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Exact propagator

  1. Feb 5, 2009 #1
    I've got a question about about eqns. 13.12, 13.13, and 13.16. in Mark Srednicki's QFT book, freely previewable here:

    http://www.physics.ucsb.edu/~mark/qft.html

    (it's a good book - this is the only section I have problems with)

    I don't really get how he derives the Lehmann-Kallen form of the exact propagator. The spectral density in equation 13.11 is a function of total momentum k. In eqn 13.12 and 13.13, the spectral density is taken out of the integration over k, which seems illegal. In deriving eqn 13.16 using 13.15, once again it seems that's only legal if the spectral density is independent of k. How is this all legal?
     
  2. jcsd
  3. Feb 6, 2009 #2

    Avodyne

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    I believe the point is that (1) the spectral density rho is a function of the four-momentum k, and not anything else (because everything else has been summed over) and (2) rho is Lorentz invariant. Therefore it can only be a function of k^2, and k^2 = -s.
     
  4. Feb 6, 2009 #3
    I get it now. Thanks. That part is a little tricky.

    I'm not used to describing a multiple particle state with a single total momentum and crazy labels for relative momenta. Usually it's something called a Fock space where you just list all the individual particles and there degeneracy. But I get why he does that. This spectral density form is also used in thermal-QFT and the Dolan-Jackiw propagator separates out nicely between vacuum part and the thermal part, just as the Lehmann-Kallen separates out free-field and higher-order interacting (multiparticle) parts.

    It's a great book, and I like how he emphasizes renormalization as needing to set expressions such as [tex]<0|\phi(x)|0>=0[/tex] for validity of the LSZ-formalism, making sure that the assumption that interacting fields go to free-fields as time goes to +-infinity is true.
     
  5. Feb 6, 2009 #4

    Avodyne

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    I agree that the book is not as clear as it could be on this point.

    It's not clear to me what the status of the Fock basis is in an interacting theory; that may be why Srednicki doesn't use it.
     
  6. Feb 6, 2009 #5
    In eqn. 62.49, there is a [tex]\mu^2 [/tex] under the log that comes from the substitution for the charge: [tex]e \rightarrow e\mu^{\frac{\epsilon}{2}} [/tex]. However, the charge e is cubed to this order calculation. When I worked this out, it seems that the expression involving the [tex] \gamma^{\mu}[/tex] matrix should be multiplied by an overall [tex]\mu^{\frac{\epsilon}{2}} [/tex], and the expression involving [tex] N^{\mu}[/tex] should be multiplied overall by [tex]\mu^{\frac{3\epsilon}{2}} [/tex].

    Am I correct in thinking that these terms (which go to 1 as [tex]\epsilon[/tex] goes to zero) are unimportant except in those cases you want dimensionless expressions, so that you just ignore them when everything is dimensionally correct? The last term involving [tex] N^{\mu}[/tex] doesn't involve the mass parameter at all the way it's written, but it should get multiplied by it since it's multiplied by [tex]e^3[/tex]
     
  7. Feb 9, 2009 #6

    Avodyne

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    I believe the whole expression in 62.49 (the one-loop correction to the QED vertex factor) should have an extra factor of [itex]\mu^{\varepsilon/2}[/itex]. Since [itex]e[/itex] is dimensionless, this gives the expression the correct mass dimension.

    However, the expansion in powers of [itex]\varepsilon[/itex] has already begun, and after the [itex]1/\varepsilon[/itex] is cancelled by [itex]Z_1-1[/itex], we can take [itex]\varepsilon\to 0[/itex], and hence [itex]\mu^{\varepsilon/2}\to 1[/itex].
     
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