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  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 ? 

asked Aug 5 in Theoretical Physics by anonymous [ revision history ]

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 $Q|0\rangle\neq 0$

I know that. What I'm asking is : If there are no zero eigenvalues , Why can't we construct a spectrum with equal number of bosons and fermions with the same mass ? Consider this as a mathematical problem. We use Q and Q dagger as raising and lowering operators starting from the clifford vaccum , Now we shouldn't be able to get the spectrum that is mentioned at the beginning of all SUSY Books if there are no common zero eigenvalues for all SUSY generators right ?

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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