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I am trying to derive $$ j^\mu(z):e^{ik\cdot X(0,0)}: \;\sim \frac{k^\mu}{2z}:e^{ik\cdot X(0,0)} \tag{2.3.14a} $$ from Polchinski's String Theory vol.1 equation (2.3.14a). using $j^{\mu}=\frac{i}{\alpha}\partial_aX^\mu$.

My attempt:

$$ j^\mu(z):e^{ik\cdot X(0,0)}: \; =\frac{i}{\alpha'}\partial_a X^\mu(z):\sum_{n=0}^\infty\frac{i^n}{n!}\left(k\cdot X(0,0)\right)^n: $$

Now the first term of the summation can be ignored since its non-singular, then we get $$ = \frac{i}{\alpha'}\partial_a X^\mu :i k^\nu X_\nu(0,0)\left(1+\frac{i}{2}(k\cdot X(0,0))+\frac{i^2}{3!}(k\cdot X(0,0))^2+\ldots\right): $$ I know that the contraction will give a $1/z$ term but my problem is with re-expressing the exponential. I don't quite get how the series is again the same exponential.

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We have \begin{align} j(z) : e^{ i k X } : &= \frac{i}{\alpha'} \partial X(z) \sum_{n=0}^n \frac{ : ( i k X )^n : }{ n! } \\ &= \frac{i}{\alpha'} \sum_{n=0}^n \frac{ ( i k)^n }{ n! } \partial X(z) : X^n : \\ &\sim \frac{i}{\alpha'} \sum_{n=1}^n \frac{ ( i k)^n }{ n! } n : X^{n-1} : \partial_z X(z) X(0,0) \\ &\sim \frac{i}{\alpha'} \sum_{n=1}^n \frac{ ( i k)^n }{ n! } n : X^{n-1} : \partial_z [ - \frac{\alpha'}{2} \log |z|^2 ] \\ &\sim \frac{k}{2z} \sum_{n=1}^n \frac{ : ( i k X )^{n-1} : }{ (n-1) ! } \\ &\sim \frac{k}{2z} : e^{ i k X } : \end{align}

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  • $\begingroup$ Thank you! Can you explain what happens from line 2 to 3? $\endgroup$ Sep 11, 2021 at 10:03
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    $\begingroup$ I contracted the fields in the usual way. For instance $X_1 : X_2 X_3 X_4 : ~=~ (X_1 X_2) :X_3 X_4: + ( X_1 X_3 ):X_2 X_4: + ( X_1 X_4 ) : X_2 X_3 : + : X_1 X_2 X_3 X_4 :$. I'm using ( ) to denote contraction. This property only holds for (generalized) free fields. $\endgroup$
    – Prahar
    Sep 11, 2021 at 10:07

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