String Theory Demystified

CHAPTER 5 Conformal Field Theory Part I
101
where ∂ f =∂ f = 0. To obtain the generators, we consider a coordinate transformation
of the form:
n
z→ z′ = z−ε z z → z′ = z −ε
z
n
+ 1 n+
1 (5.21)
n
To obtain an expression for the generators of a conformal transformation in two
dimensions, we take the derivatives of the transformed coordinates z′ , z′
and look for
terms containing the derivatives ∂ε and ∂ε
, respectively. In the fi rst case we obtain
n n
∂ ′ = ∂
n+
1 n n+
1
z ( z− εnz ) = 1− εn( n+ 1)
z −z ∂zεn
∂z
This allows us to identify the generator:
n
n =−z ∂z
+1 (5.22)
A similar procedure applied to the complex conjugate coordinate gives
n
n =−z ∂z
+1 (5.23)
In the classical case, the generators [Eqs. (5.22) and (5.23)] satisfy the Virasoro
algebra:
[ , ] = ( m−n) ⎡⎣ , ⎤⎦ = ( m−n)
(5.24)
m n m+ n m n m+ n
EXAMPLE 5.1
n+1
Show that the infi nitesimal generator n =−z ∂ satisfi es the Virasoro algebra
[ , ] = ( m− n)
+ .
m n m n
SOLUTION
We apply the generator, which is an operator, to a test function f. So we obtain
[ m, n] f = ( mn −nm)
f
+ +
=−z ∂− ( z ∂) f −( −z
∂)( −z ∂)
f
m 1 n 1 n+
1 m+
1
m+ 1 n n+ 1 2 n+ 1 m m+
1 2
=− − + ∂ − ∂ +
z [ ( n 1)
z f z f] z
= ( n+ 1)
z ∂ f + z
= − −
[ − ( m+ 1)
z ∂f −z ∂ f]
m+ n+ 1 m+ n+ 2 2 m+ n+ 1 m+ n+
2 2
m+ n+
1
( m n)[ z ∂
= ( m− n)
m+ nf
] f
∂ f − ( m+ 1)
z ∂f −z ∂ f