# Supersymmetry breaking and Goldstino

I would be glad if someone can explain to me the argument as to why supersymmetry breaking is necessarily accompanied by appearance of a massless fermion, namely the goldstino. (and also why this is a non-perturbative effect)

Let me quote here the two lines from the third volume of the QFT books by Weinberg where he tries to explain this phenomenon,

• "(when supersymmetry is broken)..any n-particle state is accompanied with 2 states of the same energy and momentum and opposite statistics, containing an additional 0-momentum goldstino of spin up or down and with another state of the same energy and momentum and the same statistics, containing two additional 0-momentum goldstinos of opposite spin."

• "In particular when supersymmetry is spontaneously broken the vacuum state has non-zero energy, so it must be paired with a fermionic state of the same energy and zero-momentum; more precisely, the vacuum and the state containing two zero-momentum goldstinos are paired with the two states of a single zero-momentum goldstino."

I am unable to understand the above two arguments and will be glad if someone can help.

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The propositions by Weinberg above are just tautologies. When SUSY is broken it means that $$Q_\alpha|0\rangle \neq 0$$ the vacuum is not annihilated by the supercharges. It means that $Q_\alpha |0\rangle$ has to be equal to something. Because $Q_\alpha$ is Grassmann-odd, it must be a state with an odd number of fermions. But the momentum is still zero because the momentum of $|0\rangle$ vanishes and $Q_\alpha$ commutes with the energy-momentum vector.
So for broken $N=1$ SUSY, one has four states, $$|0\rangle, Q_1 |0\rangle, Q_2 |0\rangle, Q_1 Q_2 |0\rangle$$ which contain 0,1,1,2 fermions, respectively. This proof is somewhat heuristic because we haven't really proved that there also exist states with a nonzero momentum goldstino. To prove that, one has to define the supergenerators dressed with a momentum. The proof is then totally analogous to the Goldstone theorem for bosonic symmetries.