Lectures on String Theory

– 123 –
B. OPE and conformal blocks
Every conformal (primary) operator enters in the Operator Product Expansion with
its conformal family (descendants). Contribution of the entire conformal family
associated to a primary operator O into the OPE is called the conformal block.
Conformal blocks are completely fixed by conformal symmetry. As an example, let
us show how to find the conformal block associated to the scalar primary operator O
of conformal dimension ∆ O , which arises in the OPE of two scalar operators A and
B of conformal dimensions ∆ A and ∆ B , respectively.
Assume the Operator Product Expansion
A(x)B(0) =
1
(x 2 ) 1 2 (∆ A+∆ B −∆ O )
∞∑
k=0
1
k! Λk (x, ∂)O(0) + . . . (B.1)
Here dots indicate the conformal families of other primary operators. We assume
that all primary operators are orthogonal w.r.t. to the two-point functions
〈O(0)O(y)〉 =
C O
.
(y 2 ) ∆ O
The three-point functions are fixed by conformal symmetry. In particular,
〈A(x)B(0)O(y)〉 =
C ABO
(x 2 ) 1 2 (∆ A+∆ B −∆ O ) (y 2 ) 1 2 (∆ B+∆ O −∆ A ) ((x − y) 2 ) 1 2 (∆ A+∆ O −∆ B ) .
Plugging the OPE into the tree-point function we get
〈A(x)B(0)O(y)〉 =
1
(x 2 ) 1 2 (∆ A+∆ B −∆ O )
∞∑
r=0
1
r! Λr (x, −∂ y )〈O(0)O(y)〉.
Thus, compatibility of the 3-point function with the OPE results into
C ABO
C 0
1
(y 2 ) 1 2 (∆ B+∆ O −∆ A ) ((y − x) 2 ) 1 2 (∆ A+∆ O −∆ B ) = ∞
∑
k=0
1
1
k! Λk (x, −∂ y ) .
(y 2 ) ∆ O
Taking into account that e −x∂ y
is the shift operator acting as
the last relation may be written as
C ABO
C 0
k=0
e −x∂ y
f(y) = f(y − x) ,
∑ 1 1
k! (y 2 ) (−x · ∂ 1 2 (∆ B+∆ O −∆ A
y) k 1
)
(y 2 ) = ∑ ∞
1
1
1
2 (∆ A+∆ O −∆ B ) k! Λk (x, −∂ y ) .
(y 2 ) ∆ O
It suggests to define
Λ k (x, ∂ y ) = C ABO
C 0
Q a,b
k
(x, ∂ y),
k=0