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?


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!


2 Answers 2


In many textbooks, the Wilsonian and old-fashioned views of renormalization are treated as totally separate, but they are actually very closely connected. I will use your notation and describe three views of renormalization as would be taught in the three semesters of a typical QFT course.

QFT I: bare perturbation theory

Bare perturbation theory is the way renormalization is first explained in many textbooks, and the way it was first worked out. For concreteness, suppose $\phi^3$ theory is relevant to our world, and we observe that the particles have physical mass $m_p$ and physical interaction strength $g_p$, by measuring cross sections. That is, naively we have a theory with Lagrangian $$\mathcal{L}_{\text{naive}} = (\partial \phi)^2 + m_p^2 \phi^2 + g_p \phi^3$$ where I suppress all numerical coefficients. Then the tree-level predictions of this theory roughly match that of observations. But this agreement is an illusion. When we go to one-loop order, we see the physical mass and interaction strength are instead predicted to be infinite.

In order to fix this problem, we must impose a cutoff $\Lambda$ and instead use the "bare" Lagrangian $$\mathcal{L}_{\text{bare}}(\Lambda) = (\partial \phi)^2 + m(\Lambda)^2 \phi^2 + g(\Lambda) \phi^3$$ where $m(\Lambda)$ and $g(\Lambda)$ are formally divergent quantities, fixed so that physical predictions are finite. Concretely, for example, we may calculate the correlator $\langle \phi \phi \phi \rangle \sim g$ as a power series in $g(\Lambda)$, giving a series expansion for $g$ in terms of $g(\Lambda)$. We then flip this around to fix $g(\Lambda)$ in terms of $g$. In this way, we get finite predictions that match experiment.

QFT II: on-shell renormalized perturbation theory

Bare perturbation theory is a bit unsatisfactory. For example, it works perturbatively in $g(\Lambda)$, a formally divergent quantity. And the logic is not in the right order: we shouldn't change the theory we're working with once we see our naive theory doesn't work, we should just work with the correct theory from the start.

Instead, in renormalized perturbation theory, we start with the correct Lagrangian and split it as $$\mathcal{L}_{\text{bare}}(\Lambda) = \mathcal{L}_{\text{naive}} + \mathcal{L}_{\text{CT}}(g_p, \Lambda).$$ Here $\mathcal{L}_{\text{naive}}$ is called the renormalized Lagrangian because it contains renormalized fields and couplings, such as $g_p$. We then perform perturbation theory in $g_p$, treating the counterterms as $O(g_p)$ and higher, which makes much more sense. The conditions used to determine the counterterms are the same as in bare perturbation theory.

To define a theory, it is sufficient to specify $\mathcal{L}_{\text{bare}}(\Lambda)$. On-shell renormalized perturbation theory, which splits this Lagrangian, is an extra layer of structure. The splitting is arbitrary, and in the case of the on-shell scheme is conceptually useful because it allows us to work with physical quantities like $m_p$ and $g_p$ throughout the calculation.

It is often said, incorrectly, that "the renormalized Lagrangian is found by adding counterterms to the bare Lagrangian". That is incorrect, because the bare Lagrangian is the whole Lagrangian; we don't add anything to it. Making this mistake causes the names to be swapped, which generates a lot of confusion.

QFT III: Wilsonian renormalization

In the Wilsonian picture, we get the best of both worlds: the naive directness of bare perturbation theory, and the proper setup of renormalized perturbation theory.

In this setup, we imagine we are performing experiments near, but below some energy $\mu$, and find particles with mass $m_p$ and interaction strength $g_p(\mu)$. We may describe these results with a Wilsonian effective action $$\mathcal{L}_{\text{eff}}(\mu) = \mathcal{L}_{\text{naive}}|_{g = g_p(\mu)}$$ which is a reasonably good description, even when all quantum/loop effects are accounted for, because the loops get cut off at the low scale $\mu$. Thus in the Wilsonian picture observed quantities get translated directly into couplings.

