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I have one more stupid question in Polchinski's string theory book. P. 46, it is said

It is convenient to take a basis of local operators that are eigenstates under rigid transformation (2.4.9) $$\mathcal{A}'(z',\bar{z}')=\zeta^{-h} \bar{\zeta}^{-\bar{h}} \mathcal{A}(z,\bar{z}). \tag{2.4.13} $$

Eq. (2.4.9) is $$z'=\zeta z.\tag{2.4.9}$$

I simply don't understand the sentence "...basis of local operators that are eigenstates under...". Which "local operator", $\mathcal{A}$ or $\mathcal{A}'$? How Eq. (2.4.13) is obtained? How to see later that the derivative $\partial_z$ increases $h$ by one, the derivative $\partial_{\bar{z}}$ increases $\tilde{h}$ by one?

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I think you mean $h$ and not $\hbar$ in the last line. – Prahar Jul 17 '13 at 2:18
Thank you. It should be $\tilde{h}$. The post has been corrected. – user26143 Jul 17 '13 at 2:21
Is your question, what a local operator is? Or how to find the set of local operators which satisfy the above condition and why they are important? – Heidar Jul 17 '13 at 2:23
I mean, what local opetator is, how to find Eq. (2.4.13)? I am very sorry for unclear expressions. The post has been improved. – user26143 Jul 17 '13 at 2:25
up vote 7 down vote accepted

String theory is a conformal field theory. This means that the quantum theory has a conformal symmetry $z \to f(z)$. Recall from representation theory, whenever a theory has any symmetry, we can choose as our basis for the Hilbert space, those states that transform under irreducible representations of the symmetry group. In a conformal field theory, one has a state-operator correspondence (as can be seen from a radial quantization of the theory), i.e. there is a one-to-one correspondence between states and operators of our theory. This implies that we can also choose a basis of operators that transform under irreducible representations of the conformal group.

Representations of the conformal group are usually constructed by considering operators that have specific transformation laws under dilation $D$ (i.e. they correspond to states that are eigenstates of $D$ under the state-operator map). These operators are called primary operators and are classified by their weights $(h, {\tilde h})$. Explicitly, under $z \to z' = \zeta z$ these operators transform as $$ {\cal A}(z, {\bar z}) \to {\cal A}'(z',{\bar z}') = \zeta^{-h} {\bar \zeta}^{-{\tilde h}} {\cal A}(z, {\bar z}) $$ The operators ${\cal A}(z, {\bar z})$ can be taken to be a basis for the set of local operators of the theory.

Why does $\partial$ increase $h$?

Consider the operator ${\cal O}(z,{\bar z}) = \partial {\cal A}(z, {\bar z})$. Under $z \to \zeta z$, this transforms as $$ {\cal O}(z,{\bar z}) \to {\cal O}'(z',{\bar z}') = \partial' {\cal A}'(z', {\bar z}') = \frac{\partial z}{\partial z'} \partial \left( \zeta^{-h} {\bar \zeta}^{\tilde h} {\cal A}(z, {\bar z}) \right) = \zeta^{-h-1} {\bar \zeta}^{-\tilde h}{\cal O}(z,{\bar z}) $$

Thus the weight of ${\cal O}(z,{\bar z})$ is $(h+1, {\tilde h})$.

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Wow, you are fast! You beat me to it! – Heidar Jul 17 '13 at 2:33
haha! If you have more insights then you should go ahead and write yours up too. – Prahar Jul 17 '13 at 2:34
I was going to write almost exactly the same as you, but with my usual style of too many confusing details and words. I think this answers the question pretty well. – Heidar Jul 17 '13 at 2:36
Thank you so much for your clear explanation! – user26143 Jul 17 '13 at 2:37
No. It's correct. More explicitly, I meant to say $z \to z' = \zeta z$. – Prahar Jul 17 '13 at 2:38

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