There's a nice way to prove that for an extended D-brane half of SUSY is preserved, from perturbative string argument with SUSY Ward identities and the doubling trick (at least, at tree-level and closed string insertions) -- basically, S-matrix argument (*). How can we generalize the argument to a more complicate worldsheet with boundaries?

For (*), in other words, from SUSY Ward identities, doubling trick and perturbative string theory (at the boundary), prove that ($i$-direction is perpendicular to the D-brane):

$ Q_\alpha + \beta_\perp \tilde{Q}_\alpha = 0 \ \ , \ \ \beta_\perp = \prod_i \beta_i$

It might be somehow very trivial with the SUSY worldsheet current, but I don't see how yet:

$j_\alpha = e^{-\phi/2} S_\alpha$

The doubling trick for $S_\alpha$:

$S_\alpha(z \rightarrow\bar{z}) = \beta_\perp S_\alpha(\bar{z})$

At the boundary (real line of the $\mathbb{C}$-plane), the doubling trick for $\int \frac{dz}{2\pi i} j_\alpha(z)$ goes back-and-ford (which is equal to 0) gives:

$\oint \frac{dz}{2 i \pi} j_\alpha(z) = \int_{\mathbb{R}} \frac{dz}{2 i \pi} e^{-\phi/2} S_\alpha(z) - \int_{\mathbb{R}} \frac{dz}{2 i \pi} e^{-\phi/2} S_\alpha (z)$

$= \int_{\mathbb{R}} \frac{dz}{2 i \pi} e^{-\phi/2} S_\alpha(z) - \int_{\mathbb{R}} \frac{d\bar{z}}{2 i \pi} e^{-\tilde{\phi}/2} \beta_\perp \tilde{S}_\alpha (\bar{z}) = 0$

Hence $Q_\alpha + \beta_\perp \tilde{Q}_\alpha$ is conserved.

This post imported from StackExchange Physics at 2015-02-28 13:42 (UTC), posted by SE-user user109798