The String Coffee Table

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

$$U = e^{iap} \quad V = e^{ibq}$$

and

$$UV = e^{i a b} VU$$

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:

$$H_\epsilon = (sin^2(\epsilon p) + sin^2(\epsilon q)) / \epsilon^2$$

which are elements in the algebra. Classically, of course, this expression does converge to $H$ as $\epsilon \to 0$. 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

$$a_\epsilon^\dagger = (sin(q\epsilon) + i sin(p\epsilon))/\epsilon$$

and similarly for $a_\epsilon$. It is not true $a^\dagger_\epsilon a_\epsilon + 1/2$ is $H_\epsilon$, but it is true to order $\epsilon^2$ if I haven’t screwed up my algebra. Now, define the operators $$b_{n,\epsilon} = \frac{1}{n!}(a_\epsilon)^n a^\dagger_\epsilon a_\epsilon (a^\dagger_\epsilon)^n$$ For the usual harmonic oscillator, we have that $b_n|0\rangle = n|0\rangle$ and so the vev of the $b_n$ is $n$. By weak continuity, we can ensure that, for $n \lt N$ and for any $\delta \gt 0$, there exists an $\epsilon_0$ such that for all $\epsilon \lt \epsilon_0$: $$|\langle 0|b_{n,\epsilon}|0\rangle - n| \lt \delta$$

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 $b_{n,\epsilon}$ are close to the integers, but ut’s not at all clear to me what these vevs have to do with the spectrum of $H_\epsilon$.

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 $Tr(\rho b_{n,\epsilon})$ is as close to $\langle b_{n,\epsilon} \rangle$ 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 $n\lt N$, $$|Tr(\rho b_{n,\epsilon}) - n| \lt \delta$$

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 $b_{n,\epsilon}$ are within a specified closeness to the integers.

Let’s grant for the moment that the vevs of these $b_{n,\epsilon}$ 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: $$\rho = \frac{1}{2}(|0\rangle \langle 0| + |1\rangle \langle 1|)$$ Then, $$Tr(\rho b_n) = \frac{1}{2}(n + (n+1)^2)$$ I can just as well produce a density matrix such that the vevs of the $b_{n,\epsilon}$ 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?