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Why can't we detect off-shell particles?

  1. Jan 12, 2012 #1
    when we draw Feynman diagrams we write external lines with onshell momentum only,,
    but those "external" lines are internal lines to an extended process(?).
    what defines the energy-momentum relation of the external states? the field equation and a(p)|p>,, so the offshell parts are due only to contraction and propagators?

    thanks a lot :)
  2. jcsd
  3. Jan 12, 2012 #2
    I think of "off-shell" as being kind of synonymous with "non-physical", as in a photon with non-zero mass. You probably know this, but such non-physical states are necessary in order to conserve energy-momentum, but they are also allowed by the energy/time version of the Heisenberg Uncertainty relation. Therefore, we can have massive photons when we need them, but only if they survive for a time short enough to satisfy the uncertainty relation, i.e. they are virtual, not observable.

    Not sure if that helps with your question, though ...
  4. Jan 13, 2012 #3


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    Energy-momentum is conseverd in Feynman diagrams via Feynman rules i.e. delta-functions in (E,p) at the vertices.

    But the question regarding external lines as internal lines in an extended process is interesting.

    Suppose we do a calculation for a scattering experiment and determine the 'cross section for an external line' i.e. an asymtotic plane wave state which is on-shell. As soon as we detect this particle, e.g. via absorption, it is no longer represented via an external line.

    I think the best one can do is to abandon the idea to construct an ontology based on Feynman diagrams.
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