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):
Qα+β⊥˜Qα=0 , β⊥=∏iβi
It might be somehow very trivial with the SUSY worldsheet current, but I don't see how yet:
jα=e−ϕ/2Sα
The doubling trick for Sα:
Sα(z→ˉz)=β⊥Sα(ˉz)
At the boundary (real line of the C-plane), the doubling trick for ∫dz2πijα(z) goes back-and-ford (which is equal to 0) gives:
∮dz2iπjα(z)=∫Rdz2iπe−ϕ/2Sα(z)−∫Rdz2iπe−ϕ/2Sα(z)
=∫Rdz2iπe−ϕ/2Sα(z)−∫Rdˉz2iπe−˜ϕ/2β⊥˜Sα(ˉz)=0
Hence Qα+β⊥˜Qα is conserved.
This post imported from StackExchange Physics at 2015-02-28 13:42 (UTC), posted by SE-user user109798