Electron - neutrino scattering effective Lagrangian The electron and neutrino can interact through an intermediary Z boson, via the Lagrangian:
$$ L= \frac{1}{2} \partial_\mu \phi_Z \partial^\mu \phi_Z - \frac{1}{2} m_Z ^2 \phi_Z ^2 -g_{\nu} \phi_Z \bar{\psi_\nu} \psi_\nu - g_e \phi_Z \bar{\psi_e} \psi_e .$$
Apparently one can come up with an expression for an effective Lagrangian using the path integral formalism by integrating over the bosonic degree of freedom and obtain the following expression:
$$L_{effective} = \frac{g_\nu g_e}{m_Z ^2} \bar{\psi_\nu} \psi_{\nu} \bar{\psi_e} \psi_e .$$
I don't know how to go about trying to prove this. I'm fairly new to the path integral formalism and I don't really have the intuition for this.
Which quantity should I start by looking at? I suspect something to do with the generating functional, but I'm not certain how to start.
EDIT: I suppose what I'm asking is this: forgetting for the moment the $e,\nu, Z$, how would one go about deriving an effective Lagrangian for a problem with 3 arbitrary particles interacting via a similar Lagrangian, at the appropriate energy scale?
 A: Okay, let's give it a try. $SU(2)$ sector of Standard Model Lagrangian is rather involved, so we will take a look at something simpler. Neutron-proton interaction comes to my mind. In low energy limit it is mediated by a massive scalar particle — a pion. We will be very qualitative about this, in reality there are a lot of details.
Lagrangian will look something like this:
$$ \mathcal{L} = \frac12 \partial^\mu \pi \partial_\mu \pi - \frac12 m_\pi^2 \pi^2 - g \overline{\psi}\pi{\psi}+ \mathcal{L}_{Dirac} $$
Basically, what you are trying to do is the following:
$$ \mathcal{Z}_{eff} = \langle \mathcal{Z} \rangle_\pi  $$
i.e., produce an expression for the partition function that would look the same as fundamental one in the low energy limit. You should remember that partition function contains an exponent of the action which basically does the job of gluing all of the Lagrangian's operators in a more complicated ones. In the end, if you expand this exponent, you will get an infinite series of all possible interactions of the theory written explicitly. We won't do that, but we will imagine.
Among them there will be the operators we're hunting for:
$$ \hat{\mathcal{O}}_1 =\overline{\psi}(x) g \pi(x) \psi(x) $$
$$ \hat{\mathcal{O}}_2 =\overline{\psi}(x) g \pi(x) \psi(x) \cdot \overline{\psi}(x') g \pi(x') \psi(x') $$ 
As we are going to average over the scalar particle, we will assign its field a zero vacuum expectation such that by itself it won't contribute:
$$ \langle \hat{\mathcal{O}}_1 \rangle_\pi = \overline{\psi}(x) g \langle\pi(x)\rangle \psi(x) \approx 0 $$
which means that these particles won't be produced. Then,
$$ \langle \hat{\mathcal{O}}_2 \rangle_\pi = g^2 \overline{\psi}(x) \psi(x) \cdot \langle \pi(x) \pi(x') \rangle \cdot \overline{\psi}(x') \psi(x') $$ 
Here we got a well-known average — the propagator. For simplicity et's go into Fourier space from now on. Here is its Fourier transform:
$$ \tilde{S}_\pi (p) = \frac{1}{p^2 -m_\pi^2}$$
As the energy scale is too small to produce a real particle, we take the lowest order contribution from this operator, which will be 
$$ \tilde{S}_\pi \approx -\frac{1}{m_\pi^2}$$
And our operator becomes
$$ \langle \hat{\mathcal{O}}_2 \rangle_\pi = -\frac{g^2}{m_\pi^2} \overline{\psi} \psi \overline{\psi} \psi $$ 
Next, by evaluating all higher-order averages of $\pi$ and rearranging the series, we can in principle gather a new exponent with an effective action.
