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The 4-dimensional $\mathcal{N} = 1$ superconformal algebra as presented in equations (2.2, 2.3, and 2.4) of the paper, "Counting chiral primaries in N = 1 d = 4 superconformal algebras (arXiv:hep-th/0510060v2)" involves fermionic generators $Q_{\alpha}$ and $S_{\dot{\alpha}}$.

First of all, decomposing operators in terms of $SU(2)_l \times SU(2)_r \times U(1) \times U(1)_R$, we have

$Q_\alpha$ belongs to $(2,1)_{-\frac{1}{2},-1}$

$Q_{\dot{\alpha}}$ belongs to $(1,2)_{\frac{1}{2},1}$

$S_\alpha$ belongs to $(2,1)_{\frac{1}{2},-1}$

$S_{\dot{\alpha}}$ belongs to $(1,2)_{-\frac{1}{2},1}$

where the subscripts denote the $U(1)$ and $U(1)_R$ charges [a related question about what $\frac{1}{2}$ means in the context of $U(1)$ is here.]

Normally in supersymmetry, $Q_{\dot{\alpha}}$ is the complex conjugate of $Q_\alpha$. And complex conjugation interchanges left- and right-handed spinors.

An anticommutator that appears in the main reference paper cited above (i.e. arXiv:hep-th/0510060v2)

$$\{Q_\alpha, (Q_\beta)^\dagger\} = \delta_\alpha^\beta\left(H - \frac{3}{2}J\right) - 4 {{\sigma_{(P)}^i}^\beta}_\alpha J_i$$

where $J$ is the $U(1)_R$ symmetry generator, $H$ is the Hamiltonian, and $\sigma^i_{(P)}$ are the Pauli matrices.

The way this equation is written does not make sense: shouldn't conjugation convert $\beta$ into $\dot{\beta}$, that is, a dotted index? Isn't $(Q_\alpha)^\dagger$ the same as $Q_{\dot{\alpha}}$?

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    $\begingroup$ I haven't seen the paper, but since there is a Hamiltonian in the equation, it follows that a frame of reference has been chosen. It then implies that the spinor indices are now viewed from the point of view of the spatial SO(3) symmetry, for which there are no dotted indices. Instead, the complex conjugate of a 3d spinor is a dual spinor (i.e. with the raised or lowered index so that you can form $\delta^\beta_\alpha$). The 4d spinors here simply restrict to 3d spinors, in other words. $\endgroup$ – Peter Kravchuk Sep 12 '16 at 6:00
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    $\begingroup$ Oh, sorry, it seems I misinterpreted the question (I didn't notice the "conformal"). What happens is that hermitian conjugate of $Q$ is not a $Q$ but an $S$,and usually people have a notation with hermitian-conjugate pairs $Q,S$ and $\bar Q,\bar S$. Here the author just uses $Q^\dagger$ instead of $S$ and dotted $Q$ for $\bar Q$. The reason no dot appears is that you are working with representations of the $SO(4)$, not $SO(3,1)$, which are unitary, so complex conjugation gives you a dual representation (raises or lowers the index). The reason why $Q^\dagger$ is $S$ is similar to $P^\dagger=K$ $\endgroup$ – Peter Kravchuk Sep 12 '16 at 17:29
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    $\begingroup$ That said, $K$ and $P$ are hermitian conjugate in radial quantization, which is by far the most popular quantization to use in Euclidean signature, so yes, we normally think $P^\dagger=K$. This is because hermitian conjugation in radial quantization is equivalent, in a sense, to an inversion $I$, and $K=IPI$. $\endgroup$ – Peter Kravchuk Sep 13 '16 at 16:42
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    $\begingroup$ Conjugation depends on your Hilbert space, which depends on your quantization. To quantize, you foliate your space with constant time surfaces. In Minkowski you want to have them spacelike, so it is most natural to work with flat slices. In that case $P^\dagger=P$. In Euclidean case one often foliates by spheres, so the conjugation rule is different. $\endgroup$ – Peter Kravchuk Sep 13 '16 at 17:59
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    $\begingroup$ See e.g. hep-th/9712074. $\endgroup$ – Peter Kravchuk Sep 13 '16 at 18:30

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