String theory - OPE and primary operators First, a disclaimer: I am new to Physics SE, and I am primarily a mathematician, not a physicist. I apologise in advance for the possibly poor quality of the question, any and thank you for your patience.
I am currently trying to understand some basics of String Theory, based on the script by D. Tong, available at: http://www.damtp.cam.ac.uk/user/tong/string.html .
I am badly confused about the OPE and some related issues. (For definition, see the mentioned script, pages 69 onwards; I don't quite know enough to know what to mention here). I do understand that a number of "operators" $O(z)$ are supposed to be "inserted" at various points $z$ of the complex plane, and the underlying physics is somehow supposed to be encoded in the singular parts of expressions $O_1(z) O_2(w)$ with $w \simeq z$. It is not quite clear to me how it can be that the singular parts somehow seem to be the only thing that matters, but this is probably too philosophical.
I would like for some explanation of the so called primary operators (page 76). 
Firstly, what is the intuition behind those? Is there some physical entity that they represent?
The definition says that $O(z)$ is primary, if it has the OPE with the stress tensor $T(z)$ of the form:
$$  T(z)O(w) ~=~ \frac{h}{(z-w)^2}O(w) + \frac{\partial O(w)}{z-w}+\ldots$$  
At the same time, it says that this is just saying that the OPE terminates at the second order, so it would sound as if it is always the case that if OPE terminates at the second order, the OPE has this particular form. Is this the case? In particular, it would seem that if $O_1(z)$ is primary with $h_1$ and $O_2(z)$ is primary with $h_2$, then $(O_1+O_2)(z)$ has the pole of order at most $2$, but does not have OPE of this form (or am I getting it wrong?).
 A: If you consider the $T(z)O(0)$ OPE, you want to write down all the singular terms. First, there may be singular terms that are more singular than $1/z^2$. If they're there, it means that $O(0)$ isn't a "tensor field". For example, $T(z)$ itself isn't a tensor field in CFTs with $c\neq 0$ because there is a $c/z^4$ term in the OPE.
However, even if $O(0)$ is a tensor field and the $1/z^2$ and $1/z$ are the only ones that appear in the OPE, it doesn't mean that $O(0)$ is a primary operator. Quite likely, it is not one. What the primary operator Ansatz requires that the term going like $1/z^2$ is a multiple of the original operator $O(0)$, the same one!
So the primary operator is an "eigenstate" of the stress-energy tensor, in a sense. Most general superpositions of primary operators won't be primary operators. If you translate the primary operator to a state in the Hilbert space by the state-operator correspondence, it will be an eigenstate of $L_0$ and the highest state vector (a vector in a representation of the Virasoro algebra with the minimum possible eigenvalue of $L_0$ among the vectors in the representation). The absence of $1/z^3$ and higher singularities is equivalent to the corresponding state's being annihilated by $L_n$ for positive values of $n$; and then there's the extra "eigenstate" condition under $L_0$, one that can be seen in the coefficient of the $1/z^2$ term of the OPE.
In some sense, it is unnatural to combine primary operators with different dimensions $h$ into superpositions: it violates the "dimensional analysis" because these operators have the units of ${\rm  mass}^h$.
