Why can we not define asymptotic states in CFTs? I have known that we can't define asymptotic states in CFTs, because we can't use Fock spaces to describe CFTs. But is that right and why? I want to know some details about it. 
 A: The problem with defining asymptotic states in a CFT is the same as in any quantum field theory with massless particles: operators don't decouple fast enough at large distances. Let me explain:
In a QFT with massive particles, the Fock space construction tells you that you can use the one-particle state
$$
\left| {\bf k} \right\rangle
$$
describing a particle with momentum $\bf{k}$,
to construct the most generic states of your theory, for instance a two-particle state
$$
\left| {\bf k}_1 \, {\bf k}_2 \right\rangle = \left| {\bf k}_1 \right\rangle \oplus \left| {\bf k}_2 \right\rangle.
$$
This construction gives you a complete basis of states. There is no double-counting since states with different particle numbers are orthogonal:
$$
\left\langle {\bf k}_1 \right|\left. {\bf k}_2 {\bf k}_3 \right\rangle = 0.
$$
(note that this is not an S-matrix element, but the overlap of two "in" or two "out" states.)
In a CFT there are no particles, but you can still do the following construction: consider one operator $\phi(x)$ acting on the vacuum, take its Fourier transform  with momentum $k$ satisfying $k^2 = 0$, and here you go, you've got a "one-particle" state
$$
\left| \phi({\bf k}) \right\rangle = \int d^dx \, e^{i | {\bf k} | x^0}
e^{i {\bf k} \cdot {\bf x}} \phi(x) \left| 0 \right\rangle.
$$
If your CFT is a free theory, then this state is exactly the same as $\left| {\bf k} \right\rangle$. 
"Multi-particle" states can be defined in a similar fashion:
$$
\left| \phi({\bf k}_1) \, \phi({\bf k}_2) \right\rangle
= \int d^dx_1 \int d^dx_2 \, 
e^{i | {\bf k}_1 | x_1^0} e^{i {\bf k}_1 \cdot {\bf x}_1}
e^{i | {\bf k}_2 | x_2^0} e^{i {\bf k}_2 \cdot {\bf x}_1}
\phi(x_1) \phi(x_2) \left| 0 \right\rangle.
$$
There is nothing wrong with using these states (they are actually closely related to the states used in the conformal collider setup of Hofman and Maldacena, if you are familiar with the CFT literature).
But unlike the Fock space construction in QFT, they do not form a basis of states in your CFT: for instance, you can't distinguish a one-particle states from say a 3-particle state, as their overlap is generically non-zero
$$
\left\langle \phi({\bf k}_1) \right|\left. \phi({\bf k}_2) \, \phi({\bf k}_3) \, \phi({\bf k}_4)  \right\rangle \neq 0,
$$
unless you are precisely in a free theory.
So these states cannot be used to define an S-matrix, for instance.
In summary, I think that the more precise statement in your question is that in a CFT it is impossible to build a complete basis of asymptotic states.
