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In this video (from 57:38 to 58:31) Aurélien Barrau talked about a simple obstruction for the existence of quantum gravity, stated in an unusual way (to me):

Original in French:

Ce qui m'intéresse ici c'est vraiment de regarder les aspects littéralement liés à la gravitation quantique. Alors pourquoi c'est si dur d'ailleurs? Bien sûr il y a des arguments techniques de non-renormalisabilité que vous connaissez sans doute, mais conceptuellement on pourrait aussi le dire de façon simple: Quand je fais la quantification canonique d'un système, je remplace x par un opérateur mais je ne remplace pas t par un opérateur. Vous voyez que la mécanique quantique distingue fondamentalement l'espace du temps, alors que en relativité générale l'espace et le temps c'est la même chose. Vous voyez que c'est trés compliqué d'avoir d'une part une théorie qui peut mélanger l'espace et le temps, et d'autre part une théorie qui ne traite pas l'espace et le temps sur un pied d'égalité. C'est une des raisons (parmi d'autres) conceptuelles qui rendent très difficile l'émergence d'une théorie de gravitation quantique; on peut dire aussi que le principe d'incertitude d'Heisenberg n'est pas géométrisable.

English translation:

What interests me here is really looking at the aspects literally related to quantum gravity. So why is it so hard? Of course there are technical arguments of non-renormalizability that you probably know, but conceptually we could also say it in a simple way: When I make the canonical quantization of a system, I replace x by an operator but I do not replace t by an operator. You see that quantum mechanics fundamentally distinguishes the space from the time, whereas in general relativity space and time are the same thing. You see that it is very complicated to have on the one hand a theory that can mix space and time, and on the other hand a theory that does not treat space and time on an equal footing . This is one (among others) of the conceptual reasons that make the emergence of a quantum gravity theory very difficult; we can also say that the Heisenberg uncertainty principle is not geometrizable.

Did you ever seen this obstruction stated like that (i.e. bolded sentence) before? Where? Is it relevant? Could it be the starting point of a new approach for quantum gravity? Is there already an approach for quantum gravity where the variable t (for time) is replaced by an operator?

This post appeared first on PhysicsOverflow.

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    $\begingroup$ For what it's worth, that is a very common statement in physics, and by no means specific to quantum gravity. For example, this is pointed out in part 1, section 1 of Srednicki's standard quantum field theory text. Accordingly, loop quantum gravity, which takes this direct canonical quantization approach, has trouble incorporating time. $\endgroup$ – knzhou Dec 11 '19 at 22:08
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Did you ever seen this obstruction stated like that (i.e. bolded sentence) before? Where? Is it relevant?

As pointed out by knzhou in a comment, this is a point that comes up at the level of QFT on a flat spacetime, before we even start talking about quantum mechanics of curved spacetime. If you look at discussions of quantum gravity that are written for specialists, they do often talk about the problems of time, but they focus on the new problems of time that occur in curved spacetime, not the old ones that occur for QM on a fixed, flat background. A recent paper on this kind of thing is Anderson. See also Smolin pp. 9, 71.

Could it be the starting point of a new approach for quantum gravity? Is there already an approach for quantum gravity where the variable t (for time) is replaced by an operator?

It's not a starting point so much as an unsolved problem. We could try to fix the problem by making a theory where time is an operator, but this doesn't seem to work, and the reasons it doesn't seem to work are, as far as we know, fundamental.

At a basic, hand-wavy level:

  • What would a time operator do on a state of zero energy? Such a state has no time-dependence.

  • In general, a time operator acting on an eigenstate of energy would have to be a phase operator, but phase isn't measurable.

At a more technical level, there is a proof that you can't have a time operator, because if there were one, it would be conjugate to energy, so $[H,T]=i\hbar$. There are then results in representation theory saying that the energy spectrum can't be bounded below, and that creates severe problems in quantum mechanics.

References

Anderson, https://arxiv.org/abs/1009.2157

Smolin, https://arxiv.org/abs/hep-th/9202022

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  • $\begingroup$ Am I misunderstanding the Baez reference and, e.g., arxiv.org/pdf/1508.06951.pdf? It appears they are stating that the energy spectrum can't be bounded below unless it is also bounded above. From the link above, "if σ(H)is bounded below but not above, the time reversal symmetry cannot be unitary." Must the energy spectrum necessarily be unbounded above? $\endgroup$ – S. McGrew Dec 12 '19 at 2:44
  • $\begingroup$ Your answer is very interesting! What is the problem with energy spectrum not bounded below? Is the problem due to negative energy? Or is the problem that it makes things too hard? But if so, it is not a good reason to exclude it, because physics could be like that, independently of our ability to manage it. $\endgroup$ – Sebastien Palcoux Dec 12 '19 at 12:08
  • $\begingroup$ @SebastienPalcoux: What is the problem with energy spectrum not bounded below? You might want to ask that as a separate question, or take a look at some of the Q&A's we already have on that. Searching on "energy bounded below" turns up quite a bit. $\endgroup$ – user4552 Dec 12 '19 at 14:59
  • $\begingroup$ The spectrum of the position operator on $L^2(\mathbb{R})$ is $\mathbb{R}$ (not bounded below, ok), but on $L^2(\mathbb{S}^1)$, it is just $\mathbb{S}^1$ (which is bounded), and if I am not mistaken, for any compact space $X$, the spectrum of the position operator on $L^2(X)$ is also bounded. So before considering the issues of energy spectrum not bounded below (for the space $\mathbb{R}$), why not just replacing $\mathbb{R}$ by $\mathbb{S}^1$ (or any compact space). In other words, does energy-time uncertainty principle works well on a compact universe? $\endgroup$ – Sebastien Palcoux Dec 12 '19 at 20:20
  • $\begingroup$ @S.McGrew: I was getting some of the technical details wrong. I've deleted the Baez link and given a link to a blog that takes some care with these details. $\endgroup$ – user4552 Dec 12 '19 at 22:14

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