Current continuity, Drude obstructions and conditional diffusion

Since $h_{j,j+1}$ and $q_{j+1}$ are self-adjoint, $[h_{j,j+1},q_{j+1}{]}^{\ast }=-[h_{j,j+1},q_{j+1}]$. Thus $j_{j,j+1}=(i\mathrm{/}\mathrm{\hslash })[h_{j,j+1},q_{j+1}]$ is self-adjoint.

The Heisenberg derivative is ${\dot{q}}_j=(i\mathrm{/}\mathrm{\hslash }){\sum }_k[h_{k,k+1},q_j]$. By [A1], only the existing bonds immediately to the left and right of $j$ can contribute. The left contribution is $(i\mathrm{/}\mathrm{\hslash })[h_{j-1,j},q_j]=j_{j-1,j}$ when that bond exists, and is zero by the open-boundary convention otherwise.

For the right contribution, [A2] gives $[h_{j,j+1},q_j]=-[h_{j,j+1},q_{j+1}]$, so $(i\mathrm{/}\mathrm{\hslash })[h_{j,j+1},q_j]=-j_{j,j+1}$ when that bond exists. If it is absent at the open right endpoint, both terms are zero by convention.

Combining steps 1.2 and 1.3 proves ${\dot{q}}_j=j_{j-1,j}-j_{j,j+1}$. Summing over $j\in {\mathrm{\Lambda }}_L$ cancels every existing open-chain or periodic bond once with each sign, so ${\dot{Q}}_L=0$, equivalently $[H_L,Q_L]=0$.

The lattice-current definition fixes $J_L={\sum }_j{j}_{j,j+1}$, and the local-current lemma gives $J_L=J_L^{\ast }$. Subtracting its real mean $\omega (J_L)$ preserves self-adjointness, so $J$ is a matrix observable in $\mathcal{B}(\mathcal{H})$.

On $\mathcal{B}(\mathcal{H})$ define $(A,B{)}_{\rho }:=\mathrm{Tr}(\rho A^{\ast }B)$, linear in its second argument. It is positive definite: if $(A,A{)}_{\rho }=0$, then $\rho \ge cI$ for some $c>0$ in finite dimension, so $0=\mathrm{Tr}(\rho A^{\ast }A)\ge c\mathrm{Tr}(A^{\ast }A)$ and hence $A=0$. The invariance $[\rho ,H]=0$ gives $({\alpha }_t(A),{\alpha }_t(B){)}_{\rho }=(A,B{)}_{\rho }$ by cyclicity of the finite trace, so ${\alpha }_t$ is unitary on this finite-dimensional Hilbert space of matrices.

Write $H={\sum }_{E}E{P}_E$ over its distinct eigenvalues. Then ${\alpha }_t(J)={\sum }_{E,F}{e}^{it(E-F)\mathrm{/}\mathrm{\hslash }}{P}_{E}J{P}_F$. For $E\mathrm{\ne }F$, $\mid \frac{1}{T}{\int }_0^T{e}^{it(E-F)\mathrm{/}\mathrm{\hslash }}dt\mid \le \frac{2\mathrm{\hslash }}{T\mathrm{\mid }E-F\mathrm{\mid }}⟶0,$ while for $E=F$ the coefficient is $1$. The sum is finite, so $M_{T}J:=T^{-1}{\int }_0^T{\alpha }_t(J)dt$ converges in matrix norm to $PJ:={\sum }_E{P}_{E}J{P}_E$. This block map is the orthogonal projection onto $\mathcal{N}:=\{A:[H,A]=0\}$ in $(\cdot ,\cdot {)}_{\rho }$: $PJ\in \mathcal{N}$, and each off-diagonal block of $J-PJ$ has zero weighted trace pairing with a block-diagonal matrix because $\rho$ is block diagonal. It is the same energy-block projection as the density-matrix dephasing in [F3]; [F3] states the density-matrix version, while the explicit finite expansion here proves the observable version used.

