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When we take the hermitian conjugate of an operator in D dimensions we have: $$ \mathcal{O}_{flat}(r,\vec{n})^\dagger=\frac{1}{r^{2\Delta}}\mathcal{O}_{flat}\left(\frac{1}{r},\vec{n}\right) $$ where $\vec{n}$ the "spherical" coordinates of our radial quantisation, $\Delta$ the weight of our field and $r$ the radial coordinate.

When we consider complex coordinates in 2D CFT this becomes$$ \mathcal{O}(z,\bar{z})^\dagger=\frac{1}{z^{2\bar{h}}}\frac{1}{\bar{z}^{2h}}\mathcal{O}\left(\frac{1}{\bar{z}},\frac{1}{z}\right) $$ where $z,\bar{z}$ the complex coordinates and the weight of the field is $(h,\bar{h})$. How can I prove the 2D case from the $N$-Dimensional case?

Note: In the 2D case we use $z=re^{i\theta},r=|z|,\theta=\arg(z)$, the standard complex plane coordinates.

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  • $\begingroup$ What exactly are you having trouble with? You know exactly how $r$ and $z$ are connected. Also $h+\bar{h}=\Delta$. $\endgroup$ – octonion Feb 16 at 20:33
  • $\begingroup$ Your 2D CFT relation does not do what you would expect it to do, in particular it does not correspond to hermitian conjugation when you rotate back to Lorentzian signature. What one needs is the notion of Euclidean adjoint (Polchinski vol1 p.336). Are you sure that what you wrote is what you are interested in? $\endgroup$ – Wakabaloola Feb 16 at 20:42
  • $\begingroup$ @Wakabaloola - Why do you think it should be something different? $\endgroup$ – Prahar Feb 23 at 4:44
  • $\begingroup$ @Prahar there are a few ways of seeing something is wrong with this definition, eg, what does it mean to replace z with $1/\bar{z}$ in a heterotic string vertex operator where chiral and anti-chiral halves don’t mix in any natural way? $\endgroup$ – Wakabaloola Feb 23 at 9:25
  • $\begingroup$ @Prahar a correct way to derive the relevant notion of conjugation is to consider a path integral on a cylinder, cut it open to produce two path integrals on half cylinders with boundaries, and use operator state correspondence to replace corresponding (sum over) states with (sum over) local operators. the resulting local operators are related by the appropriate notion of conjugation (which does not mix chiral and anti-chiral halves). $\endgroup$ – Wakabaloola Feb 23 at 9:38
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The conjugation formula you have in general dimensions is true only for scalar operators. In $d=2$, this corresponds to an operator with $h={\bar h} = \frac{\Delta}{2}$. We therefore need to prove $$ {\cal O}^\dagger(z,{\bar z}) = \frac{1}{(z {\bar z})^{\Delta} } {\cal O} \left( \frac{1}{ {\bar z} } , \frac{1}{z} \right) \qquad \qquad (1) $$ To derive this, we start from the formula in general dimensions reduced down to $d=2$ $$ {\cal O}^\dagger(r ,\theta) = \frac{1}{r^{2\Delta}} {\cal O}\left( r' , \theta' \right) , \qquad r' = \frac{1}{r} , \qquad \theta' = \theta $$ We now set $z=r e ^{i\theta}$. Then, $$ z' = r' e^{i \theta'} = \frac{1}{r} e^{i\theta} = \frac{1}{r e^{-i\theta}} = \frac{1}{ {\bar z}}. $$ Then, rewriting the transformation in complex coordinates, we immediately reproduce (1).

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  • $\begingroup$ I'm sorry I know this is trivial but how do you rewrite $\theta$ in terms of $z$ ? $\endgroup$ – fielder Feb 16 at 22:33
  • $\begingroup$ I only use the fact that $z=re^{i\theta}$ and ${\bar z} = r e^{-i\theta}$. Is this what you were asking about? $\endgroup$ – Prahar Feb 16 at 22:40
  • $\begingroup$ Oh I just "saw" it. Thanks a ton! $\endgroup$ – fielder Feb 16 at 22:45

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