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I Common interaction vacuum for QED + QCD?

  1. Apr 15, 2017 #1
    Hello,

    I know QED and QCD as isolated theorys but now I thought about particle interactions with QED and QCD processes (like fpr proton-antiproton scattering). But I'm not sure how to interpret this mathematically.

    As I understood my Feynman diagrams are nothing more like pictures for the transition amplitueds (up to some orders). For this we introduce a interaction vacuum state [itex]|\Omega\rangle[/itex]. And then we are able to calculate: [tex]\langle\Omega|\phi(x_1)...\phi(x_n)|\Omega\rangle.[/tex] I thougth this means the creation of some particle at [itex]\phi[/itex] at [itex]x_n[/itex] and anihilation at some other space time point.

    But if I like to have both interactions in one diagram I need a common interaction vacuum to write such transition amplitueds? Is there a common state for QED and QED or better for the standard model? Or are they different? But how can I interpret these processes in tis case?

    Thanks for some answers. Maybe I'm a bit to confused with this whole QFT thing.
     
  2. jcsd
  3. Apr 20, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Apr 20, 2017 #3

    king vitamin

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    The vacuum for pure QED or QCD will differ from the vacuum of both of them combined (and the two vacua naturally differ from each other). Whatever problem (Hamiltonian) you're working with, you'll want to compute correlation functions with respect to the vacuum of that Hamiltonian.
     
  5. Apr 21, 2017 #4

    Demystifier

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    You need, of course, a common vacuum. But since you work with Feynman diagrams, i.e. perturbatively, you only need the perturbative vacuum which is quite simple to find. If ##|0_{\rm QED}\rangle## and ##|0_{\rm QCD}\rangle## are perturbative QED and QCD vacuums, respectively, then the full perturbative vacuum is simply ##|0_{\rm QED}\rangle \otimes |0_{\rm QCD}\rangle##.
     
  6. Apr 24, 2017 #5

    dextercioby

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    The existence of a vacuum vector for a 4D QFT is postulated by the Wightman axioms. Ironically, the set of Wightman axioms is mathematically founded, but physically it checks out only for a QFT of a scalar field in 4D Minkowski spacetime with no self-interaction, so no QED or QCD, or their union for the (presumable) theory of electromagnetic quark-quark scattering. So ##|0_{\mbox{QED}}\rangle## and its counterpart for QCD could be very well taken as ill-defined.
     
  7. Apr 25, 2017 #6

    Demystifier

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    What's wrong with a vector or spinor free (i.e. non-interacting) field in arbitrary number of dimensions?
     
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