Do Two Operators Need A Term In Their OPE Whose Weight Is Their Combined Weight? My logic is as follows. Suppose I have two operators, $O_1$ and $O_2$. I place a copy of each operator near the origin, and a copy of each operator distance $d$ away, for some large $d$. I can first combine the two sets of $O_1$, $O_2$, and get two copies of $\sum_i C_{12i}O_i$. For large $d$, this means that the four-point function scales as $d^{-2\Delta_{min}}$, where $\Delta_{min}$ is the lowest dimension of any $O_i$ in the $O_1$-$O_2$ OPE.
Alternately, I can contract the two $O_1$s with each other and the two $O_2$s with each other. This seems like it should result in a four-point function proportional to $d^{-2(\Delta_1+\Delta_2)}$. So if these two expressions are equivalent, shouldn't we have $\Delta_{min}=\Delta_1+\Delta_2$?
 A: I don't completely understand the geometry you're considering. It seems that you have in mind a limit where both pairs of $(O_1,O_2)$ are close together, so both $O_1 \times O_2$ OPEs converge rapidly. In that case naive scaling arguments based on the OPE are reliable. But if this channel converges rapidly, the other channel, based the $O_1 \times O_1$ and $O_2 \times O_2$ OPEs, converges slowly, and to reproduce the scaling you like you need to resum infinitely many terms.
To get what you want, consider a GFF with two fields $\phi$ and $\psi$ with dimensions $\Delta_\phi, \Delta_\psi$. You can construct such a theory by putting two free scalars in AdS with different masses. The full four-point function reads
$$\langle \phi(x_1) \psi(x_2) \psi(x_3) \phi(x_4) \rangle = \frac{1}{|x_2 - x_3|^{2\Delta_\psi} |x_1 - x_4|^{2\Delta_\phi}}.$$
In one channel (where $|x_1 - x_4|$ and $|x_2 - x_3|$ are small)  this is obvious. In the other channel you exchange infinitely many composite operators:
$$
\phi(x_1) \psi(x_2) \sim \sum_{n,\ell} \; [\phi \psi]_{n,\ell}
$$
where $[\phi \psi]_{n,\ell}$ has dimension $\Delta_\phi + \Delta_\psi + 2n + \ell$ and spin $\ell$. Naively, the lowest-dimension exchanged has $n = \ell = 0$ so you expect a scaling
$$\langle \phi(x_1) \psi(x_2) \psi(x_3) \phi(x_4) \rangle \sim \frac{1}{|x_2 - x_3|^{2\Delta_\phi + 2\Delta_\psi}} + \ldots.
$$
The fact that this does not reproduce the full correlator at all means that you need to include infinitely many descendants and other exchanged operators to get a reliable result. For a given configuration $(x_1,x_2,x_3,x_4)$ there are some estimates about the rate of converge, meaning how many terms you need to include to get a small error.
