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From the polchinski page 238, given \begin{equation} [x_L,p_L] =[x_R,p_R]=i\tag{8.2.14} \end{equation} and the mode expansion $$\begin{equation} \begin{split} X_L(z) = x_L -i\frac{\alpha'}{2}p_L \ln z +i(\frac{\alpha'}{2})^{1/2} \sum_{m\neq 0} \frac{\alpha_m}{mz^m} \\ X_R(\bar z) = x_R -i\frac{\alpha'}{2}p_R \ln \bar z +i(\frac{\alpha'}{2})^{1/2} \sum_{m\neq 0} \frac{\tilde \alpha_m}{m \bar z^m} \\ \end{split}\tag{8.2.16} \end{equation}$$

The OPE ought to be $$\begin{equation} \begin{split} X_L(z_1) X_L(z_2) \sim -\frac{\alpha'}{2} \ln z_{12} \\ X_R(\bar z_1) X_R(\bar z_2) \sim -\frac{\alpha'}{2} \ln \bar z_{12} \\ X_L(z_1) X_R(\bar z_2) \sim 0 \end{split}\tag{8.2.17} \end{equation}$$

However, when I tried to compute $X_L(z_1) X_L(z_2)$ OPE with the mode expansion directly, the expansion $\ln(z_1)\approx \ln z_2 +\frac{z_1-z_2}{z_2}$ and $p_Lx_L$ could not be removed.

How to calculate the OPE of the $X_L(z_1)X_L(z_2)$ in the free boson theory from the mode expansion?

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ShoutOutAndCalculate
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1 Answers1

1

Let us concentrate on the very first relation in eq. (8.2.17). A more precise version of eq. (8.2.17) is$^1$ $$\begin{align} {\cal R}(X_L(z_1) X_L(z_2)) ~-~& :X_L(z_1) X_L(z_2): \cr ~=~& -\frac{\alpha'}{2} {\bf 1}~{\rm Ln}(z_1-z_2) \end{align}\tag{8.2.17'}$$ To prove eq. (8.2.17') note that we can assume w.l.o.g. that $|z_1|>|z_2|$. Then we can remove the radial order ${\cal R}$ on the LHS. Next by choosing

  1. the mode expansion (8.2.16),

  2. a pertinent notion of normal order $::$, and

  3. commutation relations (like eq. (8.2.14)),

it is in principle straightforward to verify (8.2.17'). $\Box$

For more details, see also this & this related Phys.SE posts.

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$^1$ Note that the branch cut of the complex logarithm ${\rm Ln}$ creates jumps on the RHS.

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