Since $J=J^{\ast }$, $({\alpha }_t(J),J{)}_{\rho }=\omega ({\alpha }_t(J)J)$. Norm convergence in step 2.1 therefore gives ${\lim }_{T\to \mathrm{\infty }}\frac{1}{T}{\int }_0^T\mathrm{Re}\omega ({\alpha }_t(J)J)dt=\mathrm{Re}(PJ,J{)}_{\rho }=\mathrm{\parallel }PJ{\mathrm{\parallel }}_{\rho }^2,$ because $J-PJ$ is orthogonal to $\mathcal{N}$ and $PJ\in \mathcal{N}$.

Set $S:=\mathrm{span}\{Q_1,\dots ,Q_m\}\subseteq \mathcal{N}$, and let $P_S$ be its orthogonal projection. Then $P_{S}J=P_{S}PJ$, so $\mathrm{\parallel }PJ{\mathrm{\parallel }}_{\rho }^2\ge \mathrm{\parallel }{P}_{S}J{\mathrm{\parallel }}_{\rho }^2$. The Gram matrix $G$ is positive definite by linear independence and the positive-definite inner product. Writing $P_{S}J={\sum }_a{c}_a{Q}_a$, orthogonality gives $Gc=v$, so $c=G^{-1}v$ and $\mathrm{\parallel }{P}_{S}J{\mathrm{\parallel }}_{\rho }^2=c^{\ast }Gc=v^{\ast }{G}^{-1}v$. Together with step 3.1 this proves the bound.

In the classical alternative fix $T>0$ and use Lebesgue measure on $[0,T]$ for the Bochner integral. For each $n\ge 1$, let $J_n:[0,T]\to L^2(\mathrm{\Omega },\mathcal{F},\mathbb{P})$ take the value $J(kT\mathrm{/}n)$ on $[kT\mathrm{/}n,(k+1)T\mathrm{/}n)$ for $0\le k<n$ and the value $J(T)$ at $T$. Each $J_n$ is a measurable simple function. Uniform continuity of $J$ on $[0,T]$ gives ${\sup }_{t\in [0,T]}\mathrm{\parallel }{J}_n(t)-J(t){\mathrm{\parallel }}_2\to 0$, and hence ${\int }_0^T\mathrm{\parallel }{J}_n(t)-J(t){\mathrm{\parallel }}_{2}dt\to 0$. Thus $J$ is strongly measurable (an almost-everywhere limit of measurable simple functions), and ${\int }_0^T\mathrm{\parallel }J(t){\mathrm{\parallel }}_{2}dt<\mathrm{\infty }$ because $J$ is bounded. By the definition of the Bochner integral as the $L^1$-norm limit of integrals of simple functions, $J$ is Bochner integrable. The Bochner integrals of $J_n$ are the left-endpoint Riemann sums for $J$; they converge both to the Bochner integral by the $L^1$ approximation and to the norm-Riemann integral by [F6]. Hence these integrals agree. The expectation functional $\mathrm{\Lambda }(Y):=\mathbb{E}[Y]=⟨Y,1{⟩}_{L^2}$ is bounded since $\mathrm{\mid }\mathrm{\Lambda }(Y)\mathrm{\mid }=\mathrm{\mid }⟨Y,1{⟩}_{L^2}\mathrm{\mid }\le \mathrm{\parallel }Y{\mathrm{\parallel }}_2\mathrm{\parallel }1{\mathrm{\parallel }}_2=\mathrm{\parallel }Y{\mathrm{\parallel }}_2$. Applying the bounded-linear-map clause of [F6] gives $\mathbb{E}[X_T]={\int }_0^T\mathbb{E}[J(t)]dt=0$, so $\mathrm{Var}(X_T)=\mathbb{E}[X_T^2]$.

