Wednesday, July 26, 2006

Where have all the trackbacks gone?

I just tried to send trackback pings for the previous posts. First I realised that there are no more trackback links at the ArXiV (because of this discussion?) and then explained to me (via Haloscan)
Problem: Server said 'You are not allowed to send TrackBack pings.'

Too bad.

Mastering anomalies?

After a long and not really fruitful discussion over at Jacques' I had a look a Thomas Thiemann's (with K. Giesel) latest opus magnum (with two further parts). I did not get very far in the introduction as already on page four he mentions an interesting trick: The Master Constraint.

Before I say what it is, let be introduce a bit of the background. We are in the context of theories with gauge invariances which are as everybody knows redundancies in the degrees of freedom. One might think that in quantising the theory one should directly work with the gauge invariant observables but this is often not the case since the description with gauge invariances often has ma much simpler structure, e.g. the space of connections is affine, as discussed elsewhere.

The price one has to pay are is the gauge invariance one drags around and which one has to mod out after the quantization. This can turn out to be impossible and for chiral theories it generically is. Of course I have just described the fact that the theory is anomalous.

Let me discuss this in a concrete example, the bosonic string in which the Virasoro algebra plays the role of the gauge invariance. In this example, the modes of everything are labeled by integers rather than by continuous variables (as for example for the axial anomaly) so there are fewer pitfalls from integrals etc one has to avoid. Plus we have discussed this case in detail in our paper so I don't have to repeat myself too much (for this discussion ignore all the parts on polymer states and the LQG way of doing things, just focus on the mathematical description of what one usually does (Fock space, Gupta Bleuler etc)).

The task in quantization is to turn the classical algebra of observables (functions on phase space) into a quantum algebra with representations on Hilbert spaces. The problem is that the classical algebra is a Poisson algebra with two multiplicative structures, the usual (pointwise) product of functions and the Poisson bracket. Both are supposed to map into in single product in the quantum algebra such that the Poisson bracket becomes the commutator for that product. Already in quantum mechanics of a single degree of freedom you know that this does not work exactly but only "up to higher order h-bar terms".

What does this mean in practice? The usual procedure is to take a subset of observables (typically coordinates of the phase space, or x and p and 1, or the field and its canonical momentum) which have simple Poisson brackets and which generate the classical algebra in terms of the pointwise product. Then one 'promotes' them to operators such that the Poisson bracket goes to commutator rule holds exactly. For all the other observables, one fixes a way of wring them in terms of the simple ones (aka one fixes an operator ordering prescription) and uses this and the product in the operator algebra to define their quantum versions.

Now, what about the gauge symmetry? In the classical theory, Noether's theorem tells us, that all symmetries are inner, that is, for each symmetry transformation, there is a function of phase space which generates it via Poisson brackets. Furthermore, the group relations for the transformations map to Poisson brackets of the generators.

In the quantum theory, you now have to deal with the gauge symmetry. There are two slightly different ways of saying what goes on: The first is quite abstract and is the one we used in the LQG string paper: You take your quantum algebra as above and have your symmetry act on it by an automorphism. Now, in a representation on a Hilbert space, you demand that this automorphism is implemented by unitary operators U(S) (for a gauge transformation S). There is no direct way to obtain these, educated guessing is probably best. The property you demand is that when A is in the algebra and p is the representation you have .

This implies that the U(S) nearly implement the group law: where is a phase. Of course, the above constistency condition for U(S) does not change if you change U(S) by a phase. The question is if you can find a consistent assignment of phases for all U(S) such that all the go away. If this is impossible, you have an anomaly.

In the other approach you use your knowledge of the classical symmetry generators and quantise them as all the other functions on phase space. Often as in the case of the bosonic string, they are quadratic in the basic fields which you quantised directly. This implies that the ordering ambiguity is just a complex number (imaginary for anti-hermitean generators). Again, the difficult step is to find an assignment of these such that the group law holds in terms of commutators.

If you don't succeed you could subtract the left hand side from the right hand side of your expression of the commutator and have the physical state condition that this anomaly (a complex non-zero number) annihilates physical states. This condition of course immediately empties your physical Hilbert space and you are left with nothing.

So the upshot of all this is: In this canonical quantisation approach, the way the anomaly manifests itself is in the inability to get the quantum symmetry algebra to work.

Now we come to the Master Constraint Trick: Assume, we write all our symmetry generators as C_i for i in some index set (let's not worry for a second that this will be infinite in the examples and thus one should worry about existence of the sums). Then form for some positive a_i.

As you can see, M annihilates a state exactly iff all C_i annihilate the state. Thus this constraint contains all the other constraints! Even better, as we only have one constraint, the algebra is trivial and for obvious reasons it also holds in the quantised version.

One is of course not yet done as again the kernel of M could be empty and thus the spectrum of this positive operator could be bound away from zero. But Giesel and Thiemann instruct as what to do:
This can be cured by subtracting from the Master Constraint the minimum of the spectrum provided of course that it is finite and vanishes as so that the modified constrain still has the same classical limit as the original one. One then defines the physical Hilbert space as the (generalised) kernel of the Master Constraint,...

Great, now we finally know how to get rid of these stupid anomalies!

Tuesday, July 04, 2006

Everything solved!

While I was away for a wonderful vacation in southern France, I nearly missed gr-qc/0606121. In the first paragraph we are reminded
The issue of the dynamics is perhaps the central problem in canonical quantization approaches to totally constrained theories like quantum general relativity. There are three salient aspects of the problem that have prevented from advancing in the quantization. The first one is how to construct a space of physical states for the theory that are annihilated by the quantum constraints and that is endowed with a proper Hilbert space structure. The second issue is related to the introduction of a correspondence principle with the classical theory, in particular to check the constraint algebra at a quantum level. The third problem is how to address the ``problem of time'' that is, to introduce a satisfactory picture for the dynamics of the theory in terms of observable quantities.

Then come three pages of semi-technical stuff (finite number of degrees of freedom models, Legendre transformations) and eventually
Summarizing, the method of uniform discretizations allows to tackle
satisfactorily the three central problems of the dynamics of quantum
general relativity and provides new avenues for studying numerically
classical relativity as well.

Well done, guys! Now we can stop worrying about quantum gravity and spend all your energies to cheer up Klinsi's Jungs!

Sorry, I didn't have anything more intelligent to say.