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  Vertex operator and normal ordering

+ 3 like - 0 dislike

The two point function, or propagator for a free massless boson, $\phi$ in 2 dimensions is given by,

$$\begin{equation} \langle \phi (z,\bar{z})\phi(\omega, \bar{\omega})\rangle ~=~ -\frac{\alpha'}{2\pi}\{\text{ln}\left|\frac{z-\omega}{2R}\right|+\text{ln}\left|\frac{\bar{z}-\bar{w}}{2R}\right|\} \end{equation}$$

where $R$ is an IR cutoff.

My question is:

How to prove that $$ \text{e}^{ik\phi(x)} ~=~ :\text{e}^{ik\phi(x)}:\text{e}^{\frac{\alpha'k^2}{2\pi}\text{ln}(a/2R)}, $$ where $a$ is an UV cutoff, and $:\mathcal{O} :$ stands for normal ordering?

This post imported from StackExchange Physics at 2015-05-06 11:41 (UTC), posted by SE-user layman
asked Apr 29, 2015 in Theoretical Physics by layman (25 points) [ no revision ]
retagged May 6, 2015

This is a special case of the formula $:e^{\phi(f)}:=e^{\phi(f)}/\langle e^{\phi(f)}\rangle$ valid for any quasifree field $\phi$ and $\phi(f)=\int d\mu(x)\phi(x)f(x)$, where $\langle \cdot\rangle$ denotes vacuum expectation. It can be taken as the definition of normal ordering; functional differentiation then gives the usual rules.

1 Answer

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

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,

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


This post imported from StackExchange Physics at 2015-05-06 11:41 (UTC), posted by SE-user layman
answered May 6, 2015 by layman (25 points) [ no revision ]

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