String Theory Demystified

CHAPTER 7 RNS Superstrings 147
From this we obtain, by acting on the equation with F 0 , the relation
2
F F F
0( ψ ψ 0
0 )= = 0
⇒ L ψ = 0
But we know that ( L − a ) ψ = 0.
Hence,
0 R
0
0 = ( L − a ) ψ = L ψ − a ψ =−a
ψ
⇒ a = 0
R
0 R 0
R R
The Open String Spectrum
Now let’s examine the states of the string. We will look at states of the open string
in this chapter. We must consider the NS and R sectors independently. Working in
the NS sector fi rst, the ground state is 0,k NS and it is annihilated by the modes
i
αn 0, k = b 0, k = 0
(7.56)
NS
where n, r > 0. The zero mode α µ
as discussed in the bosonic string case is a
0
momentum operator:
i
r
NS
µ
α 0, k = 2α′ 0,
k
(7.57)
0
NS NS
It can be shown that the normal-ordering constant in the NS sector is
a NS = 1
2
Using this we can fi nd the mass of the ground state, which is
m 2 1
=−
2α
′
(7.58)
(7.59)
Once again, we have a state with m 2
< 0,
so the theory still contains a tachyon state.
We will see later that we can get rid of the tachyon state in the superstring theory.
The ground state in the NS sector is a unique spin-0 state. To fi nd massive states, we
progressively act on the state with negative mode oscillators.