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In QFT, the Stueckelberg "trick" is often used to show how one can write a fully gauge invariant Lagrangian out of one that is not. For example, if we have

$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + m^2 A^\mu A_\mu$,

the gauge invariance becomes apparent when we rewrite the massive gauge boson in terms of a new vector field and a scalar field $\phi$: $A^\mu \rightarrow A^\mu + \frac{1}{m}\partial_\mu \phi$. Then, the Lagrangian is then invariant under $\delta \phi = -m \Lambda(x)$ and $\delta A_\mu = \partial_\mu \Lambda(x)$.

My question is this: typically, we do not ever see terms present in the above Lagrangian like $\lambda (A^\mu A_\mu)^2$. Moreover, it seems like we can always continue to add terms like $(A^\mu A_\mu)^4/m^2$, which clearly seems like a problem. If we regard the Stueckelberg theory as one in which the Higgs has been integrated out and we are only left with the massless Goldstone bosons $\phi$, terms like $\lambda (A^\mu A_\mu)^2$ should become very relevant at high energies by dimensional analysis. I would love some clarification on why they are never present in the Lagrangian.

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The reason is that the Stueckelberg-Lagrangian is written down for a massive photon, a vector boson of the $U(1)$ gauge group. As photons are not self-interacting, interaction terms (i.e. $A^3$ or higher order) are not present.

The idea is that renormalizability can be restored in a theory with broken gauge invariance. The Stueckelberg formalism serves as a tool to make the Proca-Lagrangian gauge invariant by introducing an additional scalar field. When calculating perturbation series, its mass is sent to infinity in order to keep this field from influencing the physical results.

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    $\begingroup$ Thanks; how would this argument generalize to the non-Abelian case? $\endgroup$ – Nikhil Anand Sep 28 '13 at 16:48
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    $\begingroup$ You can find a paper on precisely this question here: ptp.oxfordjournals.org/content/37/2/452 $\endgroup$ – Frederic Brünner Sep 28 '13 at 17:04
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    $\begingroup$ Note: the renormalizability is only restored in the abelian case with this trick. In the non-abelian case you have a nonlinear sigma model, which has a higher cut off, but its still non-renormalizable. $\endgroup$ – DJBunk Apr 7 '14 at 16:07
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A similar term appears, e.g., in scalar electrodynamics (Klein-Gordon-Maxwell electrodynamics) as follows. As Schroedinger noted (please see the reference in my article in Int'l J. of Quantum Information - http://akhmeteli.org/akh-prepr-ws-ijqi2.pdf ), the scalar matter field in scalar electrodynamics can be made real by a gauge transform. The resulting equations of motions can be obtained from a Lagrangian containing a term that is similar to the term that you discuss (please see Eq. 14 and the reference in my above-mentioned article).

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