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I was reading a paper that studies a gas of fermions and bosons.

It states that "due to physical background", the Witten index $\Delta=Tr[(-1)^{F}e^{-\beta H}]$ equals 1. I have read that in a system in which supersymmetry cannot be spontaneously broken $$Tr[(-1)]^{F}\neq 0.$$ But, what is the physical reason for which $\Delta=1$?. I am not versed in supersymmetry, What does it mean that "supersymmetry cannot be spontaneously broken"?.

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    $\begingroup$ Where did you read the part about "Supersymmetry cannot be spontaneously broken" ? $\endgroup$ – Stratiev Jun 25 at 10:20
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The Witten index in a supersymmetric theory counts the number of bosonic ground states minus the number of fermionic ground states. The reason for this is that the trace sums over all states and for each excited bosonic state, we also get an excited fermionic state of the same energy, due to supersymmetry, which results in a cancellation.

The reason for $\Delta=1 $ is that in the paper you cite, one builds all the states in the Fock space on one bosonic vacuum state. This one bosonic vacuum state is what the index is counting.

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