Next, because we are high energy physicists, we want to use this information to find a more fundamental theory, valid up to some higher scale $\Lambda$. Let the Lagrangian for this fundamental theory be $\mathcal{L}_{\text{fund}}(\Lambda)$. Then we have $$\mathcal{L}_{\text{fund}}(\Lambda) = \mathcal{L}_{\text{eff}}(\mu) + \Delta \mathcal{L}$$ where $\Delta \mathcal{L}$ is found by integrating out degrees of freedom between $\mu$ and $\Lambda$. Now, $\mathcal{L}_{\text{CT}}$ is found by computing the exact same integrals, but between $0$ and $\Lambda$. (The difference in the lower bound is just because the naive Lagrangian accounts for no quantum effects at all, while $\mathcal{L}_{\text{eff}}(\mu)$ accounts for quantum effects up to scale $\mu$, and don't matter all that much.)

Therefore, by comparison with what we found earlier,

  • the renormalized Lagrangian is the Wilsonian effective Lagrangian
  • the bare Lagrangian is the fundamental Lagrangian
  • the counterterm is the term needed to compensate for the RG flow between them

Note that in all three versions presented above I've included a finite cutoff. If a continuum limit exists, we may take $\Lambda \to \infty$ for the bare/fundamental Lagrangian.

One final subtlety: when we can take this limit, the counterterms are finite! They are just the difference between the low-energy effective theory and some RG fixed point. We only think counterterms diverge because they diverge order by order in a series expansion. This doesn't mean the whole counterterm diverges; note that $$\lim_{x \to \infty} \exp(-x) = 0$$ but the terms in the Taylor series diverge individually.

  • $\begingroup$ You wrote: "Then the tree-level predictions of this theory match that of observations." Why don't we just stop at the tree-level predictions, since they match that of observations? Why are we bothered with one-loop order? $\endgroup$
    – Shen
    Commented Dec 7, 2018 at 11:45
  • 2
    $\begingroup$ @Shen No, they only roughly match observation. You still need loop corrections. $\endgroup$
    – knzhou
    Commented Dec 7, 2018 at 11:54

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


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


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.

  • 2
    $\begingroup$ Thanks for your answer. A couple of quick questions. Firstly I don't quite understand your renormalized Lagrangian. From what I've read, the renormalized Lagrangian is the bare one, minus the counterterms. Would you agree? Secondly, could you possibly expand your answer on the effective Lagrangian, giving me an example of that and how to calculate with it? From what I've read the result is meant to not depend on the cutoff, so how does that work if the effective Lagrangian explicitly does depend on it? Many thanks in advance! $\endgroup$ Commented Sep 30, 2013 at 15:58
  • $\begingroup$ To answer your first question: yes, I agree. See (9.8) and (9.9) on page 72 of Srednicki (I have linked it in my answer) for the definition of the counterterm Lagrangian. Regarding second question: what do you mean by result? Do you mean physical quantities like mass? As I have stated, the quantities in the Lagrangian do not represent physical ones. Furthermore, instead of reciting a whole book in my answer, I could advise you to study section 29 of Srednicki, where you can find all calculations and explanations in appropriate detail. $\endgroup$ Commented Sep 30, 2013 at 16:03
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    $\begingroup$ No, it is not true for all theories. Taking the cutoff to infinity may render some theories trivial (non-interacting). An example would be $\phi^4$ theory. In other cases, like theories with asymptotic freedom, one can take the limit without any problems. $\endgroup$ Commented Sep 30, 2013 at 20:51
  • 1
    $\begingroup$ No, this is not true. If you send the cutoff to infinity in the Wilsonian approach to $\phi^4$ theory, the interaction terms vanish. This does not coincide with what you get from the theory if you treat it in the "non-Wilsonian" approach. Don't worry, I am happy to help. Your questions are quite interesting and it is a pleasure to discuss these issues. $\endgroup$ Commented Sep 30, 2013 at 23:07
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    $\begingroup$ It should perhaps be emphasized that the $\mathcal L$'s in the first and second line (the "bare" and "renormalized" lagrangian) are just the same lagrangian, written in terms of different variables. $\endgroup$
    – pppqqq
    Commented Nov 2, 2018 at 23:49

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