Consider the spin-boson model. Letting the system Hamiltonian for this model be set to zero, we get an integrable system,

$$ H = X\otimes \sum_u \lambda_u x_u + \mathbb{1}\otimes H_B,$$

Where $H_B$ is the Hamiltonian for a collection of independent harmonic oscillators with displacements given by $x_u$. In particular, the Pauli matrix $X$ is a constant of motion, and so the total Hamiltonian is unitarily equivalent to a direct-sum of two copies of $H_B$ via a polaron transformation:

$$U^{-1}HU\simeq H_B\oplus H_B$$

For example, the thermal Green's function of the spin is equal to unity: because $[X,e^{\tau H}]=0$,

$$K(\tau-\tau')\equiv \langle X(\tau)X(\tau')\rangle_\beta~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\,$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{Z}\text{tr}(e^{-\tau H}Xe^{(\tau-\tau')H}Xe^{-(\beta-\tau')H})=\frac{1}{Z}\text{tr}(X\cdot X\cdot e^{-\beta H})=1$$

However, it is also known that integrating-out the oscillators in the Feynman-Kac representation of the equilibrium partition function (for a detailed derivation, see equations (93)-(102) of *Quantum Dissipative Systems *by F Bascones et. al.) yields an inverse-square Ising model on a one-dimensional periodic lattice (with lattice spacing equal to the trotter time-step $\tau_c$) in imaginary time:

$$\text{tr}(e^{-\beta H})=\sum_{\{X(\tau)\}}e^{-\beta S_{\tau_c}(X(\tau))},~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

$$~~~~~~~~~~~~~~~~~\beta S_{\tau_c}(X(\tau))=\sum_{\tau\neq \tau'}J(\tau-\tau')\,X(\tau)X(\tau'),~~~~~J(\tau-\tau')\underset{\tau-\tau' \to \tau_c}{\sim}\frac{\tau_c^2}{(\tau-\tau')^{2}}$$

Now, I can evaluate the thermal Green's function $K(\tau-\tau')$ for my integrable model using the effective Euclidean action above. However, **there is no way that an inverse-square Ising model has a two-point function which is identically one across all temperature ranges!** What went wrong?