Lectures on String Theory

– 62 –
Thus, our commutator takes the form
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 ∞ 1
p +
α j −m−n
n,m=1 n
α− m αi n + α i −n α− n−m αj m − α j −m α− m−n αi n − α i
−n α− −m αj m+n
| {z } | {z } | {z } | {z }
A
B
A
B
4πT X ∞
(p + ) 2 (n − 1)α [i −n αj] n .
n=1
Again we see that upon change of the summation index the A-terms partially cancel (the same is for the B-terms) and we arrive at
I ij =
+
−
√
4πT
p +
√
4πT
p +
X ∞ nX 1
− α i
−n
n=1 m=1 n
αj n−m α− m + α− −m αj m−n αi n
∞ X
n−1 X
1
n=1 m=0 n
4πT X ∞
(p + ) 2 (n − 1)α [i −n αj] n .
n=1
− α j
m−n α− −m αi n + αi −n α− m αj n−m
From here we find
I ij =
−
√
4πT X ∞ nX 1
p +
− α i
−n
n=1 m=1 n
αj 0 α− n + α− −n αj 0 αi n − αj −n α− 0 αi n + αi −n α− 0 αj n
0
1
4πT X ∞ n−1 X n − m
@
A
(p + ) 2 α [i −n
n=1 m=1 n
αj] n −
4πT X ∞
(p + ) 2 (n − 1)α [i −n αj] n .
n=1
| {z }
1
2 (n−1)
Thus, the commutator of the internal spin components we are interested in acquires the form
[S i− , S j− ] =
√
4πT
p +
∞X
−
n=1
X ∞ 1
− α j −n
n=1 n
α− 0 αi n + αj −n αi 0 α− n + αi −n α− 0 αj n − α− −n αi 0 αj n
− α i
−n αj 0 α− n + α− −n αj 0 αi n − αj −n α− 0 αi n + αi −n α− 0 αj n
!
4πT
f(n)
(p + 2(n − 1) −
) 2 n 2 α [i −n αj] n .
The final step consists in commuting the factor α − 0
We thus find
[S i− , S j− ] = 2 2√ πT α − 0
p +
+ 1
p +
−
∞∑
n=1
∞
∑
n=1
1
n
on the left to compare with eq.(4.15).
∞
∑
1
n α[i −nαn j] +
n=1
]
[(α −nα j n − − α−nα − n)p j i − (α−nα i n − − α−nα − n)p i j
( )
4πT f(n)
2n − α
(p + )
2
n
−nα [i j]
2 n .
Finally, adding this expression to eq.(4.15) we arrive at
(
[J i− , J j− ] = 2 p − − 2 √ ) 1
πT α0
−
+ 4πT
(p + ) 2 ∞
∑
n=1
∞
∑
p +
n=1
( [d − 2
12 − 2 ]n + 1 n
1
n α[i −nα j]
n +
[
2a − d − 2
12
] ) α [i −nα j]
n + (α n → ᾱ n ).