# code 128 font vb.net NS SECTOR ALGEBRA in Java Generation QR-Code in Java NS SECTOR ALGEBRA

NS SECTOR ALGEBRA
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For the NS sector, the following relations are satis ed: [ Ln , Lm ] = ( n m ) Ln + m + c 3 ( n n ) n +m ,0 12 (7.42) (7.43) (7.44)
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1 [ Ln , Gr ] = ( n 2r )Gn +r 2 c {Gr , Gs } = 2 Lr +s + ( 4r 2 1) r +s ,0 12
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The central charge is related to the space-time dimension by c = D + D /2. Let be a physical state in the NS sector. The NS sector super-Virasoro constraints are ( L0 aNS ) = 0 Ln = 0 n > 0 Gr = 0 r > 0 (7.45) (7.46) (7.47)
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Here, following the quantization of the bosonic string, aNS is a normal-ordering constant. The open string mass formula is taken by setting L0 = aNS , which gives m2 = Where the number operator is N = n n +
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1 ( N aNS )
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(7.48)
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r =1/ 2
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br
(7.49)
String Theory Demysti ed
R SECTOR ALGEBRA
In the R sector, the commutation and anticommutation relations are [ Lm , Ln ] = ( m n ) Lm + n + m [ Lm , Fn ] = n Fm +n 2 {Fm , Fn } = 2 Lm +n + The conditions on the physical states are ( L0 a R ) = 0 Ln = 0 n > 0 Fm = 0 m 0 Here, aR is the normal-ordering constant for the R sector. EXAMPLE 7.6 Deduce that aR = 0. SOLUTION We start with the anticommutation relation satis ed by the Fm : {Fm , Fn } = 2 Lm +n + Notice that if m = n = 0, we obtain {F0 , F0 } = F0 F0 + F0 F0 = 2 F02 = 2 L0 L0 = F02 The Fm annihilate physical states . Therefore, F0 = 0 D 2 m m +n ,0 2 (7.53) (7.54) (7.55) D 2 m m +n ,0 2 D 3 m m +n ,0 8 (7.50) (7.51) (7.52)
CHAPTER 7 RNS Superstrings
From this we obtain, by acting on the equation with F0 , the relation F0 F0
)= F
=0
L0 = 0 But we know that ( L0 aR ) = 0. Hence, 0 = ( L0 a R ) = L0 a R = a R aR = 0
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 rst, the ground state is 0,k NS and it is annihilated by the modes
i n 0, k NS
= bri 0, k
(7.56)
where n, r > 0. The zero mode 0 as discussed in the bosonic string case is a momentum operator: 0 0, k
= 2 0, k
(7.57)
It can be shown that the normal-ordering constant in the NS sector is aNS = 1 2 (7.58)
Using this we can nd the mass of the ground state, which is m2 = 1 2 (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 nd massive states, we progressively act on the state with negative mode oscillators.
String Theory Demysti ed
Next we consider the R sector, which describes space-time fermions in the open string case. The ground state is annihilated by
m 0, k = dm 0, k
(7.60)
for m > 0. The zero mode d0 is actually a Dirac operator. That is,
d0 =
(7.61)
We will see below that the critical space-time dimension is 10, so the states in the R sector are 10-dimensional spinors. The ground state satis es the massless Dirac wave equation. In our notation, this is written in the following way, recalling that the momentum operator is 0 :