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Weinberg QFT I: Lorentz Transformation with interaction

  1. Aug 22, 2011 #1

    a few lines below equation (3.1.5) Weinberg writes:
    "The transformation rule (3.1.1) is only possible for particles that for one reason or another are not interacting."

    I thought a lot about it, but dont see any possible reason. Can you help please?

    After a few lines he defines the states [tex] \Phi_{\alpha} [/tex]
    then these one should transform like (3.1.1). But why?

    Thanks a lot
  2. jcsd
  3. Aug 22, 2011 #2


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    I believe he gives a reason for this in the very next sentence: "Eq (3.1.1) requires among other things that the state has an energy equal to the sum of the one-particle energies with no interaction terms that would involve more than one particle at a time."
  4. Aug 22, 2011 #3
    Thank you for the answer, I'm not quite sure, if I understand this argument. So I will rephrase it. In a theory with no interactions we expect that the energy of a multiparticle state is the sum of the single energies. Whereas in an interacting theory this should be different. But I dont understand why this should be the case?

  5. Aug 22, 2011 #4
    But this is the *definition* of interaction! In an interacting theory the Hamiltonian has the form (3.1.8), where H_0 is the sum of single particle energies and V is the interaction potential energy.

  6. Aug 23, 2011 #5
    Ok this makes sense. But then [itex] \Psi_{\alpha}^{\pm} [/itex] is not a direct product state of one particle states otherwise it would transform like (3.1.1). How can we then define something like the index [itex] \alpha [/itex] for [itex] \Psi_{\alpha}^{\pm} [/itex]?

    Beside: Where can I read more about this? I am completly confused.

  7. Aug 23, 2011 #6

    In interacting theory the total boost operator (3.3.20) is different from the non-interacting boost operator K_0. Therefore, boost transformation laws of n-particle states are different from (3.1.1) in the interacting case. For example, a 1-particle state may transform into a n-particle state under interacting boost.

  8. Aug 25, 2011 #7
    I think you can view (3.1.13) as the definition of in/out states.
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