 Radial superpartner in three-dimensional supersymmetric quantum mechanics

Let $\chi =\chi (r)$, $r=|x|$, and let $f(r)|0⟩$ be a normalized eigenstate of $H_0$ with $E>0$, where $|0⟩$ is the Clifford vacuum. Acting with the supercharge differentiates only its wavefunction:The normalized partner is $E^{-1/2}(f^{\prime }-{\chi }^{\prime }f){\psi }_r^{†}|0⟩$. Indeed $[H,Q]=0$, and $\Vert Qf|0⟩{\Vert }^2=E\Vert f|0⟩{\Vert }^2$; 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.