Lectures on String Theory

– 51 –
Assuming for definiteness that m > 0 it is not difficult to find 12
[L m , V − ] = 1
2 √ πT
∞∑
p=1
[L m , : V 0 :] =: V 0 : (α mk)
√
πT
[L m , V + ] = 1
2 √ πT
∑−1
p=−∞
e ipτ [
V − (kα m−p ) + (kα m−p )V −
]
e ipτ [
V + (kα m−p ) + (kα m−p )V +
]
=
∞∑
p=1
e −ipτ
√
πT
: V + (kα m+p ) :
Here the first commutator is of particular importance because it still contains the
terms which are not normal-ordered. Indeed, it can be written in the form
[L m , V − ] = 1 √
πT
∞
∑
p=1,p≠m
e ipτ : (kα m−p )V − :
+ 1 √
πT
e imτ V − (kα 0 ) + 1
2 √ πT
m−1
∑
p=1
e ipτ [kα m−p , V − ] .
Further we get
[kα m−p , V − ] =
k2
√
πT
e i(m−p)τ V −
This leads to
[L m , V − ] = 1 √
πT
∞
∑
p=1,p≠m
e ipτ : (kα m−p )V − :
+ 1 √
πT
e imτ V − (kα 0 ) + k2
2πT
m−1
∑
e imτ V − .
p=1
} {{ }
(m−1)e imτ V −
12 The calculation is as follows:
[L m, V − ] = 1 +∞ X
[α µ m−n
2
αn,µ, V −] = 1 +∞ X
[α µ m−n
n=−∞
2
, V −]α n,µ + α µ m−n [αn,µ, V −] .
n=−∞
The two terms on the r.h.s. are computed separately, for instance
1
+∞ X
[α µ m−n
2
, V −]α n,µ = 1 m−1 X
[α µ 1 P
m−n
n=−∞
2
, e √ ∞p=1
kν α ν −p
e πT p
ipτ 1
∞X m−1
]α n,µ =
n=−∞
2 √ X
[α µ m−n
πT
, αν −p ]
k ν
p=1 n=−∞ | {z } p eipτ V − α n,µ
n=m−p
=
1
∞X
2 √ e ipτ V − (k µ α m−p,µ ) .
πT p=1
The other terms are computed in a similar way.