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Production of bound states of slow fermions- Peskin 5.3

  1. Jan 8, 2013 #1
    Hi all,

    I've been reading section 5.3 of Peskin and Schroeder, in which the authors discuss the production of a bound state of a muon-antimuon pair close to threshold in electron-positron collisions.

    Here [itex]\xi,\xi'[/itex] are the Weyl spinors used to construct the Dirac spinors for the muon and anti-muon, respectively.

    Why does the matrix [itex]\Gamma[/itex] supposedly depend on the momentum of the nonrelativistic fermions? In their earlier analysis, this matrix was determined by the spins of the inital electron-positron pair, and the momentum of the final state muons dropped out- even their Dirac spinors didn't depend on these momenta, only on the muon mass, which is why the scattering is isotropic.

    Thanks in advance.
     
  2. jcsd
  3. Jan 9, 2013 #2

    fzero

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    The expression (5.37) takes an especially simple form because the electrons have been taken to be relativistic. Since the mass of the electron is neglected, the factor of [itex]E^2[/itex] in (5.34) cancels against the factor of [itex]1/q^2[/itex] in the expression for the matrix element (under (5.33)). In a more general case, [itex]\Gamma[/itex] has a more complicated dependence on the 4-momentum of the initial state. Energy conservation can be used to write it as a function of the final state momentum.
     
  4. Jan 9, 2013 #3
    Thanks for your reply fzero. The electrons necessarily have to be relativistic, in order to produce the muons; would this dependence on the final state momenta only become non-trivial if we were considering the scattering of fermions of comparable masses? (If the initial state particles were much heavier than those in the final state, then the process would obviously produce particles moving at relativistic velocities.)
     
  5. Jan 9, 2013 #4

    fzero

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    No. Sure, the electrons must have an energy that is large compared to their mass, but the limit in which we ignore the mass completely is an approximation. If we include corrections of order [itex]m_e/E[/itex], then they will necessarily depend on the momentum. Of course, at that level of precision, other corrections (loops) might be of comparable order too. But the point is that the book result is only independent of momentum because of the approximation that we're using.
     
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