Lemma: Uniform finite-band estimate

Suppose a symmetric operator $A$ preserves ${\mathcal{D}}_{\mathrm{fin}}$, changes particle number by at most two, and satisfies

Then $A$ is essentially self-adjoint on ${\mathcal{D}}_{\mathrm{fin}}$. If also $\Vert A\Psi \Vert \le C\Vert (\mathsf{N}+2)\Psi \Vert$, its closure is defined on $\mathrm{Dom}\mathsf{N}$ and obeys that estimate there.

Proof. Iteration gives

Thus the power series $U(t)\Psi ={\sum }_j(it{)}^j{A}^j\Psi /j!$ converges for $2C|t|<1$, with a radius independent of $n$.

We give the analytic-vector argument explicitly. In a smaller interval, the differentiated double series for $⟨U(t)\Psi ,U(t)\mathrm{X}⟩$ is absolutely convergent. Symmetry cancels its derivative term by term, so $U(t)$ is an isometry on the dense finite-particle space and extends to an isometry on $\mathcal{F}$. For $2C(|s|+|t|)<1$, the double series for $U(s)U(t)\Psi$ converges absolutely by the preceding bound and the binomial identity. It equals $U(s+t)\Psi$. Hence $U(t)$ is unitary for sufficiently small $t$, with inverse $U(-t)$.

If $A^{\ast }h=ih$, then symmetry and the series give

Density implies $U(t{)}^{\ast }h=e^{t}h$, impossible for a nonzero $h$ and a nonzero small real $t$. The same argument excludes $A^{\ast }h=-ih$. Therefore both deficiency spaces vanish. Indeed, the ranges of the closure of $A\pm i$ are closed by ${\Vert (A\pm i)\Psi \Vert }^2={\Vert A\Psi \Vert }^2+{\Vert \Psi \Vert }^2$; their orthogonal complements are those deficiency spaces. Surjectivity of both ranges proves self-adjointness of the closure.

Finally, number truncations converge in the number graph norm. The assumed graph bound makes their $A$-images Cauchy. Closedness proves the last assertion.