 Solution

Write the interaction matrix as $\mathsf{J}$ to distinguish it from its scalar strength. The printed pair sum contains both orientations and has no factor $1/2$. For a positive-definite matrix $\mathsf{J}$, completion of the square gives the Hubbard–Stratonovich transformationsince $m^T{\mathsf{J}}^{-1}m-2m^T\sigma =(m-\mathsf{J}\sigma {)}^T{\mathsf{J}}^{-1}(m-\mathsf{J}\sigma )-{\sigma }^T\mathsf{J}\sigma$. For the exponential kernel with $J>0$ and $\kappa >0$, positivity follows from the positive Fourier symbol derived in part (b), or from its finite open-chain covariance matrix.

Insert this identity into the Ising model partition sum and interchange the finite spin sum with the convergent Gaussian integral. The spin variables now factor independently:It follows thatwith $C=({\pi }^N\det \mathsf{J}{)}^{-1/2}$ for the measure used here. The factor two in the spin-field coupling is required by the double pair sum; replacing it by one would decouple a statistical Hamiltonian with a different normalization. The diagonal $J_{ii}{\sigma }_i^2=J$ is a spin-independent energy. Including it as printed makes the positive Gaussian identity directly applicable; if diagonal self-couplings are removed they should first be restored as an irrelevant constant before this real auxiliary-field decoupling.