Physical-time KMS strip equilibrium condition

The KMS convention in [F2] uses the same closed strip when $b=\beta \mathrm{\hslash }$; the following estimate only needs the stated $b>0$ strip hypotheses. Define $G(w):=F(ibw)$ on $0\le \mathrm{Re}w\le 1$. If $w=x+iy$, then $ibw=-by+ibx$, so the lines $\mathrm{Re}w=0$ and $\mathrm{Re}w=1$ map to the lower and upper edges of $S_b$. The linear map is holomorphic and maps the open strip to $S_b^{\circ }$, so $G$ is bounded and continuous on the closed unit strip and holomorphic inside.

For $0<s<b$, put $x=s\mathrm{/}b\in (0,1)$. The vertical line $\mathrm{Re}w=x$ maps onto $\{t+is:t\in \mathbb{R}\}$, while the boundary suprema of $G$ are exactly $M_0$ and $M_1$. Applying [F1] gives the asserted interpolation inequality. Both exponents $1-x$ and $x$ are positive, so the zero-supremum cases use the defined positive powers and do not invoke $0^0$.

If $M_0,M_1\le C$, step 2.1 gives $\mathrm{\mid }F(t+is)\mathrm{\mid }\le C$ for each $0<s<b$; the same bound holds on both edges by definition of $M_0,M_1$. Hence it holds on all of $S_b$.

Let $F$ and $H$ be two such functions with identical lower edges. Their difference $D=F-H$ is bounded, continuous on $S_b$, holomorphic inside, has lower supremum $0$, and has a finite upper supremum. For every $0<s<b$, step 2.1 applied to $D$ gives ${\sup }_t\mathrm{\mid }D(t+is)\mathrm{\mid }\le 0^{1-s\mathrm{/}b}{M}_1^{s\mathrm{/}b}=0$. Thus $D$ vanishes in the open strip; continuity then gives zero on the upper edge as well as the lower edge.

The finite spectral calculus [F6] defines $Z_n={\mathrm{Tr}}_{{\mathrm{\Lambda }}_n}(e^{-\beta H_{{\mathrm{\Lambda }}_n}})>0$ and the positive trace-one matrix ${\rho }_n=e^{-\beta H_{{\mathrm{\Lambda }}_n}}\mathrm{/}{Z}_n$. For any finite $X\subseteq \mathrm{\Gamma }$, use the tensor factorization of [F1] on $X\cup {\mathrm{\Lambda }}_n$ and set ${\widehat{\rho }}_X^{(n)}:={\mathrm{Tr}}_{(X\cup {\mathrm{\Lambda }}_n)\setminus X}\text{\hspace{0.17em}⁣}({\rho }_n\otimes q^{-\mathrm{\mid }X\setminus {\mathrm{\Lambda }}_n\mathrm{\mid }}{I}_{X\setminus {\mathrm{\Lambda }}_n})$, where ${\rho }_n$ is placed on ${\mathrm{\Lambda }}_n$ and each added site has density $q^{-1}I$. Tensor trace factorization shows that if $X\subseteq Y$, then tracing ${\widehat{\rho }}_Y^{(n)}$ over $Y\setminus X$ gives ${\widehat{\rho }}_X^{(n)}$: each removed added site contributes $\mathrm{Tr}(q^{-1}I)=1$, while any removed site in ${\mathrm{\Lambda }}_n$ is traced from the same ${\rho }_n$ before or after adjoining the identity densities. By [F3], these local algebras sit in the norm-completed $\mathcal{A}$; thus the consistent family [F8] gives an algebraic state ${\omega }_n$ whose restriction to ${\mathcal{A}}_{{\mathrm{\Lambda }}_n}$ is the finite Gibbs state.

Applying [F9] to $({\omega }_n)$ gives at least one pointwise-convergent subsequence. Now fix any pointwise-convergent subsequence, write it as $({\omega }_{n_k})$, and denote its state limit by $\omega$. The rest of the proof applies to this arbitrary convergent subsequence. This extraction is conditional on the exact interface [A2], not on the draft supplier's status label.

