiemann on February 7, 2004 3:58 PM | Permalink | Reply to this

### Splitting the constraints

Now the string is curious in the sense that in the usual rep. we do have an anomaly and yet get a consistent theory at least when adding supersymmetry.

No, supersymmetry has nothing to do with this issue. Even though the algebra of constraints receives a central extension, it is possible to ensure that the matrix elements of the constraints vanish between physical states, by imposing the constraints weakly:

(1)$\begin{array}{rcl}{L}_{n}\phantom{\rule{thinmathspace}{0ex}}\mid \text{phys}〉& =& 0,\phantom{\rule{1em}{0ex}}n>0\\ 〈\text{phys}\mid \phantom{\rule{thinmathspace}{0ex}}{L}_{n}& =& 0,\phantom{\rule{1em}{0ex}}n<0\\ \left({L}_{0}-a\right)\phantom{\rule{thinmathspace}{0ex}}\mid \text{phys}〉& =& 〈\text{phys}\mid \phantom{\rule{thinmathspace}{0ex}}\left({L}_{0}-a\right)=0\end{array}$

This is hardly an unusual situation. Indeed, in “most” quantum field-theoretic instances of constrained quantization, one can only impose the constraints weakly,

(2)$〈\text{phys}\mid \phantom{\rule{thinmathspace}{0ex}}\pi \left(C\right)\phantom{\rule{thinmathspace}{0ex}}\mid \text{phys}\prime 〉=0$

It almost never happens that one can impose the stronger version, $\pi \left(C\right)\phantom{\rule{thinmathspace}{0ex}}\mid \text{phys}〉=0$.

(Which is why your procedure of attempting to impose the exponentiated constraints strongly almost never works in field theory.)

Posted by: Jacques Distler on February 7, 2004 5:42 PM | Permalink | Reply to this

### Re: Anomaly Monopoly

Unfortunately you are not correct because
nobody succeeded in finding a representation
in which the Hamiltonian or stress energy tensor are well-defined for non Abelean
Yang Mills theory. There is no conclusion.

In any case, whether or not you should have an exact or projective rep. of a symmetry
depends on the physical system under study and hence must be decided ultimately by experiment. In case of the string everything is allowed until we have quantum gravity experiments.

Bzzzzt! Do not even throw the dice.

Posted by: Thomas Thiemann on February 7, 2004 3:41 PM | Permalink | Reply to this

### “Easy” problems are not worth doing

Unfortunately you are not correct because nobody succeeded in finding a representation in which the Hamiltonian or stress energy tensor are well-defined for non Abelean Yang Mills theory.

There is no technical issue that arises in the quantization of Yang-Mills theory that does not arise in much more difficult form in the quantization of gravity.

Yang Mills theory is an infinitely easier problem.

Indeed, the cutoff theory can be perfectly-rigourously defined, say by Lattice Gauge Theory. There is no analytical proof that the Lattice Gauge Theorists can take the cutoff away, and get to the continuum limit. But there is powerful numerical evidence that they can. Indeed, they can, nowadays, compute the spectrum of glueball masses (verifying, numerically, the existence of a mass gap) to within a few percent.

In any case, whether or not you should have an exact or projective rep. of a symmetry depends on the physical system under study and hence must be decided ultimately by experiment. In case of the string everything is allowed until we have quantum gravity experiments.

No, no, and no! It is mathematically inconsistent to assume that the Virasoro constraints do not pick up a central extension in the bosonic string (as Urs proved to you above). It is mathematically inconsistent to assume that dilatation symmetry is unbroken in quantum Yang-Mills.

And “Anything goes, until we have experimental evidence proving that our mathematically-inconsistent manipulations are ruled out by Mother Nature.” is not a recipe for doing good science. (OK, that last one was a religious statement, which you are welcome to disagree with.)

Posted by: Jacques Distler on February 7, 2004 6:04 PM | Permalink | Reply to this
Read the post Three Card Monte
Weblog: Musings
Excerpt: I've been pondering why I find the discussion of Thomas Thiemann's recent paper over at the String Coffee Table so disturbing. Finally, Thiemann's latest comment made it all fall into place for me (emphasis added):
Tracked: February 8, 2004 4:09 AM

### Baby & Bathwater

So now, the party line is that Thiemann’s quantization is some clever new method of quantization, completely unrelated to canonical quantization, that no one has thought of before.

This is not only my interpretation, but Thomas Thiemann himself says that the procedure, sketched above, for dealing with the constraints, should be compared to experiment to see if nature favors it over standard Dirac/Gupta-Bleuler quantization.

