= Radial superpartner in three-dimensional supersymmetric quantum mechanics
Let $\chi=\chi(r)$, $r=|x|$, and let $f(r)|0\rangle$ be a normalized <eigenstate> of $H_0$ with $E>0$, where $|0\rangle$ is the <Clifford vacuum>. Acting with the <supercharge> differentiates only its <wavefunction>:
$$
Qf(r)|0\rangle=(f'-\chi'f)\psi_r^\dagger|0\rangle,
\qquad \psi_r^\dagger=\sum_a\frac{x_a}{r}\psi_a^\dagger.
$$
The normalized partner is $E^{-1/2}(f'-\chi'f)\psi_r^\dagger|0\rangle$. Indeed $[H,Q]=0$, and $\|Qf|0\rangle\|^2=E\|f|0\rangle\|^2$; acting with $Q$ again returns $\sqrt E$ times the original state. The expression at $r=0$ is interpreted by the regular extension of the original state. Although its coefficient depends only on $r$, its degree-one components contain the radial unit vector.
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