When we have a correlation function of operators in a CFT we can express for example the 3-point function as $$ \langle\mathcal{O_1(x)}\mathcal{O_2(0)}\mathcal{O_3(z)}\rangle=\sum C_{12O} C_{O}(x,\partial y)\langle\mathcal{O}(y)|_{y=0}\mathcal{O_3}(z)\rangle $$ Now since we know that $$ \langle\mathcal{O}(y)\mathcal{O_3}(z)\rangle=\frac{1}{|y-z|^{2\Delta_3}} $$ (assuming a normalisation where the coefficient of the two-point function is 1) and knowing the three-point function form we can arrive at $$ \frac{1}{|x|^{\Delta_1+\Delta_2-\Delta_3}|z|^{\Delta_2+\Delta_3-\Delta_1}|x-z|^{\Delta_1+\Delta_3-\Delta_2}}=C_O(x,\partial y)\frac{1}{|y-z|^{2\Delta_3}}\rvert _{y=0} $$

at leading order(when $x\rightarrow 0$) it's trivial from here that $$ C_O(x,\partial y) = \frac{1}{|x|^{\Delta_1+\Delta_2-\Delta_3}} $$ I want to derive $C_O(x,\partial y)$ up to $\mathcal{O}(x^2)$ but I don't understand how to expand the equations.

The function $C_O(x,\partial y)$ is a power series in $\partial y$ so I expand that as $$ a_o + a_1 \partial_y + a_2 \partial_y^2 + \cdots $$ but I'm not sure how to expand the $|x-z|^{-(\Delta_1+\Delta_2-\Delta_3)}$ term.


1 Answer 1


If you take the multi-variable Taylor expansion \begin{equation} f(x) = f(0) + x^\mu \partial_\mu f(0) + \frac{1}{2} x^\mu x^\nu \partial_\mu \partial_\nu f(0) + \dots \end{equation} and apply it to the function above, you get \begin{align} \frac{1}{|z - x|^{\Delta_1 + \Delta_3 - \Delta_2}} = \frac{1}{|z|^{\Delta_1 + \Delta_3 - \Delta_2}} + (\Delta_1 + \Delta_3 - \Delta_2) \frac{z \cdot x}{|z|^{\Delta_1 + \Delta_3 - \Delta_2 + 2}} + \frac{\Delta_1 + \Delta_3 - \Delta_2}{2} \frac{(\Delta_1 + \Delta_3 - \Delta_2 + 2)(z \cdot x)^2 - z^2 x^2}{|z|^{\Delta_1 + \Delta_3 - \Delta_2 + 4}} + \dots \end{align} if I haven't made a mistake. Now turning to the differential operator in the OPE, \begin{equation} C_O \left ( x, \frac{\partial}{\partial y} \right ) = \frac{1}{|x|^{\Delta_1 + \Delta_2 - \Delta_3}} \left [ 1 + a_1 x^\mu \frac{\partial}{\partial y^\mu} + a_2 x^\mu x^\nu \frac{\partial}{\partial y^\mu} \frac{\partial}{\partial y^\nu} + b_2 x^2 \frac{\partial}{\partial y} \cdot \frac{\partial}{\partial y} + \dots \right ] \end{equation} is the most general second order ansatz. Acting on $|z - y|^{-2\Delta_3}$ and comparing to the above, we can eventually read off \begin{align} a_1 &= \frac{\Delta_1 + \Delta_3 - \Delta_2}{2\Delta_3} \\ a_2 &= \frac{(\Delta_1 + \Delta_3 - \Delta_2)(\Delta_1 + \Delta_3 - \Delta_2 + 2)}{8\Delta_3(\Delta_3 + 1)} \\ b_2 &= \frac{4\Delta_3 a_2 - \Delta_1 - \Delta_3 + \Delta_2}{4\Delta_3 (2\Delta_3 + 2 - d)}. \end{align}

Notice that it is much easier to continue to arbitrarily high order if we are working in one (holomorphic) dimension so that the points don't carry indices. In this case, the solution \begin{equation} C_O(x, \partial_y) = \sum_{m = 0}^\infty \frac{(\Delta_1 + \Delta_3 - \Delta_2)_m}{m!(2\Delta_3)_m} \frac{\partial_y^m}{x^{\Delta_1 + \Delta_2 - \Delta_3 - m}} \end{equation} appears in some early papers. In the harder case however, Dolan and Osborn were able to learn enough about the solution to obtain the first expression for a conformal block in $d = 4$. After one of their later papers, it became much more popular to find conformal blocks by solving the eigenvalue equation for the conformal Casimir. Interestingly, techniques such as the one you're describing, which start by fixing the OPE using covariance, have recently been resurrected.


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