Fix local $A\in {\mathcal{A}}_X$ and $B\in {\mathcal{A}}_Y$. If $X=\mathrm{\varnothing }$ or $Y=\mathrm{\varnothing }$, the corresponding local algebra is $\mathbb{C}I$ by [F1], so embed that scalar observable in any one-site local algebra; the identity-preserving inclusion leaves both correlation functions unchanged. We may therefore take $X,Y\mathrm{\ne }\mathrm{\varnothing }$. Since $R<\mathrm{\infty }$, the set $X^{+R}=\{y:{\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})\le R\}$ is finite, so for all sufficiently large $k$, $X\cup Y\cup X^{+R}\subseteq {\mathrm{\Lambda }}_{n_k}$. For those $k$, define the finite-volume entire function $F_k(z):={\omega }_{n_k}\text{\hspace{0.17em}⁣}(A{\alpha }_z^{{\mathrm{\Lambda }}_{n_k}}(B))$, where ${\alpha }_z^{\mathrm{\Lambda }}(C):=e^{iz{H}_{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}C{e}^{-iz{H}_{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$. Because the restrictions of ${\omega }_{n_k}$ to ${\mathcal{A}}_{{\mathrm{\Lambda }}_{n_k}}$ are the Gibbs states, [F7] gives both KMS boundary identities. On either boundary the finite trace state has norm one by [A2], so $\mathrm{\mid }{F}_k\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$ there. For any finite $\mathrm{\Lambda }$, put $H=H_{\mathrm{\Lambda }}=H^{\ast }$; diagonalizing this finite matrix shows that for $C\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and $z=t+is\in S_b$, $0\le s\le b$, $\mathrm{\parallel }{\alpha }_z^{\mathrm{\Lambda }}(C)\mathrm{\parallel }\le \mathrm{\parallel }{e}^{izH\mathrm{/}\mathrm{\hslash }}\mathrm{\parallel }\mathrm{\parallel }C\mathrm{\parallel }\mathrm{\parallel }{e}^{-izH\mathrm{/}\mathrm{\hslash }}\mathrm{\parallel }\le e^{2s\mathrm{\parallel }H\mathrm{\parallel }\mathrm{/}\mathrm{\hslash }}\mathrm{\parallel }C\mathrm{\parallel }\le e^{2\beta \mathrm{\parallel }H\mathrm{\parallel }}\mathrm{\parallel }C\mathrm{\parallel }$. Consequently each $F_k$ is bounded on $S_b$, with the crude bound $e^{2\beta \mathrm{\parallel }{H}_{{\mathrm{\Lambda }}_{n_k}}\mathrm{\parallel }}\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$, which need not be uniform in $k$. This establishes the bounded-strip hypothesis for [F10]; its three-lines estimate uses the uniform boundary bounds and gives $\mathrm{\mid }{F}_k(z)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$ throughout $S_b$.

On every compact real-time interval, [F4] gives ${\alpha }_t^{{\mathrm{\Lambda }}_{n_k}}(B)\to {\alpha }_t(B)$ uniformly in norm. The orbit $K_T=\{A{\alpha }_t(B):\mathrm{\mid }t\mathrm{\mid }\le T\}$ is compact because $t\mapsto {\alpha }_t(B)$ is norm-continuous. For any $\epsilon >0$, cover $K_T$ by finitely many norm-balls of radius $\epsilon \mathrm{/}4$, with centers $C_1,\dots ,C_m\in K_T$. Pointwise convergence in [F9] makes ${\max }_j\mathrm{\mid }{\omega }_{n_k}(C_j)-\omega (C_j)\mathrm{\mid }<\epsilon \mathrm{/}2$ for large $k$; the norm-one bounds in [A2] make the difference at any point of $K_T$ at most this maximum plus $2(\epsilon \mathrm{/}4)$. Combining this with uniform convergence of the dynamics proves $F_k(t)\to \omega (A{\alpha }_t(B))$ uniformly for $\mathrm{\mid }t\mathrm{\mid }\le T$. The same finite-net estimate applied to $\{{\alpha }_t(B)A:\mathrm{\mid }t\mathrm{\mid }\le T\}$ proves uniform convergence of the upper edges to $\omega ({\alpha }_t(B)A)$.