It is well-known that if one is willing to abandon locality, one has great lattitude to “cancel” the anomalies which arise in local QFT. A charitable interpretation of Thiemann’s procedure is that it correponds precisely to such a nonlocal modification of local field theory.

There are reasons to reject nonlocal modification of the worldsheet theory of the bosonic string — to do with getting consistent string interaction, a problem on which Thiemann is clueless, as he has, at best, made a failed attempt to construct the free bosonic string.

However, it is quite clear why Thiemann does not wish to apply his methods to Quantum Field Theories people care about, like Yang-Mills Theory. There, we know quite clearly whose side Mother Nature has come down on.

Posted by: Jacques Distler on February 12, 2004 4:29 PM | Permalink | Reply to this

### Re: Baby & Bathwater

Demian Cho has kindly pointed me to a paper that seems to be (indeed, claims to be) relevant to the issues that we have been discussing here:

A. Ashtekar, S. Fairhurst and J. Willis, Quantum gravity, shadow states and quantum mechanics, 2002

I have so far only read the first dozen pages or so, unfortunately. In the introduction it says that the purpose of the paper is to discuss the peculiarities of LQG-inspired quantization in the framework of a very simple toy example - the single nonrelativistic particle in 1-dimension.

Unfortunately this is not a theory with constraints, so I am worried that maybe a key aspect of our previous discussion is not dealt with in this paper. Nevertheless, many technical aspects clearly are as in Thomas Thiemann’s paper.

With only ten pages read I cannot yet comment on more than the following point:

On p.8 the well-known fact is emphasized that there are inequivalent representations of the Weyl-Heisenberg algebra, some of which are weakly continuous and give the usual quantum theory, while others are not and hence give the LQG-like quantum theory (called the polymer representation in that paper). Just for reference, let me note that in the weakly continuous case the Weyl-Heisenberg algebra is that generated by the exponential operators $U\left(a\right)=\mathrm{exp}\left(\mathrm{ia}\stackrel{̂}{x}\right)$ and $V\left(a\right)=\mathrm{exp}\left(ia\stackrel{̂}{p}\right)$.

Now one point of the LQG approach is that there are representations of this algebra which are not weakly continuous (which means that matrix products of operators are not continuous in the parameter $a$). This implies that for instance $V\left(a\right)$ exists as an operator, but the exponent $\stackrel{̂}{p}$ does not.

Recall that this was a crucial aspect of Thomas Thiemann’s quantization of the string: The worldsheet oscillators and Virasoro constraints themselves were not represented on the Hilbert space, only their classical exponentiations were. This was the very reason why no anomaly appeared, since the classically exponentiated constraints were promoted to operators directly. This was also the point of the main criticism of Thomas Thiemann’s quantization, so let’s see what the above paper has to say about this point.

To see this clearly, however, I’ll have to read the rest of the paper which has to wait until tomorrow, in my case. Let me only note that in the introduction an interesting hint is given:

There it says that the non-weakly-continuous representation does indeed describe ‘physics’ fundamentally very different from the ordinary one. But it also says that the non-standard representation (the ‘classical exponentiation’) can reproduce the usual results pertaining to the usual Schroedinger quantization in some limit (or so), using a concept called ‘shadow states’ .

Let me quote from the first paragraph on p.4:

At the mathematical level, the two descriptions [ordinary quantization and non-weakly-continuous representation] are quite distinct and, indeed, appear to be disparate. Yet, we will show that states in the standard Schroedinger Hilbert space define elements of the analog of ${\mathrm{Cyl}}^{*}$. As in quantum geometry, the polymer particle ${\mathrm{Cyl}}^{*}$ does not admit a natural inner product. Nonetheless, as indicated in [1], we can extract the relevant physics from elements of ${\mathrm{Cyl}}^{*}$ by examining their shadows, which belong to the polymer particle Hilbert space ${ℋ}_{\mathrm{Poly}}$. This physics is indistinguishable from that contained in Schroedinger quantum mechanics in its domain of applicability.

This really makes me curious, because it sounds like there might be a way to obtain the usual quantization of the string from Thomas Thiemann’s quantization, maybe in some limit or something.

If anyone (Demian Cho, Joshua Willis, Thomas Thiemann, for instance) can explain this, I would be very grateful.

