Operator product expansion says that, the product of two primary fields(of same dimension in this case) can be expanded as sum of primaries and their descendants

$$\phi_1(x)\phi_2(0) = {\Large \Sigma_\mathcal{O}}\lambda_\mathcal{O}C_\mathcal{O}(x,\partial_y)\mathcal{O}(y)|_{y=0} $$ where the summation $\Sigma_\mathcal{O}$ is over primaries. Descents appear when acted upon by the derivatives in $C_\mathcal{O}(x,\partial_y)$. Considering the three point function and since we know two point functions are diagonal we get,

\begin{equation}\langle\phi_1(x)\phi_2(0)\Phi(z)\rangle = \lambda_\Phi C_\Phi(x,\partial_y)\langle\Phi(y)|_{y=0}\Phi(z)\rangle \hspace{0.2cm} (1) \end{equation}

Now using known forms of two and three point functions below $$\langle\phi_1(x)\phi_2(x_2)\Phi(x_3)\rangle = \frac{\lambda_\Phi }{|x_{12}|^{\Delta_1+\Delta_2-\Delta_3}|x_{23}|^{\Delta_2+\Delta_3-\Delta_1}|x_{13}|^{\Delta_1+\Delta_3-\Delta_2}} $$ $$ \langle \Phi(y)\Phi(z)\rangle = \frac{1}{|y-z|^{2\Delta_\Phi}} $$

one is supposed to fix the the constants $\alpha, \beta$ in $C_\Phi(x,\partial_y)$ by assuming a form

$$ C_\Phi(x,\partial_y) = \frac{1}{|x|^{2\Delta -\Delta_\Phi}}\Big[1+ \frac{1}{2}x^\mu\partial_\mu + \alpha x^\mu x^\nu \partial_\mu\partial_\nu + \beta x^2\partial^2 + ...\Big] $$ Here, the dimensions of $\phi_1$ and $\phi_2$ are each $\Delta$ and that of $\Phi$ is $\Delta_\Phi$. Now I can see using the three point function with insertions at $x,0,z$ on the LHS of (1) is

$$\langle\phi_1(x)\phi_2(0)\Phi(z)\rangle = \frac{\lambda_\Phi }{|x|^{2\Delta -\Delta_\phi}|z|^{\Delta_\Phi}|z-x|^{\Delta_\Phi}} $$ the leading term when expanding about $x$, $\frac{\lambda_\Phi }{|x|^{2\Delta -\Delta_\phi}|z|^{2\Delta_{\Phi}}} $ matches with the RHS of eq. (1) but can't figure out how to find the the coefficients of higher order terms. I find myself trying to evaluate a binomial expansion of $|z-x|^{\Delta_\Phi}$ where now the points are in $D$ dimensional space and that I am not able to do. I am only trying to get till two orders. Any help is appreciated.

  • 2
    $\begingroup$ look up Gegenbauer or ultraspherical polynomials which are needed for your Taylor expansion in arbitrary $D$ and for arbitrary scaling dimension $\Delta_{\Phi}$. $\endgroup$ – Abdelmalek Abdesselam Apr 8 '17 at 18:56

This is how you evaluate the series for $$ \frac{1}{|z-x|^{\Delta_\phi}} $$ for small $x$. First, you pull out $|z|$, $$ \frac{1}{|z|^{\Delta_\phi}|e-\xi|^{\Delta_\phi}}, $$ where $e=\frac{z}{|z|}$, write $\xi=\frac{x}{|z|}$ and now work with $$ \frac{1}{|e-\xi|^{\Delta_\phi}}=\left[\frac{1}{(e-\xi)^2}\right]^{\Delta_\phi/2}, $$ Using $(e-\xi)^2=1-2(e\cdot\xi)+\xi^2$ and replacing $\xi\to\epsilon\xi$ to track the order you get just $$ \left[\frac{1}{1-2\epsilon(e\cdot\xi)+\epsilon^2\xi^2}\right]^{\Delta_\phi/2}, $$ which is now just a usual function of scalar argument $\epsilon$, which you can expand out in powers of $\epsilon$ by hand or using Mathematica. To get the general answer, set $\epsilon=t/|\xi|$ and get $$ \left[\frac{1}{1-2t\frac{(e\cdot\xi)}{|\xi|}+t^2}\right]^{\Delta_\phi/2}=\sum_{j=0}^\infty C^{(\Delta_\phi/2)}_j\left(\frac{e\cdot\xi}{|\xi|}\right)t^j=\sum_{j=0}^\infty C^{(\Delta_\phi/2)}_j\left(\frac{e\cdot\xi}{|\xi|}\right)|\xi|^j\epsilon^j, $$ by definition of the Gegenbauer polynomials, as noted by Abdelmalek Abdesselam.


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