For $B\in {\mathcal{A}}_Y$, [F5] gives the local generator value $\delta (B)$. Every interaction term meeting $Y$ is supported in the finite neighborhood $Y^{+R}$, so when ${\mathrm{\Lambda }}_{n_k}$ contains $Y^{+R}$ the finite-volume generator satisfies ${\delta }_{n_k}(B)=\frac{i}{\mathrm{\hslash }}[H_{{\mathrm{\Lambda }}_{n_k}},B]=\delta (B)$. Differentiating the finite matrix exponential then gives $F_k^{\mathrm{\prime }}(z)={\omega }_{n_k}\text{\hspace{0.17em}⁣}(A{\alpha }_z^{{\mathrm{\Lambda }}_{n_k}}(\delta (B)))$. Taking $C=\delta (B)$ in the finite-volume estimate of step 2.2 shows that for each $k$, $F_k^{\mathrm{\prime }}$ is bounded on $S_b$ by $e^{2\beta \mathrm{\parallel }{H}_{{\mathrm{\Lambda }}_{n_k}}\mathrm{\parallel }}\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }\delta (B)\mathrm{\parallel }$; this crude bound is finite but need not be uniform in $k$. The same finite Gibbs boundary identity [F7] applies to the pair $A,\delta (B)$. Its two real-time boundaries have modulus at most $\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }\delta (B)\mathrm{\parallel }$ by the finite state norm bound [A2]. Since the required boundedness is now established, [F10] gives the uniform estimate $\mathrm{\mid }{F}_k^{\mathrm{\prime }}(z)\mathrm{\mid }\le L:=\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }\delta (B)\mathrm{\parallel }$ on $S_b$. Integrating along the straight segment from $z$ to $w$ (which stays in the convex strip) yields $\mathrm{\mid }{F}_k(w)-F_k(z)\mathrm{\mid }\le L\mathrm{\mid }w-z\mathrm{\mid }$. Thus the functions are uniformly Lipschitz on the whole strip; if $L=0$, they are constant there.

For $m\ge 1$ and $r\ge 1$, let $G_{m,r}:=\{-m+\frac{2mj}{2^r}+i\frac{b\mathrm{\ell }}{2^r}:0\le j,\mathrm{\ell }\le 2^r\}\subset [-m,m]+i[0,b]$. The union of these grids over $m,r$ is countable and dense in the closed strip: for any point in a fixed rectangle, choosing $r$ large makes each coordinate mesh $2m\mathrm{/}{2}^r$ and $b\mathrm{/}{2}^r$ arbitrarily small. Enumerate the grid points by increasing $m+r+j+\mathrm{\ell }$, breaking ties lexicographically. For a bounded complex sequence, enclose its values in a fixed square $[-M,M{]}^2$ with $M>0$; subdivide the current square into four closed subsquares in a fixed lexicographic order and take the first containing infinitely many terms. Choose at each stage the least natural index larger than the previous one whose value lies in that square. At least one subsquare contains infinitely many terms because there are only four. The two coordinate intervals of the nested squares are closed, nested, and have lengths tending to zero; [F11] gives their unique common coordinates, and the selected values converge to that complex point. This defines a deterministic subsequence rule $E$ for each bounded complex sequence. Starting with $i_0(n)=n$, at grid point $g_j$ apply $E$ to $F_{i_{j-1}(n)}(g_j)$, obtaining a strictly increasing index map $s_j$, and set $i_j=i_{j-1}\circ s_j$. The deterministic recursion [F12] defines all $i_j$. The diagonal indices $k_j:=i_j(j)$ are strictly increasing; for each fixed $r$, the tail $(k_j{)}_{j\ge r}$ is a subsequence of $i_r$, so $F_{k_j}(g_r)$ converges. Thus one subsequence converges at every grid point without an additional choice principle.

