312
15 Closed String Field Theory
This can be seen by writing φ r as a 4-vector and K rs as a 4 × 4-matrix:
K rs =
1
2
⎛
⎜
⎜
⎝
φ ↓↓,r |
φ ↓↑,r |
φ ↑↓,r |
φ ↑↑,r |
⎞
⎟
⎟
⎠
t ⎛
⎜
⎜
⎝
c 0 ¯
c 0 L
+
0 0 0 0
0
0 0 0
0
0 0 0
0
0 0 0
⎞
⎟
⎟
⎠
⎛
⎜
⎜
⎝
|φ ↓↓,s
|φ ↓↑,s
|φ ↑↓,s
|φ ↑↑,s
⎞
⎟
⎟
⎠ .
(15.13)
The matrix is mostly empty because the states φ x,r with different x =↓↓, ↑↓
, ↓↑, ↑↑ are orthogonal (no non-diagonal terms) and the states with x =↓↓ are
annihilated by c 0 or ¯
c 0 . The same consideration applies for the delta function: if the
field does not satisfy L
−
0 = 0, then the kinetic operator is non-invertible.
To summarize, the string field must satisfy three conditions in order to have an
invertible kinetic term
L
−
0 | = 0,
b
−
0 | = 0,
b
+
0 | = 0.
(15.14)
This means that the string field is expanded on the H 0 ∩ ker L
−
0 Hilbert space:
| =
r
ψ ↓↓,r |φ ↓↓,r .
(15.15)
Ill-defined kinetic terms are expected in the presence of a gauge symmetry: this
was already discussed in Sects. 10.1.4 and 10.5 for the free theory, and this will be
discussed further later in this chapter for the interacting case.
Computation
Let us check that K rs is correctly the inverse of rs when is restricted to
H 0 :
K rs st = =φ r | c
−
0 c
+
0 L
+
0 δ L
−
0 ,0 |φ s φ
c
s | b
+
0 b
−
0
1
L
+
0
δ L
−
0 ,0 |φ
c
t
= =φ r | c
−
0 c
+
0 L
+
0 δ L
−
0 ,0 b
+
0 b
−
0
1
L
+
0
δ L
−
0 ,0 |φ
c
t
= =φ r | {c
−
0 , b
−
0 }{c
+
0 , b
+
0 } |φ
c
t
= =φ r |φ
c
t = δ rt .
The second equality follows from the resolution of the identity (11.36): due to
the zero-mode insertions, the resolution of the identity collapses to a sum over
the ↓↓ states
1 =
r
|φ r φ
c
r | =
r
|φ ↓↓,r φ
c
↓↓,r | .
(15.16)