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S-matrix in Weinberg book

  1. May 19, 2012 #1
    Hi,
    I've read a lot of posts about how Weinberg describes the S-matrix invariance in his book, but none of theme answered my questions.
    At page 116, sec 3.3 - "Lorentz Invariance" of Quantum theory of fields vol.1 Weinberg says:
    "Since the operator [itex]U(\Lambda, a)[/itex] is unitary we may write
    [tex]
    S_{\beta\,\alpha}=\langle\Psi_{\beta}^-\mid\Psi_{\alpha}^+\rangle
    =\langle\Psi_{\beta}^- \mid U^{\dagger}U\mid \Psi_{\alpha}^+\rangle
    [/tex]
    From this equation he gets some conditions that the S-matrix has to fulfill.
    But if the operator [itex]U(\Lambda, a)[/itex] is unitary, then shouldn't be
    [itex]U^{\dagger}U=1[/itex]?
    And so the equation above is always satisfied no matter the form of the S matrix!
     
  2. jcsd
  3. May 19, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    Yes, this equation is an obvious identity. Lorentz invariance is the next assertion - that when you apply U to Ψ+ and Ψ- they transform as representations of the inhomogeneous Lorentz group (Eq. 3.1.1), leading to Eq 3.3.1.
     
  4. May 21, 2012 #3
    Uhm... ok, but now another question arises...
    [itex]U^{\dagger}U=1[/itex], but [itex]\Psi_{\alpha}[/itex] and [itex]\Psi_{\beta}[/itex]
    are two different states, so they can transform with two different irreducible representations.
    In that case [itex]U^{\dagger}[/itex] and [itex]U[/itex] are two matrices of different kind, so how can i say that [itex]U^{\dagger}U=1[/itex]?
     
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