Fermion-number sectors in supersymmetric quantum mechanics (${\mathcal{H}}_n=L^2({\mathbb{R}}^d)\otimes {\Lambda }^n{\mathbb{C}}^d$)

For $d$ fermionic creation operators with the canonical anticommutation relations, the Hilbert space $L^2({\mathbb{R}}^d)\otimes {\Lambda }^{\bullet }{\mathbb{C}}^d$ splits into ${\mathcal{H}}_n=L^2({\mathbb{R}}^d)\otimes {\Lambda }^n{\mathbb{C}}^d$, $0\le n\le d$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$; its internal multiplicity is $\left(\genfrac{}{}{0ex}{}{d}{n}\right)$. An odd supercharge $Q=q+q^{†}$, with $q$ raising $n$ and $q^2=0$, gives $H=\{q,q^{†}\}$ and preserves these sectors.

On an eigenspace with $E>0$, $Q/\sqrt{E}$ is an odd involution pairing bosonic and fermionic states. More precisely,gives an orthogonal decomposition of each state into the images of $q$ and $q^{†}$: their inner product vanishes because $(q^{†}{)}^2=0$. If $u\in \mathrm{im}{q}^{†}$, then $q^{†}u=0$, $q^{†}qu=Eu$ and $\Vert qu{\Vert }^2=E\Vert u{\Vert }^2$. Thus $q/\sqrt{E}$ pairs such a state in degree $n$ with one in degree $n+1$, and $q^{†}/\sqrt{E}$ is the inverse. States with zero energy are killed by both operators and may remain unpaired. The pairing applies to positive spectral subspaces as well; discreteness is not implied.

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.