Posted by: Urs Schreiber on February 16, 2004 9:16 PM | Permalink | Reply to this

### Re: Baby & Bathwater

This morning I continued reading A. Ashtekar, S. Fairhurst and J. Willis, Quantum gravity, shadow states and quantum mechanics, 2002.

According to Demain Cho’s comments here this paper should contain the key peculiarities of LQG-like quantization in a tractable toy example.

The basic idea is to see what happens when the correspondence principle of elementary quantum mechanics is violated. Taking the risk of boring everybody let me recall that this principle says, in its most naive form, that classical canonical coordinates and momenta are promoted to self-adjoint operators on some Hilbert space in the quantum theory with CCR commutator $\left[\stackrel{̂}{x},\stackrel{̂}{p}\right]=\mathrm{ih}$.

One observes that this commutation relation may be exponentiated by defining $\stackrel{̂}{U}\left(a\right)=\mathrm{exp}\left(-\mathrm{ia}\stackrel{̂}{p}/h\right)$ and $\stackrel{̂}{V}\left(a\right)=\mathrm{exp}\left(-\mathrm{ia}\stackrel{̂}{x}/h\right)$, which gives the Weyl form of the CCR:

(1)$\stackrel{̂}{U}\left(a\right)\stackrel{̂}{V}\left(b\right)={e}^{\mathrm{iab}/h}\stackrel{̂}{V}\left(b\right)\stackrel{̂}{U}\left(a\right)\phantom{\rule{thinmathspace}{0ex}}.$

One central idea of the LQG-like quantization is to modify the correspondence principle to the effect that instead of demanding $\stackrel{̂}{x}$ and $\stackrel{̂}{p}$ to be operators on some Hilbert space satisfying the CCR, one demands $\stackrel{̂}{U}\left(a\right)$ and $\stackrel{̂}{V}\left(a\right)$ to be operators on some Hilbert space and that the Weyl form of the CCR holds.

The crucial point is that there are representations of the Weyl algebra, namely those which are not weakly continuous (expectation values of $\stackrel{̂}{V}\left(a\right)$ of $\stackrel{̂}{U}\left(a\right)$ are not continuous in $a$), which are not unitarily equivalent to that obtained by exponentiating operators $\stackrel{̂}{x}$ and $\stackrel{̂}{p}$. LQG-like quantization wants to work with these ‘exotic’ representations of the Weyl algebra, that’s the program.

To my mind this program has the following problem, which, in different guises, has been discussed here a lot already: The problem is that classical equations of motion, classical constraints, the Schroedinger equation, etc., are usually expressed in terms of $x$ and $p$. But now not both of these are available as operators $\stackrel{̂}{x}$ and $\stackrel{̂}{p}$. So what is the quantization prescription then? Are we to do the exponentiation classically, in the Poisson algebra and then promote the result to an operator? By which rules to we specify the commutation relations of the resulting operators? Again by the classical theory?

I was hoping to find an answer to this important question in the above mentioned paper. It is page 14 of that paper where an aspect of this questions is discussed. There, the task is to find a Weyl-algebra analog of the definition of coherent states

(2)$\stackrel{̂}{a}\mid {\psi }_{\zeta }〉=\zeta \mid {\psi }_{\zeta }〉\phantom{\rule{thinmathspace}{0ex}},$

where $\stackrel{̂}{a}$ is the lowering operator of ordinary 1d nonrelativistic QM. In this form this equation is not available in LQG-like QM, because there $\stackrel{̂}{p}$, which enters the definition of $\stackrel{̂}{a}$, is not defined. Therefore one has to find an exponentiated version of this equation.

Now comes the interesting point: The exponentiated version of the above equation in this paper is modeled after the respective exponentiated equation in in the usual Schroedinger quantization, using the Baker-Campbell-Hausdorff formula for the ordinary CCR algebra operators! (Please see page 14 of this paper for details.)

If this does not sound surprising,, recall how a similar step was done in Thomas Thiemann’s paper: There the question was how to represent the Virasoro constraints in exponentiated form, since the constraints themselves were not represented as operators in Thomas Thiemann’s LQG-like quantization of the string. Following the above paper by Ashtekar, Fairhurst and Willis one might have expected that this was done modeled after the usual quantum theory, which, as Jacques Distler has emphasized would still see the anomaly, of course. Instead, what Thomas Thiemann does in his paper is to use the classical Poisson-algebra of the exponentiated Virasoro constraints and represents this in terms of operators on some Hilbert space.

