Rotor interacting Hamiltonian with exact domain

In the joint basis, $H_0$ multiplies coefficients by the real nonnegative number $h(\mathbf{n})={\lambda }_E{\sum }_e{n}_e^2$. If $\varphi \in D(H_0^{\ast })$ with $H_0^{\ast }\varphi =\eta$, testing against each basis vector gives ${\eta }_{\mathbf{n},\mathbf{m}}=h(\mathbf{n}){\varphi }_{\mathbf{n},\mathbf{m}}$, so $\eta \in {\mathcal{H}}_{\mathrm{k}\mathrm{i}\mathrm{n}}$ forces $\sum h(\mathbf{n}{)}^2\mathrm{\mid }{\varphi }_{\mathbf{n},\mathbf{m}}{\mathrm{\mid }}^2<\mathrm{\infty }$ and $\varphi \in D(H_0)$. Conversely that weighted square summability and Cauchy–Schwarz give the adjoint identity. Hence $H_0^{\ast }=H_0$. Truncating a vector to finite subsets of the joint basis converges both in norm and in the $H_0$ graph norm, so finite sums are a graph core.

Every $W_p$ is unitary, so $\mathrm{\parallel }{W}_p+W_p^{†}\mathrm{\parallel }\le 2$ and $W_p+W_p^{†}$ is self-adjoint. Each CAR operator has norm at most one by the conditional supplier in [F1]; every displayed fermion/link term is a finite sum of products of bounded operators, and pairing each term with its adjoint makes $V_F$ self-adjoint. Since the graph, plaquette list, site modes, and matrices are all finite-dimensional or finite in number, $V$ is bounded and self-adjoint.

Let $\beta >\mathrm{\parallel }V\mathrm{\parallel }$ and $z=\pm i\beta$. The diagonal resolvent $R_0(z)=(H_0-z{)}^{-1}$ satisfies $\mathrm{\parallel }{R}_0(z)\mathrm{\parallel }\le 1\mathrm{/}\beta$, and $H-z=(I+V{R}_0(z))(H_0-z)\mathrm{.}$ The Neumann series ${\sum }_{k=0}^{\mathrm{\infty }}(-V{R}_0(z){)}^k$ converges in operator norm because $\mathrm{\parallel }V{R}_0(z)\mathrm{\parallel }<1$, so $I+V{R}_0(z)$ is onto and invertible. Thus both $H-i\beta$ and $H+i\beta$ are onto ${\mathcal{H}}_{\mathrm{k}\mathrm{i}\mathrm{n}}$.

On $D(H_0)$, $H$ is densely defined and symmetric because $H_0$ and $V$ are. To prove self-adjointness directly, take $y\in D(H^{\ast })$ and choose $x\in D(H)$ with $(H+i\beta )x=(H^{\ast }+i\beta )y$, using the surjectivity from step 3.1. Then $y-x\in \ker (H^{\ast }+i\beta )=\mathrm{Ran}(H-i\beta {)}^{\perp }=\{0\}$, again by step 3.1. Therefore $D(H^{\ast })\subseteq D(H)$; symmetry gives the reverse inclusion and $H=H^{\ast }$. Its domain is exactly the original $D(H_0)$ because the sum with bounded $V$ was defined there.

Each $U(\alpha )$ is diagonal in the same Fourier/occupation basis as $H_0$, so it preserves $D(H_0)$ and commutes with $H_0$. The link relation $[E_e,U_e]=U_e$ and the CAR relation $[n_x,c_{y\beta }]=-{\delta }_{xy}{c}_{y\beta }$ give $U(\alpha )U_{e}U(\alpha {)}^{†}=e^{i({\alpha }_{s(e)}-{\alpha }_{t(e)})}{U}_e$ and $U(\alpha )c_{x\beta }U(\alpha {)}^{†}=e^{i{\alpha }_x}{c}_{x\beta }$. A cycle's link phases telescope to $1$; in a hopping term the phases multiply to $e^{-i{\alpha }_{s(e)}}{e}^{i({\alpha }_{s(e)}-{\alpha }_{t(e)})}{e}^{i{\alpha }_{t(e)}}=1$, and each onsite bilinear has phase $e^{-i{\alpha }_x}{e}^{i{\alpha }_x}=1$. Therefore $U(\alpha )VU(\alpha {)}^{†}=V$ and $U(\alpha )HU(\alpha {)}^{†}=H$ on $D(H_0)$.