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In John Terning, Modern Supersymmetry, $R$-symmetry is introduced as below: "The SUSY algebra is invariant under a multiplication of the supercharges by a phase, so in general there is a linear combination of U(1) charges called $R$-charge, that doesn't commute with the supercharges."

He the proceeds to write down the following commutator relations: $$[Q_{\alpha}, R] = Q_{\alpha}, $$ $$[Q_{\alpha}^{\dagger}, R] = - Q_{\alpha}^{\dagger}$$

How are the above relations derived?

Michael Williams
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  • That's pretty much a convention: we say an operator $\mathcal O$ has charge $q$ under the $\mathrm U(1)$ symmetry generated by $T$ iff $[\mathcal O,T]=q\mathcal O$. In other words, this algebra defines what it means to be charged under a $\mathrm U(1)$ symmetry. See also this PSE post. – AccidentalFourierTransform May 21 '18 at 16:19
  • ^ what AFT said. If that's confusing, have a look at how time evolution acts on operators in Heisenberg picture -- it's by commutator. In general the finite action is by a conjugation, which infinitesimally becomes a commutator. – knzhou May 21 '18 at 16:21
  • @AccidentalFourierTransform your answer in the post was helpful, got it, thanks. – Michael Williams May 21 '18 at 16:47
  • @knzhou i went through the post AFT linked, the example which you give now follows from the general case. Thanks. – Michael Williams May 21 '18 at 16:49

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