Diagonal self-adjoint operator model

Let $I$ be a finite or countably infinite set with a specified enumeration, let $E_i\in \mathbb{R}$, and let ${\mathrm{\ell }}^2(I):=\{x:I\to \mathbb{C}:{\sum }_i\mathrm{\mid }{x}_i{\mathrm{\mid }}^2<\mathrm{\infty }\}$ with inner product $⟨x,y⟩={\sum }_i{x}_i\overline{y_i}$, linear in the first variable. Write $u_i$ for the coordinate unit vector at $i$. The diagonal multiplication operator $H_E$ is the linear operator with domain $D(H_E):=\{x\in {\mathrm{\ell }}^2(I):{\sum }_i\mathrm{\mid }{E}_i{x}_i{\mathrm{\mid }}^2<\mathrm{\infty }\}$ and action $(H_{E}x{)}_i:=E_i{x}_i$. Completeness and linear domain of the diagonal sequence model ↗ proves completeness of the coefficient space, linearity and density of this domain, and linearity of the coordinate action before the Hilbert-space operator interpretation is used. Its domain is part of the operator data in the sense of Unbounded linear operators: domain, graph and extension.

The coefficient-space and domain construction above requires no choice assumption. Under Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), Diagonal multiplication is self-adjoint ↗ proves that $H_E$ equals its Hilbert-space adjoint, including equality of domains; this is the self-adjointness referred to in the title. The choice assumption belongs to the imported adjoint convention used by that lemma.

The definition introduces no claim about a physical Hamiltonian or about a trace of $e^{-\beta H_E}$.