Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer 1

up vote 5 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 $(h+1, {\tilde h})$.

share|improve this answer
    
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
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.