7.1 Free Scalar
151
Computation: Equation (7.44)
Using (6.160b), one has
∂X(z)V k (w, ¯
w) ∼ ik ∂X(z)X(w) V k (w, ¯
w) ∼ ik
−
2
2
1
z − w
V k (w, ¯
w).
Computation: Equation (7.45)
T (z)V k (w, ¯
w) ∼ −
2 :∂X(z)∂X(z): :e
ikX(w, ¯
w)
:
∼
ik
2
1
z − w
∂X(z) :e
ikX(w, ¯
w)
: −
2 ∂X(z) :∂X(z)e
ikX(w, ¯
w)
:
∼
ik
2
1
z − w
:∂X(z) e
ikX(w, ¯
w)
: + ∂X(z) :e
ikX(w, ¯
w)
:
+
ik
2
:∂X(z)e ikX(w, ¯
w) :
z − w
∼
k 2 2
4
V k (w, ¯
w)
(z − w) 2 + ik
:∂X(w)e ikX(w, ¯
w) :
z − w
.
In the first line, we consider a single contraction (hence, there is no factor
of 2): the reason is that considering the contractions symmetrically and not
successively counts twice the first term of the last line. Indeed, there is only
one way to generate this term. It is also possible to achieve the same result by
expanding the exponential.
Computation: Equation (7.47)
Using (6.160c) and keeping only the leading term, one has
V k (z, ¯
z)V k (w, ¯
w) ∼ exp
− kk
X(z, ¯
z)X(w, ¯
w)
:e
ikX(z,¯ z) e
ik X(w, ¯
w)
:
∼ (z − w)
kk 2 /2 V k+k (w, ¯
w).