# How one goes to derive equation (2.4.19) from Polchinski's String Theory Volume 1?

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?

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).
• I have tried that but I am not getting the $\sum_i m_i$ part of the weights right. – Slayer147 Sep 30 at 17:14
• $\uparrow$ Good. – Qmechanic Oct 1 at 6:10