Why does requiring δψ = 0 give the BPS equation?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
1 reply · 4K views
ismaili
Messages
150
Reaction score
0
Supposed we are given a set of SUSY transformation law, the way to get the BPS equation is by requiring that
[tex]\delta \psi = 0[/tex]
where [tex]\psi[/tex] is a fermion field.

Could somebody explain why this is the BPS equation?
Thanks!
 

Discussion

Physics news on Phys.org
Thank You! Exactly this question I have, too!

I know vaguely, that the supersymmetry algebra can be extended by a central charge term, that effectively break some of the supercharges and thus truncate the supermultiplets. There is a bound, called the BPS bound and states that satisfy this bound are BPS states. Can't remember right now much about this bound. Anyways, I guess that the name BPS refers to the fact that you break some of the supersymmetry?

Let me know if you know more by now.