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The action for the free Maxwell theory is given by $$S=\int d^dx\sqrt{-g}\bigg(-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}\bigg)$$ The theory is invariant under conformal transformations $g_{\mu\nu}\to\Omega^2(x)g_{\mu\nu}$ only in $d=4$ as can be recognized by looking at the trace of the energy-momentum tensor of the theory, or more directly by recognizing that under such a transformation,

  • $F^{\mu\nu}F_{\mu\nu}=F_{\mu\nu}F_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}\to \Omega^{-4}F_{\mu\nu}F_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}=\Omega^{-4}F^{\mu\nu}F_{\mu\nu}$
  • $g=e^{\text{Tr}(\ln(g_{\mu\nu}))}\to e^{\text{Tr}(\ln(\Omega^2g_{\mu\nu}))}=e^{\text{Tr}(2\ln(\Omega))}g=e^{2d\ln(\Omega)}g=\Omega^{2d}g$

and thus, for $-\frac{1}{4}\sqrt{-g}F^{\mu\nu}F_{\mu\nu}$ to be invariant, $\frac{\Omega^{d}}{\Omega^4}=1$ which is the case only in $d=4$.

This means that the free Maxwell theory is not conformally invariant except in $d=4$. However, the definition of theory is the same in all dimensions and doesn't involve any dimensionful parameter, so I am confused as to what sets the scale of the problem in $d\neq 4$ when the theory is not conformally invariant.

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    $\begingroup$ To me, the question is unclear. In $d=4$, free Maxwell theory is conformally invariant so there are no meaningful scales to speak of. It's not uncommon for conformal invariance to strongly depend on the dimension of the space(-time). $\endgroup$
    – Andrew
    Nov 17, 2020 at 7:40
  • $\begingroup$ What is the question? What is the conformal invariant theory in arbitrary d? $\endgroup$
    – Noone
    Nov 17, 2020 at 7:44
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    $\begingroup$ @Andrew I realize that conformal invariance of theories depends on the dimension of spacetime but my question is that if the theory is not invariant then there should be some parameters of the theory which give me a characteristic scale. In an interacting theory, this might be coupling constants, in a free theory, this might be the mass. But I am not sure how I would get a characteristic scale of a massless free theory. $\endgroup$
    – user87745
    Nov 17, 2020 at 9:01
  • $\begingroup$ @ApolloRa Thanks for the edit and the answer. Kindly see my above comment addressed at Andrew. I will have a look at the paper you linked to in your answer, but quickly, why do we have a $k/4$ here? The kinetic term would simply be a quadratic term, why introduce a parameter in the exponent? Apologies if these are naive objections. $\endgroup$
    – user87745
    Nov 17, 2020 at 9:03
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    $\begingroup$ Oh I see, so the question is, in $d \neq 4$, what is the scale for Maxwell theory? In that case the answer is that Maxwell's theory is scale invariant in any $d$, but is not conformally invariant (except in $d=4$). So, there is no physical scale, but also no conformal invariance. $\endgroup$
    – Andrew
    Nov 17, 2020 at 9:07

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Maxwell's theory is scale invariant in any 𝑑, but is not conformally invariant (except in 𝑑=4). So, there is no physical scale, but also no conformal invariance. Scale invariance does not imply conformal invariance.

One way to see this is to note that if Ω is simply a scale transformation and not the full conformal transformation, then even if it is not canceled out in the action (which is the case in 𝑑≠4), it would simply be a constant factor which would not participate in the differentiation and would leave the EOM invariant. Another approach is to choose the scaling dimension of the field so that the kinetic term is invariant.

In QFT it is very hard to have scale invariance at all, since essentially any regularization scheme will introduce a scale. (See dimensional transmutation in QCD). So the theories that do end up preserving scale invariance quantum mechanically have to be very special and symmetric, and often just end up being fully conformally invariant (eg 𝑁=4 super yang mills).

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