How does the renormalization factor affect the propagator?

[RedX]

In Kaku's book, the self-energy in a $$\phi^4$$ scalar theory is expanded in a Taylor series as:

$$\Sigma(p^2)=\Sigma (m^2)+\Sigma'(m^2)(p^2-m^2)+\tilde_{\Sigma}(p^2)$$

where $$\tilde_{\Sigma}(p^2)$$ is finite and m is arbitrary (but finite).

The full propagator is then:

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

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

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

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

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

where $$\Sigma_{1}(p^2) =Z_\phi\tilde_{\Sigma}(p^2)$$

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

Kaku defines the renormalized propagator $$\tilde{\Delta}(p)$$ as:

$$\Delta(p)=Z_\phi \tilde{\Delta}(p)$$

which gets rid of $$Z_\phi$$ in the numerator, but not the denominator.

 

[RedX]

Okay, I think I got it. With this expression:

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

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

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

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

Anyways, that's how you get:

$$Z_\phi$$=(1+$$\alpha$$*infinity+$$\alpha^2$$*infinity^2+...)

So we had the original expression:

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

but now expand the Z in the denominator:

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

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

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

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.