242
GAUGE FIELDS AND STRINGS
It seems at first that the bosonic tachyon has appeared, since we have
obtained its vertex operator :e'^ *: on the r.h.s. of (9.399). However, its
appearance is illusory. Indeed, the fusion of two tachyons of two
fermionic strings (9.398) with
= 1/2 would produce a bosonic
tachyon with
= 1, according to (9.399), if
— k} - Xjl and
(p + Ar)^ = 1. That implies that p A : = 0 and hence the residue of the
corresponding pole in all amplitudes is zero. Hence, the true ground
state of the open fermionic string has p^ = 1/2.
It is possible in a consistent manner to eliminate even this tachyon.
Let us distinguish vertex operators by their parity under iA a * ^
(we are talking about open strings for simplicity). The tachyon state
(9.398) is odd. The fusion of two tachyons will give even states. In order
to find them we have to investigate the next terms in (9.399): the
simplest one will be the supersymmetric extension of the
operator lii^x^e*^ *: of the bosonic string theory, which describes a
massless vector state of the open string. This extension is easy to find by
using superfields. Let us consider the expression:
r,(p) = dz d0
*:
= dz d0 :i(i/^^ + 0
-ip-\|#0):
(9.400)
This vertex operator describes a vector massless state again, since the
dimensionality of the integrand is given by:
A =p^ + 1 = l;p^ = 0
(9.401)
By its construction, our vertex is supersymmetric. It is even under
reflection
Under fusion of two such particles the tachyon
(9.398) cannot appear because it is odd, and the “bosonic” tachyon
:e*^ *: cannot appear since it is not supersymmetric. To be more precise,
in the operator product of two operators (9.400), the operator :e‘^ *:
does appear, but on mass shell the coefficient in front of it is equal to
zero, just as happens in (9.399).
We come to the conclusion that the fermionic string when restricted
to the sector even under (/^-reflection has no tachyons for ^ = 10, and
its ground state is the zero mass vector particle.
In the closed string case the vertex operator is given by a direct
product of two from (9.400), one depending on z and the other on z.
It describes a massless tensor particle, which we discuss a little later.
GAUGE FIELDS AND STRINGS
It seems at first that the bosonic tachyon has appeared, since we have
obtained its vertex operator :e'^ *: on the r.h.s. of (9.399). However, its
appearance is illusory. Indeed, the fusion of two tachyons of two
fermionic strings (9.398) with
= 1/2 would produce a bosonic
tachyon with
= 1, according to (9.399), if
— k} - Xjl and
(p + Ar)^ = 1. That implies that p A : = 0 and hence the residue of the
corresponding pole in all amplitudes is zero. Hence, the true ground
state of the open fermionic string has p^ = 1/2.
It is possible in a consistent manner to eliminate even this tachyon.
Let us distinguish vertex operators by their parity under iA a * ^
(we are talking about open strings for simplicity). The tachyon state
(9.398) is odd. The fusion of two tachyons will give even states. In order
to find them we have to investigate the next terms in (9.399): the
simplest one will be the supersymmetric extension of the
operator lii^x^e*^ *: of the bosonic string theory, which describes a
massless vector state of the open string. This extension is easy to find by
using superfields. Let us consider the expression:
r,(p) = dz d0
*:
= dz d0 :i(i/^^ + 0
-ip-\|#0):
(9.400)
This vertex operator describes a vector massless state again, since the
dimensionality of the integrand is given by:
A =p^ + 1 = l;p^ = 0
(9.401)
By its construction, our vertex is supersymmetric. It is even under
reflection
Under fusion of two such particles the tachyon
(9.398) cannot appear because it is odd, and the “bosonic” tachyon
:e*^ *: cannot appear since it is not supersymmetric. To be more precise,
in the operator product of two operators (9.400), the operator :e‘^ *:
does appear, but on mass shell the coefficient in front of it is equal to
zero, just as happens in (9.399).
We come to the conclusion that the fermionic string when restricted
to the sector even under (/^-reflection has no tachyons for ^ = 10, and
its ground state is the zero mass vector particle.
In the closed string case the vertex operator is given by a direct
product of two from (9.400), one depending on z and the other on z.
It describes a massless tensor particle, which we discuss a little later.