# Goldstino wave function

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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 of $D_a$'s VEV is not zero. The thing I don't get is why vector $\tilde{G}$ is proportional to goldstino wave function. I guess that it means that after diagonalization of $m_F$ there is a field $\Psi = \sum_{i} \tilde{G}_i \, \tilde{\psi}_i$, that corresponds to zero eigenvalue of $m_F$. Is it correct? If so, could you please give any hints of proof of this statement.

This post imported from StackExchange Physics at 2014-12-17 16:11 (UTC), posted by SE-user user43283
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