Can scattering amplitudes be simplified with 1PI diagrams?

Question

I have been teaching myself quantum field theory, and need a little help connecting different pieces together. Specifically, I'm rather unsure how to tie in renormalization, functional methods, and the calculation of scattering amplitudes.

Now, I understand how the LSZ formalism is used to link scattering amplitudes (i.e. $\langle f\vert i\rangle$) to the calculation of time ordered products, which can be calculated in two different ways:

1. In the interaction picture, with Wick's theorem and the Feynman rules, by expanding $T \exp(i\int dt \ H_I)$

2. Path Integrals methods, that is, taking functional derivatives of the generating functional $Z[J] = \int \mathcal{D}\phi \ e^{iS[\phi,J]}$ (collective index on $\phi ^a$).

In particular, we may define a generating functional for 1PI diagrams by defining $Z[J] = \exp{W[J]}$, and then taking a Legendre transform to obtain the effective action $\Gamma [J]$. We then see that all the connected correlation functions can be built out of the 1PI correlation functions by constructing Legendre trees. But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams? I cannot see how to do this myself.

The other question, regarding renormalization, is similarly motivated. Now, I understand that, when considering higher corrections to scattering amplitudes, divergent integrals often arise, which can be regularized in a host of different methods (dimensional regularization, introducing a cutoff, etc.). If the the theory is renormalizable, we find that, with the addition of a finite number of counterterms to the Lagrangian (corresponding to the terms already present), we may "redefine" our fields and coupling constants so that everything is under control. This much I am comfortable with. What I don't understand is how this fits in with the general procedure of calculating scattering amplitudes. For instance, suppose I've calculated the amplitude for a typical, tree-level process for the exchange of one photon between two electrons. In the higher correction, I put a photon "wiggle" over an external electron line (that is, before the exchange process, an electron emits and reabsorbs a proton). Suppose I already regulated that electron propagator - which is 1PI - when it had that wiggle. For my higher order diagram, can I simply (in momentum space) multiply the original diagram with my regulated electron propagator?

I do hope you can help me resolve these issues. Thanks in advance.

Answer 1

But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams?

I think that you are missing the perturbation theory in your picture. -- If you know the correlation functions, then you get the amplitudes by LSZ. But in order to get the correlation function you need to calculate some diagrams -- usually representing terms in some perturbation theory.

I put a photon "wiggle" over an external electron line.
Can I simply multiply the original diagram with my regulated electron propagator?

I thought that is exactly what LSZ is about -- the scattering amplitude is a correlation function times a bunch of $\frac{1}{\sqrt{Z}} (p^2-m^2)$factors: $$\langle f|i\rangle \sim \frac{p^2-m^2_1}{\sqrt{Z_1}} \cdot \frac{q^2-m^2_2}{\sqrt{Z_2}} \cdot \dots \Gamma(p,q,\dots)$$ Which on the language of Feynman diagrams means that you need to "amputate" everything that hangs on the external lines of the correlation function $\Gamma$: $$\Gamma(p,q,\dots) = \frac{Z_1}{p^2-m^2_1}\cdot\frac{Z_2}{p^2-m^2_2}\dots\Gamma_{amp}(p,q,\dots)$$ Or, finally: $$\langle f|i\rangle \sim \sqrt{Z_1}\cdot\sqrt{Z_2}\dots\Gamma_{amp}(p,q,\dots)$$ I hope that is what you are asking...

Answer 2

Some parts of Feynman diagrams can indeed be replaced with already calculated functions, but generally not because of different topology and you have to honestly perform calculations.

External line modifications can be reduced to the usual external lines with real masses, if you speak of renormalizations. However it is not all one needs. One has to add plenty of outgoing photon lines in the electron line to take into account radiation of soft photons. It is necessary in order to obtain an inclusive cross section - the only thing different from zero. So the true external electron line is always a "hairy" one.