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One can represent a quantum system by the Weyl algebra (which is a C*-algebra). For instance, a 1 degree of freedom system can be represented by the algebra generated by $e^{\imath t Q}, e^{\imath s P}$. My guess for modeling a quantum field would be to look at C*-algebras with an uncountably infinite number of generators, but in AQFT this is done thanks to a net of C*-algebras. Is there a link between the approach I think of as being intuitive and the AQFT one?

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The Weyl algebra construction can be done abstractly for any real vector space (even infinite-dimensional) endowed with an antisymmetric bilinear form, thanks to B. Blackadar's universal C*-algebra construction using generators and relations ("Shape theory for C∗-algebras", Math. Scand. 56 (1985) 249–275). However, since you asked for a concrete correspondence, here is a more explicit description. In the case of (free) real scalar fields $\phi$ in a globally hyperbolic space-time $(M,g)$ subject to the Klein-Gordon equation $$ \Box_g\phi+m^2\phi=0\ ,$$ the Weyl unitaries are concretely given by the functionals $$ \mathscr{C}^\infty(M,\mathbb{R})\ni\phi\mapsto W_f(\phi)=e^{i\int_M f\phi\ d\mu_g}\ ,\quad f\in\mathscr{C}^\infty_c(M,\mathbb{R})\ ,$$ where $d\mu_g$ is the volume form associated to the Lorentzian metric $g$. One can understand the test functions $f$ as "component indices", just like we do in the case of the $2n$-dimensional symplectic (phase) space generated by $n$ positions $x_1,\ldots,x_n$ and $n$ momenta $p_1,\ldots,p_n$. Given any non-void open subset $O\subset M$ of the space-time manifold $M$, let $\tilde{\mathfrak{A}}(O)$ be the unital $*$-algebra generated by the Weyl unitaries $W_f$ as $f$ runs over the real-valued smooth functions compactly supported in $O$ (so we say that such $W_f$'s are localized in $O$), once we endow such functionals with the following operations:

  • Vector space operations: $(\alpha W_f+\beta W_{f'})(\phi)=\alpha W_f(\phi)+\beta W_{f'}(\phi)$, $\alpha,\beta\in\mathbb{C}$;
  • Involution: $W^*_f=W_{-f}$;
  • Identity: $\mathbb{1}=W_0$;
  • Weyl product: $W_f W_{f'}=e^{\frac{i}{2}\Delta_{m,g}(f\otimes f')}W_{f+f'}$, where $\Delta_{m,g}$ is the distributional kernel of the difference between the retarded and the advanced fundamental solutions of the Klein-Gordon operator $\Box_g+m^2$. We recall that $\Delta_{m,g}$ provides our antisymmetric bilinear form, for $$\Delta_{m,g}(f\otimes f')=-\Delta_{m,g}(f'\otimes f)\ .$$

The above operations are then extended to general elements in the usual way, and they entail that the Weyl unitaries are worthy of their name, since we clearly have $W^*_f W_f=\mathbb{1}$. This $*$-algebra has non-trivial $*$-representations (namely, Fock representations associated to quasi-free states), hence it admits a minimal nontrivial C$*$-norm $\|\cdot\|$. The Weyl algebra $\mathfrak{A}(O)$ associated to $O$ is the C$*$-completion of $\tilde{\mathfrak{A}}(O)$ with respect to this C$*$-norm, and the correspondence $O\mapsto\mathfrak{A}(O)$ is obviously an isotonous net of C$*$-algebras. Moreover, due to the causal support of $\Delta_{m,g}$ entailed by the hyperbolicity of the Klein-Gordon operator, this net is also causal (i.e. elements localized in causally disjoint regions commute). Finally, uniqueness of retarded and advanced fundamental solutions for this operator guarantees that the isometry group of $(M,g)$ acts on this net as it should - namely, for any isometry $\psi$ of $(M,g)$, the action $$ \alpha_\psi(W_f)(\phi)=W_f(\psi^*\phi)=W_{\psi_*f}(\phi) $$ uniquely extends to unit-preserving *-isomorphisms satisfying $$\alpha_\psi\circ\alpha_{\psi'}=\alpha_{\psi\circ\psi'}\ ,\quad\alpha_\psi(\mathfrak{A}(O))=\mathfrak{A}(\psi(O))\ .$$ In the above formula, $\psi^*\phi=\phi\circ\psi$ is the pullback of the field configuration $\phi$, and $\psi_*f=f\circ\psi^{-1}$ is the pushforward of the test function $f$. To summarize, we have obtained a Haag-Kastler net of C$*$-algebras.

