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What are the postulates of QFT?

  1. Mar 6, 2012 #1
    I like the way quantum mechanics can be expressed as a set of five or six axioms, like in Daniel T. Gillespie's A Quantum Mechanics Primer or David McMahon's Quantum Mechanics Demystified.

    Is there a similar set of axioms for quantum field theory?
     
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  3. Mar 7, 2012 #2

    dextercioby

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    An example would be Wightman's axioms which are pretty well-known. See the books by Bogolubov et.al. 1975 or Lopuszanski. There's also an axiomatic formulation due to Haag in terms of (nets of) operator algebras, but this harder to grasp, unless you're also reformulating normal quantum mechanics (the one mentioned in the OP) through operator algebras and their representations.
     
    Last edited: Mar 7, 2012
  4. Mar 8, 2012 #3
    Thank you.
     
  5. Mar 8, 2012 #4

    A. Neumaier

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    Unfdrtunately, while various axiom systems for QM indeed cover all quantum mechanics, the Wightman axioms are known not to cover any of the fundamental quantum fields theories (QED, QCD, and the standard models), as these are gauge theories and the Wightman axioms are not applicable to these. There have been attempts to repair this (notably by Strocchi), but with partial success only.

    The sad truth therefore is that we currently have no adequate system of axioms for QFT, only a number of ideas how it could possibly look like.

    Closest to the standard model are in fact not the nonperturbative Wightman axioms but the perturbative Epstein-Glaser approach to quantum field theory; see
    http://en.wikipedia.org/wiki/Causal_perturbation_theory (and much more by querying scholar.google.com).
     
  6. Mar 8, 2012 #5
    How strange!
     
  7. Mar 8, 2012 #6

    kith

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    When we talk about the axioms of QM, I think of the basic axioms (states in Hilbert space, observables as self-adjoint operators, von Neumann measurements, Schrödinger equation) and not the specific axioms of non-relativistic single particle QM.

    Do the basic axioms really need to be modified for QFT? I thought the problems would occur in in constructing the operators and calculating transition amplitudes.
     
  8. Mar 9, 2012 #7
    That's what I've been wondering for a long time. According to McMahon, in QFT "fields are promoted to operators". I don't know what this means.

    I understand the idea of a physical state being represented by a vector and an observable by an operator and how all that works, but if the field is an operator, upon what does it operate? What kind of mathematical object, and what does this object represent? This seems like a completely different formulation to me.
     
  9. Mar 9, 2012 #8

    kith

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    Also on the physical state.

    Let's look at the electron field for example (this is probably flawed but will show the basic features). A simple physical state is a number state |n>, where n is the number of electrons you have. The operators of the electron field are ψ(x) and ψ+(x). The first one destroys an electron at space time point x, the second one creates one. The fields themself are not observables, but auxiliary operators from which the observables (like the Hamiltonian) can be constructed. So in general, the field equations (like the Dirac equation) are neither Schrödinger equations (dynamics of the states) nor Heisenberg equations (dynamics of the observables), but dynamical equations for these auxiliary quantities which have no direct physical significance.
     
  10. Mar 9, 2012 #9

    A. Neumaier

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    It means that insterad of the classical electromagnetic field $E(x)$ you have operator-valued fields. Thus each component E_k(x) is an operator. More precisely, it is an operator-valued distribution, whch means that to get an operator you must multiply by a nice function and integrate over x. This operator operates as usual on a Hilbert space of wave functions in infinitely many variables.
     
  11. Mar 9, 2012 #10

    atyy

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    Why aren't the Wightman axioms applicable to gauge theory? I know they haven't been shown to be applicable, but have they been shown to be inapplicable?
     
  12. Mar 11, 2012 #11

    A. Neumaier

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    This was discussed at length in
    https://www.physicsforums.com/showthread.php?t=468492
    and threadss citedthere.
     
  13. Mar 13, 2012 #12
    Thank you kith and A. Neumaier, I think I understand a tiny bit now.

    A nice function? What nice function?
     
  14. Mar 13, 2012 #13

    A. Neumaier

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    In order that the product of a function and a distribution is integrable, the function must be sufficiently well-behaved. An infinitely often differentiable function with compact support will always do, and hence is nice enough. But for many distributions, much less is sufficient. Saying nice allows me to hide all these technical details.
     
  15. Mar 13, 2012 #14
    I'm sorry, I meant - what functions are you referring to?
     
  16. Mar 14, 2012 #15

    A. Neumaier

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    I had answered that already: To get an operator rather than a distiribution, multiply by an arbitrary nice function (e.g., a C^inf function with compact support) and integrate over x.. (i.e., rather than integrating over a bounded domain of interest, one typically integrates over a C^inf approximation of its charactersistic function.)

    There isn't any more to the informal notion of ''nice''.
     
  17. Mar 14, 2012 #16
    Oh ok, I figured there was a specific one that depended on the physical state, or something.
     
  18. Apr 3, 2012 #17
    Is this the reason for the phrase "second quantization"?
     
  19. Apr 13, 2012 #18

    A. Neumaier

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    No. Second quantization means that you quantize a classical field theory that itself can be considered as the quantization of a classical particle. - Thus the ''second''.
     
  20. Apr 14, 2012 #19
    How can a classical field theory be considered as a quantization of a classical particle?
     
  21. Apr 15, 2012 #20

    DarMM

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    In a formal sense. Many classical field theories are mathematically no different from the
    quantum mechanical theory of a single particle. For instance the equations for a free
    massive scalar classical field and a single relativistic quantum mechanical boson are identical as they are both the Klein-Gordon equation.

    However Second Quantisation is an out dated piece of terminology. It original arose because you would quantised a single particle and get an PDE for its quantum mechanical version. The wavefunction appearing in this PDE behaved as a classical field, which itself could then be quantised. However it's better nowadays to think of simply quantising some given classical field, as it is just a coincidence that the equations of some classical fields are identical to the quantisation of single particles.
     
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