Supersymmetry feels like a discrete symmetry to me, since the fermions are turning into bosons, and vice versa. I understand there is an infinitesimal parameter involved in the transformations, but I don't know what it actually determines physically.

  • $\begingroup$ they can continuously transform into each other. E.g. pure fermion can pick up a delta that is proportional to the boson and vice versa $\endgroup$ – Kosm May 24 at 17:13
  • $\begingroup$ What does this mean physically? I mean, how can a fermion transform into some percentage of a boson? $\phi \rightarrow \phi +\delta \phi=\phi+\bar{\epsilon}\chi$ $\endgroup$ – user45757 May 25 at 13:20
  • $\begingroup$ What does it mean "physically" for a proton to transform into some percentage of a neutron? $\endgroup$ – Cosmas Zachos May 25 at 19:16
  • $\begingroup$ @user45757 physics doesn't change w.r.t. such transformations, that's why it is called a symmetry (if it is unbroken of course). But requiring any symmetry puts restrictions on the action. $\endgroup$ – Kosm May 26 at 5:14
  1. A super-Poincare algebra is a Lie superalgebra, whose elements are generators for a Lie supergroup, and therefore formally corresponds to a continuous symmetry.

    Of course the elephant in the room is the elusive nature of Grassmann-odd numbers, cf. this related Phys.SE post and links therein.

  2. A super-charge $Q$ belongs to the super-Poincare algebra and takes bosons into fermions, and vice-versa.

    Very oversimplified (i.e. ignoring the Grassmann-nature), $Q$ acts like a raising or lowering operator, which gives it a discrete feel, cf. OP's question. Think e.g. of $su(2)$-irreps with a discrete $m$ quantum number, such as in the isospin symmetry, which is also a continuous symmetry.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.