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

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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.

asked Mar 11, 2016 in Theoretical Physics by Weather Report (210 points) [ no revision ]

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.

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