# Why is supersymmetry a continuous symmetry?

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.

• 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 – Kosm May 24 at 17:13
• What does this mean physically? I mean, how can a fermion transform into some percentage of a boson? $\phi \rightarrow \phi +\delta \phi=\phi+\bar{\epsilon}\chi$ – user45757 May 25 at 13:20
• What does it mean "physically" for a proton to transform into some percentage of a neutron? – Cosmas Zachos May 25 at 19:16
• @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. – Kosm May 26 at 5:14

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.