Conformal Field Theory: Evaluating the Vertex Operator on the Vacuum

[Identity]

Homework Statement

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.

Homework 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:
$$: e^{i k \cdot X(z)}: = e^{ik X_<(z)} e^{ik\cdot q} z^{k \cdot p} e^{ik \cdot X_>(z)}$$

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?

 

[mark726]

First, let's simplify the expression a bit by using the definition of the vertex operator and the properties of the oscillator modes:

\begin{align*}
:e^{ik \cdot X(z)}:|0\rangle &= e^{ik X_<(z)} e^{ik\cdot q} z^{k \cdot p} e^{ik \cdot X_>(z)}|0\rangle \\
&= e^{ik X_<(z)} e^{ik\cdot q} z^{k \cdot p} |0\rangle \\
&= e^{ik X_<(z)} e^{ik\cdot q} |0\rangle \\
&= e^{ik X_<(z)} |0\rangle \\
&= \exp \left( \sum_{n>0} \frac{i}{n}\alpha_n z^{-n} \right) |0\rangle
\end{align*}

Now, we can see that the limit as z goes to zero will depend on the value of k. If k is non-zero, then the exponential term will go to 1, since all the terms in the sum will go to zero. However, if k is zero then the exponential term will be undefined.

So, in general, the limit will depend on the value of k.