1.2 String Theory
9
This state is massless:
m
2
= 0
(1.12)
and since it transforms as a Lorentz vector (spin 1), it is identified with a U(1) gauge
boson. Writing a superposition of such states
|A =
d
D k A μ (k) α
μ
−1 |k ,
(1.13)
the coefficient A μ (k) of the Fourier expansion is interpreted as the spacetime field
for the gauge boson. Reparametrization invariance is equivalent to the equation of
motion
k
2 A μ = 0 .
(1.14)
One can prove that the field obeys the Lorentz gauge condition
k
μ A μ = 0 ,
(1.15)
which results from gauge fixing the U(1) gauge invariance
A μ −→ A μ + k μ λ .
(1.16)
It can also be checked that the low-energy action reproduces the Maxwell action.
The first level of the closed string is obtained by applying both α −1 and ¯
α −1 (this
is the only way to match N = ¯
N at this level)
α
μ
−1 ¯
α
ν
−1 |k
(1.17)
and the corresponding states are massless
m
2
= 0 .
(1.18)
These states can be decomposed into irreducible representations of the Lorentz
group
α
μ
−1 ¯
α
ν
−1 + α
ν
−1 ¯
α
μ
−1 −
1
D
η
μν α −1 · ¯
α −1
|p ,
α
μ
−1 ¯
α
ν
−1 − α
ν
−1 ¯
α
μ
−1
|p ,
1
D
η μν α
μ
−1 ¯
α
ν
−1 |p
(1.19)
which are respectively associated to the spacetime fields G μν (metric, spin 2), B μν
(Kalb–Ramond 2-form) and (dilaton, spin 0). The appearance of a massless spin