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Z factor in renormalization disappeared?

  1. Aug 13, 2014 #1
    Suppose you have λø4 theory and calculate the bare 4-point function:

    $$
    \Gamma_0(s,t,u)=\lambda_0+\lambda_0^2f(s,t,u)\\
    =\left[\lambda_0+\lambda_0^2f(0,0,0)\right]+\lambda_0^2\left[f(s,t,u)-f(0,0,0)\right]
    $$

    We then take a measurement at (s,t,u)=(0,0,0) and call the result λR. Then

    $$
    \Gamma_0(s,t,u)=\lambda_R+\lambda_R^2\left[f(s,t,u)-f(0,0,0)\right]
    $$

    But isn't this also the finite, renormalized [itex]\Gamma_R(s,t,u)[/itex]? We didn't use any Z factors to convert bare to renormalized, instead we just plugged in a measurement.

    Shouldn't the relation be [itex]\Gamma_0(s,t,u)=\frac{1}{Z^2}\Gamma_R (s,t,u)[/itex]
     
  2. jcsd
  3. Aug 13, 2014 #2
    Never mind. When you make a measurement at (s,t,u)=(0,0,0), you're really measuring [itex]Z^2 \Gamma_0(0,0,0)[/itex], since the pole of each propagator contributes Z and each external line has [itex]\frac{1}{\sqrt{Z}} [/itex]. So [itex]\lambda_R=Z^2 \Gamma_0(0,0,0) [/itex], leading to

    $$\Gamma_0(s,t,u)=\frac{1}{Z^2}\lambda_R+\lambda_0^2\left[f(s,t,u)-f(0,0,0)\right]\\
    \frac{1}{Z^2}\lambda_R+\lambda_R^2\left[f(s,t,u)-f(0,0,0)\right]\\
    =\frac{1}{Z^2}\Gamma_R(s,t,u)
    $$

    which is correct to order λ2.
     
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