String Theory and M-Theory

290 The heterotic string
(ii)
(iii)
There are 324 possible left-moving states:
˜α I −1 ˜α J −1|0〉L, ˜α I −2|0〉L, ˜α i −1 ˜α j
1 |0〉L, ˜α i −2|0〉L, ˜α i −1 ˜α I −1|0〉L
NR = 1, NL = 1,
16
(p I ) 2 = 2
I=1
In this case there are 24 × 480 possible left-moving states:
˜α I −1|p J 16
, (p J ) 2 = 2〉L, ˜α i −1|p I 16
,
J=1
NR = 1, NL = 0,
I=1
16
(p I ) 2 = 4
I=1
(p J ) 2 = 2〉L.
In this case there are 129 × 480 possible left-moving states:
|p I 16
,
I=1
(p I ) 2 = 4〉L.
The total number of left-moving states is 73 764. In each case the rightmovers
have NR = 1, so these are the 256 states
α i −1|j〉R, α i −1|a〉R, ; S a −1|i〉R, S a −1|b〉R.
The spectrum of the heterotic string at this mass level is given by the tensor
product of the left-movers and the right-movers, a total of almost 20 000 000
states. ✷
Appendix: The Poisson resummation formula
Let A be a positive definite m × m symmetric matrix and define
f(A) =
exp −πM T AM . (7.122)
{M}
Here M represents a vector made of m integers M1, M2, . . . , Mm each of
which is summed from −∞ to +∞. The Poisson resummation formula is
f(A) =
1
√ det A f(A −1 ). (7.123)