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


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.


1 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.

  • $\begingroup$ How could you define the $Q_\alpha$ so it satisfies the anticommutation relation for $t = 2$? $\endgroup$
    – Luke
    Jun 14, 2019 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.