The spatially normalized current convention is fixed by [F1]. In the quantum alternative, [A1] and [F2] apply to the exact ${\alpha }_t$, invariant state $\varphi$, and centered self-adjoint $V$ in the Statement, so the supplier gives the variance identity and limit. The evenness asserted in the Statement follows from $C(-t)=\mathrm{Re}\varphi (V{\alpha }_{-t}(V))=\mathrm{Re}\varphi ({\alpha }_t(V)V)=\mathrm{Re}\overline{\varphi (V{\alpha }_t(V))}=C(t)$, using invariance and self-adjointness. In the classical alternative, continuity of the inner product makes $(s,t)\mapsto \mathbb{E}[J(s)J(t)]=C(t-s)$ continuous on $[0,T{]}^2$. Riemann sums for $X_T$ converge in $L^2$, so continuity of the inner product makes the expectations of their squared norms converge to $\mathbb{E}[X_T^2]$; each squared-sum expectation is a product Riemann sum for this continuous kernel. Thus [F3] gives $\mathbb{E}[X_T^2]={\int }_0^T{\int }_0^{T}C(t-s)dsdt$, which equals $\mathrm{Var}(X_T)$ by step 1.1. The integrand is symmetric because $C$ is even. The diagonal has zero area: it lies in strips of width $\delta$ and area at most $2T\delta +{\delta }^2$, tending to zero, so a bounded continuous integrand contributes zero there. Off the diagonal the two triangular restrictions are continuous; on a grid meeting the diagonal their upper-lower gap is bounded by the kernel oscillation away from the diagonal plus twice its supremum times the diagonal-strip area. Hence the product of the triangular indicator with the kernel is Riemann integrable. Each section has at most one jump and is therefore Riemann integrable. For the upper triangle, $g(t):={\int }_0^{t}C(t-s)ds$ satisfies $\mathrm{\mid }g(t)-g(t\mathrm{\prime })\mathrm{\mid }\le T{\omega }_C(\mathrm{\mid }t-t\mathrm{\prime }\mathrm{\mid })+\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }t-t\mathrm{\prime }\mathrm{\mid }$, where ${\omega }_C(\delta ):={\sup }_{\mathrm{\mid }u-v\mathrm{\mid }\le \delta }\mathrm{\mid }C(u)-C(v)\mathrm{\mid }\to 0$; hence its outer integral is continuous. The lower triangle has the same estimate with reversed limits. Splitting the square into its two triangles and applying [F3] to each gives $2{\int }_0^T{\int }_0^{t}C(t-s)dsdt$.

For the classical alternative, apply [F4] to $\phi (s)=t-s$ on $[0,t]$. It gives ${\int }_t^{0}C(u)du=-{\int }_0^{t}C(t-s)ds$; reversing the oriented endpoints yields ${\int }_0^{t}C(t-s)ds=F(t)$, where $F(t):={\int }_0^{t}C(u)du$. By [F7], $F^{\mathrm{\prime }}=C$; the product rule gives $(tF(t){)}^{\mathrm{\prime }}=F(t)+tC(t)$, a continuous derivative. Applying [F5] to $tF(t)$ and using $F(T)={\int }_0^{T}C(u)du$ yields ${\int }_0^{T}F(t)dt={\int }_0^T(T-u)C(u)du$. Thus step 2.1 gives $\mathrm{Var}(X_T)\mathrm{/}(2T)={\int }_0^T(1-u\mathrm{/}T)C(u)du$.

In the classical alternative, put $K:={\int }_0^{\mathrm{\infty }}\mathrm{\mid }C(u)\mathrm{\mid }du<\mathrm{\infty }$. For $T>R>0$, the difference between step 3.1 and the improper integral is bounded by $\mid {\int }_0^T(1-u\mathrm{/}T)C(u)du-{\int }_0^{\mathrm{\infty }}C(u)du\mid \le \frac{R}{T}K+2{\int }_R^{\mathrm{\infty }}\mathrm{\mid }C(u)\mathrm{\mid }du\mathrm{.}$ First choose $R$ so the tail is arbitrarily small, then let $T\to \mathrm{\infty }$; this proves the classical limit. The quantum limit is supplied by [F2].

In the classical alternative every finite-time variance is nonnegative, so its limit is nonnegative; in the quantum alternative nonnegativity is part of [F2]. The limit therefore has the stated sign in either case.