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I am reading David Tong's lecture notes chapter 4 http://www.damtp.cam.ac.uk/user/tong/string.html

On the top of page 82 in the eq. before eq. (4.27), we are computing the OPE between $T$ and $e^{ikX}$ using Wick's Theorem, it says

I wonder why the first term does not have an extra coefficient of 2?

Since there are two $\partial X$ in the energy momentum tensor $T$, isn't there two ways of doing 2-contractions of $T$ and $e^{ikX}$, just like the second term?

So why doesn't the first term have a 2 like the second term?

Qmechanic
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pig2000
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2 Answers2

2

TL;DR: No, the order of performing the double-contraction$^1$ should be moded out, i.e. one should only count the number of pairs of double-contractions.

Tip: To not make combinatoric mistakes, one might want to consider a monomial first instead of the full exponential/vertex operator. For instance$^2$ $${\cal R}( :X(z)^n::X(w)^m:)$$ leads to $$ n (n-1) \times m (m-1) \times \frac{1}{2!} \text{ double contractions}, $$ and so forth.

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$^1$ Note to the reader: In Tong's normalization, each contraction comes with a factor $\frac{\alpha^{\prime}}{2}$, cf. 2nd last formula on p. 80.

$^2$ Here ${\cal R}$ denotes the often implicitly written radial ordering. The OPE calculation consists of evaluating a nested Wick's theorem between radial ordering ${\cal R}$ and normal ordering $::$, cf. my Phys.SE answer here.

Qmechanic
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2

The contractions are given by:

$$ :A::B:=\exp{\int -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z_1-z_2|^2\,\delta_{X^{\mu}(z_1,\bar{z}_1)}^{(A)}\delta^{(B)}_{X^{\nu}(z_2,\bar{z}_2)}} :AB: $$

In the case where $B=e^{ikX(z,z)}$ note that $B$ is an eigenfunctional of $\delta^{B}_{X^{\nu}(z_2,\bar{z}_2)}$:

$$ \delta^{B}_{X^{\nu}(z_2,\bar{z}_2)}e^{ikX(z,z)}=ik_{\nu}\delta^2(z-z_2)e^{ikX(z,\bar{z})} $$

so we just need to do $\delta^{B}_{X^{\nu}(z_2,\bar{z}_2)}\rightarrow ik_{\nu}\delta^2(z-z_2)$, getting

$$ :A::e^{ikX(z,\bar{z})}:=\exp{\int -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z_1-z_2|^2\,\delta_{X^{\mu}(z_1,\bar{z}_1)}^{(A)}(ik_{\nu}\delta^2(z-z_2)}) :Ae^{ikX(z,\bar{z})}: $$

and then the contraction will be just in $A$, doing

$$ X(z',\bar{z}')^{\mu}\rightarrow \frac{\alpha'}{2}ik^{\mu}\ln|z'-z|^2 $$

In your case where $A=\partial X(z,\bar{z}).\partial X(z,\bar{z})$ we have $2$ ways of contracting just one of the $X$'s and only one way of contracting both the $X$'s

Nogueira
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