The above argument is a somewhat modernized version of a construction due to J. Dimock ("Algebras of local observables on a manifold", Comm. Math. Phys. 77 (1980), no. 3, 219–228). See also C. Bär, N. Ginoux and F. Pfäffle, "Wave equations on Lorentzian manifolds and quantization", ESI Lectures in Mathematics and Physics, European Mathematical Society (2007), arXiv:0806.1036 [math]. One can perform a similar construction for Dirac fields (see for instance C. Dappiaggi, T.-P. Hack and N. Pinamonti, "The extended algebra of observables for Dirac fields and the trace anomaly of their stress-energy tensor", Rev. Math. Phys. 21 (2009) 1241-1312, arXiv:0904.0612 [math-ph]), with the simplification that for the CAR algebra it's not necessary to exponentiate the smeared fields to get C*-algebra elements.

For interacting fields, if one wants to keep close to what physicists do and steer clear of constructive methods a la Glimm-Jaffe (which are of course a better choice from the viewpoint of rigor but severely limit the models one may study in their present state of the art), one has to recourse to formal perturbation theory, which means one has to work with formal power series in the coupling constant and Planck's constant. This also means abandoning C$*$-algebras and working with more general *-algebras. Once one accepts this, perturbative renormalization can be dealt with in a rigorous way, using a language close to the one adopted above. See for instance R. Brunetti, M. Dütsch and K. Fredenhagen, "Perturbative algebraic quantum field theory and the renormalization groups", Adv. Theor. Math. Phys. 13 (2009) 1541–1599, arXiv:0901.2038 [math-ph].

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Thanks Pedro for this detailed answer. –  Issam Ibnouhsein Aug 18 '13 at 11:39
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At a crude intuitive level, yes, something like that --- you want to replace each classical degree of freedom with a one-dimensional quantum particle. Quantum fields have infinitely many degrees of freedom, okay (but only countably many, sorry). The point about nets of C*-algebras is that for each space-time region you quantize the portion of a field that lies in that region, and the resulting C*-algebras cohere in the manner described by the net. This really becomes important when you take relativity into account. If you're quantizing a nonrelativistic field I don't think you need to worry about nets.

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to "Quantum fields have infinitely many degrees of freedom, okay (but only countably many, sorry)" : How do you count them ? –  jjcale Aug 16 '13 at 17:53
    
In the case of infinitely extended field theories (i.e. thermodynamic limits), the net structure becomes important even in the non-relativistic case. A typical example is provided by spin systems. –  Pedro Lauridsen Ribeiro Aug 16 '13 at 18:11
    
@jjcale: e.g., in one of the simplest examples, you quantize the radiation field in a box by Fourier analyzing it. The nodes correspond to points of $\mathbb{Z}^3$, so, countable. In general the classical field is always modeled on a space which is separable in some appropriate sense. –  Nik Weaver Aug 16 '13 at 20:14
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I would avoid to talk about countable or uncountable degrees of freedom in this case. In the QM case one quantizes $\mathbb{C}^2$ which is clearly finite, but in the free QFT case one quantizes (for nets a subspace of) the separable Hilbert space, which has a countable basis, but also uncountable inproper basis. Taking localization into account I would say there are uncountable many degrees of freedom, in particular the local algebras are type III$_1$ factors which I would call uncountable degrees of freedom. –  Marcel Bischoff Aug 30 '13 at 14:10
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One could see the QFT as an countable product of let's say oscillators, but in this picture one loses the important information about the localization, encoded in the net and one would end up with a type I$_\infty$ factor namely all $B(H)$ on the Fock space which one can see as a QM mechanical system with countable many degrees of freedom. But this algebra does not contain information about the localization nor about that it describes a QFT. –  Marcel Bischoff Aug 30 '13 at 14:13
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