2
$\begingroup$

I have recently asked this question in Physics Stackexchange, but as there was no success there, a friend pointed out that I might have a better shot here.

Consider the Ising Model in the $d$-dimensional discrete torus with side lengh $L$, denoted by $\mathbb{T}_L $, with nearest neighbors interaction (with interaction parameter $J$, no magnetic field, and inverse of temperature $\beta$

The Gaussian Domination Bound/ Infrared Bound states that for every non-zero $ p \in \mathbb{T}_L^*:=(\frac{2\pi}{L} \mathbb{Z}^d) \cap (-\pi,\pi]$, we have:

$$ \sum_{x \in \mathbb{T}_L} e^{ip \cdot x} \langle \sigma_0 \sigma_x \rangle_{L,\beta} \le \frac{1}{2\beta E(p)}, $$

where $\langle \cdot \rangle^{0}_{L,\beta} $ denotes the expected value of a random variable with respect to finite volume Gibbs Measure in $\mathbb{T}_L$; $i$ is the imaginary unit; $ip \cdot x$ denotes the inner product of $ip$ and $x$; and $$ E(p):= J \sum_{x: \|x\|_1=1} (1-e^{ip\cdot x}). $$

Multiple articles point out that the following statement is equivalent to that bound, $\forall (v_x) \in \mathbb{C}^{\mathbb{T}_L}$, we have $$ \sum_{x,y \in \mathbb{T}_L: \|x-y\|_1=1 }v_x \bar{v}_y \langle \sigma_x \sigma_y \rangle_{L,\beta} \le \frac{1}{2\beta}\sum_{x,y \in \mathbb{T}_L: \|x-y\|_1=1 } v_x \bar{v}_y G_L(x,y) + \frac{1}{L^d} \big| \sum_{x \in \mathbb{T}_L} v_x \big|^2 \sum_{x \in \mathbb{T}_L} \langle \sigma_0 \sigma_x \rangle_{L,\beta}, $$ where $$ G_L(x,y) := \sum_{p \in \mathbb{T}_L^* \backslash \{0\} } \frac{1}{L^d} \frac{e^{ip \cdot (x-y)}}{E(p)}. $$

Well, I tried to use the usual proof as in the Chapter 10 of the book Statistical Mechanics of Lattice Systems:a Concrete Mathematical Introduction. Following the proof in page 465 until the moment they choose the values of $\alpha_i$, I believe that this equivalence will come choosing new values for $\alpha_i$, but to this moment, I don't know what to choose.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.