Fix $m$ and $\epsilon >0$. If $L=0$, convergence at a single grid point and constancy give uniform convergence on the rectangle $K_m=[-m,m]+i[0,b]$. If $L>0$, choose $r$ so that every point of $K_m$ is within $\epsilon \mathrm{/}(8L)$ of some point of $G_{m,r}$. The grid is finite, so the diagonal subsequence is uniformly Cauchy on that grid: for sufficiently large $p,q$, the values $F_{k_p},F_{k_q}$ differ by less than $\epsilon \mathrm{/}2$ at every grid point. At an arbitrary $z\in K_m$, choose a grid point $g$ within $\epsilon \mathrm{/}(8L)$; the two Lipschitz estimates from step 3.2 and the grid difference give $\mathrm{\mid }{F}_{k_p}(z)-F_{k_q}(z)\mathrm{\mid }\le L\mathrm{\mid }z-g\mathrm{\mid }+\mathrm{\mid }{F}_{k_p}(g)-F_{k_q}(g)\mathrm{\mid }+L\mathrm{\mid }g-z\mathrm{\mid }<\epsilon$. Thus the subsequence is uniformly Cauchy on each $K_m$. Completeness of $\mathbb{C}$ gives a limit $F_{A,B}$ locally uniformly on the closed strip; step 2.2 gives $\mathrm{\mid }{F}_{A,B}(z)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$.

Each $F_k$ is entire, and step 4.1 gives locally uniform convergence on every compact subset of $S_b^{\circ }$. By [F13], $F_{A,B}$ is holomorphic there; uniform convergence on compact rectangles including their horizontal edges makes it continuous on $S_b$. Step 3.1 gives uniform convergence of the two boundary sequences on every compact real interval, so the lower and upper boundaries of $F_{A,B}$ are respectively $\omega (A{\alpha }_t(B))$ and $\omega ({\alpha }_t(B)A)$ for every real $t$. Hence the stated local-pair strip conclusion holds.

Let $F:=F_{I,B}$ be the strip function from [F1]. Since every ${\alpha }_t$ is unital, its two boundary values coincide: $F(t)=\omega ({\alpha }_t(B))=F(t+ib)$ for every $t\in \mathbb{R}$. Denote this common continuous function of $t$ by $g(t)$.

For each integer $k$, define $\tilde{F}(t+iy):=F(t+i(y-kb))$ when $kb\le y\le (k+1)b$. At a shared seam the lower and upper definitions agree by step 1.1, so the definition is single-valued. It is continuous on each closed translated strip and across every seam, is $ib$-periodic, and is bounded on $\mathbb{C}$ by the bound for $F$ on $S_b$. It is holomorphic away from the horizontal seams because there it is a translate of $F$ on its open strip.

Fix a point on a seam and choose a disk about it that meets no other seam. Let $T$ be any filled triangle in this disk. If $T$ crosses the seam, cut it along that horizontal line; the resulting at most two polygons can each be triangulated by joining one vertex to the other vertices. Every resulting small triangle lies in one of the two closed half-disks. If a small triangle touches the seam, translate it a distance $ϵ>0$ into that half's open strip; for all sufficiently small $ϵ$, its translate remains inside that strip and the translated contour integral is zero by [F2]. Uniform continuity of $\tilde{F}$ on a compact neighborhood of the original contour implies the translated integrals converge to the original integral as $ϵ\downarrow 0$. Hence each small triangle has zero integral. Summing their boundary integrals cancels shared edges with opposite orientations and yields ${\int }_{\mathrm{\partial }T}\tilde{F}(z)dz=0$. Triangles wholly on one side also have zero integral by [F2].

Step 3.1 applies to every triangle in the disk. The function $\tilde{F}$ is continuous there by step 2.1, so [F3] makes it holomorphic across the seam in that disk. Away from seams it was already holomorphic by step 2.1; since the point on the seam was arbitrary, $\tilde{F}$ is entire on $\mathbb{C}$.

