4
$\begingroup$

This is a question related to chapter 2 in Polchinski's string theory book. On page 43 Polchinski calculates the Noether current from spacetime translations and then calculates its OPE with the tachyon vertex, see equations (2.3.13) and (2.3.14)

$$j_a^{\mu} = \frac{i}{\alpha'}\partial_a X^{\mu}, \tag{2.3.13}$$ $$ j^{\mu}(z) :e^{i k\cdot X(0,0)}:\quad \sim\ \frac{k^{\mu}}{2 z} :e^{i k\cdot X(0,0)}:\tag{2.3.14} $$

I wanted to do a similar calculation but for spacetime Lorentz transformations. First I calculated the Noether current, I get $$ L^{\mu\nu}(z)~=~ :X^{\mu} \partial X^{\nu}: ~-~ (\mu \leftrightarrow \nu).$$ Next I calculated the OPE using Wick's formula (in the form of equation 2.2.10). My result is $$ L^{\mu\nu}(z) :e^{i k\cdot X(0)}: \quad \sim\ -\frac{\alpha'}{2} \ln |z|^2\ i k^{\mu} :\partial X^{\nu} e^{i k\cdot X(0)}: ~-~\frac{\alpha'}{2} \frac{1}{z}\ i k^{\nu} :X^{\mu} e^{i k\cdot X(0)}: ~-~ (\mu \leftrightarrow \nu).$$ I think this answer is incorrect because of the logarithm in the right hand side. So my questions are

  1. Is $ L^{\mu\nu}(z)$ defined above indeed the Noether current from spacetime Lorentz transformations?

  2. Is the OPE $ L^{\mu\nu}(z) :e^{i k\cdot X(0)}:$ above correct?

  3. Is there a link where this calculation is performed so that I can check my result?

$\endgroup$
0
$\begingroup$

I do not know if you have not written, or didn't do the double contractions. But apart from that, it is correct.

I haven't checked the calculation of $L$, but notice that your $L^{\mu\nu}$ is not holomorphic. And so the logarithmic term is expected.

$\endgroup$
  • $\begingroup$ I did do the double contractions. $\endgroup$ – wiskundeliefhebber Mar 11 '17 at 18:45
  • $\begingroup$ How if there X's in our answer? They cancel? If so, it is everything ok. Please, note that $: exp(i k. x): $ is not a scalar field, so it is fine to transform under rotations. $\endgroup$ – OkThen Mar 11 '17 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.