Transverse Ising Majorana reduction and bulk gap closing

Each displayed Pauli matrix is an everywhere-defined endomorphism of ${\mathbb{C}}^2$. Its tensor embedding $X_j,Y_j,Z_j$ is therefore an everywhere-defined endomorphism of the finite-dimensional product space ${\mathcal{H}}_N$, using the fixed factor order in [F1].

Every $S_{j-1}$ and ${\gamma }_a$ is a finite product of such endomorphisms, and both sums defining $H_N$ are finite (the second is empty when $N=1$). Thus all displayed operators are well-defined matrices on ${\mathcal{H}}_N$; the only bond terms are exactly those indexed by $j<N$.

The string $S_{j-1}=Z_1\cdots Z_{j-1}$ is Hermitian, has square $I$, and commutes with $X_j$ and $Y_j$ because it acts on different tensor factors. Hence ${\gamma }_{2j-1}$ and ${\gamma }_{2j}$ are Hermitian, and each squares to $I$.

On one site, ${\gamma }_{2j-1}{\gamma }_{2j}=S_{j-1}^2{X}_j{Y}_j=i{Z}_j$ and ${\gamma }_{2j}{\gamma }_{2j-1}=S_{j-1}^2{Y}_j{X}_j=-i{Z}_j$. Their sum is zero, so the two distinct operators at the same site anticommute.

Let $j<\mathrm{\ell }$ and choose either $X_j$ or $Y_j$ at site $j$ and either $X_{\mathrm{\ell }}$ or $Y_{\mathrm{\ell }}$ at site $\mathrm{\ell }$. The string in the later operator contains exactly one $Z_j$; all other factors commute across distinct sites, while both $X_j$ and $Y_j$ anticommute with $Z_j$ by [F2]. Moving the earlier site matrix past the later string therefore contributes exactly one minus sign. Thus every pair of operators at distinct sites anticommutes. Together with the squares from step 1.1 and the same-site computation in step 2.1, this gives the stated CAR for all indices.

Regard $B=iA$ as a complex matrix on ${\mathbb{C}}^{2N}$. Since $A^{\mathsf{T}}=-A$ and $A$ is real, $B^{\ast }=(-i)A^{\mathsf{T}}=iA=B$. By [F1], $B$ has an orthonormal eigenbasis with real eigenvalues. Complex conjugation sends an eigenvector of eigenvalue $ϵ>0$ to an eigenvector of eigenvalue $-ϵ$, because conjugating $iAz=ϵz$ gives $-iA\bar{z}=ϵ\bar{z}$. Thus nonzero eigenvalues occur in opposite pairs.

Fix a unit eigenvector $z=u+iv$ for $ϵ>0$, with real vectors $u,v$. The vectors $z$ and $\bar{z}$ belong to distinct eigenspaces of the Hermitian matrix $B$, so $⟨z,\bar{z}⟩=0$. Expanding this equality gives $\mathrm{\parallel }u{\mathrm{\parallel }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2$ and $u\cdot v=0$; $\mathrm{\parallel }z\mathrm{\parallel }=1$ then gives $\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }v\mathrm{\parallel }=1\mathrm{/}\sqrt{2}$. From $Az=-iϵz$, comparison of real and imaginary parts gives $Au=ϵv$ and $Av=-ϵu$. Therefore $e_1=\sqrt{2}u$ and $e_2=-\sqrt{2}v$ are orthonormal and $A{e}_1=-ϵe_2$, $A{e}_2=ϵe_1$, which is the displayed block in this basis.

Choose an orthonormal basis in each positive eigenspace of $B$; [F1] supplies these finite-dimensional bases. Together with their complex conjugates, the chosen vectors are orthonormal: conjugation preserves orthonormality, and every positive eigenspace is orthogonal to every negative eigenspace of the Hermitian matrix $B$. Hence the real planes from step 2.1 are mutually orthogonal. On the zero eigenspace, $Az=0$, so the real and imaginary parts of each vector lie in the real kernel of $A$. The real and complex kernels have equal dimension because row reduction of the real matrix $A$ has the same pivots over $\mathbb{R}$ and $\mathbb{C}$; this dimension is $2N$ minus twice the total number of positive eigenvectors, hence even. Choose a real basis of $\ker A$ and successively subtract projections onto earlier vectors and normalize each nonzero residual; this gives an orthonormal real basis. Pairing consecutive vectors gives zero blocks. The resulting real orthonormal basis gives the required $O$.

Since $O$ is orthogonal, $(OA{O}^{\mathsf{T}}{)}^{\mathsf{T}}(OA{O}^{\mathsf{T}})=O{A}^{\mathsf{T}}A{O}^{\mathsf{T}}={⨁}_{r=1}^N{ϵ}_r^2{I}_2$. Thus the singular-value multiset from the Statement has two entries ${ϵ}_r$ for each block $r$, including its two zero entries when ${ϵ}_r=0$. This identifies every positive block value once per real two-plane and completes the multiplicity assertion.

For every site, ${\gamma }_{2j-1}{\gamma }_{2j}=X_j{Y}_j=i{Z}_j$, so $ih{\gamma }_{2j-1}{\gamma }_{2j}=-h{Z}_j$. For $j<N$, ${\gamma }_{2j}{\gamma }_{2j+1}=Y_j{Z}_j{X}_{j+1}=i{X}_j{X}_{j+1}$, so $iJ{\gamma }_{2j}{\gamma }_{2j+1}=-J{X}_j{X}_{j+1}$. Summing exactly these terms gives the open-chain $H_N$ of [F1], with no $j=N$ bond.

