Translation-invariant Ising Majorana symbol and bulk gap

Let $S_{+}f$ have coordinates $(S_{+}f{)}_j=f_{j+1}$ and let $S_{-}f$ have $(S_{-}f{)}_j=f_{j-1}$. Reindexing gives $\mathrm{\parallel }{S}_{\pm }f\mathrm{\parallel }=\mathrm{\parallel }f\mathrm{\parallel }$ and $⟨S_{-}f,g⟩=⟨f,S_{+}g⟩$, so $S_{-}^{\ast }=S_{+}$ and $S_{+}^{\ast }=S_{-}$ by [F6]. The Fourier-unitarity fact [F2] makes $\mathcal{K}$ a Hilbert space, so the bounded shifts and their bounded linear combinations have Hilbert adjoints under [F6]. With $B=2hI-2J{S}_{-}$, $B^{\ast }=2hI-2J{S}_{+}$; the block formula $A_{\mathrm{\infty }}=\left(\begin{array}{cc}0& B\\ -B^{\ast }& 0\end{array}\right)$ then gives $A_{\mathrm{\infty }}^{\ast }=-A_{\mathrm{\infty }}$. By [F3] and the triangle inequality, both $B$ and $B^{\ast }$ have bound $2(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }J\mathrm{\mid })$, whence $\mathrm{\parallel }{A}_{\mathrm{\infty }}x{\mathrm{\parallel }}^2=\mathrm{\parallel }Bv{\mathrm{\parallel }}^2+\mathrm{\parallel }{B}^{\ast }u{\mathrm{\parallel }}^2\le 4(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }J\mathrm{\mid }{)}^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$. Thus $A_{\mathrm{\infty }}$ and $K_{\mathrm{\infty }}=i{A}_{\mathrm{\infty }}$ are everywhere-defined bounded operators by [F4]. Finally $⟨i{A}_{\mathrm{\infty }}x,y⟩=i⟨x,A_{\mathrm{\infty }}^{\ast }y⟩=⟨x,-i{A}_{\mathrm{\infty }}^{\ast }y⟩$, so $K_{\mathrm{\infty }}^{\ast }=-i{A}_{\mathrm{\infty }}^{\ast }=i{A}_{\mathrm{\infty }}=K_{\mathrm{\infty }}$ by [F6].

For a finitely supported sequence, $\mathcal{F}(v_{j-1})(k)=e^{ik}\widehat{v}(k)$ and $\mathcal{F}(u_{j+1})(k)=e^{-ik}\widehat{u}(k)$ by reindexing the sums. Therefore $\mathcal{F}{A}_{\mathrm{\infty }}{\mathcal{F}}^{-1}$ is multiplication by $\left(\begin{array}{cc}0& 2(h-J{e}^{ik})\\ -2(h-J{e}^{-ik})& 0\end{array}\right)$; boundedness and density extend the identity to all of ${\mathrm{\ell }}^2$. Multiplication by $i$ gives the displayed $M(k)$.

Put $a(k)=2i(h-J{e}^{ik})$. Then $M(k)=\left(\begin{array}{cc}0& a(k)\\ \overline{a(k)}& 0\end{array}\right)$ and $M(k{)}^2=\mathrm{\mid }a(k){\mathrm{\mid }}^{2}I$. Its characteristic polynomial is ${\lambda }^2-\mathrm{\mid }a(k){\mathrm{\mid }}^2$, so its two eigenvalues are $\pm \mathrm{\mid }a(k)\mathrm{\mid }=\pm 2\sqrt{h^2+J^2-2hJ\cos k}$, including the double zero when $a(k)=0$.

