Why is conformal invariance only possible for massless theories? A usual mantra in field theories is the assertion that only massless theories can be conformally invariant. By a theory I mean an action $$ S = \int \mathcal{L} \, \mathrm{dVol}, $$ where $\mathcal{L}$ is the Lagrangian density, and the integral is taken over a 4-dimensional Lorentzian manifold with metric $g$. By conformal invariance I mean the statement that under the conformal rescaling of the metric $$ \hat{g} = \Omega^2g, $$ the Lagrangian transforms as $\hat{\mathcal{L}} = \Omega^{-4} \mathcal{L}$. Then, as the volume form transforms as $\widehat{\mathrm{dVol}} = \Omega^{4} \mathrm{dVol}$, the action $S$ is invariant, and the theory is said to be conformally invariant.
The usual physics explanation given is that "if a theory is supposed to be conformally, or scale, invariant, then there cannot exist an intrinsic scale to it, such as mass or a Compton wavelength". Of course, this is a load of hand waving. I guess I don't strictly know what I mean by a massless theory. Maxwell's equations, for example, are a massless conformally invariant theory. My best guess would be that the mass of a theory is its ADM mass, say $m$. So, if $m \neq0$, why must conformal invariance fail?
 A: This question is old, but let me answer it for the record. I'll use the viewpoint of group theory. The mass $m$ of a state $| \Psi \rangle$ is its eigenvalue under the action of $P_\mu^2$:
$$ P^2 | \Psi \rangle = m^2 | \Psi \rangle\,.$$
In a conformal theory, there also exists a generator $D$ of scale transformations obeying $[D,P_\mu] = -i P_\mu$.
Now suppose that there's a normalizable state $| \Psi \rangle$ in your Hilbert space with mass $m \neq 0$. We will arrive at a contradiction by computing $ \langle \Psi | [D,P^2] | \Psi \rangle $ in two ways. On the one hand, using $[D,P^2] = -2i P^2$ we find that
$$\langle \Psi | [D,P^2] | \Psi \rangle= -2i m^2 \langle \Psi | \Psi \rangle\,.$$
At the same time
$$\langle \Psi | [D,P^2] | \Psi \rangle= \langle \Psi | D P^2 - P^2 D | \Psi \rangle=  \langle \Psi | D m^2 - m^2 D | \Psi \rangle= 0$$
since $P^2$ is Hermitian. By comparing both expressions, only $m^2 = 0$ is consistent.
By the way, this proof is far from original, and you will find roughly this computation in various physics papers from the 1960s-1970s.
