Lectures on String Theory

– 61 –
2. Analogously, we compute (m > 0)
√ √
[S i− , α − 4πT
4πT
−m ] = i p + (α− −m αi 0 − αi −m α− 0 ) − i X ∞ m
p +
α i
−n
n=1 n
α− n−m − α− −m−n αi n
4πT m(m − 1)
+ i
(p + ) 2 − f(m)
!
α i −m
2
m
.
In fact these formula is also obtained from eq.(4.17) by simply substituting m → −m.
3. The next step consists in finding the commutator
[S i− , α j −m ] = −i ∞ X
= −i
1
n=1 n
√
4πT
p +
α i −n [α− n , αj −m ] − [α− −n , αj −m ]αi n − α− −n δij δ n−m
| {z }
0 as i≠j
X ∞ m
n=1 n
α i
−n αj n−m − αj −n−m αi n .
4. Finally we compute the commutator
[S i− , α j m ] = i √
4πT
p +
X ∞ m
n=1 n
α i
−n αj m+n − αj m−n αi n .
Substituting all our findings into the commutator [S i− , S j− ] we obtain
[S i− , S j− ] =
√
4πT
p +
∞ X
1
n=1 n
∞X
−
n=1
√
4πT X ∞ 1
+
p +
m,n=1 n
− α j
−n α− 0 αi n + αj −n αi 0 α− n + αi −n α− 0 αj n − α− −n αi 0 αj n
4πT n − 1
(p + ) 2 − f(n)
!
2 n 2 α [i −n αj] n
α j −m (αi −n α− m+n − α− m−n αi n ) − (αi −n αj n−m − αj −m−n αi n )α− m
+ (α i −n α− n−m − α− −n−m αi n )αj m − α− −m (αi −n αj m+n − αj m−n αi n )
.
We first analyze the last two lines of the equation above, which we write as follows
I ij =
√
4πT X ∞ 1
p +
α j −m
m,n=1 n
αi −n α− m+n −(α i −n αj n−m
−α j −m−n αi n )α− m
| {z } | {z }
A
A
+ α i −n α− n−m αj m − αj −m α− m−n αi n − α− −n−m αi n αj m −α − −m (αi −n αj m+n
| {z } | {z }
B
B
−α j m−n αi n )
.
Upon changing the summation variables the A-terms can be partially cancelled, the same is for the B-terms. We therefore obtain
I ij =
+
√
4πT X ∞ nX 1
n
p + −
n=1 m=1 n αi −n αj n−m α− m +
X 1
m=1 n α− −m αj m−n αi n
√
4πT X ∞
p +
α j
−m−n αi n α− m + αi −n α− n−m αj m − αj −m α− m−n αi n − α− −m αi −n αj m+n
.
n,m=1
One can recognize that in the second line of the equation above the first and the last terms are not normal-ordered. We consider the
first sum, which is not normal-ordered, and try to bring it to the normal-ordered form:
∞X 1
∞
n,m=1 n αj −m−n αi n α− m = X 1
∞
n,m=1 n αj −m−n α− m αi n +
X 1
n,m=1 n αj −m−n [αi n , α− m ]
√
∞X 1
4πT
=
n,m=1 n αj −m−n α− m αi n + X ∞
p +
α j −m−n αi n+m
n,m=1
√
∞X 1
4πT
=
n,m=1 n αj −m−n α− m αi n + X ∞
p + (k − 1)α j −k αi k ,
k=2
where in the last sum we made a substitution k = m + n and then summed over m, n with the condition m + n = k kept fixed; this
resulted in the factor k − 1. Analogously, we achieve the normal-ordering of the second sum
∞X 1
∞ √
n,m=1 n α− −m αi −n αj m+n =
X 1
4πT
n,m=1 n αi −n α− −m αj m+n + X ∞
p + (k − 1)α i −k αj k .
k=2