0
$\begingroup$

You may find Basic Concepts in String Theory by Blumenhagen et. al. here. In page 37, they display the propagators of a closed string.

I also did calculation to verify their results, but my results show that the propagators should contain Heaviside step functions. For example,

$\langle X_L^\mu(\bar z)X_L^\nu(\bar w)\rangle=[\frac{\alpha'}{4}\eta^{\mu\nu}\ln \bar z-\frac{\alpha'}{2}\eta^{\mu\nu}\ln(\bar z-\bar w)]\theta(\tau-\tau')+[\frac{\alpha'}{4}\eta^{\mu\nu}\ln \bar w-\frac{\alpha'}{2}\eta^{\mu\nu}\ln(\bar w-\bar z)]\theta(\tau'-\tau),$

$\langle X_R^\mu(z) X_L^\nu(\bar w)\rangle=-\frac{\alpha'}{4}\eta^{\mu\nu}\ln\bar w\,\theta(\tau'-\tau)-\frac{\alpha'}{4}\eta^{\mu\nu}\ln z\,\theta(\tau-\tau')$.

Here, $\tau$ is the time associated with $z$ and $\tau'$ associated with $w$. The reason that there are Heaviside functions is simply that the time ordered product by definition contains Heaviside functions. Please look at my results carefully, and you will find out that if we assume $\tau>\tau'$, my results agree with Blumenhagen et. al..

So what's going wrong with my calculation?

$\endgroup$
  • $\begingroup$ Can you describe the procedures to calculate above propagators, or give some references about the calculations about these propagators? Thanks! $\endgroup$ – phys_student Aug 7 '17 at 4:51
  • $\begingroup$ I found on page 47 of this lecture notes the propagators are derived, which maybe helpful for you. stringworld.ru/files/Arutyunov_G._Lectures_on_string_theory.pdf $\endgroup$ – phys_student Aug 7 '17 at 5:40
  • $\begingroup$ @phys_student Thanks! It's been a long time ago. I just worked out details very carefully, using very conventional treatment. Nothing fancy. But I will look into the lecture notes. $\endgroup$ – Drake Marquis Aug 7 '17 at 8:07
0
$\begingroup$

Your Result is right, Blumenhagen Lüst Theisen does only look the case where $\tau>\tau'$.

$\endgroup$
  • 1
    $\begingroup$ This is not really an answer to the original question. It should probably be posted as a comment, which you can do, after you've earned some reputation. $\endgroup$ – engineer Dec 2 '16 at 22:25
  • 1
    $\begingroup$ @engineer why do you say that? the answer is to the point, but I think it's fine, right? $\endgroup$ – AccidentalFourierTransform Dec 2 '16 at 22:43
  • $\begingroup$ I somehow thought this referred to another answer. Sorry, I missed that point. $\endgroup$ – engineer Dec 2 '16 at 22:57
  • $\begingroup$ Thanks! Probably I did not read their book carefully. $\endgroup$ – Drake Marquis Dec 4 '16 at 3:26

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.