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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 $f \in S \subset L^2 (T^*Q)$ (where S is a space of solutions) such that $$W(f)W(g) = \exp (-i \sigma (f, g)) W (f + g)$$ for some (pre-)symplectic form $\sigma$.

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 $$F \twoheadrightarrow X$$ with $X$ a globally hyperbolic Lorentzian manifold of dimension $n + 1$ and a Lagrangian density $$\mathcal{L}: J^1F \rightarrow \Lambda^{n + 1} T^*X$$ such that $\mathcal{L} = Ldx$. Let $$(J^1 (F))^* = \text{Hom} (J^1F, \Lambda^{n + 1}T^*X),$$ $$\mathbb{F} \mathcal{L} : J^1F \rightarrow (J^1F)^*$$ be the Legendre transform, $\omega =- d\theta$ and $\theta$ is the canonical form on $(J^1F)^*$. Then $$\omega_L = -\mathbb{F} \mathcal{L}^* \omega = -d \theta_L$$ for $\theta_L = \mathbb{F} \mathcal{L}^* \theta$ is a a multisymplectic form of degree $n + 2$.

Given Cauchy surface $\Sigma \subset X$, it's possible to produce a (pre-)symplectic form $$\Omega_{\Sigma} = \int_{\Sigma} \omega_L$$ on the off-shell (the space of all sections).

Locally on coordinates $$\theta_L = \frac{\partial L}{\partial v^a_\mu} \wedge du^a \wedge dx_0 \wedge … \wedge \hat{dx_\mu} \wedge … \wedge dx_n + (L - \frac{\partial L}{\partial v^a_\mu}v^a_\mu )dx$$

3) Analogously in http://arxiv.org/1402.1282 and http://ncatlab.org/nlab/show/multisymplectic+geometry, one can consider the first variation $$\delta \mathcal{L} = \sum_a (EL_L)_a \wedge \delta u^a + d \theta_L$$, where $(EL_L)_a$ is the Euler-Lagrange equation and $(x_{\mu}, u^a, v^a_{\mu})$ are the local coordinates of $J^1 F$. In this context $$\omega_L = EL_L + \delta \theta_L $$ 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 $\theta_L$ in 3) coincides with the usual boundary condition $$\sum_{a, \mu} \frac{\partial L}{\partial v^a_{\mu}} \delta u^a $$ (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 $$\int _{\partial V} \sum_{a, \mu} \frac{\partial L}{\partial v^a_{\mu}} \delta u^a vol_{\partial V}= \int_V \sum_{a, \mu} d(\frac{\partial L}{\partial v^a_{\mu}} \delta u^a) dx$$ for some suitable region $V \subset X$.

Anyway the first minor question remains. In other words, is $\theta_L = \frac{\partial L}{\partial v^a_{\mu}} \delta u^a dx$ in 3) equals to $\frac{\partial L}{\partial v^a_\mu} \wedge du^a \wedge dx_0 \wedge … \wedge \hat{dx_\mu} \wedge … \wedge dx_n + (L - \frac{\partial L}{\partial v^a_\mu}v^a_\mu )dx$ ($\theta_L$ in 2))?

All the terms except $Ldx$ seems reasonable to appear in $\theta_L$ of 3).

user40276
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    A comment: the Weyl quantization is unique (up to $*$-isomorphisms) once $V={f}$ and $\sigma$ are specified (with $V$ a real vector space and $\sigma : V\times V\to \mathbb{R}$ 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 $\Omega_{\Sigma}\bigr\rvert_V\neq \sigma\bigr\rvert_V$, I don't see how the two theories should be equivalent. – yuggib Jan 07 '16 at 13:13
  • @yuggib Thanks for your comment. Do you know if $\sigma$ usually comes from the boundary condition on the first variation of the Lagrangian? This would almost solve the problem. – user40276 Jan 07 '16 at 13:16
  • 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 $L^2(\mathbb{R}^d)$ as $V$; or better said one takes the real vector space with the same elements as $L^2$ and $\mathrm{Im}\langle\cdot,\cdot\rangle_2$ as $\sigma$). Maybe since you need that additional condition, this affects the choice of $\sigma$. – yuggib Jan 07 '16 at 13:22
  • In the Klein-Gordon equation $\sigma$ can be written using the causal Green function as $\sigma (f, g) = \int_X f G(g)$ for the causal Green function $G = G^+ - G^-$, but I don't know how to relate this to the boundary condition. – user40276 Jan 07 '16 at 13:29
  • 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... – yuggib Jan 07 '16 at 13:36
  • Sadly calling via "@" other persons that haven't already participated doesn't work. – user40276 Jan 07 '16 at 13:49
  • Ah...then you can ping him in chat (with a link to this question); this would definitely work. – yuggib Jan 07 '16 at 13:58

1 Answers1

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A multisymplectic form is not a symplectic form and cannot be canonically quantized. Thus your main question doesn't make sense as made ''more precise''. (By the way, I have never seen a way to quantize a multisymplectic form.

You would have to quantize the symplectic form underlying the Peierls bracket constructed from the multisymplectic framework. This would give the same result as the quantization in point 1).

The equivalence of the various symplectic forms with the Peierls bracket is demonstrated (with varying amounts of detail) in work by Forger/Romero, by Marolf, by Esposito et al., by Khavkine. They are all constructed to be equivalent, in the hope that one of them might oneday lead to a better approach to a nonperturbative quantization.

  • The presymplectic form (in an infinite dimensional space) is produced after integrating the multisymplectic form in a Cauchy surface. Therefore I'm \textbf{not} demanding a quantization of the multisymplectic form (although I think there's such method using Poisson groupoid stuff). By the way, I've been told that all the three methods (Peierls bracket and the three methods mentioned above) coincide even for a Lagrangian depending on the full (infinity) jet space, however I've never seen a complete proof. In other words, there's a bunch of (4 possible) symplectic forms, why they coincide? – user40276 Jun 29 '16 at 18:59
  • @user40276: I added a comment in my answer. – Arnold Neumaier Jun 29 '16 at 20:30