possibility for unitary CFTs? In all notes on conformal field theory, the unitarity bounds are obtained by requiring that we have a positive definite inner product. I get how this allows for a probabilistic interpretation, and unitarity is usually associated with the existence of a positive definite inner product. 
But the important question is whether the conformal symmetry can be realized unitarily. The offending commutators from the conformal algebra that are worrying me are
$$[D,P_{\mu}]=iP_{\mu}, \quad [D,K_{\mu}]=-iK_{\mu}$$
These mean that $P_{\mu}$ and $K_{\mu}$ are ladder operators of $D$, which change the eigenvalues of $D$ by $\pm i$. No matter what then, this means that $D$ is not hermitian.
This is telling me that there exist no unitary representations of the conformal algebra. If this is not the case, I would like to understand why. 
If it is the case, how are we able to consider the conformal algebra a symmetry of a quantum system? It is my understanding that symmetries must be realized in quantum mechanics by either unitary or antiunitary transformations. The nonhermiticity of the scaling generator means that a scaling transformation can be neither.
 A: Before we even start talking about ladder operators, there is a much simpler test. Consider that $P_\mu$ is Hermitian. (You might have seen the relation $P^\dagger_\mu=K_\mu$. This is a source of a lot of confusion, and I will comment on it later. For now, let's simply agree that $P_\mu$ is the usual momentum and thus certainly $P^\dagger_\mu=P_\mu$.)
Let's assume that $D$ is also Hermitian and try to check whether this is consistent with the commutation relation
$$
[D,P_\mu]=iP_\mu.
$$
Taking Hermitian conjugation on both sides, and using $[D,P_\mu]^\dagger=[P^\dagger_\mu, D^\dagger]$, we get
$$
[P_\mu,D]=-iP_\mu.
$$
This is consistent with the original commutation relation. On the other hand, had we assumed that $D$ is anti-Hermitian, then we would get a contradiction.
Now, we can go to the argument using ladder operators. You are saying that $D$ cannot be Hermitian for the following reason. 


*

*If $|\Psi\rangle$ is a normalizable eigenstate of $D$ with eigenvalue $\lambda$, then $P_\mu|\Psi\rangle$ is an eigenstate of $D$ with eigenvalue $\lambda+i$.

*Hermitian operators have real eigenvalues, but $\lambda$ and $\lambda+i$ cannot be real simultaneously.
Here I say "normalizable" because non-normalizable states are, well, not states. This argument doesn't work here, because it assumes that $D$ has normalizable eigenstates that are in the domain of $P_\mu$, and are not annihilated by $P_\mu$. This is apparently not true, at least I am not aware of any argument to the contrary. You could try to construct such states. E.g. if $\mathcal{O}(x)$ is a Hermitian primary operator, so $[D,\mathcal{O}(x)]=\Delta \mathcal{O}(x)$ for some real $\Delta$, you could try to define
$$
|\Psi\rangle = \mathcal{O}(0)|0\rangle.
$$
However, this is not a normalizable state, because its norm is
$$
\langle\Psi|\Psi\rangle = \langle 0|\mathcal{O}(0)\mathcal{O}(0)|0\rangle = \infty.
$$
(In our context $(\mathcal{O}(x))^\dagger=\mathcal{O}(x)$, without any inversion applied to $x$. This is related to $P^\dagger \neq K$ comment above. See about that below.) You could also try to patch the argument to work on the continuous spectrum of $D$ ("non-normalizable" eigenstates), but that will fail too. (You can ask a follow-up about that.)
Let me now address the issue about $P_\mu^\dagger\neq K_\mu$ and applying inversion when taking Hermitian conjugate of $\mathcal{O}(x)$. We use $P_\mu^\dagger = K_\mu$ and inversion when taking Hermitian conjugate of $\mathcal{O}(x)$ when we work in Euclidean signature, and in radial quantization. That is, after we have Wick-rotated to Euclidean and performed some further redefinitions to land in radial quantization. Doing this makes analysis of unitarity complicated. Wick-rotation introduces some $i$'s and going to radial quantization mixes them up in a way such that while we had hermitian $P_\mu$ in the beginning, we now have $P_\mu^\dagger = K_\mu$ in the end, etc. Also, after this $D$ is not what it used to be, and is in fact anti-Hermitian and has pure imaginary eigenvalues $i\Delta$.
Finally, it is worth stressing that the algebra of Hermitian conformal generators is $SO(2,d)$ (even in Euclidean signature) and its unitary representations are well-understood and classified. So even if you are not satisfied with this answer, rest assured that unitary representations of conformal symmetry do exist.