Step 2.1 gives a global bound on $\tilde{F}$, so [F4] makes it constant. Its restriction to the real axis is $g(t)=\omega ({\alpha }_t(B))$ by step 1.1, and at $t=0$ this is $\omega (B)$. Therefore $\omega ({\alpha }_t(B))=\omega (B)$ for every real $t$. Since $B$ was arbitrary, $\omega$ is invariant.

By [A1]–[A2], choose an orthonormal basis $u_1,\dots ,u_D$ with $H{u}_i=E_i{u}_i$ and $E_i\in \mathbb{R}$; [A2] puts the finite matrix algebra in the C${}^{\ast }$-dynamical setting required by Physical-time KMS strip equilibrium condition. Let $E_{ij}$ be the matrix unit that maps $u_j$ to $u_i$ and all other basis vectors to zero. Direct multiplication gives $E_{ab}{E}_{cd}={\delta }_{bc}{E}_{ad}$, and the exponential action gives ${\alpha }_z^H(E_{ij})=e^{iz(E_i-E_j)\mathrm{/}\mathrm{\hslash }}{E}_{ij}$ for $z\in \mathbb{C}$. This formula is used only for finite matrices, where the exponentials and scalar entries are entire.

Fix $i\mathrm{\ne }j$ and apply [F1] to $A=E_{jj}$, $B=E_{ij}$. Since $E_{jj}{E}_{ij}=0$, the lower boundary of its KMS function is identically zero. By [F2] this KMS function vanishes on the whole strip. Its upper edge at $t=0$ is $\omega (E_{ij}{E}_{jj})=\omega (E_{ij})$, so $\omega (E_{ij})=0$ for every off-diagonal matrix unit.

For arbitrary $i,j$, set $\mathrm{\Delta }=E_i-E_j$ and use [F1] with $A=E_{ji}$, $B=E_{ij}$. The lower edge is $e^{it\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }}\omega (E_{jj})$. The candidate $g(z)=e^{iz\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }}\omega (E_{jj})$ is entire and bounded on $0\le \mathrm{Im}z\le \beta \mathrm{\hslash }$, since $\mathrm{\mid }{e}^{i(t+iy)\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }}\mathrm{\mid }=e^{-y\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }}$ and $y$ ranges over a compact interval. It has the same lower boundary as the KMS function, so [F2] identifies them. At the upper edge and $t=0$, the KMS identity is therefore $\omega (E_{ii})=e^{-\beta (E_i-E_j)}\omega (E_{jj})$.

The ratios in step 2.2 show that $e^{\beta E_i}\omega (E_{ii})$ is a constant $c$ independent of $i$. Since ${\sum }_i{E}_{ii}=I$ and $\omega (I)=1$, $1=c{\sum }_i{e}^{-\beta E_i}$; the sum is positive and finite, so $c=Z^{-1}$ where $Z=\mathrm{Tr}(e^{-\beta H})>0$. Thus $\omega (E_{ii})=e^{-\beta E_i}\mathrm{/}Z$. This also covers $D=1$ and degenerate eigenvalues.

Every $X\in \mathcal{B}(\mathcal{H})$ has a matrix-unit expansion $X={\sum }_{i,j}{X}_{ij}{E}_{ij}$. By linearity, step 2.1, and step 3.1, $\omega (X)={\sum }_i{X}_{ii}{e}^{-\beta E_i}\mathrm{/}Z=Z^{-1}\mathrm{Tr}(e^{-\beta H}X)$. Therefore every finite KMS state is the Gibbs trace state.

Conversely, [F3] supplies the entire bounded finite Gibbs correlation with the stated lower and upper edges, which is exactly [F1]'s finite-matrix KMS condition; [A2] ensures that this KMS definition applies to $\mathcal{B}(\mathcal{H})$. The normalized Gibbs trace state is therefore KMS, proving the reverse implication.

