# On Goldstone fermions / goldstino: SUSY breaking

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There are statements about Goldstone fermions, or goldstino, seem confusing to me.

(1) Goldstone boson requires a continuous symmetry spontaneously broken. Does Goldstone fermion imply continuous SUSY spontaneously broken? But boson B and fermion F Hilbert space can be finite dimensional and discrete. The number of SUSY charge $$\mathcal{N}$$ is also discrete. What do we mean by continuous SUSY if there is ? If not, how does the Goldstone theorem on the continuous global symmetry applies to SUSY (possibly discrete, switching between bosonic B and fermionic F sector, and finite $$\mathcal{N}$$)?

(2) In Witten 1982 NPB - Constraints on Supersymmetry Breaking, he said:

(A) the number of boson zero energy modes $$n_b=0$$ and fermion zero energy modes $$n_f=0$$ --- supersymmetry is spontaneously broken and there is a massless Goldstone fermion in the infinite volume theory.

• How and why the massless Goldstone fermion appears, only in the infinite volume limit? (only when the states of bosonic B and fermionic F sector become infinite?) How to interpret Goldstone fermion when we have a finite size but large $$L$$ system?

(B) the number of boson vs fermion zero energy modes $$n_b= n_f\neq 0$$, supersymmetry is not broken; there is no Goldstone fermion, but there are zero-energy fermionic states which we interpret as evidence that the infinite volume theory has a massless fermion (a massless fermion is not a Goldstone fermion unless it is created from the vacuum by the supersymmetry current).

• How do we know there is a massless fermion (in the infinite volume)?

• How to show Goldstone fermion is created from the vacuum by the supersymmetry current? (in the infinite volume? How to interpret it in the finite volume?)

This post imported from StackExchange Physics at 2020-12-13 12:43 (UTC), posted by SE-user annie marie heart
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