String Theory Demystified

214 String Theory Demystifi ed
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 fi 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:
( )
2
P
α ′ m = 8NR = 8 N
L + 1 (P sector)
2
A
α ′ m = 8NR = 8(
N
L −1)
(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
• For a state in the P sector, N P
L =−1.
• For a state in the A sector, N A
L =+1.
Now, in ordinary quantum mechanics, you learned that a number operator
satisfi es N ≥ 0. So, we must reject the fi rst possibility which translates into
The Spectrum
• The P sector contains no massless states.
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:
ψ = left ⊗ right (12.28)
For the left movers, we just learned that the P sector contributes nothing to a
massless state. Since N A
L =+1, this means that the state from the A sector is in the
fi rst excited state. There are two possibilities. It can be a bosonic state:
left = −
α 1 0
j
L (12.29)