# Z factor in renormalization disappeared?

1. Aug 13, 2014

### geoduck

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 $\Gamma_R(s,t,u)$? We didn't use any Z factors to convert bare to renormalized, instead we just plugged in a measurement.

Shouldn't the relation be $\Gamma_0(s,t,u)=\frac{1}{Z^2}\Gamma_R (s,t,u)$

2. Aug 13, 2014

### geoduck

Never mind. When you make a measurement at (s,t,u)=(0,0,0), you're really measuring $Z^2 \Gamma_0(0,0,0)$, since the pole of each propagator contributes Z and each external line has $\frac{1}{\sqrt{Z}}$. So $\lambda_R=Z^2 \Gamma_0(0,0,0)$, 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.