# Why does the non-zero vacuum energy means SUSY is spontaneously broken ?

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I don't quite understand why super symmetry is spontaneously broken if the vacuum state has non-zero energy. If the vacuum state has non zero energy, this means the the SUSY generators have no common eigen-state with zero eigenvalue. I think in this case, SUSY will turn a one-particle state into a state which is a linear combination of the vacuum and many particle states. Is this the reason we say that SUSY is broken if the vacuum state has non-zero energy ?

Use $QQ\sim P$ anticommutation relation and you can get something like $Q^\dagger Q|0\rangle=H|0\rangle$, so if $H|0\rangle\neq 0$ you would know $Q|0\rangle\neq 0$
Ok, I think you are trying to say $e^{i\epsilon Q} |\text{particle}\rangle$ doesn't rotate to a degenerate superpartner, but rather the original particle plus lots of Goldstones, am I understanding you correctly? If so then yes, your reasoning in the main post is right, which is essentially the same as saying $Q|0\rangle=|\text{1-Goldstone}\rangle\neq 0$. This has been well understood since Nambu solved the missing chiral partner problem in nuclear physics using the idea of spontaneous chiral symmetry breaking.
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