# 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?).

• Primary operators are the highest-weight vectors of representations the conformal group (with the corresponding weights). So in practice you never need to study non-primary operators. If $O_1$ is a primary w/ weight $h_1$ and $O_2$ with $h_2$, then $O_1 + O_2$ is not a primary operator! It doesn't transform correctly under conformal mappings. Mar 25, 2013 at 22:03
• We are only interested in the singular terms as this kind of expressions appears usually in loop integrals (once the 2D spacetime is mapped on the cylinder, integrals over a space slice become contour integrals in the complex plan, hence the focus on the singular contributions). Plus the OPE is based on the algebraic structure of the (infinite number of) operators in the theory, D. Tong says a few words about it in the same notes. And as explained by Vibert $O_1 + 0_2$ is a primary operator only for $h_1=h_2$ giving you a nicely defined OPE ;) Mar 25, 2013 at 22:47

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 ${\rm mass}^h$.
• So what happens if I take $O(0)$ to be $X^\mu(0)$. According to Polchinski's book, the OPE $T(z)X^\mu(0)$ is $\frac{\partial X(0)}{z}$. Why is there no $1/z^2$ term? Or in the notation above $O(w)$? Apr 4, 2013 at 11:29
• Dear @Afriendlyhelper, there "is" a $1/z^2$ but its coefficient is zero which means that the formal dimension of $X^\mu(0)$ is $h=0$. However, it's not a full-fledged tensor operator of dimension zero because $X(z)X(0)\sim\ln|z|^2$ which isn't a power law in $z$. Apr 4, 2013 at 15:38
• Thanks for your answer. I was just wondering because Polchinski simply state s$T(z)X^\mu(0)$ without further explanation in chapter 2. So there's a little but of an argument (aka h=0) to be made to really see the OPE, isn't there? Thanks anyways! Apr 4, 2013 at 17:48