Let $\lambda \in \mathbb{C}$ lie outside the union of the two bands in the Statement. By step 3.1, $ϵ(k)=\mathrm{\mid }a(k)\mathrm{\mid }$ for every $k$, so $d(k):=\mathrm{\mid }{\lambda }^2-\mathrm{\mid }a(k){\mathrm{\mid }}^2\mathrm{\mid }$ is continuous and strictly positive on $\mathbb{T}$. For each $k_0$, the open set $U_{k_0}:=\{k:d(k)>d(k_0)\mathrm{/}2\}$ contains $k_0$. Compactness from [F7] supplies a finite subcover $U_{k_1},\dots ,U_{k_m}$, whence $d(k)>c:={\min }_{1\le r\le m}d(k_r)\mathrm{/}2>0$ for every $k$. The matrix identity $(M(k)-\lambda I)(M(k)+\lambda I)=(\mathrm{\mid }a(k){\mathrm{\mid }}^2-{\lambda }^2)I$ therefore gives the pointwise inverse $(M(k)+\lambda I)\mathrm{/}(\mathrm{\mid }a(k){\mathrm{\mid }}^2-{\lambda }^2)$. This inverse matrix field is continuous and its operator norm is uniformly bounded because $\mathrm{\parallel }M(k){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }J\mathrm{\mid })$ and $d(k)\ge c$. Multiplication by it is an everywhere-defined bounded two-sided inverse for $\mathsf{M}-\lambda I$, and its negative is an inverse for $\lambda I-\mathsf{M}$. Conjugating by $\mathcal{F}$ puts $\lambda$ in the resolvent of $K_{\mathrm{\infty }}$ by [F5]. Thus the spectrum is contained in the two bands.

Fix $k_0\in \mathbb{T}$. When $a(k_0)\mathrm{\ne }0$, the vectors $w_{+}=(a(k_0)\mathrm{/}\mathrm{\mid }a(k_0)\mathrm{\mid },1)\mathrm{/}\sqrt{2}$ and $w_{-}=(-a(k_0)\mathrm{/}\mathrm{\mid }a(k_0)\mathrm{\mid },1)\mathrm{/}\sqrt{2}$ are unit eigenvectors of $M(k_0)$ with eigenvalues $\pm ϵ(k_0)$ by step 3.1; when $a(k_0)=0$, $M(k_0)=0$ and either standard unit vector is an eigenvector. For either eigenvalue $\lambda$, set ${\delta }_n=\min (\pi \mathrm{/}2,1\mathrm{/}n)$ and let $I_n=\{k_0+[t]:-{\delta }_n<t<{\delta }_n\}$, an open arc of angular length $2{\delta }_n$ and measure $m_{\mathbb{T}}(I_n)={\delta }_n\mathrm{/}\pi >0$ by [F7]. Put ${\psi }_n(k)={\mathbf{1}}_{I_n}(k)w\mathrm{/}\sqrt{m_{\mathbb{T}}(I_n)}$ using the corresponding fixed eigenvector $w$. Then $\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}_2=1$ and continuity of $M$ gives $\mathrm{\parallel }(\mathsf{M}-\lambda I){\psi }_n{\mathrm{\parallel }}_2\le {\sup }_{k\in I_n}\mathrm{\parallel }(M(k)-M(k_0))w\mathrm{\parallel }\to 0$. If $\mathsf{M}-\lambda I$ had a bounded inverse, these unit vectors would have residual norms bounded below by the reciprocal inverse norm, a contradiction. By [F5] and the unitary equivalence in step 2.1, every band value lies in the spectrum of $K_{\mathrm{\infty }}$.

Steps 4.1 and 4.2 give both inclusions, so the spectrum is exactly ${\bigcup }_{k\in \mathbb{T}}\{-ϵ(k),ϵ(k)\}$. Since $\cos k\in [-1,1]$, step 3.1 gives ${\min }_kϵ(k{)}^2=4(h^2+J^2-2\mathrm{\mid }hJ\mathrm{\mid })=4(\mathrm{\mid }h\mathrm{\mid }-\mathrm{\mid }J\mathrm{\mid }{)}^2$. The minimum is attained at $k=0$ when $hJ\ge 0$ and at $k=\pi$ when $hJ<0$. Taking the nonnegative square root gives ${\mathrm{\Delta }}_{\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}}=2\mid \mathrm{\mid }h\mathrm{\mid }-\mathrm{\mid }J\mathrm{\mid }\mid$, which vanishes exactly for $\mathrm{\mid }h\mathrm{\mid }=\mathrm{\mid }J\mathrm{\mid }$. This concerns the one-particle convolution operator only.