# Why is the infinitesimal SUSY variation generated by the sum of a left- and right-chiral generator

I was wondering why in many (all? e.g.https://arxiv.org/abs/hep-ph/9709356) resources on N=1 SUSY the variation of a field in the simplest free susy model is defined as $$\delta_\epsilon \phi = (\epsilon \cdot Q + \overline \epsilon \cdot \overline Q) \phi$$ and not simply as $$\delta_\epsilon \phi = (\epsilon \cdot Q) \phi$$ with a corresponding conjugated transformation. This somewhat confuses me, because I would expect that it is necessary to show that the action in invariant under action of the left-chiral generator alone.

Would it be possible to define the SUSY transformations separately for $Q$ and $\overline Q$ such that, e.g. for the free chiral model $$[Q_\alpha, \phi] \propto \chi_\alpha, \qquad \{ Q_\alpha,\chi_\beta \} =0, \\ [\overline Q_{\dot \alpha}, \phi] \propto \overline \chi_{\dot \alpha}, \qquad \{ \overline Q_{\dot \alpha},\chi_\beta\} \propto \sigma^\mu_{\beta\dot \alpha} \partial_\mu \phi ?$$

I believe I found the answer myself, see below.

A part of the supersymmetry algebra is $$\{Q_a,~{\bar Q}_{\dot b}\}~=~-2i\sigma^\mu_{a\dot b}\partial_\mu$$ which is a momentum operator $p_\mu~=~-i\partial_\mu$. The graded Lie algebra $g~=~h~+~k$ $$[h,~h]~\subset~h,~[h,~k]~\subset~k,~\{k,~k\}~\subset~h,$$ where the last of these contains the above anti-commutator. This model has chiral symmetry. It then means the left and right handed generators act on a scalar and Dirac field $$\delta_\epsilon\phi~=~(\epsilon Q~+~\bar\epsilon\bar Q)\phi~=~\epsilon\bar\psi~+~\bar\epsilon\psi$$ $$\delta_\epsilon\psi~=~\epsilon\gamma\cdot\partial\phi~+~\bar\epsilon\gamma\cdot\partial\bar\phi.$$ This is the standard SUSY model.

The Wess-Zumino model introduces as pseudoscalar field $\eta$ and the model is taken to be left or right handed chiral that is added to the Dirac field under the SUSY transformations $$\delta_\epsilon\phi~=~\bar\epsilon\bar Q\phi~=~\epsilon\bar\psi$$ $$\delta\eta~=~\bar\epsilon\gamma_5\psi$$ $$\delta_\epsilon\psi~=~\epsilon(\gamma\cdot\partial\phi~+~\gamma_5\eta).$$ The transformation generators are Majorana, and the field $\psi$ is a Majorana fermion. The Majorana fermion is its own antiparticle. The charge conjugation of $\psi$ is $C\psi~=~i\psi^*$. The appearance of $\psi$ and $C\psi$ in the Lagrangian means that the Majorana field must be electrically neutral to conserve charge. This would then be a particle such as the neutrino. We can with the charge conjugation operator transform $C\epsilon~=~\epsilon^*$ $=~\gamma^0\bar\epsilon$ and similarly $CQ~=~Q^*$ $=~\gamma^0\bar Q$ and as such we can define the two super-transformations separately this way.

As for the commutators, I would maybe take a small issue with the bottom right. The elements $\chi_a$ are contained in $k$ with $\{k,~k\}~\subset~h$ and so this I think should be an anticommutator that is $$\{{\bar Q}_{\dot a},~\chi_b\}~=~{\bar Q}_{\dot a}\chi_b~+~\chi_b{\bar Q}_{\dot a}$$ $$\propto~{\bar Q}_{\dot a}Q_b\phi ~-~{\bar Q}_{\dot a}\phi Q_b~+~\phi Q_b{\bar Q}_{\dot a}~-~Q_b\phi {\bar Q}_{\dot a}.$$ For $\phi$ a scalar field that transforms by the supergenerators we must the take care to commute this past the supergenerators $$\{{\bar Q}_{\dot a},~\chi_b\}~=~\{{\bar Q}_{\dot a},~Q_b\}\phi~-~({\bar Q}_{\dot a}\phi) Q_b~+~(Q_b\phi){\bar Q}_{\dot a}.$$ $$=~-2i\sigma^\mu_{\dot a b}\partial_\mu\phi~+~\bar\psi_{\dot a} Q_b~+~\psi_b\bar Q_{\dot a}.$$ The last two expressions in the first line above have parentheses indicating the supergenerator only acts on the field $\phi$. Now $\bar\psi_{\dot a} Q_b~=~\overline{\psi_a\bar Q_{\dot b}}$ and with the Majorana valued fermion defined $C\psi~=~i\gamma^0\bar\psi$ and similarly for the generators the occurrence of $i^2~=~-1$ means the last two terms subtract.

