# Conformal Field Theory: Evaluating the Vertex Operator on the Vacuum

1. May 3, 2012

### Identity

1. The problem statement, all variables and given/known data

Evaluate
$$\lim_{z \to 0}:e^{ik \cdot X(z)}:|0\rangle$$
where $X(z)$ is a free chiral scalar field in the complex plane.

2. Relevant equations

In Conformal Field Theory, the free chiral scalar field in the complex plane is given by:
$$\begin{array}{rcl} X(z) &=& \frac{1}{2}q - ip\log{z}+i\sum_{n \neq 0}\frac{1}{n}\alpha_n z^{-n}\\ &=& \frac{1}{2}q-ip\log{z}+X_{<}(z)+X_{>}(z)\end{array}$$

Where $q,p$ are position and momentum respectively, and $\alpha_n$ are oscillator modes with the following properties:
$$[\alpha_m^\mu,\alpha_n^\nu]=m\eta^{\mu\nu}\delta_{m+n,0}$$
$$\alpha_{-n} = \alpha_n^\dagger\ ,\ \alpha_n|0\rangle = 0\ ,\ n>0$$

The (normal-ordered) Vertex operator is defined to be:
$$: e^{i k \cdot X(z)}: = e^{ik X_<(z)} e^{ik\cdot q} z^{k \cdot p} e^{ik \cdot X_>(z)}$$

3. The attempt at a solution

So effectively we want to evaluate:

$$\lim_{z \to 0}e^{ik X_<(z)} e^{ik\cdot q} z^{k \cdot p} e^{ik \cdot X_>(z)}|0\rangle$$

We know that since $e^{ik \cdot X_>(z)}$ consists of annihilation operators, $e^{ik \cdot X_>(z)}|0\rangle = |0\rangle$

But what about the other terms? I have no idea how they act on the vacuum. Moreover, when we take the limit to zero, doesn't the whole thing go to zero because we have $z$ in the base of an exponent here?