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CHAPTER 12 Heterotic String Theory
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( L0 a P ) = 0 ( L0 a A ) = 0 (P sector) (A sector) )
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(12.23)
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The L0 are not identical in these two equations, since we have the two number operators [Eq. (12.20)]. So we have p2 P + N L aP 8 p2 A + N L aA L0 = 8 L0 =
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(P sector) (12.24) (A sector) A
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The task now is to determine the values of a P and a A . It turns out that this can be done readily given what we know about superstrings. That is A periodic boson makes a contribution of 1/24 to the normal ordering constant. A periodic fermion makes a contribution of 1/24 to the normal ordering constant. The normal ordering constant is actually formed by the sum a = bosonic contribution + fermionic contribution i Going to the light-cone gauge, there are eight transverse bosonic components. We also have the 32 fermions A. So, the total normal ordering constant for the P sector is 1 1 aP = 8 + 32 = 1 24 24 Now for the A sector, we need to know that An antiperiodic fermion contributes 1/48 to the normal ordering constant. In the A sector, the bosonic contribution is the same. Hence 1 1 aP = 8 + 32 = 1 48 24 (12.26) (12.25)
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String Theory Demysti ed
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You should be aware that the bosonic and fermionic contributions to the normal ordering constants are due to the zero point energy of the bosonic and fermionic modes. Note also that these zero energy contributions are nite. Using Eqs. (12.24), (12.25), and (12.26) and including the relation obtained for the right-moving modes, we obtain the mass formulas for the P and A sectors:
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P m 2 = 8 N R = 8 ( N L + 1
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m 2 = 8 N R
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) = 8 ( N 1)
(P sector) (A sector) (12.27)
We have made the obvious leap of faith that the mass must be the same for a given string state whether looking at the right-moving or left-moving modes. We immediately notice that A state with zero mass has N R = 0 . Put another way, a state with zero mass in heterotic string theory has the rightmoving modes in the ground state. In addition, if m = 0, then
P For a state in the P sector, N L = 1. A For a state in the A sector, N L = +1.
Now, in ordinary quantum mechanics, you learned that a number operator satis es N 0. So, we must reject the rst possibility which translates into The P sector contains no massless states.
The Spectrum
To describe the spectrum of the theory, we follow the usual procedure of constructing states which are tensor products of left-moving and right-moving modes:
= left right
(12.28)
For the left movers, we just learned that the P sector contributes nothing to a A massless state. Since N L = +1 , this means that the state from the A sector is in the rst excited state. There are two possibilities. It can be a bosonic state: left = j1 0
(12.29)
CHAPTER 12 Heterotic String Theory
Or it can be a fermionic state, since we need to consider the A as well:
A B left = 1/ 2 1/ 2 0 L
(12.30)
For right movers, with N R = 0 , we can have a bosonic state: right = i Or a fermionic state a (12.32) (12.31)
Now let s consider the case when the left movers are in the bosonic state [Eq. (12.29)]. The bosonic sector is given by the tensor product with the bosonic states of the right movers [Eq. (12.31)]:
= j1 0 L i
(12.33)
The states [Eq. (12.33)] can be summarized as follows. The particle spectrum includes: A scalar, the dilaton An antisymmetric tensor state given by j1 0 L i The graviton which is the state
j 1 R i 1 0 L
0 L i
i 1
j L
Now let s take a look at the fermionic sector for the massless states. We can get this by pairing up the bosonic states from the left movers with the fermionic states from the right movers. This is going to give us the superpartners of Eq. (12.23). The states can be written as
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