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The virtue particle is mass on-shell or off-shell?

  1. Jul 21, 2011 #1
    Please teach me this:
    I do not know whether interaction transfer boson particle(virtue particle) is mass on-shell or mass off-shell(meaning square(4-p)=square(m) or= -square(m)) or both case happening?
    Thank you very much in advance.
     
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  3. Jul 21, 2011 #2

    tiny-tim

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    hi ndung200790! :smile:

    anything with creation and annihilation operators must be on-shell

    so in the position representation, the https://www.physicsforums.com/library.php?do=view_item&itemid=287" are on-shell (and 3-momentum is conserved)

    in the momentum representation, the virtual particles are off-shell (and 4-momentum is conserved) :wink:
     
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  4. Jul 21, 2011 #3

    vanhees71

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    I don't understand this remark. Of course an creation operator wrt. the energy-momentum eigenstates adds an onshell (asymptotically) free particle with the given momentum (and spin or helicity) to the state the creation operator adds on.

    In terms of Feynman diagrams, which are nothing else than a special notation for the calculation of S-matrix elements in perturbation theory, these initial- or final-state asymptotically free particles are represented by the external legs. The "virtual particles" are represented by inner lines, connecting to vertices of the diagram. These lines stand for particle propagators. At each vertex, energy-momentum conservation holds, and usually the four momenta of internal lines are off-shell. It's in fact a problem in naive perturbation theory, when the kinematics of a process is such that an internal line's four momentum becomes on-shell since there the propagator has a pole. The reason are usually infrared of collinear divergences when massless particles are involved. These divergences have to be remedied by an appropriate resummation of many diagrams (e.g., by the Bloch-Nordsieck argument in QED).
     
  5. Jul 21, 2011 #4

    tiny-tim

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    hi vanhees71! :smile:
    but not in the position representation (coordinate-space representation), only in the momentum representation?

    apart from that, aren't you agreeing with me? :confused:
     
  6. Jul 21, 2011 #5

    vanhees71

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    I just said that I don't understand what you mean. What do you mean by "on shell" in the position representation?
     
  7. Jul 22, 2011 #6
    Thanks all of you very much.Knowing that in momentum representation the virtue particle is mass off-shell is important(it seem to me).
     
  8. Jul 23, 2011 #7
    The position and 3-momentum representation theory in normal Quantum Mechanics is already clear.But how about the position and momentum representation in Quantum Field Theory.Please give me a favour to explaint more detail.
     
  9. Jul 23, 2011 #8
    At the moment, I think that the position and momentum representation relate with each other by Fourier transformation(Fourier integral transformation).Is that correct?
     
  10. Jul 24, 2011 #9

    tiny-tim

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    hi ndung200790! :smile:
    no

    the position and momentum representations are both Fourier integrals

    the position representation (of the propagator) is a Fourier integral over d3p, see (6.2.1) at p274

    the momentum representation is obtained by the trick of extending the physical variable p to a mathematical variable q, where q = p and q0 is mathematically convenient

    the momentum representation (of the propagator) is a Fourier integral over d4q, see (6.2.18) at p277 :wink:
     
  11. Jul 25, 2011 #10
    Considering propagator,we use Fourier integral over 4-momentum.Please teach me why we consider the energy as mathematical convenient,because the energy is determined by 3-momentum then it seems that 4-momentum q is mass on-shell but not off-shell.
     
  12. Jul 25, 2011 #11

    tiny-tim

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    no, the "energy" q0 is not determined by the 3-momentum q (= p)

    q0 is defined as p0 + s, where s is a new variable which can take any value

    (and where p0 is the energy determined by the 3-momentum p)

    so q = (q0 , q) = (p0 + s , p), which is off-shell :smile:

    see p276 of Weinberg "Quantum Theory of Fields", just below (6.2.15), viewable online (it says "Volume 2", bit it's actually Volume 1 :rolleyes:) at http://books.google.co.uk/books?id=...t=book-preview-link&resnum=1&ved=0CC4QuwUwAA"

    (sorry, the references in my previous post were to the same book … i confused this with another thread, and thought you'd already mentioned you were reading Weinberg :redface:)

    btw, the mathematical convenience is that integrating over d3p is not Lorentz invariant (nothing that's 3D can be Lorentz invariant), so we invent a new 4D variable q that is Lorentz invariant, and we integrate over d4q :wink:
     
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  13. Jul 25, 2011 #12
    Then 4-q vector may be space-like vector,may be time-like vector?
     
  14. Jul 26, 2011 #13
    In QFT book of Schroder&Peskin,chapter 12.4 ''Renormalization of Local Operators'' I do not understand why they can neglect the square of 4-q vector comperision with square of massive boson,because I think that the square of 4-q is arbitrary value(4-q is mass off-shell but arbitrary).Please be pleasure to teach me this.
     
  15. Jul 26, 2011 #14

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  16. Jul 26, 2011 #15

    samalkhaiat

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    How did reach this conclusion? why should on-shell/off-shell "status" of the fields depends on the representations?
     
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  17. Jul 26, 2011 #16

    tiny-tim

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    creation and annihilation operators are only defined for on-shell particles
    "off-shell" isn't a property of the field, it's a property of the variable, q
     
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