Understanding the BPS Equation from SUSY Transformation Law

[ismaili]

Supposed we are given a set of SUSY transformation law, the way to get the BPS equation is by requiring that
\delta \psi = 0
where \psi is a fermion field.

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

 

[Lemaho]

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.