Lectures on String Theory

– 20 –
which result into
We see that we have two constraints
We find the following Poisson brackets
Ẍ µ − X ′′µ = □X µ = 0
C 1 = P µ P µ + T 2 X ′ µX ′µ , C 2 = P µ X ′µ (3.25)
{C 1 (σ), C 1 (σ ′ )} = 4T 2 ∂ σ C 2 (σ)δ(σ − σ ′ ) + 8T 2 C 2 (σ)∂ σ δ(σ − σ ′ ) , (3.26)
{C 1 (σ), C 2 (σ ′ )} = ∂ σ C 1 (σ)δ(σ − σ ′ ) + 2C 1 (σ)∂ σ δ(σ − σ ′ ) , (3.27)
{C 2 (σ), C 1 (σ ′ )} = ∂ σ C 1 (σ)δ(σ − σ ′ ) + 2C 1 (σ)∂ σ δ(σ − σ ′ ) , (3.28)
{C 2 (σ), C 2 (σ ′ )} = ∂ σ C 2 (σ)δ(σ − σ ′ ) + 2C 2 (σ)∂ σ δ(σ − σ ′ ) . (3.29)
Instead of the constraints C 1 and C 2 we can equally consider their linear combinations
Their Poisson algebra becomes
T ++ = 1
8T 2 (C 1 + 2T C 2 ) = 1
8T 2 (P µ + T X ′ µ) 2 , (3.30)
T −− = 1
8T 2 (C 1 − 2T C 2 ) = 1
8T 2 (P µ − T X ′ µ) 2 . (3.31)
{T ++ (σ), T ++ (σ ′ )} = 1 (
)
∂ σ T ++ (σ)δ(σ − σ ′ ) + 2T ++ (σ)∂ σ δ(σ − σ ′ ) ,
2T
{T −− (σ), T −− (σ ′ )} = − 1 (
)
∂ σ T −− (σ)δ(σ − σ ′ ) + 2T −− (σ)∂ σ δ(σ − σ ′ ) ,
2T
{T ++ (σ), T −− (σ ′ )} = 0 . (3.32)
Constraints T ++ and T −− Poisson commute and form two independent Poisson algebras!
Now one can easily find the evolution equations for T ++ and T −− . We have
∂ τ T ++ = {T ++ (σ), H} = ∂ σ T ++ =⇒ (∂ τ − ∂ σ )T ++ = ∂ − T ++ = 0 ,
∂ τ T −− = {T −− (σ), H} = −∂ σ T −− =⇒ (∂ τ + ∂ σ )T −− = ∂ + T −− = 0 ,
Thus, we see that evolution equations imply that
The Hamiltonian itself is
T ++ = T ++ (σ + ) , T −− = T −− (σ − ) . (3.33)
H = 2T
∫ 2π
0
dσ(T ++ + T −− ) ,
i.e. it is a sum of zero modes of the left- and right-moving constraints.