QUANTUM STRINGS AND RANDOM SURFACES
219
However, we should not simply subtract all levels given by (9.301). That
would lead to overcounting for the following reason. Among the zero
norm states on the second level, which as we said have the form:
I/> = (T-2 + |T 1 Q |A = - 1 >
(9.303)
there are some, for which the state | A = — 1 > is itself secondary. These
cases are again found from the Kac formula:
+ «/n = ^ - T (3n _ 2 m f = - 1
3« - 2m = + 7
(9.304)
which is satisfied, say, on the level 3, by n = 3, m = 1. Therefore, some of
the states (9.303) will be of the form:
l/> = (T_2 + I T l ,x r _ 3 T ...)|A = - 4 >
(9.305)
This can be considered as a degenerate state at the level 5. So, this state
should not be counted as we go to the level 5, setting n = 2 in (9.301).
Another state which should not be counted is the following:
|/ > = 7 _ J A = 0>
|A = 0> = {T_^-f .••}|A = - 4 >
because according to the Kac formula:
(9.306)
25
1
8)" = 0
On the other hand, the Kac formula predicts only one secondary state
at the level 5, and we have already counted two states. The conclusion is
that these two states are in fact the same, and that we have counted it
twice. Hence, to correct the situation, we must write instead of (9.302):
1
V (x) = (1 - X
) n 1 -X*
(9.307)
It is not hard to do the same combinatoricst at each level, using the Kac
formula repeatedly. The result of this counting is:
v(x) = h + X
.n(3n-t- l)/2 -h X'
n(3n - 1./2)
n = 1
1 \2 5
1
\2 4
Zk ,
(9.308)
t Described in the Appendix to Ch. 9.10, p. 221.