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Witten index, defined as ${\rm Tr}(-1)^F$, determines if supersymmetry is spontaneously broken or not for a given model. However, it is also known that supersymmetry can be dynamically broken. One could think of a mechanism (yet to be implemented if any) à la Nambu-Jona-Lasinio model for chiral symmetry in QCD. In this case, how does Witten index change? I think it should reflect the fact that, even if we are not requiring explicitly a selected ground state breaking symmetry, particle masses are lifted and a gap equation is satisfied.

A more general question is: if Witten index applies as well to dynamical breaking of supersymmetry, for all the models one can possibly conceive.

As usual, good references are welcome.

Thanks.

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2 Answers 2

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When ${\rm Tr}(-1)^F\neq 0$, then supersymmetry cannot be spontaneously broken. One could say that this basic fact is the very point of the Witten index.

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    $\begingroup$ Yes, I know this. I have read Witten's paper. But how does this idea translate to dynamical breaking of symmetry? And if I would have a NJL mechanism? $\endgroup$
    – Jon
    Nov 29, 2012 at 9:58
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Last time I checked, dynamical SUSY breaking was a particular kind of spontaneous SUSY breaking. It just means using certain non-perturbative effects to lift some of the flat directions in moduli space.

So Witten's index trick still works: You can prove that SUSY is unbroken by showing that the Witten index is non-zero.

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  • $\begingroup$ I think this is a right view. $\endgroup$
    – Jon
    Nov 29, 2012 at 15:42
  • $\begingroup$ It's the right answer, but I gave it -1 because it's exactly copied from my answer that was written 5 hours earlier. Why? $\endgroup$
    – Luboš Motl
    Dec 2, 2012 at 18:04
  • $\begingroup$ @LubošMotl: Jon didn't seem to realize that dynamical susy breaking is a form of spontaneous susy breaking and I thought that stating this explicitly was important. Your answer didn't directly address the point he was confused about. Anyways, no particular harm done. (And by the way, I'm the sole upvoter of your answer.) $\endgroup$
    – user1504
    Dec 3, 2012 at 4:20
  • $\begingroup$ It is unfortunate that this rather dippy question has 6 upvotes, while its two correct answers have (net) 1 each. $\endgroup$
    – user1504
    Dec 3, 2012 at 4:32

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