Lectures on String Theory

– 26 –
i.e. the total mass momentum of string coincides with p µ .
The total angular momentum of closed string in the conformal gauge is is defined as
J µν = T
∫ 2π
0
Substituting the oscillator expansion we get
dσ(X µ Ẋ ν − X ν Ẋ µ ) . (3.51)
where
J µν = x µ p ν − x ν p µ
} {{ }
l µν +S µν + ¯S µν ,
S µν = −i
¯S µν = −i
∞∑
n=1
∞∑
n=1
1 (
α
µ
n
−nαn ν − α−nα n) ν µ , (3.52)
1 (ᾱµ
n
−nᾱn ν − ᾱ−nᾱ n) ν µ . (3.53)
For the case of open string expressions are the same (again integration runs from 0 to
π) except ¯S µν is absent. Here l µν is the angular momentum of string and S µν µν
+ ¯S
is its internal spin.
Let us show that both P µ and J µν are invariant under the action of the Virasoro
algebra. We have
{L m , P µ } = {L m , p µ } = 0 (3.54)
as L m does not contain x µ . Let us separate the zero mode part 6 of L m
We first compute
L m = α ρ 0α mρ + 1 2
∑
n≠0,m
α ρ m−nα nρ .
{L m , l µ } = {α ρ 0α mρ , x µ p ν − x ν p µ } = 1 √
4πT
α mρ {p ρ , x µ p ν − x ν p µ } =
Second, since S µν = − ∑ k≠0
= α ν mα µ 0 − α µ mα ν 0 . (3.55)
∑
n,k≠0;n≠m
i
k αµ −k αν k
we have
{ 1 2 αρ m−nα nρ , − i k αµ −k αν k} = 0 ,
∑
{α0α ρ mρ , − i k αµ −k αν k} = α mα µ 0 ν − αmα ν µ 0 ,
k≠0
6 We assume here that m ≠ 0.