1
$\begingroup$

May it is very trivial, but I am stuck here, given (I have suppressed the conjugate coordinates) $$ \phi_i(x) \phi_j(y) \sim \sum_{k} c_{ijk} (x-y)^{h_k - h_i - h_j} \phi_k(y) $$

$$ \langle \phi_i(x) \phi_j(y)\rangle = \delta_{ij} \dfrac{1}{(x-y)^{2h_i}}$$

Show that $c_{ijk}$ is symmetric in three indices, (i,j) is straightforward how to go about (j,k) ?

$\endgroup$
1
$\begingroup$

Hint: The fusion rule Clebsch-Gordan-like coefficients $c_{ij}{}^k=c_{ijk}$ are related to the 3-point function $\langle \phi_i(x) \phi_j(y)\phi_k(z)\rangle$ of 3 primary fields, which in turn is totally symmetric.

$\endgroup$
3
  • $\begingroup$ Correct me if am wrong, consider $$ \langle \phi_1(z_1) \phi_2(z_2) \phi_3 (z_3) \rangle = \dfrac{c_{123}}{z_{12}^{h_2 + h_1 - h_3} z_{23}^{h_2 + h_3 - h_1} z_{13}^{h_1 + h_3 -h_2}} $$ which is same has $$ \langle \phi_1(z_1) \phi_3(z_3) \phi_2 (z_2) \rangle = \dfrac{c_{132}}{z_{13}^{h_1 + h_3 - h_2} z_{32}^{h_3 + h_2 - h_1} z_{12}^{h_1 + h_2 -h_3}} $$ owning to radial ordering, hence $ c_{123} = c_{132}$. $\endgroup$ – Jaswin Feb 15 '15 at 18:43
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Feb 15 '15 at 22:32
  • $\begingroup$ Btw the word symmetric should be interpreted in a superized sense, i.e. with appropriate Grassmann sign factors. $\endgroup$ – Qmechanic Feb 16 '15 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.