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I am deriving the four-point functions, using translation and Lorentz invariance I start with the following form:

$$ \langle \phi_1(x_1)\phi_2(x_2)\phi_3(x_3)\phi_4(x_4)\rangle=C_{1234}x_{12}^ax_{13}^bx_{14}^cx_{23}^dx_{24}^ex_{34}^f. $$ From scaling invariance I get one constraint on $a,b,c,d,e,f$. From special conformal invariance I get 4 constraints (one for each point).

My question is, assuming that the coefficients $C_{1234}$ are functions of cross-ratios $u = \frac{x_{12}x_{34}}{x_{14}x_{23}}$, $v= \frac{x_{13}x_{24}}{x_{14}x_{23}}$ (invariant under the conformal group) how can I write two additional constraints on $a,b,c,d,e,f$? Such that I can solve the system uniquely for $a,b,c,d,e,f$ and obtain the expected form (equ. 4.62 of Di Francesco's book):

$$ \langle \phi_1(x_1)\phi_2(x_2)\phi_3(x_3)\phi_4(x_4)\rangle = f(u,v)\Pi_{i<j}x_{ij}^{\Delta/3-\Delta_i-\Delta_j}.\tag{4.62} $$

For the moment I can't find the right exponent.

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  • $\begingroup$ A similar question is asked (physics.stackexchange.com/q/512322) I guess his answer uses a different approach, as he doesn't have 6 constraints. $\endgroup$
    – Amateur Physicist
    Commented Oct 14 at 22:20
  • $\begingroup$ No, it's the same approach. There is no way to get 6 constraints because you could have equally well made the numerator $u^r v^s f(u, v)$ for any $r$ and $s$. $\endgroup$
    – Connor Behan
    Commented Oct 15 at 0:46

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