Supersymmetry non-breaking $\iff$ no "Goldstone fermion"?

Question

In Supersymmetry and Morse Theory (1982) by E Witten,

Concern whether the supersymmetry is broken by checking whether $$ Q | 0 \rangle=0 $$ exists or not --- Witten said:

SUSY breaking: A solution may be shown not to exist by calculating a reliable, positive lower bound to the energy eigenvalues.

SUSY non-breaking: It may be shown that a solution does exist by showing that the theory has a mass gap so that there is no potential "Goldstone fermion".

my question is that why SUSY non-breaking has something to do with

• no potential "Goldstone fermion"?
• theory has a mass gap?

Are these if and only if conditions?

$$ \text{Supersymmetry non-breaking} \iff \text{mass gap and no "Goldstone fermion"?} $$

This post imported from StackExchange Physics at 2020-12-12 20:07 (UTC), posted by SE-user annie marie heart

Comments

A "Goldstone fermion" is a massless fermion associated to a broken fermionic symmetry (in this case, supersymmetry). This is directly analogous to Goldstone bosons, which appear when we have a broken bosonic symmetry. If a theory has a mass gap, it means that any excitation above the ground state has mass, so there can be no Goldstone boson, so supersymmetry cannot be broken. This doesn't answer your question regarding "if and only if", but should at least illuminate the relationship of these ideas.

This post imported from StackExchange Physics at 2020-12-12 20:07 (UTC), posted by SE-user George Hulsey

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Goldstone boson requires a continuous symmetry spontaneously broken. Does Goldstone fermion imply continuous SUSY? But boson and fermion Hilbert space are discrete?

This post imported from StackExchange Physics at 2020-12-12 20:07 (UTC), posted by SE-user annie marie heart

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Supersymmetry is a continuous fermionic symmetry: it depends on an infinitesimal Grassmann parameter. For example, the complex scalar of the chiral multiplet transforms under infinitesimal SUSY as $\delta_\epsilon \phi = \sqrt{2}\epsilon \psi$. There is a continuous (but fermionic) parameter associated to the transformations. So the situation is exactly analogous to the bosonic case, except the continouous parameter is Grassmann valued rather than c-number valued.

This post imported from StackExchange Physics at 2020-12-12 20:07 (UTC), posted by SE-user George Hulsey

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but SUSY quantum mechanics do not require a continuous SUSY parameters? One can just switch between bosonic ($-1^F=+1$) and fermionic ($-1^F=-1$) sector

This post imported from StackExchange Physics at 2020-12-12 20:07 (UTC), posted by SE-user annie marie heart