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Single particle state and symmetry

  1. Feb 23, 2016 #1
    Known that $$ [Q^a,p^\mu]=0 $$ where Q is a representation for the a-th generator of the algebra of group simmetry and p the 4-momentum, if we consider a single particle state, eigenstate of p

    $$ p^\mu | \psi_A (k) > = k^\mu | \psi_A (k) > $$

    then also $$ Q^a | \psi_A (k) > $$

    is eigenstate of p. My text says that If we suppose that in our theory there aren't massless particle then the result state of the action of Q on the single particle state is in general linear combination of single particle states, because otherwise if i add another particle i change the invariant mass.
    How i can derive this?
     
  2. jcsd
  3. Feb 28, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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