0
$\begingroup$

In chapter 7 of superstring theory, it is written $$ g\langle0;k_1|\zeta\cdot\alpha_1V_0(k_2)\zeta_3\cdot\alpha_{-1}|0;k_3\rangle=g\langle0;k_1|\zeta\cdot\alpha_1e^{k_2\cdot\alpha_{-1}} e^{-k_2\cdot\alpha_{1}}\zeta_3\cdot\alpha_{-1}|0;k_3\rangle. \tag{7.1.56}$$ Why is it that only $\alpha_1$ and $\alpha_{-1}$ in $X^\mu$ are relevant to the calculation?

$\endgroup$
1
$\begingroup$

Here, $V_0(k_2)$ could have been replaced by $e^{k_2 \alpha_{-1}} e^{-k_2 \alpha_1}$ while the factors from $\alpha_{\pm n}$ for $n\gt 1$ could have been neglected because $\alpha_n$ annihilates everything that appears in the matrix element on the right side from $V_0$ (because it ultimately annihilates $|0\rangle$), and similarly for $\alpha_{-n}$ that annihilates the bra $\langle 0 |$.

The exponentials of these multiples of $\alpha_{\pm 2}$ etc. are therefore effectively equal to $\exp(0)=1$, in that situation. The oscillators $\alpha_{\pm 1}$ are the only exceptions because of the extra noncommuting $\alpha_{\pm 1}$ that appear in the matrix element.

The zero mode $x_0$ hiding on $X$ was ignored by Green-Schwarz-Witten but it's mostly because their treatment was a bit sloppy or heuristic. This $x_0$ factor of the calculation produces the delta-function $\delta(k_1+k_2+k_3)$ with the right signs and perhaps powers of $(2\pi)$ which they know to be there but they sometimes omit it when they talk about "the amplitude".

This zero-mode part is one that also appears in point-like particle field theories and they only focus on the factors that are "new" in string theory relatively to point-like QFTs.

$\endgroup$
  • $\begingroup$ Sorry for the late reply, the external Javascript had been blocked... Aside from the $x_0$ term, I'm also confused about the equality between the massless vector state $|1;k_3\rangle=:\zeta_3\cdot X e^{ik_3\cdot X}:|0;0\rangle$ and $\zeta_3\cdot\alpha_{-1}|0;k_3\rangle$. Why are they equal? $\endgroup$ – Xavier May 23 '16 at 13:13
  • $\begingroup$ It's how state-operator correspondence works, with some identities in the OPEs etc. It might be that Polchinski's book covers all these things much more systematically and accurately than GSW. $\endgroup$ – Luboš Motl May 23 '16 at 16:51

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.