Exponentially coupled Ising inverse kernel ($J^{-1}(q)=(\cosh \kappa -\cos q)/(J\sinh \kappa )$)

For the infinite one-dimensional coupling matrix $J_{ij}=J{e}^{-\kappa |i-j|}$ with $J,\kappa >0$, its Fourier symbol is $J\sinh \kappa /(\cosh \kappa -\cos q)$. Its inverse is nearest-neighbor even though the original interaction is not. The inverse quadratic form equals $[2J\sinh \kappa {]}^{-1}{\sum }_j(m_j-m_{j+1}{)}^2+[\tanh (\kappa /2)/J]{\sum }_j{m}_j^2$. Finite open chains add endpoint corrections; a periodized coupling gives the same translation-invariant formula.

For $N\ge 2$, the open-chain Ising model coupling $[J_N{]}_{ij}=J{\rho }^{|i-j|}$, $0<\rho <1$, has a tridiagonal inverse. Its endpoint diagonal entries are $1/[J(1-{\rho }^2)]$, its interior diagonal entries are $(1+{\rho }^2)/[J(1-{\rho }^2)]$, and adjacent off-diagonal entries are $-\rho /[J(1-{\rho }^2)]$. Direct matrix multiplication verifies the inverse because the geometric sequences cancel in every interior row. The displayed quadratic form follows with $D=\rho /[J(1-{\rho }^2)]$, $A=(1-\rho )/[J(1+\rho )]$ and $B=\rho /[J(1+\rho )]$. For $\rho =e^{-\kappa }$, $D=(2J\sinh \kappa {)}^{-1}$ and $A=\tanh (\kappa /2)/J$. Omitting the endpoint term requires an infinite chain, a correctly periodized coupling, or a bulk thermodynamic limit.