= Fermion-number sectors in supersymmetric quantum mechanics
{title2=$\mathcal H_n=L^2(\mathbb R^d)\otimes\Lambda^n\mathbb C^d$}
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 $\mathcal H_n=L^2(\mathbb R^d)\otimes\Lambda^n\mathbb C^d$, $0\leq n\leq d$. The <fermion number operator> $N=\sum_a\psi_a^\dagger\psi_a$ has value $n$ on $\mathcal H_n$; its internal multiplicity is $\binom dn$. An odd <supercharge> $Q=q+q^\dagger$, with $q$ raising $n$ and $q^2=0$, gives $H=\{q,q^\dagger\}$ and preserves these sectors.
On an <eigenspace> with $E>0$, $Q/\sqrt E$ is an <odd involution pairing bosonic and fermionic states>. More precisely,
$$
v=E^{-1}(qq^\dagger+q^\dagger q)v
$$
gives an <orthogonal decomposition> of each state into the images of $q$ and $q^\dagger$: their <inner product> vanishes because $(q^\dagger)^2=0$. If $u\in\operatorname{im}q^\dagger$, then $q^\dagger u=0$, $q^\dagger qu=Eu$ and $\|qu\|^2=E\|u\|^2$. Thus $q/\sqrt E$ pairs such a state in degree $n$ with one in degree $n+1$, and $q^\dagger/\sqrt E$ 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.
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