Recent content by SmalltownBoy

Ok, then arbitrary vector is goldstino field...?

Unfortunately, it's not pretty clear for me because ##m_F G## is not even a matrix...

Ok, let me explain how I see it...For simplcity let's consider theory wtih only chiral superfields. We know that in some mininum of scalar potential ##V(\phi,\phi^*)## VEV's of auxillary fields ##F_i## are nonzero and vector ##(F_1,...,F_N)## is anihillated by fermion mass matrix ##m_F##. Than I...

Could you please prove this statement? In fact this is the point that I don't get... :(

Yes, basically I can prove that ##G## vector has zero eigenvalue. But obviously it's not goldstino field,because it's a constant. I assume that goldstino field is ##\sum_i \, F_i \, \psi_i + \sum_a \, D_a \, \lambda_a## but I cannot prove it.

Yet it's awful...It's rainy in Moscow and there is now snow( And my exams are approaching...

I'm a novice in SUSY and I'v got a question concerning spontaneous supersymmetry breaking and goldstinos. In Martin's review on page 68 there is a proof of a statement about existence of massless particle when one of ##F_i##'s or ##D_a##'s VEV is not zero. The thing I don't get is why vector...

Hi all! My name is Ivan and I'm undergraduate student from Russia how likes theoretical phycics :)