Supersymmetry algebra convention?

Question

As derived for instance in this review (page 23-24), the supersymmetry algebra involving grassmann valued generators $Q_\alpha$ and ${\bar Q}_{\dot\alpha}$ is given by:

$$[Q_\alpha,M^{\mu\nu}]=(\sigma^{\mu\nu})_\alpha{}^\beta Q_\beta~~~~~~,~~~~~~[Q_\alpha,P^\mu]=0~~~~~~,~~~~~~\{Q_\alpha,Q_\beta\}=0$$

and adjoint versions of the above, where $P^\mu$ is momentum (translation) generator, $M^{\mu\nu}$ is boost generator, and $(\sigma^{\mu\nu})_\alpha{}^\beta=\frac{i}{4}(\sigma^\mu{\bar\sigma}^\nu-\sigma^\nu{\bar\sigma}^\mu)_\alpha{}^\beta$ with $\sigma^\mu=(\mathbb{1}_{2\times2},\vec\sigma),~\bar\sigma^\mu=(\mathbb{1}_{2\times2},-\vec\sigma)$ and $\vec\sigma$ the three Pauli matrices. Finally, there is also the anti-commutation relation

$$\{Q_\alpha,\bar Q_{\dot\beta}\}=t(\sigma^\mu)_{\alpha\dot\beta}P_\mu$$

As stated on page 24, apparently there is no way to constrain and fix the constant $t$ to any specific value - therefore it is set to $t=2$ by convention. Considering all the algebra equations above, it is clear that this constant can be scaled to any finite value without changing the algebra by simply redefining the scale of $Q_\alpha,\bar Q_{\dot\alpha}$ operators. However, there also exists the option of just setting the constant to

$$t=0$$

Naively, this choice seems to be even more symmetric than the original choice $t=finite$, since with $t=0$ the supersymmetry generator action would be completely local without inducing any translation whatsoever. The commutation relation with the boost generator $M^{\mu\nu}$ would still couple the two sectors, properly extending the algebra as we want. So my questions are:

Was the possibility of $t=0$ considered by anyone? What are the consequences compared to the usual choice of $t=2$? Does $t=0$ introduce any consistency issues down the road? Does it make sense to think in this direction, or is there a good reason a priori to discard this possibility?

Thanks for any suggestion.

Answer 1

$t=0$ is completely consistent. It is simply too boring since it implies that in any unitary theory, $Q_\alpha |\psi\rangle = 0$ for any state $|\psi\rangle$. This is because in this case, we have $$ 0 = \langle\psi|\{ Q_\alpha , Q^\dagger_\alpha \} | \psi \rangle = \| Q^\dagger_\alpha |\psi\rangle \|^2 + \| Q_\alpha |\psi\rangle \|^2 $$ Positivity of norm means $$ Q_\alpha |\psi\rangle = Q_\alpha^\dagger | \psi \rangle = 0 $$ This means that the charges $Q_\alpha$ simply decouple from the entire theory and have no consequences whatsoever.