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I have a stupid (homework tag may be suitable =_=) question about the problem 2.7 in Polchinski's string theory volume 1.

Why the weight of operator $:e^{ik\cdot X}:$ is $(\frac{\alpha'k^2}{4}, \frac{\alpha' k^2}{4})$?

From the solution manual and a similar procedure in the answers of this post Identity of Operator Product Expansion (OPE),

$$ T(z):e^{ik\cdot X}: \sim \frac{ \alpha' k^2}{4z^2} :e^{ik\cdot X(0,0)}: +\frac{1}{z} i k_{\mu} : \partial X^{\mu} (0) e^{ik\cdot X(0,0)}: (1)$$

(skip the antiholomorphic part.)

How Eq.(1) is related to the weight $(\frac{\alpha'k^2}{4}, \frac{\alpha' k^2}{4})$ without the contribution of $1/z^2$?

For my understanding, consider $T(z) :\mathrm{operator}:$ is to use the Ward identity (2.3.11) in Polchinski's string theory

$$\mathrm{Res}_{z \rightarrow z_0} j(z)A(z_0, \bar z_0) + \overline{\mathrm{Res}}_{\bar z \rightarrow \bar z_0} \tilde{j}(z)A(z_0, \bar z_0) = \frac{1}{i \epsilon} \delta A(z_0, \bar z_0)$$ with $j(z)=iv(z)T(z)$ and $\tilde{j}(z)=iv(z)\tilde{T}(z)$ thus we relate $T(z) :\mathrm{operator}:$ to $\delta :\mathrm{operator}:$

For example, the operator $X^{\mu}$, we got $\sim \frac{1}{z} \partial X^{\mu}(0)$. For conformal transformation $z \rightarrow \zeta z$, $\partial$ increase weight 1, $1/z$ decrease 1, thus the weight is (0,0).

However, how Eq. (1) is related to $(\frac{\alpha'k^2}{4}, \frac{\alpha' k^2}{4})$ without the contribution of $1/z^2$?

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1 Answer 1

up vote 1 down vote accepted

Remember the equation

$$ T(z){\cal O}(0,0) = \frac{h}{z^2} {\cal O}(0,0) + \frac{1}{z} \partial {\cal O}(0,0) $$

Since you already have the OPE of the operator $: e^{i k \cdot X(z, {\bar z})}:$, you can use the OPE to read of the weight $h$ of the operator. The OPE can be written as $$ T(z): e^{i k \cdot X(z, {\bar z})}: = \frac{\frac{\alpha' k^2}{4} }{z^2} : e^{i k \cdot X(0,0)}: + \frac{1}{z} \partial: e^{i k \cdot X(0,0)}: $$ Thus $h = \frac{\alpha' k^2}{4} $. Similarly, we can work it out ${\tilde h}$ by evaluating the OPE with ${\tilde T}({\bar z})$.

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Thank you very much! I really appreciate your help! – user26143 Jul 18 '13 at 17:07

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