# What is classical Lagrangian? The Bare one or renormalized one? Are counterterms quantum corrections to renormalized Larangian?

EDIT

1. When one talks about the “classical Lagrangian” of a field, does one mean the renormalized Lagrangian with physical/renormalized masses and physical/renormalized couplings and without counterterms?

2. If yes, does it therefore mean that the bare Lagrangian is the "quantum corrected" Lagrangian of the theory which includes quantum corrections in the form of countertems (where $\mathcal{L}_{bare}=\mathcal{L}_{renorm}+\mathcal{L}_{counter}$)?

If not, why is that? Why is it wrong to say that "renormalized Lagrangian without the counterterm" $\mathcal{L}_{renorm}$ as the classical Lagrangian. This reference advocates the claim of question 1. This reference says, bare Lagrangian is different from classical Lagrangian. 12.11 is the bare Lagrangian. It says, below Eq. 12.13, classical Lagrangian doesn't have counterterms, and counterterms are added as quantum corrections to the classical Lagrangian. This is explained in the paragraph above eqn. 12.6. Also the footnote around Eqn. 12.14. Can you help me with this?

3. Do the counterterms have any physical effect? In particular, I have heard that quantum corrections can trigger spontaneous symmetry breaking (SSB) even if the theory is classically unbroken (say, in massless $\phi^4$ theory which is unbroken classically). Since, the counterterms, as I understand, are quantum contributions to the classical Lagrangian, I wonder, whether it is the counterterms which are responsible in some way in triggering the SSB?

Renormalisation has nothing to do with Classical vs Quantum. Any theory, classical or quantum mechanical, needs renomalisation if and only if it is non-linear. The origin of the confusion is that, in QFT, non-linear theories are treated with perturbation theory around the free theory, which is linear, which may make it look like that renomalisation is related to loops, i.e., to quantum corrections.

Take as an example the anharmonic oscillator: the true frequency is not $\omega_0$ but $\omega^2\sim\omega_0^2+\gamma^2$ instead, i.e., the frequency has to be renormalised. A more involved example is General Relativity, where the mere definition of mass is highly non-trivial. Both these examples can be treated with Classical mechanics or Quantum mechanics. In both cases, renormalisation is needed because the theory is non-linear. If you want to use the classical Lagrangian, you have to renormalise it too.

As a matter of fact, the process of renormalisation itself in the classical theory is in general different than that of the quantum theory. For example, the classical theory of the electron mass has a quadratic divergence, while the quantum theory has a log divergence. In principle it is even possible that the renormalisation constants of the classical theory are finite and those of the quantum theory diverge, or vice versa.

1. The classical Lagrangian is the bare Lagrangian.

2. (and 3.) It's not the counterterms that have a physical effect, it's renormalization itself. Renormalization in the modern Wilsonian understanding means setting a cutoff energy scale of your theory, above which the Fourier modes of your quantum fields have been integrated out. The Wilsonian effective action is the one that counts to do now QFT below this scale. Not some counterterms you have added in your favourite renormalization scheme. However, it may well be that using the counterterm model, the "renormalized Lagrangian" (that is, bare+counterterm) is a major ingredient for the Wilsonian effective action or the 1PI effective action (for the difference, see this question), as is indeed the case for the Coleman-Weinberg SSB that breaks massless $\phi^4$.

• Re. 1. I'm sure some people have in mind that the classical Lagrangian is something like a renormalized Lagrangian, as if $\hbar \to 0$ the counter-terms/loop corrections vanish. – innisfree Mar 16 '16 at 11:41
• @innisfree: That happens for the 1PI effective action: Taking $\hbar\to 0$ gives back the classical action, but the integrand of the 1PI effective action is generally not the renormalized Lagrangian of renormalized perturbation theory (although it may appear as a part of it). – ACuriousMind Mar 16 '16 at 11:50
• Ah yes, that's true, I didn't think about the effective action. I just meant if you arrange the Lagrangian as L = renormalizated + counter terms, the latter vanish if $\hbar \to 0$ leaving only the renormalized. – innisfree Mar 16 '16 at 12:01
• @innisfree "Re. 1. I'm sure some people have in mind that the classical Lagrangian is something like a renormalized Lagrangian, as if $\hbar\rightarrow 0$ the counter-terms/loop corrections vanish." Is that notion right or wrong? If wrong, why? – SRS Dec 24 '16 at 17:38
• @ACuriousMind Also can you elaborate your reply at innisfree's comment as a note in your answer? I have the same idea in mind as innisfree suggested in his first comment. – SRS Dec 24 '16 at 17:43