I've hit a stumbling block where I'm just not seeing how to get from line to line in the following calculation from David Tong's strings notes. Can someone spell out how line 1 becomes line 2 in the $\partial X :\mathrm{e}^{ikX} :$ calculation and how that result is actually being employed in the calculation for $T : \mathrm{e}^{ikX} : $?

enter image description here

  • 1
    $\begingroup$ Partial answer: Can make precisely one contraction for each $X(w)$ under the sum, so we get that $\partial X(z) : X(w)^n : \, = n :X(w)^{n-1}: \langle \partial X(x) X(w) \rangle$ to get from line 1 to line 2 $\endgroup$ – Diffycue Feb 28 '18 at 3:55
  • 1
    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ – Emilio Pisanty Feb 28 '18 at 13:43
  • $\begingroup$ @EmilioPisanty I am sensitive to that, which is why I repeated all relevant terms for searching in either the tags or what I wrote. $\endgroup$ – Diffycue Feb 28 '18 at 17:35
  • 1
    $\begingroup$ You're still expected to transcribe any and all text you quote, for the sake of readability, and so that the effort you put into your question matches the effort you're expecting answerers to put in. I'm downvoting on that account; fix it, or don't, up to you. $\endgroup$ – Emilio Pisanty Feb 28 '18 at 17:40

Contraction with exponential is actually more easy than polynomials. This is so because the exponential is the eigenfunctional of $\frac{\delta}{\delta X_{\mu}}$. The cross contractions of $(...)_A$ and $(...)_B$ are given by

$$ (...)_A(...)_B=\exp\left(-\frac{\alpha'}{2}\eta^{\mu\nu}\int d^2z_1\int d^2z_2 \ln|z_{12}|^2 \frac{\delta}{\delta X^{\mu}_A(z_1,\bar{z}_1)}\frac{\delta}{\delta X^{\nu}_B(z_2,\bar{z}_2)}\right)(...)_{AB} $$

and we have the eigenfunctional equation

$$ \frac{\delta}{\delta X^{\mu}(z_1,\bar{z}_1)}\exp(ik.X(z,\bar{z}))=ik_{\mu}\delta^{2}(z-z_1;\bar{z}-\bar{z}_1)\exp(ik.X(z,\bar{z})) $$

then if $(...)_{B}=\exp(ik.X(z_2,\bar{z}_2))$ we have

$$ (...)_A(...)_B=\exp\left(-\frac{i\alpha'k^{\mu}}{2}\int d^2z_1 \ln|z_2-z_1|^2 \frac{\delta}{\delta X^{\mu}_A(z_1,\bar{z}_1)}\right)(...)_{AB} $$

just contractions on $(...)_A$ with:

$$ -\frac{i\alpha'k^{\mu}}{2}\ln|z_2-z_1|^2 $$

for each $X^{\mu}$ in $(...)_A$. So, if $(...)_A=\partial X^{\mu}\partial X_{\mu}$, we have one $0$-contraction (0) plus two identical $1$-contraction (1) plus one $2$-contractions (2), and only this:

(0) $:\partial X^{\mu}\partial X_{\mu}(z_1)\exp(ik.X(z_2,\bar{z}_2)):$

(1) $2:\partial_1(-\frac{i\alpha'k^{\mu}}{2}\ln|z_2-z_1|^2 )\partial X_{\mu}(z_1)\exp(ik.X(z_2,\bar{z}_2)):$

(2) $:\partial_1(-\frac{i\alpha'k^{\mu}}{2}\ln|z_2-z_1|^2 )\partial_1(-\frac{i\alpha'k_{\mu}}{2}\ln|z_2-z_1|^2 )\exp(ik.X(z_2,\bar{z}_2)):$

Adding all together you get the answer.

| cite | improve this answer | |

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