# How to show that $L_n^\dagger=L_{-n}$ for the Virasoro generators in CFT?

+ 3 like - 0 dislike
312 views

It seems to be a common knowledge that for the Virasoro generators in CFT the rule of hermitian conjugation reads
$$L_n^\dagger=L_{-n}$$
There is probably more then one way to show this. I ask for a clarification of a particular one.

One first defines the scalar product on a pair of fields $A_1,A_2$ as a correlator
$$\langle A_1,A_2\rangle:=\langle A_1(\infty)A_2(0)\rangle:=\lim\limits_{z\to\infty}z^{2\Delta_1}\langle A_1(z) A_2(0)\rangle$$

The action of the Virasoro generator $L_n$ on fieds is defined via
$$L_n A_1(z):=\frac1{2\pi i}\oint_z d\zeta (\zeta-z)^{n+1} \langle T(\zeta) A_1(z)\rangle$$

Using these definitions I can show that $L_1^\dagger=L_{-1}$ in the following way. Consider
$$\langle L_1 A_1(\infty) A_2(0)\rangle=\lim\limits_{z\to\infty}\frac{z^{2\Delta_1-2}}{2\pi i}\oint_z d\zeta (\zeta-z)^2\langle T(\zeta)A_1(z)A_2(0)\rangle$$

Integral around point $z$ can be represented as the sum of integrals around $0$ and $\infty$. Using the ordinary regularity condition for the energy momentum tensor $T(\zeta)\sim\zeta^{-4}, \zeta\to\infty$ one sees that the integral around $\infty$ vanishes. In the integral around $0$ out of three terms in expansion of $(\zeta-z)^2$ only the one with $z^2$ survives in the $z\to\infty$ limit. Hence
$$\langle L_1 A_1(\infty)A_2(0)\rangle=\lim\limits_{z\to\infty}\frac{z^{2\Delta_1}}{2\pi i}\oint_0 d\zeta \langle T(\zeta)A_1(z)A_2(0)\rangle=\langle A_1(\infty) L_{-1}A_2(0)\rangle$$

However this trick does not work for me already for $n=2$. The corresponding scalar product can be represented as
$$\langle L_2A_1(\infty)A_2(0)\rangle=\lim\limits_{z\to\infty}\frac{z^{2\Delta_1-4}}{2\pi i}(\oint_0+\oint_\infty) d\zeta (\zeta-z)^3 \langle T(\zeta)A_1(z)A_2(0)\rangle$$
Here the situation is opposite. It seems to me that the integral around $0$ vanishes in the $z\to\infty$ limit, since it grows as $z^3$ and gets diminished by a factor $z^{-4}$. On the other hand, the integral around $\infty$ seems to be non-vanishing as the integrand grows faster that in preceeding case, and the $\zeta^{-4}$ fall off of $T(\zeta)$ is not enough to make it regular at infinity. But I do not see how to bring this integral to the expected form $\langle A_1(\infty)L_{-2}A_2(0)\rangle$. Any help is appreciated.

Just a comment:  the scalar product you define is only correct if $A_1$ is primary.  Thus, $L_1$ and $L_2$ should annihilate it.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.