Q:

Effective operator in four-fermion interaction

In one book, I have got the following lines which I found myself unable to understand what is effective operator? The paragraph is given below:

The weak interaction describes nuclear beta decay, and at low energy it is given
  by an effective four fermion interaction. Since the effective operator has dimension 6, the coupling constant has inverse mass-squared dimension.

A:

To make things clear for the specific case you are talking about, think about what is really happening in beta decay. It occurs via the following process: 

It involves the exchange of a W boson and so to model the true process properly using the standard model requires the three point fermion gauge boson vertices (operators): 
$$\frac{g}{\sqrt{2}} W^\mu\left(\bar\nu \, \gamma_\mu e + \bar u \gamma_\mu d \right).$$
These are interactions between two fermions and the W boson allowing us to draw the above diagram. Note that the coupling constant $g$ is dimensionless. The contribution of the exchange of the W boson is given by
$$\frac{g^2}{E^2-M_W^2}$$ 
where E is the energy transferred by the W boson.
Now at low energies which correspond to small distances the W boson only propagates for a very small distance and it can be approximated by a contact interaction between the fermions:

(Note here that we have replaced the individual quarks by $p$ and $n$ but that is not important to the argument). You see that the contribution of the $W$ is reduced to a contact interaction between four fermions! So the effective operator for this interaction now contains a term with four fermions such as
$$G_F \, \bar\nu e \bar p n.$$
At low energies $(E^2 \ll M_W^2)$ we find
$$\frac{g^2}{E^2-M_W^2} \to \frac{g^2}{M_W^2}$$
and so the new coupling constant (Fermi constant) is given by
$$\frac{G_F}{\sqrt{2}}=\frac{g^2}{8M_W^2}$$
which has dimension of inverse mass squared! (The numerical factors are not important for the argument and come out from doing the full calculation properly).
I am not sure if it was the use of the word operator in the quote that confused you. Remember that in QFT the fields are operators and so a term in the Lagrangian containing a bunch of fields is often referred to as an operator. If four fermion fields are involved (as above for the contact interaction) then it is called a four fermion operator. If the operator appears from integrating out some other fields (the $W$ in this case) then it is known as an effective operator meaning at low energies it effectively captures the correct physics.

