Finite fermion-mode occupation basis

For $k=0$, ${\xi }_{\mathrm{\varnothing }}=\mathrm{\Omega }$ is the unit vacuum. For $k\ge 1$, repeated application of the creation formula in [F2] gives ${\xi }_S=\sqrt{k!}{P}_{a,k}(e_{j_1}\otimes \cdots \otimes e_{j_k})\mathrm{.}$ The product-basis theorem in [F4], under [A1], makes elementary tensors of basis vectors a complete orthonormal basis of ${\mathcal{K}}^{\widehat{\otimes }k}$. If an index repeats, antisymmetrization is zero by pairing permutations that differ by swapping the equal factors. If all indices are distinct, their antisymmetrization is, up to the permutation sign, a scalar multiple of the vector displayed for the uniquely ordered subset $S$.

For $k\ge 1$ and distinct increasing indices, $\mathrm{\parallel }{P}_{a,k}(e_{j_1}\otimes \cdots \otimes e_{j_k}){\mathrm{\parallel }}^2=⟨e_{j_1}\otimes \cdots \otimes e_{j_k},P_{a,k}(e_{j_1}\otimes \cdots \otimes e_{j_k}){⟩}_2=1\mathrm{/}k!$, since only the identity permutation has nonzero inner product with that ordered elementary tensor. Distinct subsets have orthogonal permutation orbits in the tensor basis, so $⟨{\xi }_S,{\xi }_T{⟩}_2={\mathbf{1}}_{S=T}$. The vacuum case has norm one by step 1.1.

For each $k\ge 1$, the bounded projection $P_{a,k}$ maps the complete basis of elementary tensors onto a dense subspace of its range. Step 1.1 shows this image is spanned by the ${\xi }_S$ with $\mathrm{\mid }S\mathrm{\mid }=k$; step 2.1 shows those vectors are orthonormal. Thus they form a basis of each antisymmetric sector. If $k>\mathrm{\mid }J\mathrm{\mid }$, every elementary basis tensor repeats an index and its projection vanishes, so that sector is zero. The direct-sum construction and completeness supplied by [A1], [F1] and [F3], together with the unit vacuum in step 1.1, now show that all ${\xi }_S$ form an orthonormal basis of ${\mathcal{F}}_a(\mathcal{K})$.

Write $c_j=c(e_j)$ and $c_j^{†}=c^{†}(e_j)$. By [F3] both operators are bounded, so $N_j=c_j^{†}{c}_j$ is bounded. If $j\mathrm{\notin }S$, anticommute $c_j$ through the creation factors in ${\xi }_S$ using the mixed CAR; it reaches $\mathrm{\Omega }$ and annihilates it, so $N_j{\xi }_S=0$. If $j=j_{\mathrm{\ell }}\in S$, anticommuting it past the first $\mathrm{\ell }-1$ distinct creators gives $c_j{\xi }_S=(-1{)}^{\mathrm{\ell }-1}{\xi }_{S\setminus \{j\}}$; creating $j$ again and reordering it into its original position gives the same sign, so $N_j{\xi }_S={\xi }_S$. Thus by the ONB in step 3.1, $N_j$ is the diagonal multiplier ${\mathbf{1}}_{j\in S}$; its diagonal values are real and idempotent, so it is a self-adjoint orthogonal projection.

On every basis vector ${\xi }_S$, the finite sum $n_x={\sum }_{j\in J_x}{N}_j$ acts by $\mathrm{\mid }S\cap J_x\mathrm{\mid }$. These are real numbers between zero and $\mathrm{\mid }{J}_x\mathrm{\mid }$, so the bounded diagonal operator $n_x$ is self-adjoint. If $z\in \mathbb{C}\setminus \{0,\dots ,\mathrm{\mid }{J}_x\mathrm{\mid }\}$, the diagonal operator with entries $(\mathrm{\mid }S\cap J_x\mathrm{\mid }-z{)}^{-1}$ is bounded by $\mathrm{dist}(z,\{0,\dots ,\mathrm{\mid }{J}_x\mathrm{\mid }\}{)}^{-1}$ and is an inverse for $n_x-zI$; hence $z$ is outside its spectrum.