Lectures on String Theory

where the expression on the r.h.s. is specified by the periodicity/boundary properties
of A(τ).
Explicit example of a vertex operator
Consider the following operator acting in the Hilbert space of open string
√
V (k, τ, σ) = e 1 P ∞
kµα µ −n
πT n=1 e n inτ cos nσ e ik µ(x µ + pµ
πT τ) e − √ 1 P ∞ kµα µ n
πT n=1 n e−inτ cos nσ .
At σ = 0 this simplifies to
V (k, τ) = e 1 √
πT
P ∞
n=1
kµα µ −n
e n
inτ
e ikµ(xµ + pµ
πT
} {{ τ)
} e − 1 P
√ ∞ kµα µ n
πT n=1 n
} {{ e−inτ
} .
V − V 0 V +
} {{ }
This operator is “almost” normal-ordered and can be concisely written as
V (k, τ) = V − V 0 V +
Here only zero modes entering V 0 are not normal-ordered. Indeed, according to our
conventions : p µ x ν := x ν p µ , the operator p µ should be always on the right from x µ
which is not the case for V 0 .
The normal-ordering of V 0 can be achieved with the help of the Baker-Campbell-
Hausdorff formula 11
Thus, we have
: e ik µx µ e i k µp µ
πT τ := e ik µx µ e i k µp µ
e A e B = e A+B+ 1 2 [A,B]
πT τ = e ik µx µ +i k µp µ
πT
Recalling that [x µ , p ν ] = iη µν we therefore find that
τ− τ
2πT k µk ν [x µ ,p ν ]
e ik µx µ +i k µp µ
πT τ = e iα′ k 2 τ
: e ik µx µ e i k µp µ
πT τ := e iα′ k 2 τ
: V 0 :
Thus, we obtain completely normal-ordered vertex operator
V = e iα′ k 2τ V − : V 0 : V +
We would like to investigate the transformation properties of this operator under
conformal transformations. To this end we need to compute by using the Leibnitz
rule
e −iα′ k 2τ [L m , V ] = [L m , V − ] : V 0 : V + + V − [L m , : V 0 :]V + + V − : V 0 : [L m , V + ] .
11 Suppose you forgot an exact coefficient in front of the commutator term. Write the operator
identity in the form e A e B = e A+B+α[A,B] with the forgotten coefficient α. Rescale A and B with a
small parameter ɛ to get e ɛA e ɛB = e ɛA+ɛB+ɛ2 α[A,B] . Now expand both sides up to order ɛ 2 keeping
the order of operators. You will find that fulfilment of the operator relation at order ɛ 2 will require
to fix α = 1 2 . – 50 –