Suppose you have λø(adsbygoogle = window.adsbygoogle || []).push({}); ^{4}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 λ_{R}. 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]

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Z factor in renormalization disappeared?

**Physics Forums | Science Articles, Homework Help, Discussion**