BPS equation

  • Thread starter ismaili
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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!
 

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  • #2
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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.
 

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