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Homework Help: Conformal Field Theory: Evaluating the Vertex Operator on the Vacuum

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{z \to 0}:e^{ik \cdot X(z)}:|0\rangle [/tex]
    where [itex]X(z)[/itex] 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:
    [tex]\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}[/tex]

    Where [itex]q,p[/itex] are position and momentum respectively, and [itex]\alpha_n[/itex] are oscillator modes with the following properties:
    [tex]\alpha_{-n} = \alpha_n^\dagger\ ,\ \alpha_n|0\rangle = 0\ ,\ n>0[/tex]

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

    3. The attempt at a solution

    So effectively we want to evaluate:

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

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

    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 [itex]z[/itex] in the base of an exponent here?
  2. jcsd
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