2
$\begingroup$

I know about the crossing symmetry in conformal field theory, which is a symmetry satisfied by the primary fields' conformal block. Is there a generalized crossing symmetry relation which holds also for conformal blocks of descendant fields?

$\endgroup$

2 Answers 2

1
$\begingroup$

This can be derived by the idea of OPE. Consider the following correlator: $$< {\phi _1}|{\phi _2}({z_1}){\phi _3}({z_2})|{\phi _4} > $$ Insert a complete set of basis, it can be written as : $$\sum_i<\phi_1|\phi_2(z_1)|\phi_i><\phi_i|\phi_3(z_2)|\phi_4>$$ Then we consider the limit $z_1\rightarrow z_2$. In this limit we can first do a OPE of $\phi_2$ and $\phi_3$, and it's easy to see that:$$\phi_2(z_1)\phi_3(z_2)\sim\sum_k<\phi_k|\phi_2(z_1-z_2)|\phi_3>\phi_k(z_2)$$ Therefore, we get another expression for the original correlator: $$\sum_k<\phi_1|\phi_k(z_2)|\phi_4><\phi_k|\phi_2(z_1-z_2)|\phi_3>$$

Finally we analytically continue the expression to outside the region of $z_1\rightarrow z_2$. This completes the proof of crossing symmetry relation for any fields from $\phi_1$ to $\phi_4$. $$\sum_i<\phi_1|\phi_2(z_1)|\phi_i><\phi_i|\phi_3(z_2)|\phi_4>=\sum_k<\phi_1|\phi_k(z_2)|\phi_4><\phi_k|\phi_2(z_1-z_2)|\phi_3>$$

$\endgroup$
0
$\begingroup$

Yes, conformal blocks of descendent fields obey the same crossing symmetry relations as conformal blocks of the corresponding primary fields.

For example, in a case where a descendent's blocks are obtained by acting on a primary's blocks with a differential operator, you can apply that operator to the crossing symmetry relation involving the primary's blocks, and get the crossing symmetry relation for the descendent's blocks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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