String Theory Demystified

146 String Theory Demystifi ed
R SECTOR ALGEBRA
In the R sector, the commutation and anticommutation relations are
D 3
[ L , L ] = ( m− n) L + m δ (7.50)
m n m+ n m+ n,
0
8
⎛ m ⎞
[ L , F ] = −n
F
⎝
⎜
2 ⎠
m n ⎟ m+ n
The conditions on the physical states are
(7.51)
D 2
{ F , F} = 2L
+ m δ (7.52)
m n m+ n m+ n,
0
2
( L − a ) ψ = 0
(7.53)
0 R
L ψ = n>
0 0 (7.54)
n
F ψ = m≥
0 0 (7.55)
Here, a is the normal-ordering constant for the R sector.
R
EXAMPLE 7.6
Deduce that a R = 0.
SOLUTION
We start with the anticommutation relation satisfi ed by the F m :
Notice that if m= n=0,
we obtain
m
D 2
{ F , F} = 2L
+ m δ
m n m+ n m+ n,
2
2
{ F, F} = FF + FF = 2F= 2L
0 0 0 0 0 0 0
2
⇒ L = F
0 0
The F annihilate physical states ψ . Therefore,
m
F ψ =
0
0
0
0