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I am following Blumenhagen & Plauschinn, “Introduction to Conformal Field Theory,” §2.9. For the holomorphic piece of the vertex operator

$$ V_{\alpha}(z,\bar z)=:\!\exp\!\bigl(i\alpha\,X(z,\bar z)\bigr)\!: $$

they write (their eq. (2.93))

$$ \boxed{% V_{\alpha}(z,\bar z)= \underbrace{\exp\!\Bigl(i\alpha x_{0} -\alpha\sum_{n\le -1}\frac{j_{n}}{n}\,z^{-n}\Bigr)}_{\text{left / creation}} \; \underbrace{\exp\!\Bigl(\alpha\pi_{0}\ln z -\alpha\sum_{n\ge 1}\frac{j_{n}}{n}\,z^{-n}\Bigr)}_{\text{right / annihilation}} \;} \tag{2.93} $$

with the mode expansion

$$ X(z)=x_{0}-i\pi_{0}\ln z +i\sum_{n\neq0}\frac{j_{n}}{n}\,z^{-n},\qquad [x_{0},\pi_{0}]=i,\qquad \pi_{0}=\tfrac12\,j_{0}. $$

Question: All oscillator creation modes $j_{-n}\;(n>0)$ are placed to the left of annihilation modes $j_{+n}$. But $x_{0}$ commutes with every $j_{n}$, so naively I could move $e^{i\alpha x_{0}}$ to the right block without changing the operator. So why did they put $x_0$ in the left factor and $\pi_0$ in the right factor? Is it an arbirary choice, is it just a convention followed by everyone, or is there a physical/algebraic reason that forces us make this choice?

Any clarification would be appreciated.

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baba26
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1 Answers1

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TL;DR: The normal ordering prescription $$:px:~=~:xp:~=~xp\tag{A}$$ (sometimes called $xp$-ordering) is a convenient choice$^1$ since the $XX$ OPE then acquires a simple form$^2$ $${\cal R}X(z,\bar{z})X(w,\bar{w}) ~=~ -\kappa\hat{\bf 1}~\ln|z-w|^2~+~:X(z,\bar{z})X(w,\bar{w}): \tag{B}$$ where ${\cal R}$ denotes radial ordering.

  1. In more detail, the normal ordering prescription (A) means that the closed bosonic string separates into creation and annihilation parts as $$\begin{align} X(z,\bar{z})~=~&X_{\rm crea}(z,\bar{z})~+~X_{\rm anni}(z,\bar{z}), \cr X_{\rm crea}(z,\bar{z})~=~&x ~+~i\sum_{n\in\mathbb{N}}\frac{j_{-n}}{-n}z^n ~+~i\sum_{n\in\mathbb{N}}\frac{\tilde{j}_{-n}}{-n}\bar{z}^n, \cr X_{\rm anni}(z,\bar{z})~=~&\kappa p\ln|z|^2 ~+~i\sum_{n\in\mathbb{N}}\frac{j_{n}}{n}z^{-n}~+~i\sum_{n\in\mathbb{N}}\frac{\tilde{j}_{n}}{n}\bar{z}^{-n}. \end{align}\tag{C} $$

  2. Using the technique of my Phys.SE answer here, we decompose the $XX$ OPE as $$\begin{align} &-\kappa\hat{\bf 1}\Theta(|z|-|w|)\left(\color{red}{\ln|z|^2}+\color{green}{\ln(1-\frac{w}{z})}+\color{blue}{\ln(1-\frac{\bar{w}}{\bar{z}})} \right) ~+~ (z\leftrightarrow w)\cr ~=~&-\kappa\hat{\bf 1}\Theta(|z|-|w|)~\ln|z-w|^2 ~+~ (z\leftrightarrow w)\cr ~=~&-\kappa\hat{\bf 1}~\ln|z-w|^2\cr ~\stackrel{(B)}{=}~&{\cal R}X(z,\bar{z})X(w,\bar{w}) ~-~ :X(z,\bar{z})X(w,\bar{w}):\cr ~=~&\ldots\cr ~=~&\Theta(|z|-|w|) [X_{\rm anni}(z,\bar{z}),X_{\rm crea}(w,\bar{w})] ~+~ (z\leftrightarrow w). \end{align}\tag{D}$$

  3. Let us calculate the holomorphic oscillator part of the commutator $[X_{\rm anni}(z,\bar{z}),X_{\rm crea}(w,\bar{w})]$: $$ \begin{align} \left[i\sum_{n\in\mathbb{N}}\frac{j_{n}}{n}z^{-n},~i \sum_{m\in\mathbb{N}}\frac{j_{-m}}{-m}w^m\right] ~\stackrel{(2.83)}{=}~& \kappa\hat{\bf 1}\sum_{n\in\mathbb{N}}\frac{1}{n}\left(\frac{w}{z}\right)^n\cr ~\stackrel{|w|<|z|}{=}~&-\kappa\hat{\bf 1}~\color{green}{\ln(1-\frac{w}{z})}, \end{align}\tag{E}$$ which fits nicely into eq. (D). Here we used the oscillator algebra $$ [j_n,j_m]~=~\kappa\hat{\bf 1}~n\delta_{n+m,0}.\tag{2.83}$$ There is a similar story for the anti-holomorphic part.

  4. For eq. (D) to be satisfied, we apparently need that the center of mass (COM) sector satisfies $$\begin{align} \left[\left(x+\kappa p\ln|z|^2\right)_{\rm anni},~\left(x+\kappa p\ln|w|^2\right)_{\rm crea}\right] ~\stackrel{(D)}{=}~&-\kappa{\bf 1}\color{red}{\ln|z|^2}. \end{align}\tag{F}$$ Eq. (F) leads to the $xp$-ordering prescription (A), cf. the CCR.

References:

  1. R. Blumenhagen & E. Plauschinn, Intro to CFT, Lecture Notes in Physics 779, 2009; Subsection 2.9.1.

  2. R. Blumenhagen, D. Lust & S. Theisen, Basic Concepts of String Theory, 2012; p. 37 eq. (3.6).

$^1$ The $xp$-ordering prescription (A) corresponds to the choice of translationally invariant in-vacuum $$p|\Omega\rangle~=~0,\tag{G}$$ and an out-vacuum $$\langle\Omega|x~=~0,\tag{H}$$ with zero center of mass (COM), cf. e.g. my Phys.SE answer here.

$^2$ The XX OPE (B) is used in Wick's theorem and e.g. in calculations with vertex operators (2.93), cf. e.g. this, this & this Phys.SE posts.

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