3
$\begingroup$

We are interested in proving the Markov property for the long range Ising type model in $\mathbb{Z}^d$.

Setting:

Define $\Omega = \{-1, +1\}^{\mathbb{Z}^d}$ the space of all possible configurations in $\mathbb{Z}^d$ and let $\Lambda$ be a finite subset of $\mathbb{Z}^d$. Then we define the hamiltonian for this model as: $$H_{\Lambda, \beta}^{\omega_{\Lambda^c}}(\sigma_{\Lambda}) = - \sum_{i \in \Lambda, j \in \Lambda} J_{ij}\sigma_i\sigma_j - \sum_{i \in \Lambda, j \notin \Lambda} J_{ij}\sigma_i\omega_j,$$ where $\omega_{\Lambda^c}$ is a fixed configuration outside $\Lambda$ and $\sigma_{\Lambda}$ is the restriction of a configuration $\sigma$ to $\Lambda$ and $J_{ij} = \frac{1}{\vert\vert i - j \vert\vert_1}$.

For a fixed $\Lambda \subset \mathbb{Z}^d$, we define a mesure in $\Omega_{\Lambda}$ as $$\mu_{\Lambda, \beta}^{\omega_{\Lambda_c}}(\sigma_{\Lambda}) = \frac{1}{Z_{\Lambda}^{\omega_{\Lambda^c}}}\exp(-\beta H_{\Lambda, \beta}^{\omega_{\Lambda^c}}(\sigma_{\Lambda}) ),$$ where $Z_{\Lambda}^{\omega_{\Lambda^c}}$ is the partition function (or normalizing constant)

Problem

Prove that for all $\Delta \subset \Lambda \subset \mathbb{Z}^d$, $\Lambda$ finite subset of $\mathbb{Z}^d$, and all configurations $\eta \in \Omega$ and $\omega' \in \Omega$ such that $\omega'_i = \eta_i$, for all $i \in \Lambda^c$, $$\mu^{\eta}_{\Lambda, \beta}(\cdot \hspace{2mm} | \hspace{2mm} \sigma_i = \omega'_i, \forall i \in \Lambda \setminus \Delta) = \mu_{\Delta, \beta}^{\omega'}(\cdot).$$

My attempt

By the definition of the measure, we have that for any $w \in \Omega$,

$$\mu_{\Delta^c, \beta}^{\omega'_{\Delta_c}}(\omega) = \frac{1}{Z_{\Delta}^{\omega_{\Delta^c}}}\exp(-\beta H_{\Delta, \beta}^{\omega'_{\Delta^c}}(\sigma_{\Delta}) ) = \frac{1}{Z_{\Delta}^{\omega'_{\Delta^c}}}\exp\left(-\beta \left(- \sum_{i \in \Lambda, j \in \Lambda} J_{ij}\omega_i\omega_j - \sum_{i \in \Lambda, j \notin \Lambda} J_{ij}\omega_i\omega'_j\right)\right) = \frac{1}{Z_{\Delta}^{\omega'_{\Lambda^c}}}\exp\left(-\beta \left(- \sum_{i \in \Lambda, j \in \Lambda} J_{ij}\omega_i\omega_j - \sum_{i \in \Lambda, j \notin \Lambda} J_{ij}\omega_i\eta_j\right)\right).$$

Also,

$$\mu_{\Lambda^c, \beta}^{\eta_{\Lambda_c}}(\omega \hspace{1mm} | \hspace{1mm} \sigma_i = \omega'_i, \forall i \in \Lambda \setminus \Delta) = \frac {\mu_{\Lambda^c, \beta}^{\eta_{\Lambda_c}}(\omega \hspace{1mm} \cap \hspace{1mm} \{ \sigma_i = \omega'_i, \forall i \in \Lambda \setminus \Delta \})}{\mu_{\Lambda^c, \beta}^{\eta_{\Lambda_c}}(\{ \sigma_i = \omega'_i, \forall i \in \Lambda \setminus \Delta \})}.$$

I am having some trouble manipulating this term. I tried to partition the sum of the hamiltonian in the sets $\Delta, \Lambda \setminus \Delta$ and $\Lambda^c$, but it didn't get me anywhere.

Could someone help me? I would really appreciate!!

$\endgroup$
  • 1
    $\begingroup$ Could you explain these notations? $\endgroup$ – LoveQYG Sep 7 '18 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.