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MS renormalization

  1. Feb 11, 2009 #1
    I have a question about the MS renormalization scheme. When you choose this scheme, all sorts of strange things start happening. The mass in your Lagrangian can no longer be the physical mass. The 4-momentum of a physical particle squares to the physical mass, not the free-field mass. But what I don't get is why the creation and annihilation operators get divided by the square root of the residue of the propagator at the physical mass. How does this happen?
  2. jcsd
  3. Feb 12, 2009 #2


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    I'm not sure I know what you're asking. Very generally, if the residue of the pole in the propagator is Z, then you need a factor of Z^(-1/2) multiplying each field in the LSZ formula for the scattering amplitude. See e.g. P&S eq.(7.42) (where they have Z^(1/2) times the scattering ampliude instead of Z^(-1/2) times the field).
  4. Feb 12, 2009 #3
    I was looking at page 173 of MS's book where he talks about the MS scheme (Mark Srednecki). He writes that " [tex]Z^{\frac{-1}{2}}\phi(x)[/tex] now has unit amplitude to create a one-particle state ".

    The creation and annihilation operators are usually the Fourier coefficients of the free-field solutions, solved in terms of the field. But from the derivation of the LSZ forumula, you should really insert the renormalized field into the expression for the creation and annihilation operators, and not the bare fields. But although that might explain factors of [tex]Z_{\phi}[/tex], how does the residue R come into play?

    I'll see if I can find a preview of P&S online from google books - my library doesn't have it.
  5. Feb 12, 2009 #4
    Okay, thanks, I figured it out from those pages of PS you suggested. I like MS's interpretation better, but I only understood what MS was saying after I read the pages of PS you gave. I was able to reconcile MS's and PS's interpretation.
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