15.2 Gauge Fixed Theory
313
The third equality uses that L
+
0 commutes with the ghost modes, that φ r is
annihilated by b
±
0 , and that (δ L
−
0 ,0 ) 2 = δ L
−
0 ,0 = 1 on states with L
−
0 = 0.
Finally, we find that the kinetic term matches the classical quadratic vertex V 0,2
defined in (14.56) such that
S 0,2 =
1
2
V 0,2 ((
2 ) =
1
2
| c
−
0 c
+
0 L
+
0 δ L
−
0 ,0 | .
(15.17)
15.2.2 Interactions
The second step to build the action is to write the interaction terms from the
Feynman rules. Before proceeding to SFT, it is useful to remember how this works
for a standard QFT.
Example 15.1: Feynman Rules for a Scalar Field
Consider a scalar field with a standard kinetic term and a n-point interaction:
S =
d
D x
1
2
φ(x)(−∂
2
+ m
2 )φ(x) +
λ
n!
φ(x)
n
.
(15.18)
First, one needs to find the physical states that correspond to solutions of the
linearized equation of motion. In the current case, they are plane-waves (in
momentum representation):
φ k (x) = e
ik·x .
(15.19)
Then, the vertex (in momentum representation) V n (k 1 , . . . , k n ) is found by
replacing in the interaction each occurrence of the field by a different state, and
summing over all the different contributions. Here, this means that one considers
states φ k i (x) with different momenta:
V n (k 1 , . . . , k n ) =
λ
n!
d
D x n!
n
i=1
φ k i (x) = λ
d
D x e
i(k 1 +···+k n )x
= λ(2π)
D δ
(D) (k 1 + · · · + k n ).
(15.20)
The factor n! comes from all the permutations of the n states in the monomial
of order n. Reversing the argument, one sees how to move from the vertex
V n (k 1 , . . . , k n ) written in terms of states to the interaction in the action in terms
of the field.
Obviously, if the field has more states (e.g. if it has a spin or if it is in
a representation of a group), then one needs to consider all the different
313
The third equality uses that L
+
0 commutes with the ghost modes, that φ r is
annihilated by b
±
0 , and that (δ L
−
0 ,0 ) 2 = δ L
−
0 ,0 = 1 on states with L
−
0 = 0.
Finally, we find that the kinetic term matches the classical quadratic vertex V 0,2
defined in (14.56) such that
S 0,2 =
1
2
V 0,2 ((
2 ) =
1
2
| c
−
0 c
+
0 L
+
0 δ L
−
0 ,0 | .
(15.17)
15.2.2 Interactions
The second step to build the action is to write the interaction terms from the
Feynman rules. Before proceeding to SFT, it is useful to remember how this works
for a standard QFT.
Example 15.1: Feynman Rules for a Scalar Field
Consider a scalar field with a standard kinetic term and a n-point interaction:
S =
d
D x
1
2
φ(x)(−∂
2
+ m
2 )φ(x) +
λ
n!
φ(x)
n
.
(15.18)
First, one needs to find the physical states that correspond to solutions of the
linearized equation of motion. In the current case, they are plane-waves (in
momentum representation):
φ k (x) = e
ik·x .
(15.19)
Then, the vertex (in momentum representation) V n (k 1 , . . . , k n ) is found by
replacing in the interaction each occurrence of the field by a different state, and
summing over all the different contributions. Here, this means that one considers
states φ k i (x) with different momenta:
V n (k 1 , . . . , k n ) =
λ
n!
d
D x n!
n
i=1
φ k i (x) = λ
d
D x e
i(k 1 +···+k n )x
= λ(2π)
D δ
(D) (k 1 + · · · + k n ).
(15.20)
The factor n! comes from all the permutations of the n states in the monomial
of order n. Reversing the argument, one sees how to move from the vertex
V n (k 1 , . . . , k n ) written in terms of states to the interaction in the action in terms
of the field.
Obviously, if the field has more states (e.g. if it has a spin or if it is in
a representation of a group), then one needs to consider all the different