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Renormalizing the propagator

  1. Jun 24, 2010 #1
    In Kaku's book, the self-energy in a [tex]\phi^4[/tex] scalar theory is expanded in a Taylor series as:

    [tex]\Sigma(p^2)=\Sigma (m^2)+\Sigma'(m^2)(p^2-m^2)+\tilde_{\Sigma}(p^2) [/tex]

    where [tex]\tilde_{\Sigma}(p^2) [/tex] is finite and m is arbitrary (but finite).

    The full propagator is then:

    [tex]i\Delta(p)=\frac{i}{p^2-m_{0}^2-\Sigma (m^2)-\Sigma'(m^2)(p^2-m^2)-\tilde_{\Sigma}(p^2)+i\epsilon} [/tex]

    where m0 is the bare mass that's in the original Lagrangian. If we define [tex]m_{0}^2+\Sigma(m^2)=m^2 [/tex], i.e., the infinite bare mass cancels a divergence in a self-energy term to give something finite, then:

    [tex]i\Delta(p)=\frac{i}{(1-\Sigma'(m^2))(p^2-m^2)-\tilde_{\Sigma}(p^2)+i\epsilon} [/tex]

    Here's what I don't understand. Kaku now factors out a [tex] Z_\phi=\frac{1}{1-\Sigma'(m^2)} [/tex] to get:

    [tex] i\Delta(p)=\frac{iZ_\phi}{(p^2-m^2)-\Sigma_{1}(p^2)+i\epsilon}[/tex]

    where [tex]\Sigma_{1}(p^2) =Z_\phi\tilde_{\Sigma}(p^2)[/tex]

    The [tex]Z_\phi[/tex] in the numerator of the propagator can be absorbed by bare constants, but I'm not sure how the [tex]Z_\phi[/tex] in the denominator (through [tex]\Sigma_1(p^2) [/tex]) can be gotten rid of.

    Kaku defines the renormalized propagator [tex]\tilde{\Delta}(p)[/tex] as:

    [tex]\Delta(p)=Z_\phi \tilde{\Delta}(p) [/tex]

    which gets rid of [tex]Z_\phi[/tex] in the numerator, but not the denominator.
     
  2. jcsd
  3. Jul 11, 2010 #2
    Okay, I think I got it. With this expression:

    [tex]
    i\Delta(p)=\frac{iZ_\phi}{(p^2-m^2)-Z_\phi\tilde_{\Sigma}(p^2)+i\epsilon}
    [/tex]

    you don't renormalize by just absorbing the [tex]Z_\phi[/tex] in the numerator into bare constants. You first calculate [tex]Z_\phi[/tex]=(1+[tex]\alpha[/tex]*infinity+[tex]\alpha^2[/tex]*infinity^2+...) [where infinity represents some regulated infinity like [tex]\frac{1}{\epsilon} [/tex] in dimensional regularization, and [tex]\alpha [/tex] is a bare constant) by calculating [tex]\Sigma'(0) [/tex] using Feynman diagrams, and using the master formula: [tex]Z_\phi=\frac{1}{1-\Sigma'(m^2)} [/tex].

    We know that [tex]\Sigma'(m^2) [/tex] is infinity, but we do the insane idea that [tex]\alpha*infinity [/tex] is actually small, so instead of [tex]Z_\phi [/tex] being small because it has infinity in the denominator, [tex]Z_\phi=\frac{1}{1-\Sigma'(m^2)}=1+\Sigma'(m^2)}+... [/tex]

    because [tex]\Sigma'(m^2)[/tex]=[tex]\alpha[/tex]*infinity+[tex]\alpha^2[/tex]*infinity^2+... and the RHS is small by the logic above.

    Anyways, that's how you get:

    [tex]Z_\phi[/tex]=(1+[tex]\alpha[/tex]*infinity+[tex]\alpha^2[/tex]*infinity^2+...)

    So we had the original expression:

    [tex]

    i\Delta(p)=\frac{iZ_\phi}{(p^2-m^2)-Z_\phi\tilde_{\Sigma}(p^2)+i\epsilon}

    [/tex]

    but now expand the Z in the denominator:

    [tex] i\Delta(p)=\frac{iZ_\phi}{(p^2-m^2)-(1+\alpha*infinity+\alpha^2*infinity^2+...)\tilde_{\Sigma}(p^2)+i\epsilon}[/tex]

    But now the infinity terms in the denominator are really small since they are multiplied by alpha:

    [tex] i\Delta(p)=\frac{iZ_\phi[1+(\alpha*infinity+\alpha^2*infinity^2+...)/(p^2-m^2-\tilde_{\Sigma}(p^2))+...]}{p^2-m^2-\tilde_{\Sigma(p^2)}+i\epsilon}[/tex]

    So now we get something that is multiplicatively renormalizeable, only instead of only Z being just absorbed into the bare constants, the entire numerator is absorbed.

    I suspect that 1-loop calculations are insensitive to this extra step, but when calculating higher-order loops, the extra term in the numerator must be considered.
     
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