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

$$\begin{equation} \langle \phi (z,\bar{z})\phi(w, \bar{w})\rangle ~=~ -\frac{\alpha^{\prime}}{2\pi}\{\text{ln}\left|\frac{z-w}{2R}\right|+\text{ln}\left|\frac{\bar{z}-\bar{w}}{2R}\right|\} \tag{1} \end{equation}$$

where $R$ is an IR cutoff.

My question is:

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

Qmechanic
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layman
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2 Answers2

3

I) Recall first the $\phi\phi$-Operator Product Expansion (OPE):

$$\tag{A} {\cal R}\left\{\phi(z,\bar{z})\phi(w,\bar{w})\right\} ~-~: \phi(z,\bar{z})\phi(w,\bar{w}): ~=~C(z,\bar{z};w,\bar{w}) ~{\bf 1}, $$

where the contraction is assumed to be a $c$-number:

$$\tag{B} C(z,\bar{z};w,\bar{w})~=~ \langle 0 | {\cal R}\left\{\phi(z,\bar{z})\phi(w,\bar{w})\right\} |0\rangle ~=~ -\frac{\alpha^{\prime}}{2\pi}\ln\frac{|z-w|^2+a^2}{(2R)^2}, $$ cf. e.g. this Phys.SE post. Besides the IR cut-off $R>0$, we have added an UV regulator $a>0$ in the contraction (B). Hence in a coinciding world-sheet point, the contraction remains finite

$$\tag{C} C(z,\bar{z};z,\bar{z})~=~ -\frac{\alpha^{\prime}}{\pi}\ln\frac{a}{2R}.$$

II) The ${\cal R}$ symbol in eqs. (A)-(B) denotes radial ordering. Note that many authors don't write the radial ordering symbol ${\cal R}$ explicitly, cf. e.g. OP's eq. (1). It is often only implicitly implied in the notation. Radial order ${\cal R}$ has an interpretation as time ordering, and is necessary in order to make contact with the pertinent operator formalism and correlation functions. Note in particular that the exponential in OP's eq. (2) can be written with radial order

$$ \tag{D} {\cal R}\left\{ e^{ik\phi(z,\bar{z})}\right\}. $$

III) We next use Wick's theorem between the radial and the normal ordering:

$$\tag{E} {\cal R}\left\{{\cal F}[\phi]\right\} ~=~ \exp \left(\frac{1}{2} \int\! d^2z~d^2w~ C(z,\bar{z};w,\bar{w})\frac{\delta}{\delta\phi(z,\bar{z})} \frac{\delta}{\delta\phi(w,\bar{w})} \right) :{\cal F}[\phi]:, $$

cf. e.g. Ref. 1 and my Phys.SE answer here. Here ${\cal F}[\phi]$ denotes an arbitrary functional of the field $\phi$.

For the exponential (D), the Wick's theorem (E) becomes

$$ {\cal R}\left\{ e^{ik\phi(z,\bar{z})}\right\} ~\stackrel{(E)}{=}~\exp \left(\frac{(ik)^2}{2}C(z,\bar{z};z,\bar{z})\right) :e^{ik\phi(z,\bar{z})}: $$ $$\tag{F}~\stackrel{(C)}{=}~ \exp \left(\frac{\alpha^{\prime}k^2}{2\pi}\ln\frac{a}{2R}\right) :e^{ik\phi(z,\bar{z})}: ~=~\left(\frac{a}{2R}\right)^{\frac{\alpha^{\prime}k^2}{2\pi}} :e^{ik\phi(z,\bar{z})}, $$

which is OP's sought-for formula (2).

IV) Alternatively, assume that the field $$ \tag{G} \phi(z,\bar{z})~=~\varphi(z,\bar{z}) + \varphi(z,\bar{z})^{\dagger}, \qquad \varphi(z,\bar{z})|0\rangle~=~0,$$ can be written as a sum of an annihilation and a creation part. Then the UV-regularized OPE (A) becomes

$$ [\varphi(z,\bar{z}),\varphi(z,\bar{z})^{\dagger}] ~\stackrel{(G)}{=}~\phi(z,\bar{z})\phi(z,\bar{z}) ~-~: \phi(z,\bar{z})\phi(z,\bar{z}):$$ $$\tag{H} ~\stackrel{(A)}{=}~C(z,\bar{z};z,\bar{z}) ~{\bf 1}. $$

Moreover, the vertex-operator becomes

$$ :e^{ik\phi(z,\bar{z})}: ~\stackrel{(G)}{=}~e^{ik\varphi(z,\bar{z})^{\dagger}}e^{ik\varphi(z,\bar{z})} ~\stackrel{\text{BCH}}{=}~e^{ik\varphi(z,\bar{z})^{\dagger}+ik\varphi(z,\bar{z})+\frac{1}{2}[ik\varphi(z,\bar{z})^{\dagger},ik\varphi(z,\bar{z})]}$$ $$\tag{I}~\stackrel{(H)}{=}~ e^{ik\phi(z,\bar{z})}e^{\frac{k^2}{2}C(z,\bar{z};z,\bar{z})}~\stackrel{(C)}{=}~ e^{ik\phi(z,\bar{z})}\exp \left(-\frac{\alpha^{\prime}k^2}{2\pi}\ln\frac{a}{2R}\right) ,$$

which again leads to OP's sought-for formula (2). In eq. (I) we have used the truncated BCH formula, see e.g. this Phys.SE post.

References:

  1. J. Polchinski, String Theory, Vol. 1; p. 39, eq. (2.2.7).
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Qmechanic
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1

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. Radial 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.

layman
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