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.
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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 '19 at 17:13
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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 '19 at 13:20
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What does it mean "physically" for a proton to transform into some percentage of a neutron? – Cosmas Zachos May 25 '19 at 19:16
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@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 '19 at 05:14
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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.
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.
