quantum field theory - Is the effective Lagrangian the bare Lagrangian?

In standard (non-Wilsonian) renormalization we split the bare Lagrangian $\mathcal{L}_0$ into

• a physical Lagrangian $\mathcal{L}_p$ with measurable couplings and masses


• counterterms $\mathcal{L}_{ct}$ with divergent couplings

We then calculate with the bare Lagrangian perturbatively to a cutoff $\Lambda$ adjusting the counterterms so that it's valid to take the continuum limit.

In the Wilsonian theory we start with a Lagrangian $\mathcal{L}_w$ and a scale $\Lambda$. We suppose $\mathcal{L}_w$ is a small perturbation to some fixed point of the renormalization group flow. Using the renormalization group we deduce an effective Lagrangian $\mathcal{L}_{eff}$ at some scale $\mu<\Lambda$. We then take the limit $\Lambda \to \infty$.

What do we calculate with in the Wilsonian theory? Is it the case that $\mathcal{L}_{eff} = \mathcal{L}_0$ from the standard renormalization theory? If so, why?

Example:

In $\phi^4$ theory we have

• 
  

  $\mathcal{L}_0 = \frac{1}{2}Z_0(\partial_\mu \phi)^2-\frac{1}{2}Z_0m_0^2\phi^2+ \frac{\lambda_0Z_0^2}{4!}\phi^4$

  
  


• 
  

  $\mathcal{L}_p = \frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2 \phi^2+\frac{\lambda}{4!}\phi^4$

  
  


• 
  

  $\mathcal{L}_{ct} = \frac{1}{2}\delta_Z(\partial_\mu\phi)^2-\frac{1}{2}\delta_Z\delta_{m^2}\phi^2+ \frac{\delta_\lambda\delta_Z^2}{4!}\phi^4$

  
  

I have no idea what $\mathcal{L}_w$ and $\mathcal{L}_{eff}$ are though? And no book (Peskin and Schroeder, Deligne and Witten etc.) seems to explain it. My guess is as follows

$$\mathcal{L}_w = \frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m_\Lambda^2 \phi^2+\textrm{all interactions}$$

and that at fixed $\mu$ we have $\mathcal{L}_{eff}\to\mathcal{L}_0$ as $\Lambda\to\infty$. I have no real reason to think this though, and still less idea how to prove it!

Answer

The short answer to the question in the title is "No". As there is a lot of confusion on the terminology, I will try to give a clear picture on what kind of different Lagrangians we encounter in this discussion.

Bare Lagrangian

This is the Lagrangian which encodes the basic field content and form of the theory, defines kinetic and interaction terms. For $\phi^3$ theory (I am using the example from Srednicki, where you can find much more detail on the whole issue), it is given by

$$\mathcal{L}=-\frac{1}{2}\partial^\mu\phi_0\partial_\mu\phi_0-\frac{1}{2}m_0^2\phi_0^2+\frac16g_0\phi_0^3+Y_0\phi_0.$$

The quantities appearing in this expression are referred to as bare fields and bare parameters.

Renormalized Lagrangian

After the procedure of renormalization (in the $\overline{MS}$ scheme), we end up with a an expression of the form

$$\mathcal{L}=-\frac12Z_\phi\partial^\mu\phi\partial_\mu\phi-\frac12Z_mm^2\phi^2+\frac16Z_gg\tilde{\mu}^{\epsilon/2}\phi^3+Y\phi.$$

The fields and parameters appearing are now referred to as renormalized. Note that these parameters are not to be interpreted as physically observable quantities. The physical mass of the particle for example does not correspond to the $m$ in the Lagrangian, but is given by the location of the pole of the propagator. The renormalization group tells us how the parameters in the Lagrangian evolve with the energy scale of the theory. Furthermore, note that the counterterm Lagrangian is implicitely present the above expression.

Effective Lagrangian

The effective Lagrangian is still a renormalized Lagrangian, but the parameters will now have an additional dependence: they are not just a function of the energy scale, but also of the cutoff. The action appearing in the path integral of such an effective field theory is called the Wilsonian effective action.