3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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, 2015 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, 2015 at 16:50

2 Answers 2

1
$\begingroup$

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.)

$\endgroup$
0
$\begingroup$

I hope that 8 years later, you're still interested in hearing a different version of an answer. (@Heterotic, unfortunately I couldn't fully understand your explanation.)

Let me start by warning that $\epsilon$ has to be strictly holomorphic, not meromorphic (such as $x\mapsto\frac{1}{x-z}$).

Now, the argument to pass from the Noether charge identity (2.21) to the Noether current identity (2.22) -the unintegrated form- is in complete analogy to the respective step on page 19, namely: $$\oint\frac{dz}{2\pi i}\epsilon(z)RT(z)\Phi(w)=h\partial\epsilon(w)\Phi(w)+\epsilon(w)\partial\Phi(w) \iff RT(z)\Phi(w)=\frac{h}{(z-w)^2}\Phi(w)+\frac{\partial\Phi(w)}{z-w}+\operatorname{reg}_z$$ ($\operatorname{reg}_z$ and $\epsilon(z)$ mean arbitrary holomorphic functions in $z$, $\epsilon$ small.) Here is how I think this implication is argued for:

  1. The reverse direction "$\Longleftarrow$" follows directly from Cauchy's differentiation formula, $$f\,\text{holomorphic}\implies f'(w)=\oint\frac{dz}{2\pi i}\frac{f(z)}{(z-w)^2}\quad\&\quad f(w)=\oint\frac{dz}{2\pi i}\frac{f(z)}{z-w},$$ used for $f=\epsilon$. (Here, a meromorphic $\epsilon$ could totally screw up the argument.)
  2. "$\Longrightarrow$" I cannot make as rigorous right now. When I read Ginsparg's notes, I was satisfied by the fact that $\epsilon$ is arbitrary among small holomorphic functions (an infinite dimensional algebra), maybe this suffices for you as well.

If you believe this "$\Longleftrightarrow$" then "(2.21)$\iff$(2.22)" is immediate as already noted ($f=\epsilon$ again, only a $\sum_{j=1}^n$ has to added).


Let me add that the important step in this section 2.4 on conformal Ward identities you're talking about is not this passing from charge to current equation, I think. Instead, it's that we learn that products of primaries $\phi_1\phi_2\dots\phi_n$ transform in a Leibniz rule fashion. The cool thing is that we are able to infer this by means of results from contour integration of meromorphic functions - the first step in (2.21).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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