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Technical question about loop corrections

  1. May 28, 2015 #1
    Does anyone know a simple explanation for the following statement:

    Gauge invariance ⇒ $Πμνϒϒ(0) = ΠμνϒZ(0) = 0$

    Where ΠVV' is the V to V' one loop correction, ϒ is the photon field and Z is the Z-boson field. The argument of Π is the incoming momentum q2 = 0
     
  2. jcsd
  3. May 29, 2015 #2

    vanhees71

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    Your photon propagator is
    $$D_{\mu \nu}(q)=-\frac{1}{q^2-q^2 \Pi(q^2)+\mathrm{i} 0^+}[g_{\mu \nu}-q_{\mu} q_{\nu}] + A(q^2) q^{\mu} q^{\nu},$$
    where ##A(q^2)## is a gauge-dependent non-interacting piece, which doesn't enter any physical result.

    Now the photon has strictly 0 mass. Together with the Ward-Takahashi identity of the photon polarization tensor, which makes it purely 4-transverse, this implies that
    $$\Pi_{\mu \nu}=q^2 \Pi(q^2) (g_{\mu \nu}-q_{\mu} q_{\nu}).$$
    ##\Pi## is a logarithmically divergent scalar. Now to make the residuum of the photon propgator 1 at ##q^2=0##, you impose the renormalization condition
    $$\Pi(q^2=0)=0.$$
    The same argument holds for the ##\gamma##-Z mixing piece too.

    Note that the above renormalization condition is dangerous with regard to infrared divergences, which must be resummed. For this purpose it's better to choose the renormalization point in the space-like.
     
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