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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?

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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

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    $\begingroup$ So this is the proper way of doing contractions using $e^{ikX}$ as a whole. Thanks a lot for your answer and providing a different perspective other than tearing $e^ikX$ apart! $\endgroup$ – pig2000 Apr 8 '18 at 11:32
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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.

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  • $\begingroup$ I see now, my previous mistake was due to treating $e^{ikX}$ as a single operator and...in fact you can't do 2-contractions in this way. Thank you very much! $\endgroup$ – pig2000 Apr 7 '18 at 18:54
  • $\begingroup$ Note: One can indeed do the contraction treating $e^{ikX}$ as a single opertor instead of a monomial...it's just my way of doing it before was wrong. Readers may want to check out the answer of Nogueira regarding this. I'd like to thank both of the authors again for their great answers from two different perspective! $\endgroup$ – pig2000 Apr 8 '18 at 11:37

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