String Theory Demystified

84 String Theory Demystifi ed
If the following relations are satisfi ed by a state ψ :
( L − aδ ) ψ = 0 ( L − aδ
) ψ = 0 (4.28)
m m, 0 m m,
0
then the state ψ is a physical state. That is, if m > 0 , then the Virasoro operator
annihilates the physical state Lm ψ = 0, Lm
ψ = 0.
The condition satisfi ed when
m = 0 is the “mass shell” condition, ( L0 − a) ψ = 0,( L0 − a)
ψ = 0.
Once again, it
can be shown that negative norm states can be avoided in the theory provided that
a= 1, D=
26.
The Virasoro operators L0 and L0can
be written in terms of the
number operators as follows:
L
1 µ
1 µ
= p pµ + N L = p pµ + N
8πT8πT
0 L 0
R
The sum and difference of L0 and L0annihilate
the physical states:
( L + L − 2a) ψ = 0 ( L − L ) ψ = 0
0 0 0 0
The constraint ( L0 − L0)
ψ = 0 is called level matching. Using the Einstein
relations, we can arrive at an expression for a mass operator:
2 µ 2
M =− p pµ = ( NL + NR
−2)
(4.29)
α ′
The ground state of the closed bosonic string is 0, k , found when NL = NR
=0.
This is a tachyon that satisfi es
M 2
4
=−
α ′
The next case to consider is NL = NR
=1, which is the fi rst excited state. Here
we get hints that string theory is a unifi ed theory. The fi rst excited states are massless,
so M 2
= 0.
They are derived from the ground state in the following way:
µ ν
ε ( k) α α , k
µν
−1 −1
0
The object ε µν ( k) is a tensor which can be decomposed into symmetric ε ( µν ) ( k)
k parts (see Relativity Demystifi ed if you aren’t sure about
and antisymmetric ε ( µν ) ( )