Lectures on String Theory

– 32 –
The physical Hamiltonian and the Poisson brackets look the same as the ones in the
conformal gauge, however, the important difference is that now they involve 2(d − 2)
physical fields only plus two additional degrees of freedom (x − , p + ). First, from
eq.(3.65) we find that the zero mode of X − evolves as
ẋ − = 1
p + H =⇒ x− (τ) = x − + H p + τ .
It is this τ-independent mode x − which is conjugate to p + . Second, equations
∂ ± X − = 2πT
p + (∂ ±X i ) 2 (3.70)
can be solved for the longitudinal oscillators αn − , ᾱn
− with n ≠ 0 by substituting an
expansion
X − (τ, σ) = x − + p−
}{{} 2πT
τ + i
p + H
We find p − = 2πT
p + H and
√
4πT
∑
n≠0
1
(
αn − e −inσ− + ᾱn − e −inσ+) . (3.71)
n
α − n =
ᾱ − n =
√
πT
p +
√
πT
p +
∞
∑
m=−∞
∑ ∞
m=−∞
α i n−mα i m , n ≠ 0, (3.72)
ᾱ i n−mᾱ i m , n ≠ 0 . (3.73)
These formulae give a complete solution for X − .
Thus, the light-cone gauge allows for the explicit solution of the Virasoro constraints.
The variables (P i , X i ) are physical excitations while X ± , P + were removed by the
light-cone gauge choice and by solving the constraints. The variable P − plays the
role of the Hamiltonian for physical excitations! Equations of motion for physical
fields are the same as before
Ẍ i − X ′′i = 0 , i = 1, . . . , d − 2.
The variables (P i , X i ), where i = 1, . . . , d − 2, are called transversal, while X ± , P ±
are longitudinal. The only constraint which we were not able to solve explicitly is
the level-matching constraint V = 0. It is easy to check that
{X i , V} = ∂ σ X i , {P i , V} = ∂ σ P i , (3.74)
i.e. V generates the rigid σ-rotations. We also have the evolution equations
{X i , H} = 1 T P i = ∂ τ X i , {P i , H} = T ∂ 2 σX i = ∂ τ P i . (3.75)