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State space of QFT, CCR and quantization, and the spectrum of a field?

  1. Mar 11, 2013 #1
    State space of QFT,CCR and quantization,spectrum of a field operator?

    In the canonical quantization of fields, CCR is postulated as (for scalar boson field ):
    [ϕ(x),π(y)]=iδ(x−y) ------ (1)
    in analogy with the ordinary QM commutation relation:
    [xi,pj]=iδij ------ (2)
    However, using (2) we could demo the continuum feature of the spectrum of xi, while (1) raises the issue of δ(0) for the searching of spectrum of ϕ(x), so what would be its spectrum?

    I guess that the configure space in QFT is the set of all functions of x in [itex]R^3[/itex], so the QFT version of ⟨x′|x⟩=δ(x′−x) would be ⟨f(x)|g(x)⟩=δ[f(x),g(x)],
    but what does δ[f(x),g(x)] mean?

    If you will say it means ∫Dg(x)F[g(x)]δ[f(x),g(x)]=F[f(x)], then how is the measure Dg(x) defined?

    And what is the cardinality of the set {g(x)}? Is the state space of QFT a separable Hilbert sapce also? Then are field operators well defined on this space?

    Actually if you choose to quantize in an [itex]L^3[/itex] box, many issues will not emerge, but many symmetries cannot be studied in this approximation, such as translation and rotation, so that would not be the standard route,
    so I wonder how the rigor is preserved in the formalism in the whole space rather than in a box or cylinder model?

    I'm now beginning learning QFT, and know little about the mathematical formulation of QFT, so please help me with these conceptual issues.
     
    Last edited: Mar 11, 2013
  2. jcsd
  3. Mar 11, 2013 #2

    strangerep

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    Science Advisor

    In QFT the fields are operator-valued distributions. So ##\phi(x)## is not an operator on any Hilbert space, but $$\phi_f ~:=~ \int\! dx\, \phi(x) f(x)$$ is.

    In the above, f(x) is an arbitrary Schwartz test function.
    Cf. http://en.wikipedia.org/wiki/Schwartz_space .

    So the CCR above involving a Dirac delta must be understood by integrating both sides with an arbitrary test function.

    For free theories, the usual Fock space is separable. http://en.wikipedia.org/wiki/Fock_space

    For interacting theories, the situation is less clear. Afaik, there are still no rigorous theories of interacting QFT in 4D.

    In such cases (i.e., infinite degrees of freedom), the Stone-von Neumann unitary-equivalence theorem breaks down.
    http://en.wikipedia.org/wiki/Stone–von_Neumann_theorem
    In various particular situations, one can work with so-called "Bogoliubov transformations" to pass between separable subspaces of a much larger nonseparable space. http://en.wikipedia.org/wiki/Bogoliubov_transformation
    This is a large and difficult subject in general and that Wiki page is a bit dubious in places.
    Also try Umezawa's textbook on "Advanced Field Theory".
    https://www.amazon.com/gp/product/1...d_t=201&pf_rd_p=1278548962&pf_rd_i=1563960818

    (BTW, this might also give a little more insight into the question in your other thread about degenerate vacua.)

    When comparing textbooks on practical QFT (using perturbation theory, renormalization, etc) and textbooks on mathematically rigorous approaches to QFT, one might be forgiven for wondering whether they're really about the same subject.
     
    Last edited by a moderator: May 6, 2017
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