## September 2, 2006

### The Harmonic Oscillator in LQG

#### Posted by Aaron I’ve been trying to understand Thomas Thiemann’s riposte to the papers of Nicolai, Peeters and Zamaklar, Nicolai and Peeters and Helling and Policastro. I’m fairly busy right now with a paper of my own and moving, so I’ll concentrate on the part where he describes how LQG-quantization replicates the usual quantization of the harmonic oscillator. Maybe later, I can get to trying to understand the master constraint program.

I’ve already posted some comments at Christine Dantas’s blog, but I thought I might also try to post them here. I really would love to see some sort of discussion on these points. One of the things I think any scientist should be able to do is to get in front of a chalkboard and be able to communicate a pretty good idea about what’s going on with their work. Think of this as a long distance chalkboard, and I’m the skeptical visitor.

Anyways, I will try in this post to summarize my understanding of the construction in Thiemann. I hope people will correct me if I get it wrong. (And I hope I get the algebra correct….)

In LQG-quantization, we start with a ${C}^{*}$-algebra, in this case the algebra given by the exponentiated position and momentum operators. This is often used because $p$ and $q$ are unbounded operators. In particular, we have

$U={e}^{\mathrm{iap}}\phantom{\rule{1em}{0ex}}V={e}^{\mathrm{ibq}}$

and

$\mathrm{UV}={e}^{iab}\mathrm{VU}$

The Stone-von Neumann theorem tells us that ordinarily all unitary representations of this algebra are unitarily equivalent. In LQG, however, we have weakly continuous representations, so we can’t appeal to this any more. Unfortunately, the usual harmonic oscillator Hamiltonian

$H={p}^{2}+{q}^{2}$

isn’t an element in this ${C}^{*}$-algebra. However, we can define a one parameter family of operators:

${H}_{ϵ}=\left({\mathrm{sin}}^{2}\left(ϵp\right)+{\mathrm{sin}}^{2}\left(ϵq\right)\right)/{ϵ}^{2}$

which are elements in the algebra. Classically, of course, this expression does converge to $H$ as $ϵ\to 0$. It would be nice now to determine the spectrum of this operator, but apparently for the usual LQG representation, we cannot. Thus, let us define the raising and lowing operators

${a}_{ϵ}^{†}=\left(\mathrm{sin}\left(qϵ\right)+i\mathrm{sin}\left(pϵ\right)\right)/ϵ$

and similarly for ${a}_{ϵ}$. It is not true ${a}_{ϵ}^{†}{a}_{ϵ}+1/2$ is ${H}_{ϵ}$, but it is true to order ${ϵ}^{2}$ if I haven’t screwed up my algebra. Now, define the operators${b}_{n,ϵ}=\frac{1}{n!}\left({a}_{ϵ}{\right)}^{n}{a}_{ϵ}^{†}{a}_{ϵ}\left({a}_{ϵ}^{†}{\right)}^{n}$For the usual harmonic oscillator, we have that ${b}_{n}\mid 0⟩=n\mid 0⟩$ and so the vev of the ${b}_{n}$ is $n$. By weak continuity, we can ensure that, for $n and for any $\delta >0$, there exists an ${ϵ}_{0}$ such that for all $ϵ<{ϵ}_{0}$:$\mid ⟨0\mid {b}_{n,ϵ}\mid 0⟩-n\mid <\delta$

We’re still nowhere near the LQG harmonic oscillator, though. To summarize what we’ve done so far, we have defined a one parameter family of operators that exist in the algebra generated by $U$ and $V$ that, such that if we interpret them as in the algebra with the $p$ and $q$, they would go to the usual Hamiltonian and raising and lowering operators. There, the vevs of the ${b}_{n}$ give exactly the spectrum of the Hamiltonian. We would have ensired that the vevs of a finite number of the ${b}_{n,ϵ}$ are close to the integers, but ut’s not at all clear to me what these vevs have to do with the spectrum of ${H}_{ϵ}$.

Regardless, we don’t necessarily have the harmonic oscillator vacuum floating around. It is a theorem (apparently) that the space of traceclass operators on our (GNS-)Hilbert space is dense in the space of states on the ${C}^{*}$-algebra. Thus, we can find a traceclass operator, $\rho$, such $\mathrm{Tr}\left(\rho {b}_{n,ϵ}\right)$ is as close to $⟨{b}_{n,ϵ}⟩$ in the Harmonic oscillator vaccum as we desire. Since that’s as close to the integers as we specify, we’ve produced a state (actually, an infinite number of them) such that, for $n,$\mid \mathrm{Tr}\left(\rho {b}_{n,ϵ}\right)-n\mid <\delta$

So, what have we proven? We’ve produced an infinite number of mixed states for the GNS-representation provided by LQG-quantization. In each of these mixed states, the vevs of the ${b}_{n,ϵ}$ are within a specified closeness to the integers.

