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:

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

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&#039;(0) using Feynman diagrams, and using the master formula: Z_\phi=\frac{1}{1-\Sigma&#039;(m^2)}.

We know that \Sigma&#039;(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&#039;(m^2)}=1+\Sigma&#039;(m^2)}+...

because \Sigma&#039;(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 /> <br /> i\Delta(p)=\frac{iZ_\phi}{(p^2-m^2)-Z_\phi\tilde_{\Sigma}(p^2)+i\epsilon}<br /> <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.