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When I read Mukanov's book in "Quantum effects in gravity", I found the following interesting remark at p. 57.

Remark: Lorentz invariance.To quantize a field theory, we use the Hamiltonian formalism which explicitly separates the time coordinate $t$ from the spatial coordinate $x$. However, if the classical theory is relativistic (Lorentz-invariant), the resulting quantum theory is also relativistic.

This is a statement. I would like to ask: is there any proof, or any simple way to see that it is true? Namely, why by a recipe that is not relativistically invariant (at least it looks so), one can be sure that all the consequences are relativistically variant.

I can't see why it should be true.

PS: This question hits my scar. When I asked exactly that question in my university, I could not get any answer, neither from the lecturer, nor from any book I found. I felt so bad that I even abandoned the quantum field theory course after that.

Qmechanic
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cnguyen
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  • What is the question exactly? – gented Jul 29 '16 at 11:10
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    Just look at the eventual results for that what is physically relevant: Expectation values, S-matrix elements. They all are properly covariant. – ACuriousMind Jul 29 '16 at 11:10
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    @ACuriousMind but that is the question: why starting from some recipes that is not covariant, one gets the covariant expectation values, S-matrix elements... Or asking in another way, how to prove that there is no consequence from the recipes that violate relativity (expectation, S-matrix do not cover all consequences, I suppose) – cnguyen Jul 29 '16 at 11:16
  • @GennaroTedesco please see my comments above, I'll try to edit the question to make it clearer. – cnguyen Jul 29 '16 at 11:18
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    Well, the "proof" is that everything you can conceive of that is physically meaningful is covariant, even when computed with the Hamiltonian recipes. Are you also bothered by the ADM formalism giving covariant results? If we are free to choose coordinates how we want, then choosing specific ones must be allowed. I'm not sure what exactly bothers you about that. – ACuriousMind Jul 29 '16 at 11:19
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    @ACuriousMind But how do we know that we have conceived everything we can? I am not familiar with the ADM formalism, but I suppose it can be proved to be equivalent to other covariant forms. Likewise, I have no problem with going from Lagrangian to Hamiltonian in relativistic mechanics too. But quantisation is rather an extrapolation, it is not derived from the classical field equation. – cnguyen Jul 29 '16 at 11:30
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    Maybe have a look at http://physics.stackexchange.com/q/14481/50583 – ACuriousMind Jul 29 '16 at 11:45
  • @ophelia Why would you say that quantisation is not relativistically invariant? The standard way to quantise things is to take the path integral against a relativistically invariant measure. If the original Lagrangian is invariant, so is the path integral. Hamiltonian quantisation is, in general, not manifestly invariant and checks must be performed eventually on the physical quantities (as already remarked by @ACuriousMind). Usually, you do not want to choose the latter method exactly for this reason. – gented Jul 29 '16 at 14:31
  • @GennaroTedesco in the question, I meant "canonical quantisation" (subjecting fields and momenta to non-zero commutation) which probably means "Hamiltonian quantisation" in your comments, and not functional integral quantisation. Then I'd take your answer that there is indeed a problem with "Hamiltonian quantisation" :-) – cnguyen Jul 29 '16 at 16:09

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A good/intuitive discussion of how to take a classical field theory and obtain a quantum field theory in the Heisenberg picture is discussed in Bjorken and Drell [BD] section 11.3. A discussion on how classical symmetries like Lorentz covariance are to be expressed on a quantum level is provided in BD 11.4, directly motivating the above-mentioned material in Srednicki. Discussions of Lorentz covariance of specific quantum field theories are provided in BD 12.1, 13.5, 14.3 and in 15.3. Covariance of e.g. classical Bosonic string theory only goes to the quantum level for D = 26, as one checks analogously to the BD references given, so it's not as straightforward as the quoted passage makes it seem.

The way this is done is to take $U(a,b) \psi_r(x) U^{-1}(a,b) = S^{-1}_{rs}(a) \psi_s(ax+b)$ (motivated in [11.4]) to first order for translations and Lorentz transformations, i.e. $i[\hat{P}^{\mu},\psi_r(x)] = \partial^{\mu} \psi_r(x)$ and $i[\hat{M}^{\mu \nu},\psi_r(x)] = x^{\mu} \partial^{\nu} \psi_r(x) - ...$ and to show that these transformation laws, which are expressions of the classical displacement/Lorentz invariance of the classical theory [11.4], do not upset the canonical commutation relations, or rather, the theory with the commutation relations imposed is still Lorentz invariant, which is shown by showing the commutation relations still hold for transformed fields, as is shown in e.g. [13.5] explicitly.

bolbteppa
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I would like to refer to the presentation of canonical quantization of scalar fields in chapter 3 of Quantum Field Theory, Srednicki (2007). To clarify this covers canonical quantization in standard Quantum Field Theory (not Quantum Gravity) which I understand is the question.

Srednicki is very careful to use a Lorentz invariant differential so you can see clearly that Lorentz invariance is maintained throughout the quantization procedure (moving from classical variables to quantum operators).

The "proof" then (for the free theory) is to show at the end that once you have created the Fock space of particle states then these all transform as expected under Lorentz transformation. For example for a one-particle state $|k\rangle$ you can prove that

$$U(\Lambda)|k\rangle = |\Lambda k\rangle$$

where $U(\Lambda)$ is the unitary operator associated with the classical Lorentz transformation $\Lambda$.

Problem 3.3 works you through this.

Bosoneando
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isometry
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