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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.

StupidQuestionsIGuess
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  • 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. – QuantumFieldMedalist Jun 24 '23 at 13:39
  • @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? – StupidQuestionsIGuess Jun 24 '23 at 13:53
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    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. – emir sezik Jun 24 '23 at 15:21
  • 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"? – QuantumFieldMedalist Jun 24 '23 at 16:32
  • @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? – StupidQuestionsIGuess Jun 25 '23 at 11:19
  • In QFT in curved spacetime, the choice of mode functions is independent of the choice of chart but the construction of the Fock space of the same mode functions I believe is dependent of the choice of chart (of course tho the two Fock spaces of the same mode functions in different charts are canonically isomorphic, while different mode functions even in the same chart, as we know, can be not-canonically isomorphic). – StupidQuestionsIGuess Jun 25 '23 at 11:22
  • 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. – StupidQuestionsIGuess Jun 25 '23 at 11:33
  • The mode functions are obtained by solving the equations of motions in a given chart. The corresponding annihilation and creation operators are defined relative to this choice. If one were to choose a different chart, they would obtain different mode functions and different quantizations. As the quantization is of the same theory, they must be related by a transformation. For example, in curved space, the Bogolyubov transformations are the relevant ones. – QuantumFieldMedalist Jun 25 '23 at 13:07
  • Minor note: I think you are making an unclear distinction between geometric and algebraic transformations. The isometry group relevant for quantization is of course dependent on the spacetime manifold and its metric. These isometries are geometric. As the fields are operators on Hilbert space, they will have an algebraic representation when analyzing the behavior of the fields under geometric transformations. – QuantumFieldMedalist Jun 25 '23 at 13:09
  • @QuantumFieldMedalist I didn't know that choosing different charts necessarily brings to different mode functions, good to know. Ok so you agree with me I guess: different charts implies different quantization. But, for geometric reasons we expect this different choices to be related via a transformation which we model via, for example, unitary representations of the Lorentz Group. – StupidQuestionsIGuess Jun 25 '23 at 13:24
  • @QuantumFieldMedalist I think that what I am saying is: consider a scalar field on a manifold $\phi: M\to V$ where V is a vector space. What we usually quantize is $\phi\circ\psi^{-1}$ (with $\psi:U\subset M\to \tilde{U}\subset\mathbb{R}^n$). Of course we could have chosen another chart with another map and quantize with respect to these. Of course, geometrically we expect a connection between the two. Thus at the level of Fock space we introduce the various transformations. Right? – StupidQuestionsIGuess Jun 25 '23 at 13:39
  • Sounds right to me. – QuantumFieldMedalist Jun 25 '23 at 13:42

1 Answers1

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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.

ACuriousMind
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    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 – LolloBoldo Jun 24 '23 at 14:37
  • @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. – ACuriousMind Jun 24 '23 at 15:46
  • 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 :) – LolloBoldo Jun 24 '23 at 16:06
  • I've updated the question, so maybe it's clearer. Anyway the answer of @ACuriousMind was very helpful! – StupidQuestionsIGuess Jun 25 '23 at 11:40
  • @StupidQuestionsIGuess I don't really understand your edit: For most (i.e. interacting) theories, the Fock spaces live in the asymptotic past/future of the theory, not on specific charts. I also do not understand the distinction between "geometric" and "algebraic" you seem to be making here. – ACuriousMind Jun 25 '23 at 12:13
  • @ACuriousMind For example: the transformation law for a tensor field. In QFT you view the transformation rule as a representation of the Lorentz Group. In a more geometric way the transformation comes from the fact that decomposing the tensor in coordinates using the local trivialization of the tangent bundle and then changing chart you change also the basis of the tangent bundle locally given by the chart and the change of basis is modulated via the Jacobian Matrix, but the tensor is the same and the transformation rule is natural, no need for unitary representations etc – StupidQuestionsIGuess Jun 25 '23 at 12:26
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    @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 – ACuriousMind Jun 25 '23 at 12:32
  • @ACuriousMind I don't think that they really answer my question. If quantization was really chart independent, then how a field transforms in QFT should not be a definition. On a Manifold I have no definitions on how things transform. They do it naturally, geometrically. While in QFT you define the transformations. For what I understand in QFT is seems almost like we completely forget that a field is a smooth section of a vector/spinor bundle on a manifold. It's clear that I'm not competent enough to formulate clearly my questions. – StupidQuestionsIGuess Jun 25 '23 at 12:56
  • @ACuriousMind I think that what I am saying is: consider a scalar field on a manifold $\phi: M\to V$ where V is a vector space. What we usually quantize is $\phi\circ\psi^{-1}$ (with $\psi:U\subset M\to \tilde{U}\subset\mathbb{R}^n$). Of course we could have chosen another chart with another map and quantize with respect to these. Of course, geometrically we expect a connection between the two. Thus at the level of Fock space we introduce the various transformations. Right? (Of course for sections of tangent or spinor bundles is more interesting, but this way is easier) – StupidQuestionsIGuess Jun 25 '23 at 13:45
  • @StupidQuestionsIGuess The standard version of QFT as taught in intro textbooks (Fock spaces, canonical quantization, LSZ formula etc.) is formulated not on some general manifold, but specifically on Minkowksi space $\mathbb{R}^4$. That's what the Standard Model is, and it is what most people mean when they say "QFT". How to formulate QFTs on general manifolds is an entirely different question from this kind of QFT, and indeed the domain of AQFT, TQFT, FQFT and various other approaches. – ACuriousMind Jun 25 '23 at 14:28