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I have a question regarding equation (2.22) in Ginsparg's lecture notes on CFTs. Equation (2.22) is $$ \langle T(z) \phi_1(w_1, {\bar w}_1) \cdots \rangle = \sum_{i=1}^n \left( \frac{h_i}{(z-w_i)^2} + \frac{1}{z-w_i} \frac{\partial}{ \partial w_i} \right) \langle \phi_1(w_1, {\bar w}_1) \cdots \rangle $$ Here, $T(z)$ is the stress tensor of the CFT and $\phi_i$ is a primary operator of weight $(h_i,0)$ which transforms under conformal transformations as $$ \delta_\epsilon \phi_i = \left( h_i \partial \epsilon + \epsilon \partial \right) \phi_i $$ He derives (2.22) from (2.21) which reads $$ \langle \oint \frac{dz}{2\pi i} \epsilon(z) T(z)\phi_1(w_1, {\bar w}_1) \cdots \rangle = \sum_{i=1}^n \langle \phi_1(w_1, {\bar w}_1) \cdots \delta_\epsilon\phi_i(w_i, {\bar w}_i) \cdots \rangle $$ by setting $\epsilon(x) = \frac{1}{x-z}$.

My question is - Is (2.22) correct?

Here are my reasons to believe that it is not -

  1. I believe he derives (2.22) from (2.21) by setting $\epsilon(x) = \frac{1}{x-z}$ in (2.21). (2.22) is then derived if the following holds $$ \langle \oint \frac{dx}{2\pi i} \frac{T(x)}{x-z} \phi_1(w_1, {\bar w}_1) \cdots \rangle = \langle T(z)\phi_1(w_1, {\bar w}_1) \cdots \rangle $$ This would be true if the integrand on the LHS had only a pole at $x-z$. However, it has also has poles at each $x = w_i$, but those contributions aren't considered.

  2. I can try and derive (2.22) in a different way - namely via contractions. I start with the LHS of (2.22) and contract $T(z)$ with each $\phi_i$. Each contraction is replaced with the operator product $$ T(z) \phi_i(w_i {\bar w}_i) = \frac{h_i \phi_i(w_i {\bar w}_i) }{ ( z - w_i )^2 } + \frac{ \partial \phi_i(w_i {\bar w}_i) }{ z - w_i } + : T(z) \phi_i(w_i {\bar w}_i) : $$ Again, if I only consider the singular terms, I reproduce the RHS of (2.22). But what about $: T(z) \phi_i(w_i {\bar w}_i) :$?? In a general CFT, conformal normal ordering $:~:$ is not equivalent to creation-annihilation normal ordering ${}^\circ_\circ~{}^\circ_\circ$. The latter would vanish in a correlation function, but not the former. So, I believe in general there would be extra terms on the right of (2.22).

What am I misunderstanding?

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  • $\begingroup$ I don't think you should be setting $\epsilon=\frac{1}{x-z}$. Have a look at Tong's string theory notes page 76 and see if it helps... $\endgroup$ – Heterotic Jan 15 '15 at 10:56
  • $\begingroup$ @Heterotic - I don't think that helps. For instance, I'm still concerned about the "+ non-singular" terms in the first equation on page 76 of the notes. Naively, I would think they give non-trivial contributions to the correlation function. $\endgroup$ – Prahar Jan 15 '15 at 16:50
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The integral will give you the residue, which is the coefficient of the $\frac1{z-w}$ term. Nothing else will contribute (this is known form complex analysis). So we need to tag carefully the $\frac1{z-w}$ term of $\epsilon(z) T(z)\phi_1(w_1, {\bar w}_1)$.

Using $\epsilon(z)=\epsilon(w)+\epsilon'(w)(z-w)+\cdots$

and $$T(z) \phi_i(w_i {\bar w}_i) = \frac{h_i \phi_i(w_i {\bar w}_i) }{ ( z - w_i )^2 } + \frac{ \partial \phi_i(w_i {\bar w}_i) }{ z - w_i } + : T(z) \phi_i(w_i {\bar w}_i) :$$

we see that

$$\epsilon(z) T(z)\phi_1(w_1, {\bar w}_1)=\cdots+\frac{\epsilon(w) \partial \phi_1(w_1, {\bar w}_1) }{ z - w_1 }+ \frac{\epsilon'(w) h_1 \phi_1(w_1, {\bar w}_1) }{ z - w_1 } +\cdots$$ where dots denote other terms that behave differently than $\frac1{z-w}$.

Then, doing the integral we obtain $$\oint \frac{dz}{2\pi i} \epsilon(z) T(z)\phi_1(w_1, {\bar w}_1)=\epsilon(w) \partial \phi_1(w_1, {\bar w}_1)+\epsilon'(w) h_1 \phi_1(w_1, {\bar w}_1) =\delta_\epsilon \phi_1.$$

"Putting the integral in brackets" and by re-instating the other (spectator) $\phi_i$'s we get (2.22).


Well in this last step we also have to use that the integral actually breaks as $$\oint_{around\ everything}=\oint_{around\ w_1}+\cdots+\oint_{around\ w_n}$$ and when doing the i-th integral the only poles in that region will be $\frac{1}{z-w_i}$ and nothing more, (ie the other w's do not contribute. We can always choose the contours this way.)

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