In a physics article I read recently, the author introduces the notion of supersymmetry by saying basically that the system described by the Hamiltonian $H$ is supersymmetric if $H$ can be decomposed as $H=QQ^\dagger + Q^\dagger Q$ with $Q^2=0$. This condition then leads to a nice splitting of the underlying Hilbert space, which seems to help a lot when working with $H$. Even from the point of view of simple linear algebra, this property is obviously interesting.

Then, I heard that the symmetry is said to be broken or unbroken depending on whether or not $H$ has any ground states, or in other words if the null space of $H$ is non-trivial. I can understand this as a definition, but why would this be interesting? In particular, what consequences are there to supersymmetry being (un)broken (preferably at the low level)? Which systems are "better"?

  • $\begingroup$ Do we observe e.g. a scalar electron at 0.5 MeV? The consequence is a phenomenologically viable mass spectrum of supersymmetric particles, heavy enough to have escaped detection. $\endgroup$ – innisfree Oct 25 '13 at 10:02
  • 1
    $\begingroup$ @innisfree I believe the OP is talking about supersymmetric quantum mechanics as opposed to field theory, in which case I am unsure about the practical import, but it would not have to do with particle phenomenology. :) Of course there are natural analogies between mechanics and field theory. In particular breaking of SUSY QM would lead to a split spectrum as it does in particle physics. I just don't know why one would care about it. $\endgroup$ – Michael Brown Oct 25 '13 at 10:11
  • $\begingroup$ @MichaelBrown oops yes you are right, sorry OP my comment is a little flippant :-) $\endgroup$ – innisfree Oct 25 '13 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.