# Two-point function of operators with different scaling dimension on a disk

This paper's equation (2.39) says $$\left\langle \sigma(z, \overline{z}) O(y)\right\rangle_{\text{disk}} \propto \frac{(1-|z|^2)^{h_b - 2h}}{|1-z\overline{y}|^{2 h_b}}$$ where $$h, h_b$$ are scaling dimensions of $$\sigma$$ and $$O$$ respectively. Here the operator $$O(y)$$ is on the boundary of the disk.

I've been trying to prove it, but have been unsuccessful so far. Here is my attempt.

Steps.

1. Conformally transform the above to a function on UHP.
2. Use doubling trick for UHP function to get it on $$\mathbb{C}$$

Step 1. Jacobian factors

Conformal transformation modifies [See Ginsparg for example] correlation functions as $$\langle \phi_1(x_1) \phi_2(x_2) \rangle = \left| \frac{\partial x'}{\partial x} \right|_{x_1}^{\Delta_1/d} \left| \frac{\partial x'}{\partial x} \right|_{x_2}^{\Delta_2/d} \langle \phi_1 (x_1')\phi_2(x_2') \rangle$$ where the prefactors are Jacobians. Here $$x \to z$$ and $$x'\to w$$. Then $$\langle \sigma(z_1) O(z_2) \rangle = \left| \frac{\partial w}{\partial z} \right|_{z_1}^{h/2} \left| \frac{\partial w}{\partial z} \right|_{z_2}^{h_b/2} \langle \sigma (w_1) O(w_2) \rangle$$ Here $$z_1, z_2, w_1, w_2$$ are actually doublets, like $$z_1, \overline{z}_1$$, etc.

We need the Jacobian of transformation $$z, \overline{z} \to w, \overline{w}$$: $$\left|\frac{\partial w}{\partial z} \right| = \frac{\partial w}{\partial z} \frac{\partial \overline{w}}{\partial \overline{z}}$$ I've verified that the map from UHP to disk is, $$z= \frac{i - w}{i+w}$$ where $$w, z$$ are coordinates for plane and disk respectively, which gives $$w = i \left(\frac{1-z}{1+z}\right)$$ $$\frac{\partial w}{\partial z}= i \left(- \frac{2}{\left(z + 1\right)^{2}}\right), \quad \frac{\partial \overline{w}}{\partial \overline{z}}= -i \left(- \frac{2}{\left(\overline{z} + 1\right)^{2}}\right)$$ so Jacobian is $$\frac{4}{|z+1|^2}$$ and $$\langle \sigma(z_1) O(z_2) \rangle = \left(\frac{2}{|z_1+1|}\right)^{h} \left(\frac{2}{|z_2+1|}\right)^{h_b} \langle \sigma (w_1) O(w_2) \rangle_{UHP}$$ Step 2.

Let's place $$w_2$$ on the boundary, i.e. $$\overline{w}_2 = w_2$$, and we get $$\left\langle \sigma(w_1) O(w_2)\right\rangle_{\text{UHP}} = \left\langle \sigma(w_1) O(w_2) \sigma(\overline{w}_1)\right\rangle_{\mathbb{C}}$$ $$= \frac{(w_1 - \overline{w}_1)^{h_b -2 h}}{[(w_1 - w_2) (\overline{w}_1 - w_2)]^{h_b}}$$ Here I'm using the three point formula while assuming I don't need to take mod signs because this is a "chiral theory" (?) Next, I need to substitute $$w_k = i \left( \frac{1 - z_k}{ 1 + z_k}\right) , \quad k =1,2$$ and then $$z_2 \to y$$. Substituting gives

$$\left(\frac{i{\left(- 2 z \overline{z} + 2 \right)}}{|1+z|^2}\right)^{- 2 h + h_{b}} \left(- \frac{4 y^{2} \overline{z} - 4 y z \overline{z} - 4 y + 4 z}{\left(y + 1\right)^{2} |1+z|^2}\right)^{- h_{b}}$$ $$\propto { {\left(1-|z|^2\right)}}^{- 2 h + h_{b}} \left( \frac{ y^{2} \overline{z} - y z \overline{z} - y + z}{\left(y + 1\right)^{2}}\right)^{- h_{b}}$$ So $$\langle \sigma(z_1) O(y) \rangle \propto { {\left(1-|z|^2\right)}}^{- 2 h + h_{b}} \left( \frac{\left(y - z\right) \left(y \overline{z} - 1\right)}{\left(y + 1\right)^{2}}\right)^{- h_{b}} \left(\frac{1}{|z+1|}\right)^{h} \left(\frac{1}{|y+1|}\right)^{h_b}$$ $$= \frac{{\left(1-|z|^2\right)}^{- 2 h + h_{b}}}{( y \overline{z} - 1)^{h_b}} \left( \frac{\left(y + 1\right)} {\left(y - z\right) }\right)^{ h_{b}} \left(\frac{1}{|z+1|}\right)^{h} \left(\frac{y+1}{|y+1|}\right)^{h_b}$$ $$\propto \frac{{\left(1-|z|^2\right)}^{- 2 h + h_{b}}}{( y \overline{z} - 1)^{h_b}} \left( \frac{y + 1} {\left(y - z\right) }\right)^{ h_{b}} \left(\frac{1}{|z+1|}\right)^{h}$$ I've dropped the phase factor but I think this is wrong because there is a $$|z+1|$$ with power $$h$$.

Why isn't the result matching with theirs?