for the case of the ordinary bosonic oscillator.

We begin with the construction of the mutually conjugate nilpotent supercharge operators $Q^+$ and $Q^-$, $Q^+=(Q^-)^\dagger$, in the simplest possible form, linear in the bosonic operators $a^\pm$, but also containing a dependence on the Klein operator $K$: $$Q^{+}=\frac{1}{2}a^{+}(\alpha+\beta K)+\frac{1}{2}a^{-}(\gamma+\delta K).$$ The nilpotency condition $Q^{\pm 2}=0$ leads to the restriction on the complex number coefficients: $\beta=\epsilon\alpha$, $\delta=\epsilon\gamma$. Therefore, we find that there are two possibilities for choosing operator $Q^+$: $$Q^{+}_{\epsilon}=(\alpha a^{+}+\gamma a^{-})\Pi_\epsilon,
\quad \epsilon=\pm,$$ where $\Pi_\pm$ are the projector operators (\[proj\]). The anticommutator $\{Q^{+}_{\epsilon},Q^{-}_{\epsilon}\}$ has here the form: $$\{Q^{+}_{\epsilon},Q^{-}_{\epsilon}\}=
a^{+2}\alpha\gamma^{*}+a^{-2}\alpha^{*}\gamma
+\frac{1}{2}\{a^{+},a^{-}\}(\gamma\gamma^{*}+\alpha\alpha^{*})
-\frac{1}{2}\epsilon K
[a^{-},a^{+}](\gamma\gamma^{*}-\alpha\alpha^{*}).$$ Whence we conclude that if we choose the parameters in such a way that $\alpha\gamma^{*}=0$, the anticommutator will commute with the number operator $N$, and, as a consequence, the spectra of the corresponding Hamiltonians, $H_\epsilon
=\{Q^+_\epsilon,Q^-_\epsilon\}$, $\epsilon=\pm$, will have the simplest possible form. Let us put $\alpha=0$ and normalize the second parameter as $\gamma=e^{i\varphi}$. Since we can remove this phase factor by the unitary transformation of the operators $a^{\pm}$, we arrive at the nilpotent operators in the very compact form: $$Q^{+}_{\epsilon}=a^{-}\Pi_{\epsilon},\quad
Q^{-}_{\epsilon}=a^{+}\Pi_{-\epsilon}.
\label{qexp}$$ They together with the operator $$H_{\epsilon}=\frac{1}{2}\{a^{+},a^{-}\}
-\frac{1}{2}\epsilon K [a^{-},a^{+}]
\label{hexp1}$$ form the $N=2$ superalgebra, which we 