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  Canonical second quantization vs canonical quantization with multisymplectic form in AQFT

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First of all, I'm a mathematician that knows less than the basics of QFT, so forgive me if this question is trivial. Please, keep in my mind that my background in physics is very poor.

1) The usual process of quantization of a free scalar field on Q is by using Weyl operators W(f) for each fSL2(TQ) (where S is a space of solutions) such that W(f)W(g)=exp(iσ(f,g))W(f+g) for some (pre-)symplectic form σ.

This process is explained for the Klein-Gordon equation here The most general procedure for quantization

2) On the other side, in http://arxiv.org/abs/physics/9801019 and http://arxiv.org/abs/math-ph/0408008, given a configuration bundle FX with X a globally hyperbolic Lorentzian manifold of dimension n+1 and a Lagrangian density L:J1FΛn+1TX such that L=Ldx. Let (J1(F))=Hom(J1F,Λn+1TX), FL:J1F(J1F) be the Legendre transform, ω=dθ and θ is the canonical form on (J1F). Then ωL=FLω=dθL for θL=FLθ is a a multisymplectic form of degree n+2.

Given Cauchy surface ΣX, it's possible to produce a (pre-)symplectic form ΩΣ=ΣωL on the off-shell (the space of all sections).

Locally on coordinates θL=Lvaμduadx0^dxμdxn+(LLvaμvaμ)dx

3) Analogously in http://arxiv.org/1402.1282 and http://ncatlab.org/nlab/show/multisymplectic+geometry, one can consider the first variation δL=a(ELL)aδua+dθL, where (ELL)a is the Euler-Lagrange equation and (xμ,ua,vaμ) are the local coordinates of J1F. In this context ωL=ELL+δθL is a n+2-form which is (pre-) multisymplectic.

In a lot of references, it's said that the variational principle 3) coincides with 2). It's said too that θL in 3) coincides with the usual boundary condition a,μLvaμδua (which seems impossible). I would like an answer to these two questions. These are two minor questions that are probably because of a miscalculation of my part or the authors.

Now my main question is : How can one relate 1) to 3)?

More precisely, are the symplectic forms the same in these two approaches? If not, do canonical quantization of 3) (via Weyl CCR or non-exponeniated CCR) produce an equivalent QFT?


EDIT

The second of my minor question is obviously true, because of the Stokes formula. The boundary condition can be written as Va,μLvaμδuavolV=Va,μd(Lvaμδua)dx for some suitable region VX.

Anyway the second minor question remain. In other words, is θL=Lvaμδuadx in 3) equals to Lvaμduadx0^dxμdxn+(LLvaμvaμ)dx (θL in 2))?

All the terms except Ldx seems reasonable to appear in θL of 3).


This post imported from StackExchange Physics at 2016-01-08 10:49 (UTC), posted by SE-user user40276

asked Jan 7, 2016 in Theoretical Physics by user40276 (140 points) [ revision history ]
edited Jan 8, 2016 by Dilaton
A comment: the Weyl quantization is unique (up to -isomorphisms) once V={f} and σ are specified (with V a real vector space and σ:V×VR bilinear, non-degenerate, and skew-symmetric). This is a result of Slawny 1971 (a bicharacter instead of a symplectic form is sufficient for uniqueness IIRC). Therefore, if ΩΣVσV, I don't see how the two theories should be equivalent.

This post imported from StackExchange Physics at 2016-01-08 10:49 (UTC), posted by SE-user yuggib
@yuggib Thanks for your comment. Do you know if σ usually comes from the boundary condition on the first variation of the Lagrangian? This would almost solve the problem.

This post imported from StackExchange Physics at 2016-01-08 10:49 (UTC), posted by SE-user user40276
Sorry, I don't know exactly about that. In my experience, it is not strictly necessary that the one-particle space is the space of solutions of some equation (in non-relativistic QFT, one takes the whole L2(Rd) as V; or better said one takes the real vector space with the same elements as L2 and Im,2 as σ). Maybe since you need that additional condition, this affects the choice of σ.

This post imported from StackExchange Physics at 2016-01-08 10:49 (UTC), posted by SE-user yuggib
In the Klein-Gordon equation σ can be written using the causal Green function as σ(f,g)=XfG(g) for the causal Green function G=G+G, but I don't know how to relate this to the boundary condition.

This post imported from StackExchange Physics at 2016-01-08 10:49 (UTC), posted by SE-user user40276
I can't help you on that; @ValterMoretti should probably be able to give you much more insight. Maybe with this ping we will attract his attention...

This post imported from StackExchange Physics at 2016-01-08 10:49 (UTC), posted by SE-user yuggib

In the absence of gauge invariance, the canonical, Lagrangian, and multisymplectic formulations are fully equivalent. In particular, this holds for free fields. The resulting quantum theory is the same. People explore different classical formalisms in the hope that one of them may give an improved handle on the quantization of interactive field theories, but this didn't materialize so far.

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