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Cluster decomposition of the S-matrix

  1. Dec 7, 2009 #1
    Hi,

    I am reading through Weinberg's "Quantum Theory of Fields" (vol. 1)
    and I am somewhat confused about the signs in the cluster decomposition
    of the S-matrix. Specifically, referring to eq. 4.3.2, lets say the term
    coming from the partition
    [itex]\alpha \to \alpha_1\alpha_2[/itex],[tex]\beta \to \beta_1\beta_2[/tex]
    would be
    [tex]+S^C_{\beta_1,\alpha_1}S^C_{\beta_2,\alpha_2}[/tex].
    Lets now assume that permuting [tex]\alpha_1,\alpha_2[/tex] gives a sign, while permuting [tex]\beta_1,\beta_2[/tex] doesn't,
    then the equivalent partition
    [itex]\alpha \to \alpha_2\alpha_1[/itex],[tex]\beta \to \beta_2\beta_1[/tex]
    would give rise to the term
    [tex]-S^C_{\beta_2,\alpha_2}S^C_{\beta_1,\alpha_1}[/tex],
    i.e. minus what I had before. This clearly cannot be right, but I'm not sure where
    the flaw in my reasoning is. Does it have something to do with conservation
    of [tex](-1)^F[/tex], where [tex]F[/tex] is the number of Fermions?

    I would be grateful for any hints. And no, this is not homework :smile:, the course I'm
    following is using Peskin & Schroeder.

    Regards,
    Asger
     
  2. jcsd
  3. Dec 8, 2009 #2

    tiny-tim

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    Science Advisor
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    Welcome to PF!

    Hi Asger! Welcome to PF! :smile:

    (have an alpha: α and a beta: β and try using the X2 tag just above the Reply box :wink:)

    Since αi and βi together form a separate experiment, for each i, the number of fermions in αi will be odd only if the number in βi is.

    So swapping two experiments will give the same number of fermion interchanges for α as for β. :wink:
     
  4. Dec 9, 2009 #3
    Hi Tim!

    Thanks for the answer. I'm sure Weinberg thought that it too obvious to mention.

    Regards,
    Asger
     
  5. Dec 9, 2009 #4
    That is Weinberg's style :) If it is too simple, he does not mention it ;)
     
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