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

Question

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 $$[Q_\alpha, \phi] \propto \chi_\alpha, \qquad \{ Q_\alpha,\chi_\beta \} =0, \\ [\overline Q_{\dot \alpha}, \phi] \propto \overline \chi_{\dot \alpha}, \qquad \{ \overline Q_{\dot \alpha},\chi_\beta\} \propto \sigma^\mu_{\beta\dot \alpha} \partial_\mu \phi ?$$

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I believe I found the answer myself, see below.

Answer 1

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.

Answer 2

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