String Theory Demystified

CHAPTER 6 BRST Quantization 117
It is assumed that Q= Q
† . We say that the BRST operator is nilpotent of degree
two. This means that if we square the operator we get zero:
Q 2
= 0
(6.6)
Notice that Eq. (6.6) can also be expressed as { Q, Q } = 0. We label the BRST
operator with a Q to imply that this is a conserved charge of the system, we often
call this the BRST charge. A second operator that is composed solely of ghost fi elds
is called the ghost number operator U. This is given by
U c b
i
= (6.7)
(Again note the Einstein summation convention is being used.) This operator has
integer eigenvalues. If the dimension of the Lie algebra is n, then the eigenvalues of
U are the integers 0,...,n. A state ψ has ghost number m if U ψ = m ψ .
EXAMPLE 6.1
Show that Q raises the ghost number by 1.
SOLUTION
Using the anticommutation relations for the ghost fi elds [Eq. (6.3)], notice that
i
r i
r i i
Uc K = c b c K = c −c
b K
i
∑ r i ∑ δr
r
r
i i r
= cK −ccbK
i
r
r
i
( )
i i r i i
= cK + cK cb = cK + cKU
i
∑
i
∑
Now consider some state ψ with ghost number m, that is U ψ = m ψ . Then
⎛ i 1 k i j
U( Q ψ ) = U c Ki − fij cc b
⎞
k ψ
⎝
⎜
⎠
⎟
2
⎛ i ⎛ r ⎞ 1 k i j ⎞
= Uc Ki − ∑ c br
⎝
⎜
⎠
⎟
⎝
⎜
fij cc bk
2 ⎠
⎟
⎛ i i 1
= cKi + cKU i − fij
⎝
⎜
2
r
∑
r
i
r
ψ
k r i j
cbccb r k
r
⎞
⎠
⎟ ψ
i
r
i
i
i i i
( Use Uc K = c K + c K U )
i
i
i