Let’s grant for the moment that the vevs of these ${b}_{n,ϵ}$ operators are really related to what we measure for the Harmonic oscillator energy states. Given a particular choice of mixed state, there is a prediction that at some accuracy, these values will differ from the standard prediction. Presumably, something similar will hold for any comparison of LQG-quantization and standard quantization.

What we have not furnished at this point is a first principles method for determining the correct mixed state. I don’t see how we have any predictivity without this.

What’s more, there are plenty of other traceclass operators. For example, take the density matrix for the harmonic oscillator:$\rho =\frac{1}{2}\left(\mid 0⟩⟨0\mid +\mid 1⟩⟨1\mid \right)$Then,$\mathrm{Tr}\left(\rho {b}_{n}\right)=\frac{1}{2}\left(n+\left(n+1{\right)}^{2}\right)$I can just as well produce a density matrix such that the vevs of the ${b}_{n,ϵ}$ give these numbers to the specified accuracy. Can I distinguish from first principles why I should choose one of the above density matrices that give us ‘good’ answers rather than one of the ones that give these values?

Given this wide variety of ‘energy levels’ for the Harmonic oscillator, it seems to me that LQG has not predicted (or retrodicted in this case) anything. Worse, no matter how accurate traditional quantization appears to be, we can always find a state that reproduces our measurement to the needed accuracy, so LQG does not predict when we should see deviations from standard quantization.

This isn’t completely awful in situations like the Harmonic oscillator where traditional quantum mechanics tells us the answer to look for. But what happens in quantum gravity, then, when we don’t have a traditional quantization to guide us? Does this surfeit of mixed states cease to exist? How do we get any predictions at all?

Posted at September 2, 2006 3:11 AM UTC

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### Re: The Harmonic Oscillator in LQG

> a first principles method for determining the correct mixed state.

In the standard case the ‘first principle’ which selects the correct states results from the requirement of stationarity, which hinges on the availibility of a ‘preferred’ time parameter.

I assume that such a preferred time parameter is not available in a ‘background independent’ quantization a la LQG.
Therefore one would probably have to introduce some sort of measuerement device or observer reference frame first and can then study the interaction of both. This should then tell one what energy spectrum such an observer measures.

Just a guess and I am aware that this comment might be pure nonsense 8-)

Posted by: wolfgang on September 3, 2006 1:18 AM | Permalink | Reply to this

### Re: The Harmonic Oscillator in LQG

First, I think, there’s a typo: The polymer state is not weakly continious.

Further down, the construction of the mixed state in the polymer Hilbert space hinges on the fact that Thiemann cuts off the Hilbert space at some level (the non-relativisitic states as he wants to call them). It is crucial to restrict to a finite dimensional Hilbert space and thus to a finite number of observables. Otherwise the construction does not work.

And let me iterate again, that for the construction of that mixed state you have to know in advance which observables to measure as it will depend on this choice. So you don’t get close to all integers, only to the first N!

Posted by: Robert on September 4, 2006 12:59 PM | Permalink | Reply to this

### Re: The Harmonic Oscillator in LQG

I don’t think I used weak continuity for the polymer state, only the usual harmonic oscillator vacuum.

I mentioned the $n condition a few times, but maybe I should have emphasized it more. I spent a little time wondering if I could get any set of values I want from the expression

$⟨{b}_{n}⟩=\sum _{i}{a}_{i}\left(i+n\right)\left(\left(\genfrac{}{}{0}{}{i+n}{n}\right)\right)$

with ${a}_{i}>0$ and

$\sum {a}_{i}=1$

but I figured it wasn’t worth too much effort.

Posted by: Aaron Bergman on September 4, 2006 5:44 PM | Permalink | Reply to this
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### Re: The Harmonic Oscillator in LQG

The desire to work with bounded operators is natural. Passing from $q$ to $\mathrm{exp}\left(\mathrm{iq}\right)$ is natural.

What is not so natural is passing from $H$ to ${H}_{ϵ}$. That’s full of arbitrariness (and it’s not just the choice of $ϵ$).

The canonical choice seems to be rather $\mathrm{exp}\left(\mathrm{iH}\right)$.

Posted by: urs on September 4, 2006 8:42 PM | Permalink | Reply to this

### Re: The Harmonic Oscillator in LQG

I would like to bring to the attention to the reader of this blog, interested in the LQG version of the harmonic oscillator, of a recent preprint:
http://arxiv.org/abs/gr-qc/0610072
where the system is considered from the perpective of renormalization. A continuum limit is found that coincides precisely with the usual Schroedinger representation.

Posted by: A Corichi on October 18, 2006 6:17 PM | Permalink | Reply to this

### Re: The Harmonic Oscillator in LQG

Another interesting source about the LQG version of the harmonic oscillator is “Rovelli C 1998 Loop quantum gravity Living Rev. Rel. 1 1”.

Posted by: The Meta Search Engine on March 5, 2007 10:54 AM | Permalink | Reply to this

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