Here

$\begin{equation}
\langle ik\phi(x)ik\phi(x)\rangle = \frac{\alpha'k^2}{\pi} \text{ln}(a/2R),
\end{equation}$

where $a$ is an UV cutoff.

Now we can write (as all the $\phi$'s are located at $x$ i.e. Time ordered $\{\phi^n(x)\}=\phi^{n}(x)~$)

$\begin{align}
\{ik\phi\}^{n}(x) ~&=~ :\{ik\phi\}^n(x): +\sum_{\text{all contractions}} \\
&=~ :\{ik\phi\}^n(x): + ~ ^nC_2 \left(\frac{\alpha'k^2}{\pi} \text{ln}(a/2R)\right) :\{ik\phi\}^{n-2}(x):+\frac{^nC_2~ ^{n-2}C_2}{2} \left(\frac{\alpha'k^2}{\pi} \text{ln}(a/2R)\right)^2:\{ik\phi\}^{n-4}(x): +\cdots \\
&=~ :\{ik\phi\}^n(x):+n(n-1) \left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right) :\{ik\phi\}^{n-2}(x):+ \frac{n(n-1)(n-2)(n-3)}{2!} \left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right)^2 :\{ik\phi\}^{n-4}(x):+\cdots
\end{align}$

We expand the vertex operator,

$\begin{align}
\text{e}^{ik\phi(x)} &= \sum_{n=0}^\infty \frac{(ik\phi)^n(x)}{n!} \\
&= \sum_{n=0}^\infty \frac{:\{ik\phi\}^n(x):}{n!} + \left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right)\sum_{n=2}^\infty \frac{:\{ik\phi\}^{n-2}(x):}{(n-2)!} +\frac{1}{2!}\left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right)^2\sum_{n=4}^\infty \frac{:\{ik\phi\}^{n-4}(x):}{(n-4)!} +\cdots \\
&= \sum_{n=0}^\infty \frac{:\{ik\phi\}^n(x):}{n!} \left[1+\left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right) +\frac{1}{2!}\left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right)^2 +\cdots \right] \\
&=~ :\text{e}^{ik\phi(x)}: \text{e}^{\left(\frac{\alpha'k^2}{2\pi} \text{ln}(a/2R)\right)}.
\end{align}$

Q.E.D.

This post imported from StackExchange Physics at 2015-05-06 11:41 (UTC), posted by SE-user layman