80
3 Worldsheet Path Integral: Scattering Amplitudes
Computation of the Amplitude
In this section, we compute the 2-point amplitude from (3.33)
A 0,2 (k, k
) =
C S 2
Vol K 0,2
V k (∞, ∞)V k (0, 0) S 2 .
(3.35)
The volume of K 0,2 reads (by writing a measure invariant under rotations and
dilatations, but not translations nor special conformal transformations) [4, 5]
Vol K 0,2 =
d 2 z
|z| 2 = 2
2π
0
dσ
∞
0
dr
r
,
(3.36)
by doing the change of variables z = re iσ . Since the volume is infinite, it must be
regularized. A first possibility is to cut off a small circle of radius around r = 0
and r = ∞ (corresponding to removing the two punctures at z = 0, ∞). A second
possibility consists in performing the change of variables r = e τ and to add an
imaginary exponential
Vol K 0,2 = 4π
∞
0
dr
r
= 4π
∞
−∞
dτ = 4π lim
ε→0
∞
−∞
dτ e
iετ
= 4π × 2π lim
ε→0
δ(ε),
(3.37)
such that the regularized volume reads
Vol ε K 0,2 = 8π
2 δ(ε).
(3.38)
In fact, τ can be interpreted as the Euclidean worldsheet time on the cylinder since
r corresponds to the radial direction of the complex plane.
Since the worldsheet is an embedding into the target spacetime, both must have
the same signature. As a consequence, for the worldsheet to be also Lorentzian, the
formula (3.37) must be analytically continued as ε = −iE and τ = it such that
Vol M,E K 0,2 = 8π
2 i δ(E),
(3.39)
where the subscript M reminds that one considers the Lorentzian signature.
Inserting this expression in (3.34) and taking the limit E → 0, it looks like the
two δ(0) will cancel. However, we need to be careful about the dimensions. Indeed,
the worldsheet time τ and energy E are dimensionless, while the spacetime time
and energy are not. Thus, it is not quite correct to cancel directly both δ(0) since
they do not have the same dimensions. In order to find the correct relation between
the integrals in (3.37) and of the zero-mode in (3.31), we can look at the mode
expansion for the scalar field (removing the useless oscillators)
X
0 (z, ¯
z) = x
0
+
i
2
α
k
0 ln |z|
2
= x
0
+ iα
k
0 τ,
(3.40)
3 Worldsheet Path Integral: Scattering Amplitudes
Computation of the Amplitude
In this section, we compute the 2-point amplitude from (3.33)
A 0,2 (k, k
) =
C S 2
Vol K 0,2
V k (∞, ∞)V k (0, 0) S 2 .
(3.35)
The volume of K 0,2 reads (by writing a measure invariant under rotations and
dilatations, but not translations nor special conformal transformations) [4, 5]
Vol K 0,2 =
d 2 z
|z| 2 = 2
2π
0
dσ
∞
0
dr
r
,
(3.36)
by doing the change of variables z = re iσ . Since the volume is infinite, it must be
regularized. A first possibility is to cut off a small circle of radius around r = 0
and r = ∞ (corresponding to removing the two punctures at z = 0, ∞). A second
possibility consists in performing the change of variables r = e τ and to add an
imaginary exponential
Vol K 0,2 = 4π
∞
0
dr
r
= 4π
∞
−∞
dτ = 4π lim
ε→0
∞
−∞
dτ e
iετ
= 4π × 2π lim
ε→0
δ(ε),
(3.37)
such that the regularized volume reads
Vol ε K 0,2 = 8π
2 δ(ε).
(3.38)
In fact, τ can be interpreted as the Euclidean worldsheet time on the cylinder since
r corresponds to the radial direction of the complex plane.
Since the worldsheet is an embedding into the target spacetime, both must have
the same signature. As a consequence, for the worldsheet to be also Lorentzian, the
formula (3.37) must be analytically continued as ε = −iE and τ = it such that
Vol M,E K 0,2 = 8π
2 i δ(E),
(3.39)
where the subscript M reminds that one considers the Lorentzian signature.
Inserting this expression in (3.34) and taking the limit E → 0, it looks like the
two δ(0) will cancel. However, we need to be careful about the dimensions. Indeed,
the worldsheet time τ and energy E are dimensionless, while the spacetime time
and energy are not. Thus, it is not quite correct to cancel directly both δ(0) since
they do not have the same dimensions. In order to find the correct relation between
the integrals in (3.37) and of the zero-mode in (3.31), we can look at the mode
expansion for the scalar field (removing the useless oscillators)
X
0 (z, ¯
z) = x
0
+
i
2
α
k
0 ln |z|
2
= x
0
+ iα
k
0 τ,
(3.40)