For any paired indices $a<b$ with coefficient $c$, setting $(A_N{)}_{ab}=2c$ and $(A_N{)}_{ba}=-2c$ contributes $(i\mathrm{/}4)(2c{\gamma }_a{\gamma }_b-2c{\gamma }_b{\gamma }_a)=ic{\gamma }_a{\gamma }_b$ by [F2]. Applying this to $c=h$ on $(2j-1,2j)$ and $c=J$ on $(2j,2j+1)$ yields $H_N=(i\mathrm{/}4){\gamma }^{\mathsf{T}}{A}_N\gamma$.

By [F3], choose a real orthogonal $O$ with $O{A}_N{O}^{\mathsf{T}}$ in the displayed $2\times 2$ blocks, and put $\eta =O\gamma$. Since $O$ is real and orthogonal, [F2] gives $\{{\eta }_a,{\eta }_b\}=2{\delta }_{ab}I$. Define $b_r=({\eta }_{2r-1}+i{\eta }_{2r})\mathrm{/}2$. Direct expansion using these anticommutators gives $\{b_r,b_s^{\ast }\}={\delta }_{rs}I$, $\{b_r,b_s\}=0$, and $b_r^{\ast }{b}_r=\frac{1}{2}I+\frac{i}{2}{\eta }_{2r-1}{\eta }_{2r}$. Using the matrix expression from step 2.1, substitution into the block expression gives $H_N={\sum }_r{ϵ}_r(b_r^{\ast }{b}_r-\frac{1}{2}I)$.

The projections $n_r=b_r^{\ast }{b}_r$ commute by the CAR in step 3.1 and are simultaneously diagonalizable: diagonalize $n_1$ using [F5], restrict the remaining commuting Hermitian matrices to its invariant eigenspaces, and continue through the finite list. Their joint eigenvalues lie in $\{0,1{\}}^N$. If a joint eigenspace has occupancy $0$ in coordinate $r$, $b_r^{\ast }$ maps it isometrically into the eigenspace with that bit changed to $1$, since $\mathrm{\parallel }{b}_r^{\ast }\psi {\mathrm{\parallel }}^2=⟨\psi ,(I-n_r)\psi ⟩=\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$; when the bit is $1$, $b_r$ is the inverse isometry. Hence all $2^N$ joint eigenspaces have the same positive dimension. Their orthogonal sum is ${\mathcal{H}}_N$, whose dimension is $2^N$, so each eigenspace is one-dimensional. The resulting ordered occupation basis identifies the $b_r$ representation unitarily with the finite fermionic Fock occupation basis.

In that occupation basis, ${\sum }_r{ϵ}_r{b}_r^{\ast }{b}_r$ is the diagonal finite-mode Hamiltonian with energies ${ϵ}_r$. Apply [F4] with mode set $\{1,\dots ,N\}$, chemical potential $0$, and $q_r=e^{-\beta {ϵ}_r}$ to obtain trace ${\prod }_r(1+e^{-\beta {ϵ}_r})$. The scalar shift $-\frac{1}{2}{\sum }_r{ϵ}_{r}I$ multiplies this trace by $e^{\beta {\sum }_r{ϵ}_r\mathrm{/}2}$, and [F5] permits the unitary change back to ${\mathcal{H}}_N$. Thus $Z_N={\prod }_r(e^{\beta {ϵ}_r\mathrm{/}2}+e^{-\beta {ϵ}_r\mathrm{/}2})={\prod }_{r}2\cosh (\beta {ϵ}_r\mathrm{/}2)$. If ${ϵ}_r=0$, its factor is $2$.

For finitely supported $x,y$, expanding the finite sums and using ${\int }_{\mathbb{T}}{e}^{i(j-\mathrm{\ell })k}dk\mathrm{/}(2\pi )={\delta }_{j\mathrm{\ell }}$ gives $⟨{\mathcal{F}}_{0}x,{\mathcal{F}}_{0}y{⟩}_{L^2}={\sum }_{j,\mathrm{\ell }}⟨x_j,y_{\mathrm{\ell }}{⟩}_{{\mathbb{C}}^2}{\int }_{\mathbb{T}}{e}^{i(j-\mathrm{\ell })k}\frac{dk}{2\pi }={\sum }_j⟨x_j,y_j{⟩}_{{\mathbb{C}}^2}=⟨x,y{⟩}_{{\mathrm{\ell }}^2}\mathrm{.}$ Thus ${\mathcal{F}}_0$ is an isometry on its finite-support domain.

By [F2] under the explicit Countable Choice hypothesis [A1], the scalar Fourier-coefficient map is a surjective linear isometry from the complete space $L^2(\mathbb{T})$ of [F3] onto scalar ${\mathrm{\ell }}^2(\mathbb{Z};\mathbb{C})$, with inverse given by the finite-subset-net Fourier series. This proves scalar ${\mathrm{\ell }}^2$ completeness: a Cauchy sequence pulls back by that inverse to a Cauchy sequence in $L^2$, whose limit maps to its ${\mathrm{\ell }}^2$ limit. Applying this to each of the two coordinates proves completeness of ${\mathrm{\ell }}^2(\mathbb{Z};{\mathbb{C}}^2)$ with [F1]'s norm. The inverse Fourier map applied coordinatewise is a unitary and agrees with ${\mathcal{F}}_0$ on finitely supported sequences by its finite polynomial formula. Those sequences are dense: for any $\epsilon >0$, the finite-subset tail property in [F1] gives a finite $F\subset \mathbb{Z}$ with ${\sum }_{j\mathrm{\notin }F}\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2<{\epsilon }^2$; choosing $M$ with $F\subseteq [-M,M]$ makes the squared truncation error at most that tail, hence below ${\epsilon }^2$. Thus $c_{00}$ is dense and the continuous extension is unique.

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.