In quantum field theory I've came across splitting up a Lagrangian (which has an interaction part) into renormalized parts with counterterms. Are there any lagrangians for which renormalisation is not required? When do you know you need to introduce counterterms?
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$\begingroup$ Related: physics.stackexchange.com/q/26948/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jan 4, 2021 at 18:25
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Insertion of counterterms is needed, when you have certain non-vanishing diverging diagrams. For instance, the self-energy diagram in QED (Quantum Electrodynamics) for electrons leading to the mass renormalization, or polarization loop diagram for photon, which is logarithmically divergent, and renormalizes the charge.
One doesn't need counteterms definitely for the case of free theories, there is nothing to renormalize. It is a trivial example.
Less trivial case are the supersymmetric theories, where non-renormalization follows from the fact, that bosonic and fermionic contributions annihilate each other. One of the famous examples in non-renormalization of F-terms. Terms of form: $$ \int d^4 \theta \ W (\Phi) $$ Where $W(\Phi)$ is some superpotential of the superfield $\Phi$ do no receive perturbative corrections. This can be proven via supergraphs, but the most beautiful argument due to N.Seiberg uses the power of holomorphy - https://arxiv.org/abs/hep-th/9408013v1.
There is a nice review by S.Weinberg - https://arxiv.org/abs/hep-th/9803099v1
answered Jan 4, 2021 at 17:58
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$\begingroup$ Even the free scalar and fermion theories do actually require counter terms to cancel the vacuum divergence. This is trivial enough that we normally ignore it, but if one doesn't remove the constant from the action, one doesn't get a well-defined partition function. $\endgroup$– user1504Commented Jan 4, 2021 at 19:54
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$\begingroup$ @user1504 do you mean the subtraction of the infinite constant in normal ordering? $\frac{1}{2} \sum_k \omega_k (a_k a_k^{\dagger} + a_k^{\dagger} a_k) = \sum_k \omega_k a_k^{\dagger} a_k + c$. Yes, the partition function will contrain this infinity, and in will cancel only in connected correlation functions. $\endgroup$ Commented Jan 4, 2021 at 20:07