Speaking of quantum field theories. Is one of the following implications correct?

gauge theory (gauge invariant) => renormalizable

renormalizable => gauge theory (gauge invariant)

If yes do you have any reference? If not have "gauge invariance / gauge theories" something to do with renormalizability?

  • 2
    $\begingroup$ What did you try to search for yourself? Did you bother to google "nonrenormalizable gauge theories"? $\endgroup$
    – Danu
    May 26, 2015 at 21:39
  • $\begingroup$ I did, but so far I haven't found any satisfactory answer. I read a couple of forum and the arguments seemed rather confused to me... I also tried to get access to the article: "Derivation of Gauge Invariance from High-Energy Unitarity Bounds on the S Matrix" which should have the answer in it... But I can't find it. $\endgroup$
    – Worldsheep
    May 26, 2015 at 21:44
  • $\begingroup$ Related: physics.stackexchange.com/q/110402/2451 , physics.stackexchange.com/q/128053/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 16, 2016 at 9:00

1 Answer 1


The first statement is correct to some extent, the second isn't.

  1. Take the case of vector gauge theories, like the ones in the Standard Model. These theories have a massless vector field, which can be described by two degrees of freedom (2 polarisations) while the classical field itself, $A_\mu$, is described by 4 components. Gauge invariance is related to this mismatch.

    At the quantum level, the interacting theory has processes which, diagrammatically, involve the so-called loops. These loops are often (computationally) infinite and the theory only makes sense if there is a systematic way of subtracting these infinities at each order in perturbation theory, this is what we mean by a Renormalisable theory.

    Now, this in terms mean that some different diagrams will need to be related in order for these infinities to cancel at each order in perturbation theory (or more precisely for the Lagrangian bare quantities to be related at each order in perturbation theory through the coefficients Z). It so happens that if one demands the gauge invariance to be a symmetry of the quantum theory (and not only of the classical level) one find the conservation law is the so-called Ward-Takahashi identities (and not Noether's theorem) state: for a gauge invariant theory one has to have gauge invariant observables, then we should have under a gauge transformation

    $$\delta_\epsilon \langle O(x) \rangle = \delta_\epsilon \int d\phi O(x) e^{i S(\phi,\partial \phi)} = 0 \Rightarrow \langle \delta O(x) \rangle - i \int d^4x \epsilon(x) \langle O \partial_\mu J^\mu \rangle = 0$$ where $O(x)$ is some local operator, say for example a n-point function. What this is that for some n-point functions you will be able to relate some diagrams, and ultimately you will find equivalences between renormalisation of different parameters of the theory, which are exactly the correspondences you would need to employ in order to guarantee renormalisablitly.

    As the theory is renormalisable one can show that in the end we need only one counter-term for each bare quantity, such that we can just write down a counter-term Lagrangian which has the same form of the bare quantities.

    When does this fail?: Gravity. Gravity is a Gauge theory, but the gauge group is not a (semi) simple Lie group as in the Standard Model. Gravity's gauge group is the one of diffeomorphisms of space-time, and this is not a Lie group (in the sense it does not have a finite dimensional Lie algebra). Furthermore, Gravity Lagragian has negative mass dimension couplings $(G_N \propto 1/M^2_{Pl})$, and by inspecting the superficial degree of divergency (or dimensional analysis, following some texts) one can show that the number of required counter-terms increases with perturbation theory. This last notice is a general comment: negative mass dimension couplings always lead to non-renormalisable theories in 4 dimensions!

  2. A theory can be renormalisable without being a gauge theory. The usual $\lambda \phi^4$ scalar theory in 4 dimensions is renormalisable. This means that the number of required counter-terms does not increase with the order of perturbation theory and so the divergencies are kept under-controlled at each order of perturbation theory, in the sense they can be systematically subtracted from the theory.




Literally any QFT book with renormalisation chapter. Namely, to understand the counter-term and bare quantities discussion follow a book with diagrammatic description of QFT (Peskin, etc), while for just a functional methods discussion follow something like Zee (conceptual, but good introduction).

  • $\begingroup$ No worries, glad it helped even with a confusing writing. $\endgroup$
    – romanovzky
    May 29, 2015 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.