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I have a bit of confusion because when doing QFT and QFT in curved spaces this particular issue seems to be avoided.

I have this feeling that when we quantize a theory, we somehow choose a chart and we stick to it. This feeling comes from, for example, the way we deal with Lorentz transformations in QFT, namely via unitary representations. In my head, change of coordinates is something more geometrical rather than algebraic as is done in QFT.

I also asked a professor of mine and he told me that the usual way of quantizing things is chart-dependent and then suggested I read TQFT and AQFT papers for which I'm not ready yet. Can someone help me understand? I am searching for a mathematically rigorous construction of the quantization process (in canonical quantization) and if it can be done in a coordinate free way.

Hope my question does make sense.

EDIT:

I think my question was misunderstood: I do believe that, of course, the physics in QFT is Lorentz invariant. But in my understanding of the process of quantization what we are doing mathematically is the following: pick a chart, construct Fock space/Quantize and then model in that Fock space Lorentz transformations via Unitary transformations. In this process, if I take another chart I construct a different (but canonically isomorphic) Fock space. So you see: in QFT (I believe) you don't treat change of chart in a more geometric way, but you model it algebraically.

I think that what I'm saying can be seen in Wightman axioms: there is no reference on the spacetime manifold (of course the choice of the Lorentz Group comes from the isometries of the Minkowski metric but one can avoid talking about the metric completely) , it's purely algebraic. So are Lorentz transformations.

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  • $\begingroup$ A change of coordinates is indeed something geometrical, but the fields $\psi$ are functions on spacetime. So in a chart, they are coordinate-dependent. If you change the coordinates, the fields will transform accordingly as well. $\endgroup$ Jun 24, 2023 at 13:39
  • $\begingroup$ @QuantumFieldMedalist Yes indeed. But thats the point of my question: I feel a tension between what I did in my exam of differential geometry and the way I have done things in my QFT e QFT in curved spacetime. So, why? $\endgroup$ Jun 24, 2023 at 13:53
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    $\begingroup$ I would also like to add that quantisation of fields (specifically definition of a particle) in non-stationary spacetimes becomes ambiguous as you cannot define positive and negative frequency solutions in an unambiguous way. This is because you cannot define time using a killing vector. This causes different observers to have different definition of vacuum and particles. $\endgroup$
    – emir sezik
    Jun 24, 2023 at 15:21
  • $\begingroup$ The theory should be independent of the choice of coordinates used. This is why there must be transformations at the level of the theory in a chart that allow it to know about the theory in a different chart. The theory restricted to a chart can predict very different things depending on this chart (cf. Unruh effect). Can you elaborate on the "tension"? $\endgroup$ Jun 24, 2023 at 16:32
  • $\begingroup$ @QuantumFieldMedalist I think my question was misunderstood: I do believe that, of course, the physics in QFT is Lorentz invariant. But in my understanding of the process of quantization what we are doing mathematically is the following: pick a chart, construct Fock space/Quantize and then model in that Fock space Lorentz transformations via Unitary transformations. In this process, if I take another chart I construct a different (but canonically isomorphic) Fock space. So you see: in QFT (I believe) you don't treat change of chart in a more geometric way, but you model it algebraically.Right? $\endgroup$ Jun 25, 2023 at 11:19

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A fully rigorous formulation of the quantization process in QFT is as of yet unknown. Or, rather, to the extent that we have rigorous formulations of QFT we don't quite know how to apply them to QFT as practiced, and to the extent that we have QFT in practice (such as the Standard Model), it is not rigorous. The problem of formulating QFT in 4 dimensions rigorously so that it can cover e.g. the Standard Model is the unsolved Yang-Mills millenium problem.

It is nevertheless true that during "canonical quantization" in QFT, we usually choose a fixed time coordinate along the way - so indeed, a priori this is a "chart-dependent" process. Physics texts usually phrase this as the loss of "manifest Lorentz invariance", and that phrase already contains the solution: This is merely the loss of manifest independence from the choice of coordinates, but not the loss of actual independence. Specifically, the QFT scattering amplitudes that the canonical derivation ends up with have no dependence on the coordinate choice, and neither do any of the other results that we actually want to use.

Think of this like any other computation in differential geometry: Sometimes you need to choose nice coordinates (like Riemann normal coordinates) to do a computation; this does not necessarily mean the result of that computation is somehow only valid for that particular choice of coordinates - proper tensors that are equal in one chart are equal in all charts, after all.

Other approaches to QFT. like the path integral formulation, do not choose particular coordinates in such an obvious manner, yet end up with the same results, another indication that the loss of manifest Lorentz invariance should not worry us all too much.

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    $\begingroup$ That is not completely true: the Unruh effect comes from a scattering amplitude but the function of the thermal bath is frame-dependent. I think it's more correct to state that QFT is invariant only for inertial observers, while for non inertial ones it get a bit more subtle $\endgroup$
    – LolloBoldo
    Jun 24, 2023 at 14:37
  • $\begingroup$ @LolloBoldo That depends on what we mean by "dependent" here, really: Of course observations are dependent on the frame of the observer, this is true even without going to curved spacetime. I took the question to mean "dependent" in the sense that different choices of chart lead to inequivalent theories; the Unruh effect is not a demonstration of inequivalent quantization, but that in curved spacetime "the" vacuum state is no longer unique. $\endgroup$
    – ACuriousMind
    Jun 24, 2023 at 15:46
  • $\begingroup$ Oh ok yeah, i agree with that. The procedure is the same, i took the question as the dependence of the modes on the chart instead :) $\endgroup$
    – LolloBoldo
    Jun 24, 2023 at 16:06
  • $\begingroup$ I've updated the question, so maybe it's clearer. Anyway the answer of @ACuriousMind was very helpful! $\endgroup$ Jun 25, 2023 at 11:40
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    $\begingroup$ @StupidQuestionsIGuess QFT has two representations of the Lorentz group: One on the target space of the fields, which comes from the coordinate transformations, and one on the space of states, which doesn't. The two are related by assumption/Wightman axiom, see also this answer of mine and this answer by Valter Moretti $\endgroup$
    – ACuriousMind
    Jun 25, 2023 at 12:32

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