# The integrable spin-boson and 1D inverse-square Ising models: thermal two-point functions

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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?

asked Jun 26, 2017
edited Jun 27, 2017

## 1 Answer

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The integrable spin-boson model is not an inverse-square Ising model. For that equivalence to hold, one has to include terms that do not commute with $X\otimes \mathbb{1}$.

Let's re-examine the Feynman-Kac representation for the integral kernel of the partition function: when $[H,X]=0$, Kronecker deltas appear which massively simplify the partition function:

$$Z[\tau_c]~~\underset{\tau_c\to 0}{\sim} \sum_{\{X(\tau)\},\{x_u(\tau)\}}\cdots \langle X(\tau_{i-1}),x_u(\tau_{i-1})|e^{-\tau_c\cdot X(\tau_i)\sum_u \lambda_u x_u(\tau_i)}e^{-\tau_c H_B}|X(\tau_i),x_u(\tau_i)\rangle\cdots$$

$$~~~~~~~= \sum_{\{X(\tau)\},\{x_u(\tau)\}}\cdots \delta(X(\tau_{i-1}),X(\tau_i))\cdot \langle x_u(\tau_{i-1})|e^{-\tau_c\cdot X(\tau_i)\sum_u \lambda_u x_u(\tau_i)}e^{-\tau_c H_B}|x_u(\tau_i)\rangle\cdots$$

To justify the intermediate step, since $H_B$ does not affect the system degrees of freedom, $X(\tau_i)$ is just an eigenvalue, and so the bra $\langle X(\tau_{i-1})|$ meets the ket $|X(\tau_i)\rangle$ directly, inducing a prefactor. These Kronecker deltas in the integral kernel are precisely the reason why the thermal Green's function $\langle X(\tau)X(\tau')\rangle_\beta$ for the integrable spin-boson model is identically one (as you correctly deduced by a direct calculation sans path integral techniques).

Of course, if we turned on a nontrivial system Hamiltonian $H_S$ (e.g. add a term proportional to $Z\otimes \mathbb{1}$), then there would be a possibility to connect different eigenspaces of $X$ at different imaginary times, and these Kronecker-deltas would disappear (this is the starting point of the paper which you cited). But then the integrability of the model would be lost. Either way, the model is either solvable via a polaron transformation, or is equivalent to an inverse-square Ising model; both cannot be true simultaneously.

answered Jun 28, 2017 by (135 points)
edited Jun 28, 2017

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