It seems to me that by modifying the usual correspondence principle (which incidentally also means giving up the path integral) one arives at a proposal for a new form of quantization which is not uniquely well defined. Of course a similar statement is true for the standard form of Schroedinger-like quantization, where ordering ambiguities in expressions like $\mathrm{xp}$ have to be dealt with. But the ambiguity in the LQG-like quantization seems to be much more severe.

If in Thomas Thiemann’s paper one were to follow the prescription indicated on page 14 of the Ashtekar,Fairhurst&Willis paper, one would find the anomaly. If one instead uses the classical algebra one misses it.

How are we supposed to deal with the ambiguities that arise as soon as the usual correspondence principle based on the CCR is replaced by one based on the Weyl-form of the CCR?

One way is indicated by Ashtekar,Fairhurst&Willis in their simple QM example: If one takes care that the usual relations are correctly translated to the new formalism (as is done on their page 14) then one finds the same results as in the usual quantum theory, essentially. In this case, however, one might wonder what the LQG-like formalism buys us.

The other way is to model the Weyl-CCR operator relations after the classical algebra, as done by Thomas Thiemann for the LQG string and in general in LQG for the spatial diffeomorphism constraints of gravity. This approach however has very little in common with what one usually calls ‘quantization’. And it is also doubtful that an argument as in Ashtekar,Fairhurst&Willis can recover the usual theory in this case, I think.

Posted by: Urs Schreiber on February 17, 2004 11:32 AM | Permalink | Reply to this

### Re: Baby & Bathwater

Following the above paper by Ashtekar, Fairhurst and Willis one might have expected that this was done modeled after the usual quantum theory, which, as Jacques Distler has emphasized would still see the anomaly, of course. Instead, what Thomas Thiemann does in his paper is to use the classical Poisson-algebra of the exponentiated Virasoro constraints and represents this in terms of operators on some Hilbert space.

Is that even consistent with “quantizing” the Weyl form of the of the CCR’s? Are the (exponentiated) Virasoro constraints even expressible in terms of the $\stackrel{̂}{U}\left(a\right)$’s and $\stackrel{̂}{V}\left(a\right)$’s?

Another, unrelated, question is just how non-weakly continuous a representation one can abide. In the presence of several $x$’s and $p$’s, one should generalize the above equation to

(1)$\stackrel{̂}{U}\left(\stackrel{⇀}{a}\right)\stackrel{̂}{V}\left(\stackrel{⇀}{b}\right)={e}^{i\stackrel{⇀}{a}\cdot \stackrel{⇀}{b}/h}\stackrel{̂}{V}\left(\stackrel{⇀}{a}\right)\stackrel{̂}{U}\left(\stackrel{⇀}{b}\right)$

and demand that there exist a rotation operator $\stackrel{̂}{W}\left(R\right)$ such that

(2)$\stackrel{̂}{W}\left(R\right)\stackrel{̂}{U}\left(\stackrel{⇀}{a}\right)\stackrel{̂}{W}\left(R{\right)}^{-1}=\stackrel{̂}{U}\left(R\circ \stackrel{⇀}{a}\right),\phantom{\rule{1em}{0ex}}\stackrel{̂}{W}\left(R\right)\stackrel{̂}{V}\left(\stackrel{⇀}{a}\right)\stackrel{̂}{W}\left(R{\right)}^{-1}=\stackrel{̂}{V}\left(R\circ \stackrel{⇀}{a}\right)$

The Poincaré group is generated by $\stackrel{̂}{W}$ and $\stackrel{̂}{U}$. Surely, for there to be any sense to this, we must demand that the Poincaré group is represented weakly-continuously.

Is it possible to get a non-weakly-continuous representation of the Weyl relation above, while still having Poincaré represented weakly-continuously?

Posted by: Jacques Distler on February 17, 2004 3:40 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

K.-H. Rehren, who has worked on Pohlmeyer invariants, was so kind to answer to my mail. I have posted the reply here.

Among many other things he says:

Hence implementation of the constraints with c=0 is possible.

Posted by: Urs Schreiber on February 17, 2004 2:38 PM | Permalink | Reply to this

### Re: Thiemann’s quantization of the Nambu-Goto action

The discussion seems to have moved to sci.physics.research .

Posted by: Urs Schreiber on March 12, 2004 4:54 PM | Permalink | PGP Sig | Reply to this
Read the post Thiemann's quantization of the Nambu-Goto action
Weblog: The String Coffee Table
Excerpt: Aspects of the recent paper by Th. Thiemann on the 'LQG-string' are discussed.
Tracked: September 2, 2006 6:12 AM