The local algebra ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is norm dense in $\mathcal{A}$ by [F2]. Fix arbitrary $A,B\in \mathcal{A}$. Choose sequences $A_j,B_j\in {\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ with $A_j\to A$ and $B_j\to B$ in norm. They are bounded sequences, so set $M_A:={\sup }_j\mathrm{\parallel }{A}_j\mathrm{\parallel }<\mathrm{\infty }$ and $M_B:={\sup }_j\mathrm{\parallel }{B}_j\mathrm{\parallel }<\mathrm{\infty }$.

For each $j$, [F4] gives a bounded continuous strip function $F_j$ holomorphic in $S_b^{\circ }$, $b=\beta \mathrm{\hslash }$, with lower and upper boundaries $\omega (A_j{\alpha }_t(B_j))$ and $\omega ({\alpha }_t(B_j)A_j)$; the finite-volume Gibbs boundary input used by [F4] is the exact identity in [F7]. On both boundaries, the state norm and the isometry of ${\alpha }_t$ from [A2]–[F3] give modulus at most $\mathrm{\parallel }{A}_j\mathrm{\parallel }\mathrm{\parallel }{B}_j\mathrm{\parallel }$. Since $F_j$ is already bounded by [F4], applying [F5] to $F_j$ gives $\mathrm{\mid }{F}_j(z)\mathrm{\mid }\le \mathrm{\parallel }{A}_j\mathrm{\parallel }\mathrm{\parallel }{B}_j\mathrm{\parallel }\le M_A{M}_B$ throughout $S_b$. For $j,k$, each boundary difference is bounded uniformly in real $t$ by ${\eta }_{jk}:=\mathrm{\parallel }{A}_j-A_k\mathrm{\parallel }{M}_B+M_A\mathrm{\parallel }{B}_j-B_k\mathrm{\parallel }$, because $\mathrm{\parallel }\omega \mathrm{\parallel }=1$ by [A2] and each ${\alpha }_t$ is isometric by [F3]. Applying the three-lines estimate of [F5] to the bounded function $F_j-F_k$ gives ${\sup }_{z\in S_b}\mathrm{\mid }{F}_j(z)-F_k(z)\mathrm{\mid }\le {\eta }_{jk}$. Since $(A_j)$ and $(B_j)$ are norm-Cauchy, ${\eta }_{jk}\to 0$ as $j,k\to \mathrm{\infty }$. Thus $(F_j)$ is uniformly Cauchy on the whole closed strip.

The uniform limit $F_{A,B}:={\lim }_j{F}_j$ exists on $S_b$ because $\mathbb{C}$ is complete. It is bounded by $M_A{M}_B$ and continuous on $S_b$ as a uniform limit of bounded continuous functions. The convergence is locally uniform in the open strip, so [F6] makes $F_{A,B}$ holomorphic on $S_b^{\circ }$.

For fixed real $t$, the lower boundaries converge to $\omega (A{\alpha }_t(B))$: indeed, $\mathrm{\mid }\omega (A_j{\alpha }_t(B_j))-\omega (A{\alpha }_t(B))\mathrm{\mid }\le \mathrm{\parallel }{A}_j-A\mathrm{\parallel }\mathrm{\parallel }{B}_j\mathrm{\parallel }+\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{B}_j-B\mathrm{\parallel }$, using $\mathrm{\parallel }\omega \mathrm{\parallel }=1$ and isometry of ${\alpha }_t$. The right-hand side tends to zero. The same calculation with the order reversed shows the upper boundaries converge to $\omega ({\alpha }_t(B)A)$. Since $F_j\to F_{A,B}$ uniformly on $S_b$, its two edge values are therefore the required KMS boundary values.

Steps 3.1 and 4.1 show that for every $A,B\in \mathcal{A}$ the bounded continuous function $F_{A,B}$ is holomorphic inside and has the two boundaries in Physical-time KMS strip equilibrium condition. Thus $\omega$ is a $\beta$-KMS state for $\alpha$ in physical time. Existence of at least one such subsequential $\omega$ is supplied conditionally by [F4].