It must be remembered that with $i\sigma^\mu_{\dot a b}\partial_\mu\phi$ this in fact operates on both $\phi$ and any other field or wave $$i\sigma^\mu_{\dot a b}\partial_\mu(\phi\chi)~=~i\sigma^\mu_{\dot a b}\left((\partial_\mu\phi)\chi~+~\phi\partial_\mu\chi\right)$$ and that this is an operator that acts on fields.

• Hi, thank you for your answer. I have a few more questions. Using the term Wess-Zumino Model I was referring to e.g. arxiv.org/abs/hep-ph/9709356 which seems to differ from what you say or what can be found on Wikipedia. In Martins SUSY Primer he calls a free chiral model with a Weyl fermion and a complex scalar the "Wess Zumino model" without mentioning that the fermion is Majorana or part of the complex scalar is parity odd. I've thus changed the model name above to "free chiral model" to avoid any confusion. Jul 22, 2016 at 15:55
• I believe what I mean is what you call the "standard susy model". Now my question is why in this model the variation contains both, the left and right chiral generator. Instead one could have that the left and right chiral generators act separately, but apparently this is not the case. In you answer you seem to imply that this has to do with chiral symmetry. Could you elaborate on this? Jul 22, 2016 at 16:01
• About the other comments: It could be that I need an anti-commutation for the action of Q on $\chi$. I'll check that, thanks! Also, I don't think I understand what you are saying in the last two equations. Just to be sure, I've called my scalar field Z and it transforms under the SUSY generators. (I'll probably edit it to $\phi$ just to avoid confusion) Jul 22, 2016 at 16:01
• If $Z$ is a scalar field that transforms under these generators then things appear to be more complicated. BTW I fixed a "cut and paste" error above. I will have to think about this a bit more later today and maybe update my answer. Jul 22, 2016 at 17:39
• Thanks for the update on your post. It didn't quite answer my question but it was useful for figuring it out by myself. Jul 24, 2016 at 9:40

Ok, I think I figured things out myself. I will follow the conventions of Wess & Bagger.

If one wants to construct a free, $$\mathcal N=1$$ SUSY theory with a complex scalar and a Weyl Fermion, then the possible transformations are dictated by representation theory of the Lorentz group, dimensionalities, and the requirement that we have a free theory and one obtains: $$[Q_\alpha, \phi] = c_0 \chi_\alpha \qquad [Q_\alpha, \phi^*] = c_1 \chi_\alpha \\ \{Q_\alpha,\chi_\beta \} = 0 \qquad \{Q_\alpha,\overline{\chi}_{\dot \beta} \} = c_3 \sigma^\mu_{\alpha \dot \beta} \partial_\mu \phi + c_4 \sigma^\mu_{\alpha \dot \beta} \partial_\mu \phi^*$$ plus the Hermitian conjugates of these (anti-)commutation relations. Closure of the SUSY algebra then imposes restrictions on the coefficients $$c_i$$ which can for example be solved by: $$c_0 = \sqrt{2}, \qquad c_4 = i \sqrt 2.$$ The non-zero transformations are $$[Q_\alpha, \phi] = \sqrt{2} \phi_\alpha \qquad [\overline Q_{\dot \alpha}, \phi^*] = \sqrt{2} \overline \chi_{\dot \alpha} \\ \{Q_\alpha,\overline{\chi}_{\dot \beta} \} = i \sqrt{2} \sigma^\mu_{\alpha \dot \beta} \partial_\mu \phi^* \qquad \{\overline Q_{\dot \alpha},\chi_\beta \} = i \sqrt{2} \sigma^\mu_{\beta \dot \alpha} \partial_\mu \phi$$ The point is that acting on the Lagrangian with either generator leaves the Lagrangian invariant, so @ my second question, yes it is possible to define the commutators in the way I wrote in the question.

This does not quite explain why the variation is defined the way it is. The only argument I have at the moment (which makes me sufficiently happy) is that:

1. It is possible, because variation due to the left- and right-chiral generators do not mix.
2. The generator of the variation is Hermitian