Equation (2.4.19) states that an exponential times a general product of derivatives $$:(\Pi_i \partial^{m_i}X^{\mu_i})(\Pi_j \bar{\partial}^{n_j}X^{\nu_j})\exp(i k \cdot X ):\tag{2.4.18}$$ has weight given by

$$ \bigg(\frac{\alpha' k^2}{4} + \sum_i m_i,\frac{\alpha' k^2}{4} + \sum_j n_j\bigg).\tag{2.4.19}$$

I attempted starting from equation (2.2.10), namely

$$ :\mathfrak{f}: :\mathfrak{g}:= \exp\bigg( \frac{\alpha'}{2} \int d^2z_1 d^2z_2 \ln|z_{12}| \frac{\delta}{\delta X^{\mu}_F(z_1,\bar{z}_1)} \frac{\delta}{\delta X_{G \mu}(z_2,\bar{z}_2)}\bigg) :\mathfrak{f} \mathfrak{g}: $$

but got anywhere. Could someone please tell me steps to arrive at that equation?


1 Answer 1


Hint: If we call the vertex operator (2.4.18) for ${\cal A}$ then the main idea is to use eq. (2.2.10) to calculate the weights (2.4.19) as coefficients $h$ and $\bar{h}$ of the double poles in the OPEs for $T(z){\cal A}(w,\bar{w})$ and $\bar{T}(z){\cal A}(w,\bar{w})$, respectively, cf. eq. (2.4.14).

  • $\begingroup$ I have tried that but I am not getting the $\sum_i m_i$ part of the weights right. $\endgroup$
    – Slayer147
    Sep 30, 2019 at 17:14
  • $\begingroup$ Consider showing your work. $\endgroup$
    – Qmechanic
    Sep 30, 2019 at 17:22
  • $\begingroup$ I manage to get the right answer, will post whenever I have time. $\endgroup$
    – Slayer147
    Oct 1, 2019 at 6:09
  • $\begingroup$ $\uparrow$ Good. $\endgroup$
    – Qmechanic
    Oct 1, 2019 at 6:10

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