# R-symmetry commutator

I've seen the claim made several placed; Terning's "Modern Supersymmetry" p. 5 on N=1 SUSY algebra states it as well as anyone:

The SUSY algebra is invariant under a multiplication of $Q_\alpha$ by a phase, so in general there is one linear combination of $U(1)$ charges, called the $R$-charge, that does not commute with $Q$ and $Q^\dagger$:

$[Q_\alpha,R] = Q_\alpha, \;\;\;[Q^\dagger_\dot{\alpha},R]=-Q^\dagger_\dot{\alpha}$

The first statement is straightforward to see. But

(1) Why is there is one linear combination of charges that does not commute?

(2) How do we arrive at these commutators? (I imagine that the generators can be rescaled to give the coefficient $\pm1$, but I would like a clearer explanation.)

I spoke to a peer who said that the commutation relations could be found in a very general, mathematically heavy treatment of the most general possible SUSY algebra. Is there some easier way to understand?

The point is that the SUSY algebra, \begin{align} & \left\{ Q _\alpha ,Q _\beta \right\} = \left\{ \bar{Q} _{\dot{\alpha}} , \bar{Q} _{\dot{\beta}} \right\} = 0 \\ & \left\{ Q _\alpha , \bar{Q} _{\dot{\beta}} \right\} = 2 \sigma ^\mu _{ \alpha \dot{\beta} }P ^\mu \end{align} is invariant under multiplication of $Q _\alpha$ by a phase, \begin{align} Q _\alpha & \rightarrow e ^{ - i \phi } Q _\alpha \\ \bar{Q} _{\dot{\alpha}} & \rightarrow e ^{ i \phi } \bar{Q} _{\dot{\alpha}} \end{align} This means that you can have a SUSY invariant theory but still have an additional symmetry which differentiates between bosons and fermions (since it doesn't need to commute with $Q _\alpha$).
To see this explicitly consider the effect of an $R$ symmetry on $Q _\alpha$: \begin{align} e ^{ -i R \phi } Q _\alpha e ^{ i R \phi } & = e ^{ - i \phi } Q _\alpha \\ \left( 1 - i R \phi - ... \right) Q _\alpha \left( 1 + i R \phi - ... \right) & = - i \phi Q _\alpha \\ i \left[ Q _\alpha , R \right] \phi = - i \phi Q _\alpha \\ \left[ Q _\alpha , R \right] = - Q _\alpha \end{align} Thus this phase shift symmetry implies that the commutation between the $R$ symmetry generator and $Q _\alpha$ is nontrivial and hence bosons and fermions can have a different R charge. This is what makes R symmetries so special. This discussion is likely more complicated for ${\cal N} > 1$.