Can the Lagrangian of an effective field theory have higher derivative terms?

Question

For example, the effective field theory Lagrangian with cutoff $\Lambda$ for the renormalizable $\varphi^4$ theory is $$\mathcal L_{\mathrm{eff}}(\varphi;\Lambda)=\frac{1}{2}Z(\Lambda)\partial_\mu\varphi\partial_\mu\varphi+\frac{1}{2}m^2(\Lambda)\varphi^2+\frac{1}{24}\lambda(\Lambda)\varphi^4+\sum_{d\geq6}\sum_ic_{d,i}(\Lambda)\mathcal O_{d,i}$$ The operators $\mathcal O_{d,i}$ consist of all terms that have mass dimension $d\geq6$ and respect the original symmetry of the renormalizable $\varphi^4$ theory. (Reference: Srednicki's QFT Chapter 29)

Then, terms like $\partial_\mu\varphi\partial_\mu\varphi\,\varphi^2$ are allowed in the operators $\mathcal O_{d,i}$. Can higher derivative terms like $(\partial_\mu\varphi\partial_\mu\varphi)^2\varphi^2$ be allowed? If these higher derivative terms are included, the effective field theory will be different from the original renormalizable $\varphi^4$ theory which only has second order derivative.

Answer 1

Yes, it can. In principle there isn't a reason not to include them. Of course, once we add them the theory will be different, just as it happens when you add any other type of higher dimensional operator.

However, if you're cutting the expansion at some fixed dimension, you can sometimes remove this higher derivative terms using a field redefinition, if the maximum dimension is low enough. For example, in a theory with one scalar $\varphi$ and operators up to dimension six, an operator with more than two derivatives is $(\partial^2\varphi)^2$. The field redefinition $\varphi \to \varphi + k \partial^2\varphi$ (for suitable $k$) has the effect of leaving the structure of the Lagrangian unchanged except for removing the operator in question (and producing dim. $>6$ operators, which we're ignoring).

Answer 2

Yes, in principle a Wilsonian effective action will have pretty much every term you can imagine. The idea is that when constructing a QFT you choose a family of bare Lagrangians $\mathcal{L}(\Lambda)$ indexed by the UV cutoff $\Lambda$. For every pair of cutoff values (scales essentially) $\Lambda_1\ge \Lambda_2$ one has a Wilsonian Renormalization Group map $RG_{\Lambda_1,\Lambda_2}$ which takes a Lagrangian $\mathcal{L}$ at scale $\Lambda_1$ and produces $RG_{\Lambda_1,\Lambda_2}(\mathcal{L})$, namely, the effective Lagrangian at scale $\Lambda_2$. The main goal of renormalization theory is to be able to remove cutoff, i.e., find a correct prescription for the bare Lagrangians $\mathcal{L}(\Lambda)$ as functions of $\Lambda$ such that: $$ \forall \Lambda_2, \ \ \lim_{\Lambda_1\rightarrow \infty} RG_{\Lambda_1,\Lambda_2}(\mathcal{L}(\Lambda_1))\ \ {\rm exists}. $$ The latter defines the effective theory at scale $\Lambda_2$ of the QFT in the continuum without cutoffs. Typically the bare Lagrangians $\mathcal{L}(\Lambda)$ will not contain higher derivative terms (in fact will only contain an explicit and simple finite list of terms corresponding to relevant or marginal interactions). However the effective theories $\lim_{\Lambda_1\rightarrow \infty} RG_{\Lambda_1,\Lambda_2}(\mathcal{L}(\Lambda_1))$ will typically contain higher derivatives.

Answer 3

Consider Chiral Perturbation theory (ChPT), an effective field theory for the strong interactions described from QCD. In the limit of zero and equal quark masses the QCD Lagrangian exhibits a symmetry of $SU(2) \times SU(2) $ meaning the right and left "handed" defined quarks do not mix under rotation applied from the members of the group. Since QCD in low energy cannot be treated via perturbation theory. So we construct an effective theory with deggrees of freedom the nucleons and the mesons.

The demand to have a reasonable theory based on QCD is to hold all the symmetries of the underlying interactions to the effective model. This can indeed be achieved by writing the most general Lagrangian with infinitely many terms symmetric under the whole group $SU(2) \times SU(2) $.

Note: The above answer concentrates on the two flavour QCD with only the up and down quarks. Of course there exist the proper generalization to many flavours.